Post by: hamdani yusuf on 05/10/2016 11:12:16
Photon model gains reputation in early 20th century due to photoelectric effect, especially that low frequency light cannot poduce emitted electron even with increased intensity. But newer development of laser can generate high enough intensity which can poduce emitted electron even with lower frequency than usual threshold.
Is photon model of electromagnetic wave an oversimplification?
Is it possible to explain wave-like phenomena in electromagnetism using photon model?
Is it possible to explain particle-like phenomena in electromagnetism using wave model?
Post by: hamdani yusuf on 06/10/2016 07:40:25
(https://www-zeuthen.desy.de/exps/physik_begreifen/chris/PhotomultiSkizze.jpg)
First of all, despite the name, photomultipliers don't actually multiply photons. These detectors multiply electric current produced by incident light by as much as 100 million times. Hence they are basically high gain amplifiers connected to fotosensitive detectors. When the incident light is powerful enough to kick one or more electrons from light receiver to the input of the amplifier, the signal will be amplified so it can be represented visually e.g. as a spike in oscilloscope.
The quantization of the detection is thus due to the quantization of electric current, which can not be less than single electron.
Post by: puppypower on 06/10/2016 11:31:53
For example, if we had two waves of water, 180 degrees out of phase, that cancel, there will be no wave. However, the water molecules or particles do not disappear.
A simple test of the particle and wave duality, to find the hierarchy, is to cancel two waves, while trying to induce a photoelectric affect. With the waves gone, nothing should happen if the wave rule. If something happens, particles rule.
Post by: hamdani yusuf on 08/10/2016 04:27:30
This is incompatible with the view that each spike is indicating single photon. It's more like each spike is indicating that single electron is transfered from photoreceptor to the first dynode, which is then multiplied by the next dynodes to produce detectable signal.
Hence, the time gap between one spike and the next can be interpreted as the electrons in photoreceptor are accumulating energy from incoming light (which is then used to get free), instead of caused by the lack of incoming photon.
Post by: Bored chemist on 12/10/2016 20:58:24
Detection of extremely low intensity light by photomultiplier has convinced many people that every spike shown on the oscilloscope is indicating a detection of single photon.Possibly, but more of them are convinced by the photoelectric effect measured in a simple vacuum photo-diode - and the way that the current varies with wavelength and intensity.
You would need to do better than the current model if you wanted to add anything to the field. Essentially you need to answer two questions.
What problems do you think exist?
How have you solved them?
Post by: hamdani yusuf on 13/10/2016 02:40:41
You would need to do better than the current model if you wanted to add anything to the field. Essentially you need to answer two questions.I think I have mentioned the problem in the first paragraph of this thread.
What problems do you think exist?
How have you solved them?
"particle models cannot easily account for the refraction, polarization, diffraction and birefringence of light."
I have a model which can easily explain the mechanism of reflection, polarization, interference of electromagnetic wave. But still need more work for experimental evident of refraction, which I think will be the basic for other phenomena like total internal reflection, channeling, diffraction, and birefringence.
Post by: nilak on 13/10/2016 08:09:15
Post by: Bored chemist on 14/10/2016 21:27:02
We can explain refraction etc perfectly well.
You may not be familiar with the explanation.
Post by: nilak on 14/10/2016 21:56:52
Photons are not classical particles or classical waves, They are photons.Do you consider the propagation of electric and magnetic field, at a phase difference and polarization on a certain axis, a classical wave ?
We can explain refraction etc perfectly well.
You may not be familiar with the explanation.
Post by: hamdani yusuf on 16/10/2016 10:55:40
Photons are not classical particles or classical waves, They are photons.Can you tell us where to find such explanation?
We can explain refraction etc perfectly well.
You may not be familiar with the explanation.
Post by: Bored chemist on 17/10/2016 09:33:25
etc would be a good start.
Post by: hamdani yusuf on 17/10/2016 12:17:26
https://en.wikipedia.org/wiki/Photon_polarizationThe explanation there is mainly classical, which uses wave model.
etc would be a good start.
There is no explanation how polarizers can rotate the orientation of the photons.
What about refraction?
My experiment to determine photon size using microwave has convinced me against photon model.
It is shown in this video
Post by: Bored chemist on 18/10/2016 11:09:46
Post by: hamdani yusuf on 18/10/2016 18:14:56
Photons don't have a "size"So how to determine whether or not "photon" can get through microwave oven's window?
Post by: Bored chemist on 18/10/2016 20:06:39
Post by: hamdani yusuf on 19/10/2016 02:38:29
Have you found explanation for light refraction using particle model?
Post by: Bored chemist on 19/10/2016 11:06:51
The simple version of the photon model of refraction is that the photon is absorbed and then re-emitted. The delay slows down the passage of the photon through the material. Refraction is all about the change in speed of light.
It seems to me that you are trying to belittle the photon model without even bothering to understand it.
That's not science- it's religion.
Post by: hamdani yusuf on 20/10/2016 00:00:23
Quote from: Bored chemist
Do you realise that it's your job to look for explanations, not mine?
The simple version of the photon model of refraction is that the photon is absorbed and then re-emitted. The delay slows down the passage of the photon through the material. Refraction is all about the change in speed of light.
It seems to me that you are trying to belittle the photon model without even bothering to understand it.
That's not science- it's religion.
I asked you because you are the one who claimed to have perfectly well explanation, while I can't find such a thing from the internet.
Photons are not classical particles or classical waves, They are photons.
We can explain refraction etc perfectly well.
You may not be familiar with the explanation.
Your version doesn't seem to be perfectly well.
I am the one who's doing the doubt here.
Post by: Bored chemist on 22/10/2016 14:15:49
You don't seem to have tried very hard. I just googled it and got a result; why couldn't you do that?
I asked you because you are the one who claimed to have perfectly well explanation, while I can't find such a thing from the internet.
Post by: hamdani yusuf on 22/10/2016 17:40:49
Maybe we just have different standard for a perfectly well explanation.
Or do you think that first page of google search provide best explanation?
Post by: Bored chemist on 22/10/2016 18:42:52
I just thought you had a better explanation than that.No, but I think that the first hit from Google is sufficient to prove that this
Maybe we just have different standard for a perfectly well explanation.
Or do you think that first page of google search provide best explanation?
"particle models cannot easily account for the refraction, polarization, diffraction and birefringence of light."
isn't true.
Post by: hamdani yusuf on 30/01/2017 12:57:43
Detection of extremely low intensity light by photomultiplier has convinced many people that every spike shown on the oscilloscope is indicating a detection of single photon. But there is alternative explanation that I can think of.A weight scale in a filling station reminds me of this situation. The display can only show liquid product's weight change every half kg.
(https://www-zeuthen.desy.de/exps/physik_begreifen/chris/PhotomultiSkizze.jpg)
First of all, despite the name, photomultipliers don't actually multiply photons. These detectors multiply electric current produced by incident light by as much as 100 million times. Hence they are basically high gain amplifiers connected to fotosensitive detectors. When the incident light is powerful enough to kick one or more electrons from light receiver to the input of the amplifier, the signal will be amplified so it can be represented visually e.g. as a spike in oscilloscope.
The quantization of the detection is thus due to the quantization of electric current, which can not be less than single electron.
The quantization effect is actually just precision limit of the measuring instrument. In photomultiplier, the limit is quantization of electric charge.
Post by: alancalverd on 31/01/2017 14:30:30
Post by: jeffreyH on 31/01/2017 19:46:29
All models are an oversimplification, which is why they are called models. A dead mouse is a perfect model of a live mouse, but only for an inifintesimal perod of time.
If you tie a string to its head and one each to the legs you can extend the model. A good name for this would be string theory.
Post by: hamdani yusuf on 08/02/2017 10:59:44
All models are an oversimplification, which is why they are called models. A dead mouse is a perfect model of a live mouse, but only for an inifintesimal perod of time.IMO, we use models to represent reality because we can take some advantages from them, e.g. more accessible, more controllable, etc.
Rough models can be useful in some situations, and they are usually used due to their simplicity.
Using flat earth model is usually adequate if you want to build a house, or calculating trajectory of a trebuchet.
Newton's mechanic is adequate to send people to the moon and back.
But beyond some limit, we need more refined models where the cost from deviation of model's prediction compared to measurement outweigh the benefit from its simplicity.
Unexpected results come from false assumptions.
Post by: hamdani yusuf on 09/02/2017 11:09:13
Post by: alancalverd on 22/02/2017 10:54:22
X-ray "intensifying screens" are fluorescent material that converts single incoming high energy photons into multiple visible photons that can be recorded on conventional photographic film. At very low x-ray intensities you can see "quantum mottle" where individual x-ray photons have triggered a local burst of visible photons.
Or you can use a photonuclear reaction to detect photons with energies above about 1 MeV by the products of that reaction. The consistency of yield strongly suggests that the trigger is quantised.
At 1.022 MeV the incoming photon in the vicinity of a nucleus can produce an electron-positron pair which annihilates into exactly two 511 keV photons that you can detect by all sorts of methods. I can't think of another mathematical model that would predict this.
Post by: hamdani yusuf on 28/02/2017 05:27:56
Then you can use an oldfashioned instrument!What is the minimum size of "quantum mottle" have you ever seen? IMO, it's not much less than 1 micron since it's visible to the naked eye.
X-ray "intensifying screens" are fluorescent material that converts single incoming high energy photons into multiple visible photons that can be recorded on conventional photographic film. At very low x-ray intensities you can see "quantum mottle" where individual x-ray photons have triggered a local burst of visible photons.
Or you can use a photonuclear reaction to detect photons with energies above about 1 MeV by the products of that reaction. The consistency of yield strongly suggests that the trigger is quantised.
At 1.022 MeV the incoming photon in the vicinity of a nucleus can produce an electron-positron pair which annihilates into exactly two 511 keV photons that you can detect by all sorts of methods. I can't think of another mathematical model that would predict this.
Can you name one method to detect 511 keV photons which doesn't involve electron activity?
Post by: hamdani yusuf on 09/03/2017 04:25:23
In popcorn case, the energy is supplied more or less continuously, but the popping events occur discretely. Analogous with detector screen, we can think that the screen receive electromagnetic wave continuously, but the events of change in screen color/brightness occur discretely.
Post by: timey on 09/03/2017 13:52:02
Post by: alancalverd on 09/03/2017 14:53:38
Then you can use an oldfashioned instrument!What is the minimum size of "quantum mottle" have you ever seen? IMO, it's not much less than 1 micron since it's visible to the naked eye.
X-ray "intensifying screens" are fluorescent material that converts single incoming high energy photons into multiple visible photons that can be recorded on conventional photographic film. At very low x-ray intensities you can see "quantum mottle" where individual x-ray photons have triggered a local burst of visible photons.
Or you can use a photonuclear reaction to detect photons with energies above about 1 MeV by the products of that reaction. The consistency of yield strongly suggests that the trigger is quantised.
At 1.022 MeV the incoming photon in the vicinity of a nucleus can produce an electron-positron pair which annihilates into exactly two 511 keV photons that you can detect by all sorts of methods. I can't think of another mathematical model that would predict this.
Can you name one method to detect 511 keV photons which doesn't involve electron activity?
If you want to be pedantic, everything that happens outside a nucleus involves electron activity. But it's quite easy* to measure photon flux with a calorimeter.
*it only took me 13 years
Post by: timey on 09/03/2017 19:57:50
I hesitate to interject my notions into your thread, but I'll just do so once as an observation.
(btw, I've tried you YouTube vids, but the sound doesn't come through so well on my phone, so I can't hear the commentary so good. I don't know if this is a problem when not watching on a phone)
My theory of time states time as being a function of energy, and wavelength as being a function of time.
Under this remit you can have both the photon model, and the wave model united into one model, where you will always know the photons flux, or where an electron is located and how fast it is moving, by understanding that the timing/frequency of a particle, and the time/energy of a field that a particle moves through, is a function of the energy of that particle, or field.
wavelength = h/p
where:
p = h*vbar
and:
vbar = v/a
and
h = Planck's h constant.
h can be considered a time function describing field on it own, and then p derived via h*vbar can be considered as inclusive of a time function for a particle.
What one timing will record of another will be proportional to the difference in timing. i:e: discrete.
Post by: hamdani yusuf on 22/03/2017 04:30:44
I don't want to be pedantic. I'm just a simple man with a simple mind, and I prefer simple explanation.Then you can use an oldfashioned instrument!What is the minimum size of "quantum mottle" have you ever seen? IMO, it's not much less than 1 micron since it's visible to the naked eye.
X-ray "intensifying screens" are fluorescent material that converts single incoming high energy photons into multiple visible photons that can be recorded on conventional photographic film. At very low x-ray intensities you can see "quantum mottle" where individual x-ray photons have triggered a local burst of visible photons.
Or you can use a photonuclear reaction to detect photons with energies above about 1 MeV by the products of that reaction. The consistency of yield strongly suggests that the trigger is quantised.
At 1.022 MeV the incoming photon in the vicinity of a nucleus can produce an electron-positron pair which annihilates into exactly two 511 keV photons that you can detect by all sorts of methods. I can't think of another mathematical model that would predict this.
Can you name one method to detect 511 keV photons which doesn't involve electron activity?
If you want to be pedantic, everything that happens outside a nucleus involves electron activity. But it's quite easy* to measure photon flux with a calorimeter.
*it only took me 13 years
IMO the mottle size is too large to be caused by single photon, which is even smaller than electron, which is already very small.
My point is, discrete phenomena related to electromagnetic wave are more likely to be caused by discrete nature of its source, which is charged particle, e.g. proton and electron, rather than discrete nature of electromagnetic wave itself, which is thought as energy packet called photon.
How does calorimeter measure photon flux without involving electron activity?
Post by: hamdani yusuf on 22/03/2017 04:46:54
Humandi - I like your analogy because it is compatible to my time theory.Thanks for your appreciation.
I hesitate to interject my notions into your thread, but I'll just do so once as an observation.
(btw, I've tried you YouTube vids, but the sound doesn't come through so well on my phone, so I can't hear the commentary so good. I don't know if this is a problem when not watching on a phone)
My theory of time states time as being a function of energy, and wavelength as being a function of time.
Under this remit you can have both the photon model, and the wave model united into one model, where you will always know the photons flux, or where an electron is located and how fast it is moving, by understanding that the timing/frequency of a particle, and the time/energy of a field that a particle moves through, is a function of the energy of that particle, or field.
wavelength = h/p
where:
p = h*vbar
and:
vbar = v/a
and
h = Planck's h constant.
h can be considered a time function describing field on it own, and then p derived via h*vbar can be considered as inclusive of a time function for a particle.
What one timing will record of another will be proportional to the difference in timing. i:e: discrete.
I'm sorry that my videos have low quality sound. That's because I only use cellphone to record them.
Some even has unexpected cut outs, especially when changing slide. That happened when I convert the power point file into MP4.
I checked my source file and found no problem, but some things were missing in the video file.
I planned to put the script in the description when I have time.
Post by: yor_on on 24/03/2017 20:05:34
Post by: hamdani yusuf on 26/03/2017 03:40:29
Hamdani, have you looked at the origins of 'photons'?AFAIK, it was Einstein's interpretation to Planck's result, E=n.h.f, which describe quantification of thermal radiation.
This interpretation then gain support for explaining photoelectric effect.
But IMO, it can also be interpreted as quantification of electric charge as the source of electromagnetic wave.
Post by: yor_on on 28/03/2017 17:04:36
https://en.wikipedia.org/wiki/Planck%27s_law
Post by: hamdani yusuf on 28/03/2017 17:16:30
Post by: yor_on on 28/03/2017 17:30:01
Post by: Colin2B on 28/03/2017 18:32:39
Then the question becomes. What exactly is it that 'bumps' into a 'black body'?The variation of the em field. We interpret the interaction as a bump - transfer of energy/momentum - and so can say that it has at least one of the properties of a particle, but that doesn't make it a little hard piece of mass.
Post by: jeffreyH on 28/03/2017 22:12:03
Post by: Colin2B on 28/03/2017 22:36:54
The photon cannot be described physically. Only modelled.I'm pretty sure the same is true of the wave side as well, too easy to think like water or sound waves.
Post by: yor_on on 04/04/2017 12:44:47
Post by: hamdani yusuf on 06/11/2017 12:27:53
I though I had a good idea about quantum mechanics until I started studying it more in depth and understanding its strangeness. The photon cannot be described physically. Only modelled.I think many strangeness in quantum mechanics are related to result of double slit experiment, when the light is interpreted as photon particle.
I’ve shown in my experiments about diffraction of light that double slit aperture is simply an obstacle with four edges. How the photon model explains the effect of removing the outer edges (which makes it a thin wire diffraction, which according to Babinet’s principle, should produce the same pattern as single slit aperture)?
How the photon model explains the result of circular pattern when the diffraction grating is vertically tilted?
Post by: Bill S on 06/11/2017 15:11:55
My very inexpert comment would be that you are dealing with quons (sensu Herbert), and therefore may find that, not only are our models all oversimplifications, but that this is especially the case when we start looking closely at QM.
Maybe, one day it will be discovered that QM gives us a “window” to infinity. Should this be the case, all we can hope for in our 3+1 dimensional “reality” is improved modelling.
Post by: hamdani yusuf on 21/11/2017 04:20:22
Referring to photomultiplier tubes handbook from Hamamatsu Photonics, we find that single photon detection is not as simple as depicted in popular documentary movies. In my opinion, the existence of dark current and thermionic emission reduce credibility of single photon detection. Instead, they show electron detections from earlier stages of the dynode.
https://www.hamamatsu.com/resources/pdf/etd/PMT_handbook_v3aE.pdf
Post by: Bill S on 21/11/2017 11:39:23
Post by: hamdani yusuf on 21/11/2017 13:24:31
Interesting looking link Hamdani. I've read the intro, but there is no way I'll have time to wade through the lot. If you could point me in the direction of the part(s) that cover the difficulties of single photon detection, I'd be obliged.you can find link to a chapter about single photon detection mode, and see a graph representing output signal. You will find a statement about a threshold to differentiate which spike counts as photon reading, which counts as noise.
Post by: Bill S on 22/11/2017 18:07:55
Post by: hamdani yusuf on 17/02/2021 13:46:02
Quote
We know that light behaves as a wave AND a particle... but can we treat sound in exactly the same way? And what about this exciting new particle known as the "dance particle"?
Hey everyone, in this video I wanted to discuss how physicists like to take complicated physics phenomena and model them as atoms in order to hugely simplify the mathematics used to solve problems. The main example I'll be discussing is the treatment of sound waves as particles known as phonons. Now technically, these particles are QUASIparticles because they're not real, and are just a mathematical reimagination of the sound wave. This is slightly different to light, because light particles (photons) DO actually exist. We have plenty of evidence to show us that the wave model of light is not enough to explain certain things we observe, such as the photoelectric effect, black body radiation (ultraviolet catastrophe) and so on. In fact, light needs both the wave model and the particle model to explain all the phenomena we observe. However a sound quasiparticle (phonon) is not like this - it's a mathematical trick.
The reason we work with quasiparticles at all, becomes very evident when we consider the reality of a sound wave moving through a solid, for example. The sound wave is formed by the oscillation of atoms in the solid. The bonds between atoms allow energy to be transferred from one region in the solid to another, and as soon as the disturbance has passed, the atoms return to their original positions. Now if we want to study how sound waves move through the solid in any detail, then we have to consider the position of each affected atom, as well as the strength of the bonds between atoms. We can treat these bonds like springs, though this is a big simplification. But even then we have to consider the "stiffness" of the springs, which is dependent on the type of atoms forming the bond. And then we may have to think about the disturbances to each atom at every instant in time, if there are multiple sound waves moving through the solid in different directions. Things become complicated very quickly. Instead, if we treat the atoms in the solid as a "background", and the sound wave as a sound particle, then we only need to consider the phonon representing each wave rather than the (likely) tens of thousands of atoms in the solid, and how their positions change over time.
Another example we will discuss is when electrons in a solid can escape from their energy levels, leaving behind a vacancy. This vacancy can be filled with nearby electrons, making it seem as if the vacancy is moving around within the solid. This is treated as its own quasiparticle, known as a "hole", and is much easier to deal with than considering all the electrons filling any nearby vacancy.
More information about the electron hole quasiparticle: https://en.wikipedia.org/wiki/Electro...
Timestamps:
0:00 - The DANCE particle + how physicists work with quasiparticles
0:38 - How we deal with light - waves and particles (photons)
1:06 - Sound waves: oscillations in air (+ other gases liquids and solids)
1:59 - Sound wave in a solid: atomic structure and bonds transmit energy
3:53 - Treating sound waves as particles (phonons) - quasiparticles
5:37 - Why phonons are useful (multiple sound waves and phonon-phonon interactions)
6:15 - Electron hole quasiparticles (vacancy vs electron motion)
What convinced us that photon is any more physical than phonon?
Post by: alancalverd on 17/02/2021 16:30:15
We know that light behaves as a wave AND a particle.No. Light behaves as light. We model it as a wave or a particle.
Post by: hamdani yusuf on 18/02/2021 05:33:54
The sentence was quoted from the description of the video.We know that light behaves as a wave AND a particle.No. Light behaves as light. We model it as a wave or a particle.
In many cases light behaves similar to other kind of waves. In other cases it behaves similar to other kind of particles.
The question is, is it possible to explain wavelike behavior of light using particle model?
Is it possible to explain particle-like behavior of light using wave model?
What additional assumptions are required to achieve those tasks?
Post by: alancalverd on 18/02/2021 11:31:56
The question is, is it possible to explain wavelike behavior of light using particle model?No. The only connection is E = hν.
Is it possible to explain particle-like behavior of light using wave model?
The photon-phonon parallel comes from Planck's original concept of electromagnetic "waves in a box" which leads to the quantification of black body radiation and the explanation of the black body spectrum. If you want to describe nuclear motion in a regular solid you can use the same fundamental concept and end up with quantised collaborative excitations which explain some properties of solids, particularly charge transport in semiconductors.
Post by: hamdani yusuf on 19/02/2021 14:09:30
No. Light behaves as light. We model it as a wave or a particle.Is it still valid if the word light is replaced by sound?
Post by: alancalverd on 19/02/2021 15:00:04
Post by: hamdani yusuf on 21/02/2021 12:08:54
No. Sound is compression waves in a medium. The particles of the medium certainly move but do so cyclically - there is no gross motion in the direction of propagation, and sound does not propagate in vacuo.What do you think about AC electric power transmission through metal wires? Is it similar to sound, or light? Is it a completely different type of phenomena?
Post by: hamdani yusuf on 06/07/2021 01:15:59
Post by: wolfekeeper on 06/07/2021 02:50:54
Photons and how they interact with electrons are extremely well described in modern physics by quantum electrodynamics:
https://en.wikipedia.org/wiki/Quantum_electrodynamics
Post by: Kryptid on 06/07/2021 05:46:51
What convinced us that photon is any more physical than phonon?
Because, unlike phonons, photons don't rely on a medium to transmit them.
Post by: hamdani yusuf on 06/07/2021 12:42:34
Phonons are quantum mechanical phenomena, just like photons in fact.The video shows that imagining photons like extremely small particles produces more questions than answers. In many situations they behave like waves.
Photons and how they interact with electrons are extremely well described in modern physics by quantum electrodynamics:
https://en.wikipedia.org/wiki/Quantum_electrodynamics
Post by: hamdani yusuf on 06/07/2021 12:47:17
We can hypothesize that light propagates in the fabric of space-time, just like gravitational wave.What convinced us that photon is any more physical than phonon?
Because, unlike phonons, photons don't rely on a medium to transmit them.
Post by: Kryptid on 06/07/2021 14:53:25
We can hypothesize that light propagates in the fabric of space-time, just like gravitational wave.
Then what would make it different from a gravitational wave?
Post by: Eternal Student on 06/07/2021 16:48:15
Quote from: hamdani yusuf on Today at 12:47:17
We can hypothesize that light propagates in the fabric of space-time, just like gravitational wave.
Then what would make it different from a gravitational wave?
Now that's a good question: Can we observe diffraction, refraction, reflection and interference in gravitational waves?
Post by: hamdani yusuf on 06/07/2021 20:01:14
The objects that interact with it.We can hypothesize that light propagates in the fabric of space-time, just like gravitational wave.
Then what would make it different from a gravitational wave?
In case of gravity, it's objects with momentum.
In case of light, objects with electric charge.
Post by: evan_au on 06/07/2021 23:13:56
Quote from: Kryptid
Then what would make (light) different from a gravitational wave?One difference is that light is generated by an electric dipole,
- while gravitational waves are generated by a mass quadrupole.
So if you had a macroscopic (ie much bigger than atom-sized) negative charge rotating around a positive charge at 100Hz, it would generate electromagnetic waves at 100Hz.
- As I understand it, if you had a macroscopic (ie much bigger than earth-sized) positive mass rotating around another positive mass at 100Hz, it would generate gravitational waves at 200Hz.
See: https://en.wikipedia.org/wiki/Gravitational_wave#Sources
Post by: Colin2B on 06/07/2021 23:24:17
One difference is that light is generated by an electric dipole,You beat me to it Evan. This difference also means that the polarisations of the light and gravitational waves are very different as is the effect on any matter they interact with.
- while gravitational waves are generated by a mass quadrupole.
As with all waves you would expect refraction, diffraction etc, but I would need to think what arrangements of mass would cause those in the case of gravitational waves. I suspect the effects would be small.
Post by: Kryptid on 06/07/2021 23:25:48
The objects that interact with it.
In case of gravity, it's objects with momentum.
In case of light, objects with electric charge.
And how does a wave through space-time "know" which one to interact with?
Post by: hamdani yusuf on 07/07/2021 00:57:11
It's still a hypothesis, which is more or less speculative, until more evidence is found. Perhaps the space-time is multidimensional, and the objects in it only interact with some of its aspects which are relevant to them.The objects that interact with it.
In case of gravity, it's objects with momentum.
In case of light, objects with electric charge.
And how does a wave through space-time "know" which one to interact with?
As mentioned in the end of the video, we can interpret photons in some different ways. But interpreting them as point like particles moving through space is not very helpful.
Post by: wolfekeeper on 07/07/2021 01:40:47
That's also true of electrons. That's pure Quantum Mechanics baby! That's how it works.Phonons are quantum mechanical phenomena, just like photons in fact.The video shows that imagining photons like extremely small particles produces more questions than answers. In many situations they behave like waves.
Photons and how they interact with electrons are extremely well described in modern physics by quantum electrodynamics:
https://en.wikipedia.org/wiki/Quantum_electrodynamics
Post by: hamdani yusuf on 07/07/2021 03:12:22
At least electron has quantified charge and mass.That's also true of electrons. That's pure Quantum Mechanics baby! That's how it works.Phonons are quantum mechanical phenomena, just like photons in fact.The video shows that imagining photons like extremely small particles produces more questions than answers. In many situations they behave like waves.
Photons and how they interact with electrons are extremely well described in modern physics by quantum electrodynamics:
https://en.wikipedia.org/wiki/Quantum_electrodynamics
What's the most convincing experiment showing that light behaves like particles and can't be explained by wave analogy?
On the other hand, what's the most convincing experiment showing that electron behaves like waves and can't be explained by particle analogy?
Post by: evan_au on 07/07/2021 09:33:43
Quote from: hamdani yusuf
Perhaps the space-time is multidimensionalThat it is: 3 of space + 1 of time; that much is confirmed.
There are more speculative ideas about additional dimensions (eg string theory), but they have not been confirmed at this point in time.
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Perhaps the space-time is multidimensional, and the objects in it only interact with some of its aspects which are relevant to them.That is true. But the fact that light and gravity both follow an inverse-square rule to incredible accuracy at the scale of the Solar System implies that there are 3 dimensions of space (ie the density of phonons/gravitons hitting a given area reduces by a factor of 4 if you double the distance).
There are some thoughts that maybe gravity could be influenced by additional dimensions (eg string theory), and that hypothesis is being tested by measuring gravity on millimeter scales and below.
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What's the most convincing experiment showing that light behaves like particles and can't be explained by wave analogy?They both have behaviors that act like waves, and behaviors that act like particles - that's why some people dubbed them "wavicles".
On the other hand, what's the most convincing experiment showing that electron behaves like waves and can't be explained by particle analogy?
- Both electron microscopes and light microscopes have a resolution limit which depends on the energy of the particles.
Post by: Eternal Student on 07/07/2021 10:07:28
@hamdani yusuf
1. Experiments showing electrons behave like waves: I think the double slit experiment with an electron instead of a photon emitter was previously mentioned.
2. Going back to an earlier comment I made:
Now that's a good question: Can we observe diffraction, refraction, reflection and interference in gravitational waves?What I meant was - this could be something for the short video you were making where diffraction and refraction were discussed. I don't know the full answer, I'm not sure anyone does at the moment.
- - - - - - -
(About Gravitational Waves)... As with all waves you would expect refraction, diffraction etc, but I would need to think what arrangements of mass would cause those in the case of gravitational waves. I suspect the effects would be small.This looks a lot like something you would get from the superposition of two waves. Most waves follow a superposition principle. Gravitational waves are unusual because the Einstein Field Equations (unlike Maxwell's equations) are non-linear, so that two separate solutions do not (in general) add to make another valid solution.
For small oscillations (small gravitational waves) the space is assumed to be mainly empty and then the metric looks like the usual Minkowski metric plus a small perturbation. I don't know what happens when two such waves meet, possibly the resulting perturbation is like the sum of the separate perturbations to first order approximations.
Best wishes to all.
Post by: alancalverd on 07/07/2021 10:58:41
What's the most convincing experiment showing that light behaves like particles and can't be explained by wave analogy?Photon counting or "quantum mottle" with very low intensities of light or x-rays shows that interactions with a photodetector are individual transfers of fixed quantities of energy and momentum. Increasing intensity at a fixed wavelength/photon energy increases the rate of interactions but not the amplitude of an individual event.
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On the other hand, what's the most convincing experiment showing that electron behaves like waves and can't be explained by particle analogy?Electron diffraction patterns look just like x-ray diffraction patterns if you assign the appropriate wavelength. In practice, neutron diffraction is easier to demonstrate but the principle is the same.
Electron-positron pair production and annihilation is fun because it demonstrates the reversibility of E = mc^2 and you end up with two 511 keV photons hurtling in exactly opposite directions, like billiard balls rather than water waves!
Post by: hamdani yusuf on 07/07/2021 11:22:30
- As I understand it, if you had a macroscopic (ie much bigger than earth-sized) positive mass rotating around another positive mass at 100Hz, it would generate gravitational waves at 200Hz.Will the wave be sinusoid, or rectified sine wave?
(https://upload.wikimedia.org/wikipedia/commons/thumb/9/93/Full-wave_Rectified_Sine_Wave.svg/1440px-Full-wave_Rectified_Sine_Wave.svg.png)
Post by: hamdani yusuf on 07/07/2021 11:26:14
1. Experiments showing electrons behave like waves: I think the double slit experiment with an electron instead of a photon emitter was previously mentioned.What's the material that make up the slit, i.e. blocks the motion of the electrons? Is changing the material has any effect?
Post by: hamdani yusuf on 07/07/2021 11:31:45
Can you show some links to the research, pictures or videos showing those experiments?What's the most convincing experiment showing that light behaves like particles and can't be explained by wave analogy?Photon counting or "quantum mottle" with very low intensities of light or x-rays shows that interactions with a photodetector are individual transfers of fixed quantities of energy and momentum. Increasing intensity at a fixed wavelength/photon energy increases the rate of interactions but not the amplitude of an individual event.QuoteOn the other hand, what's the most convincing experiment showing that electron behaves like waves and can't be explained by particle analogy?Electron diffraction patterns look just like x-ray diffraction patterns if you assign the appropriate wavelength. In practice, neutron diffraction is easier to demonstrate but the principle is the same.
Electron-positron pair production and annihilation is fun because it demonstrates the reversibility of E = mc^2 and you end up with two 511 keV photons hurtling in exactly opposite directions, like billiard balls rather than water waves!
Post by: alancalverd on 07/07/2021 16:39:14
Electron diffraction is used in electron microscopy en.wikipedia.org/wiki/Electron_diffraction (http://en.wikipedia.org/wiki/Electron_diffraction)I used it many years ago to calibrate a small electron accelerator. Neutron diffraction has been used in crystallography particularly of organic compounds and hydrides. It's easier because neutrons have a long range in air.
Pair production and annihilation is used every day in medical imaging https://en.wikipedia.org/wiki/Positron_emission_tomography (https://en.wikipedia.org/wiki/Positron_emission_tomography)
Post by: alancalverd on 07/07/2021 16:42:17
Best to use a conductor otherwise the blocker acquires a charge that deflects the electron beam!1. Experiments showing electrons behave like waves: I think the double slit experiment with an electron instead of a photon emitter was previously mentioned.What's the material that make up the slit, i.e. blocks the motion of the electrons? Is changing the material has any effect?
Post by: hamdani yusuf on 08/07/2021 07:12:16
Is using a superconductor gives a better result?Best to use a conductor otherwise the blocker acquires a charge that deflects the electron beam!1. Experiments showing electrons behave like waves: I think the double slit experiment with an electron instead of a photon emitter was previously mentioned.What's the material that make up the slit, i.e. blocks the motion of the electrons? Is changing the material has any effect?
Post by: hamdani yusuf on 08/07/2021 07:18:13
Quantum mottle is part of the basic radiography syllabus. howradiologyworks.com/x-ray-cnr/ has some examples of a everyday phenomenon. It's the reason you can't increase the sensitvity of x-ray film beyond a certain limit, which is less for pediatrics than for adults, because you lose spatial resolution as the quantum noise exceeds the image contrast.
Why can't it be explained using wave mechanism?
Photon detector screen reminds me of popcorn analogy to explain half life of radioactive materials.
In popcorn case, the energy is supplied more or less continuously, but the popping events occur discretely. Analogous with detector screen, we can think that the screen receive electromagnetic wave continuously, but the events of change in screen color/brightness occur discretely.
Post by: evan_au on 08/07/2021 10:25:15
Quote from: hamdani yusuf
Will the (gravitational) wave be sinusoid, or rectified sine wave?The sudden change in direction of a rectified sinusoid produces lots of high-frequency harmonics.
- I don't see anything in the path of two orbiting neutron stars that would produce such high frequency components. So a gravitational wave can't be a rectified sine wave.
- The discontinuity of derivative (in fact, the reversal of derivative) in a rectified sine wave implies some unreal forces to reverse the acceleration of a neutron star twice every orbit. Another reason it can't be a rectified sine wave.
I expect that the amplitude of a gravitational wave will be more sin(x)^2, where x is the angle in the orbit.
- But remember that gravitational waves distort more than one dimension of space, and also distorting the passage of time, so a 1-dimensional function can't do it credit!
- And in all of these measures, there is an arbitrary additive constant that can move the entire graph up or down...
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Post by: alancalverd on 08/07/2021 11:20:27
If you reduce the doserate sufficiently you can detect individual photon events in real time. Easier with a geiger counter or photomultiplier. We know that x and γ radiation are the same stuff, so you can replace the x-ray source with a radionuclide which we know can only emit discrete pulses, not continuous radiation, and get the same effect.Quantum mottle is part of the basic radiography syllabus. howradiologyworks.com/x-ray-cnr/ has some examples of a everyday phenomenon. It's the reason you can't increase the sensitvity of x-ray film beyond a certain limit, which is less for pediatrics than for adults, because you lose spatial resolution as the quantum noise exceeds the image contrast.
Why can't it be explained using wave mechanism?Photon detector screen reminds me of popcorn analogy to explain half life of radioactive materials.
In popcorn case, the energy is supplied more or less continuously, but the popping events occur discretely. Analogous with detector screen, we can think that the screen receive electromagnetic wave continuously, but the events of change in screen color/brightness occur discretely.
But beware of falling into your own trap of thinking that a photon is a wave or a particle. Remember that, as with physics in general, these are mathematical models of what it does, not what it is.
Post by: hamdani yusuf on 08/07/2021 12:28:49
The sudden change in direction of a rectified sinusoid produces lots of high-frequency harmonics.
I calculated the gravitational force from two identical objects orbiting each other.
(https://www.thenakedscientists.com/forum/index.php?action=dlattach;topic=68595.0;attach=32156;image)
Post by: hamdani yusuf on 08/07/2021 13:00:15
If you reduce the doserate sufficiently you can detect individual photon events in real time. Easier with a geiger counter or photomultiplier. We know that x and γ radiation are the same stuff, so you can replace the x-ray source with a radionuclide which we know can only emit discrete pulses, not continuous radiation, and get the same effect.Waves don't have to be continuous. It can also last for a few cycles only.
But beware of falling into your own trap of thinking that a photon is a wave or a particle. Remember that, as with physics in general, these are mathematical models of what it does, not what it is.
(https://sites.suffolk.edu/jhtl/files/2019/04/Ferrara-1yhonyg.png)
Post by: Eternal Student on 08/07/2021 13:56:50
I calculated the gravitational force from two identical objects orbiting each other.I'm really not sure what that graph shows so I can't comment on it. This is a shame because it looks like you've spent a lot of time on it. Would you like to describe what it is?
Evan_au also made this comment earlier...
But remember that gravitational waves distort more than one dimension of space, and also distorting the passage of time, so a 1-dimensional function can't do it credit!I was considering replying to Evan_au earlier about this. In essence gravitational waves are not like most other waves.
Post by: hamdani yusuf on 08/07/2021 15:50:22
An observer is located at coordinate 2, 0.
The graph is a plot in time.
(x1, y1) is position of first star.
(x2, y2) is position of second star.
g1 is gravitational force from first star to observer.
g2 is gravitational force from second star to observer.
I missed something here. The y component of the forces are cancelling out. The total force should add x component only. Now The picture is corrected.
Post by: alancalverd on 08/07/2021 15:52:37
Waves don't have to be continuous. It can also last for a few cycles only.
Nice image, but if you make a fourier analysis you will see that it contains all sorts of harmonics, and a single pulse has an infinite number of harmonics, whereas a single photon only has one associated frequency 'cos it isn't a wave (or a particle).
Post by: hamdani yusuf on 08/07/2021 16:03:08
(https://www.thefouriertransform.com/pairs/gaussianplot.gif)
Waves don't have to be continuous. It can also last for a few cycles only.
Nice image, but if you make a fourier analysis you will see that it contains all sorts of harmonics, and a single pulse has an infinite number of harmonics, whereas a single photon only has one associated frequency 'cos it isn't a wave (or a particle).
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(https://www.thefouriertransform.com/pairs/endresult.gif)"TheFourierTransform.com - Fourier Transform of the Gaussian" https://www.thefouriertransform.com/pairs/gaussian.php
Equation [9] states that the Fourier Transform of the Gaussian is the Gaussian! The Fourier Transform operation returns exactly what it started with. This is a very special result in Fourier Transform theory.
Post by: Eternal Student on 08/07/2021 16:45:20
The graph is a plot in time.......... and something to do with force on an observer at a fixed poisiton (1.5, 0).
O.K. Thanks for explaining that Hamdani. Then it's a nice graph. I thought the previous discussions were about gravitational waves and that graph might have been something to represent this. Sorry for my confusion but thanks for taking the time to explain it.
Post by: Bored chemist on 08/07/2021 17:22:31
Waves don't have to be continuous. It can also last for a few cycles only.
Nice image, but if you make a fourier analysis you will see that it contains all sorts of harmonics, and a single pulse has an infinite number of harmonics, whereas a single photon only has one associated frequency 'cos it isn't a wave (or a particle).
That's why people use these
https://en.wikipedia.org/wiki/Wavelet
Post by: alancalverd on 08/07/2021 23:44:46
The simplest proof that a photon isn't a particle or a wave is to look at the diffraction pattern of a monochromatic beam. It's the same color as the original beam.
Now we know that we can generate a diffraction pattern when only one photon is present at a time (the classic double slit experiment), which is why we can model the photon as a wave..
If the single photon split and reappeared simultaneously in several positions, conservation of energy demands that the color of the diffraction pattern would be different. If not, we could focus the diffracted light back into a single beam with more energy than the incoming photon! So whilst wavelet analysis predicts the shape of the diffraction pattern, it can't predict the color.
Therefore modelling a photon as a wave proves that it is an inadequate model. But modelling it as a particle doesn't predict a diffraction pattern at all, so that is also an inadequate model..
Post by: hamdani yusuf on 08/07/2021 23:55:58
If the photon split and reappeared simultaneously in several positions, conservation of energy demands that the color of the diffraction pattern would be different. If not, we could focus the diffracted light back into a single beam with more energy than the incoming photon! So whilst wavelet analysis predicts the shape of the diffraction pattern, it can't predict the color.How do you split a single photon?
Post by: alancalverd on 09/07/2021 00:32:37
Post by: hamdani yusuf on 09/07/2021 05:25:41
Precisely the point! If it's a particle, it can't split so it must go through one slit or the other, without losing energy. If it's a wave, it can split and pass though both slits, but the energy at any point on the receiver will be less than in the primary beam. What actually happens? It appears to go though both slits and arrives with the same energy as it set out!Have you watched the video in my reply #56? Using a photon as a particle analogy, we are forced to conclude that the photon must interact with its own duplicate which has already arrived at the screen a few moments ago. Moreover, we must assume that a macroscopic beam splitter made of large number of atoms can consistently split each photon to intended directions without much loss due to scattering randomly. It seems like the results become significantly randomized in the screen only. This is in line with the popcorn analogy which I mentioned earlier.
Post by: alancalverd on 09/07/2021 09:26:23
My approach to teaching physics is to avoid analogies which I find confusing at best and positively misleading at worst. Start with what you observe and construct a purely abstract mathematical model that produces the observed result. If you need to use two models, just accept that they are both approximations. All that matters is that they should be consistent with other models (such as conservation of energy, momentum, mass....) and adequately predictive.
The "a few moments ago" is confusing because students know (or should know, if they stayed awake in the previous lecture) that a single photon is unlikely to produce a spot on a photographic film. You need two visible photons to strike a halide grain within a fairly short time in order to produce blackening. This phenomenon is known as reciprocity failure - the apparent sensitivity of film decreases at very low (and very high) photon intensities. Astronomers used to "pre-fog" their photographic plates to lift the sensitivity above the RF level, and radiographers exploit it to mask noise and produce clean white bones.
Post by: Bored chemist on 09/07/2021 09:57:44
If the single photon split and reappeared simultaneously in several positions,It doesn't.
It forms a probability density that is higher than zero, but less than one simultaneously in several positions.
It's sort of more like a quarter of the photon is in each of 4 place, but that's not the whole story either.
Post by: hamdani yusuf on 09/07/2021 10:15:17
The "a few moments ago" is confusing because students know (or should know, if they stayed awake in the previous lecture) that a single photon is unlikely to produce a spot on a photographic film. You need two visible photons to strike a halide grain within a fairly short time in order to produce blackening.What's your reference to two photons requirement?
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This breakdown in the usual tradeoff between aperture and shutter speed is known as reciprocity failure. Each different film type has a different response at low light levels. Some films are very susceptible to reciprocity failure, and others much less so. Some films that are very light sensitive at normal illumination levels and normal exposure times lose much of their sensitivity at low light levels, becoming effectively "slow" films for long exposures. Conversely some films that are "slow" under normal exposure duration retain their light sensitivity better at low light levels.https://en.wikipedia.org/wiki/Reciprocity_(photography)#Reciprocity_failure
Post by: alancalverd on 09/07/2021 10:45:34
Unlike camera film, which may stay in the camera for hours or years before being developed, "x-ray" film is processed within minutes of exposure so doesn't suffer from latent image fade but still needs at least two photons to switch on a halide grain. "X-ray" in inverted commas because photographic film (a) is actually not very sensitive to x-rays and (b) doesn't display reciprocity failure! In practice we use double-emulsion film sandwiched between fluorescent screens that generate several visible (or low energy x-ray) photons for each incoming high energy photon.
Post by: hamdani yusuf on 09/07/2021 10:55:09
Which illustrates the dangers of analogy - you end up with a menagerie of quasiparticles that travel faster than light and get randomised by a photographic film but not by a scintillation counter that does exactly the same thing, but faster.That's when you use particle analogy. But if you use wave analogy as suggested by the video, the result can be understood less surprisingly.
I have an idea to test the validity of the wave explanation. Let's make the long path of the interfering light beam much longer than the short path, so there's a significant time difference in their arrival at the screen. To balance the intensity, the short path can be reduced further using a filter.
First, the light source is turned on continuously. The single photon production is done using attenuating filter alone. As demonstrated in the video, the screen shows interference pattern.
Second the light source is turned on in short pulses. The pulse width is shorter than time difference between the paths of the light beams. The silence period between the pulses should be randomized to minimize the effect from resonance or echo. If the wave explanation is correct, then the interference pattern should not be observed significantly.
Post by: hamdani yusuf on 09/07/2021 11:06:32
Check the paragraph before the one you just quoted!You mean this one?
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At very low light levels, film is less responsive. Light can be considered to be a stream of discrete photons, and a light-sensitive emulsion is composed of discrete light-sensitive grains, usually silver halide crystals. Each grain must absorb a certain number of photons in order for the light-driven reaction to occur and the latent image to form. In particular, if the surface of the silver halide crystal has a cluster of approximately four or more reduced silver atoms, resulting from absorption of a sufficient number of photons (usually a few dozen photons are required), it is rendered developable. At low light levels, i.e. few photons per unit time, photons impinge upon each grain relatively infrequently; if the four photons required arrive over a long enough interval, the partial change due to the first one or two is not stable enough to survive before enough photons arrive to make a permanent latent image center.It doesn't specifically say that two photons is needed. It says a few dozen instead.
More modern detectors might be able to detect absorbtion of a single photon.
Post by: alancalverd on 09/07/2021 11:23:42
Post by: hamdani yusuf on 09/07/2021 12:26:28
As I said, a good photocathode will respond to a single photon, but a film won't (note "usually": x-ray film is easily fogged by natural background radiation because it is much more sensitive than ordinary camera film).
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The effectiveness of a photocathode is commonly expressed as quantum efficiency, that being the ratio of emitted electrons vs. impinging quanta (of light). The efficiency varies with construction as well, as it can be improved with a stronger electric field.https://en.wikipedia.org/wiki/Photocathode#Construction
This fact reduces the credibility of claim that each detection in photocathode represents a single photon impact. The temperature dependence of detection rate as I mentioned earlier in photomutliplier only strengthen this suspicion.
Post by: Bored chemist on 09/07/2021 13:03:06
Astronomers used to "pre-fog" their photographic plates to lift the sensitivity above the RF level,So, one presumes, do physicists monitoring low photon fluxes.
Post by: Bored chemist on 09/07/2021 13:04:05
This fact reduces the credibility of claim that each detection in photocathode represents a single photon impact.How?
The temperature dependence of detection rate as I mentioned earlier in photomutliplier only strengthen this suspicion.In what way?
Post by: hamdani yusuf on 09/07/2021 14:11:26
This fact reduces the credibility of claim that each detection in photocathode represents a single photon impact.How?The temperature dependence of detection rate as I mentioned earlier in photomutliplier only strengthen this suspicion.In what way?
Ideally, every photon impingement is followed by a detection on the screen. Other factors such as electric field and temperature shouldn't significantly affect the result.
When the detections are significantly less than the photon, then there must be some undetected photons. On the other hand, when there are more detections than the photons, then there must be some other causes that trigger detections other than photons.
Alternatively, the detections are mainly caused by something else. The energy from light source only adds the chance that the detection is triggered.
The further away from the ideal condition, the result reduces the credibility of photon model and increases the credibility of the alternatives.
Post by: Bored chemist on 09/07/2021 14:32:38
Not every photon initiates a cascade.
But most do.
The size of the pulse is pretty much identical for each photon.
It's very hard to explain that except in terms of a single electron being set free in the first place.
then there must be some other causes that trigger detections other than photons.That's never been in dispute. It's called the "dark current".
Thermal energy of electrons is occasionally high enough to get them to escape (especially for detectors sensitive to longer wavelengths in the IR.
So, you would expect that higher temperatures increase the likelihood of an electron escaping the photocathode without being hit by a (light) photon.
Also, of course things like cosmic rays will occasionally set an electron free and cause a pulse.
The fact that not every event registered by a PMT is due to a photon, does not mean that photons don't cause events.
Post by: alancalverd on 09/07/2021 14:55:43
No. A single spark can ignite an entire petrol refinery or ammunition dump, so the explosion tells you that there has been one spark. Same with most radiation detectors: you rely on one photon event initiating a cascade so you get a big output pulse for each incoming photon. The geiger counter is a particular example where the cascade causes the entire gas cloud to conduct so the output pulse amplitude is independent of the energy of the incident photon. A photomultiplier uses a sequence of photocathodes (cascodes) to produce a fixed energy-independent gain but the output pulse amplitude depends on the energy of the incident photon, which determines how many electrons are emitted from the first interaction.QuoteThe effectiveness of a photocathode is commonly expressed as quantum efficiency, that being the ratio of emitted electrons vs. impinging quanta (of light). The efficiency varies with construction as well, as it can be improved with a stronger electric field.https://en.wikipedia.org/wiki/Photocathode#Construction
This fact reduces the credibility of claim that each detection in photocathode represents a single photon impact. The temperature dependence of detection rate as I mentioned earlier in photomutliplier only strengthen this suspicion.
Temperature dependence of low-energy (photocathode) detectors is simply due to the thermal excitation of conduction band electrons in the cathode. If they have more thermal energy then it requires less energy to expel them from the surface so the gain of the target and each cascode stage increases with temperature. But you still only get one pulse per photon!
Post by: Bored chemist on 09/07/2021 17:26:28
A photomultiplier uses a sequence of photocathodes (cascodes) to produce a fixed energy-independent gain but the output pulse amplitude depends on the energy of the incident photon, which determines how many electrons are emitted from the first interaction.No.
One photo-electron per photon (or near offer).
Your head still seems to be stuck in the x-rays or something.
Post by: hamdani yusuf on 09/07/2021 22:44:33
It's very hard to explain that except in terms of a single electron being set free in the first place.I don't dispute that. In popcorn analogy, I doubt that the activation of detector means that a photon just hit it. It's explained in the end of the video I cited previously.
Post by: hamdani yusuf on 09/07/2021 23:18:19
Temperature dependence of low-energy (photocathode) detectors is simply due to the thermal excitation of conduction band electrons in the cathode. If they have more thermal energy then it requires less energy to expel them from the surface so the gain of the target and each cascode stage increases with temperature. But you still only get one pulse per photon!No. The temperature increases the rate of detection, whether or not the light source is on. The frequency required to increase the detection rate depends on the voltage between electrodes.
Post by: alancalverd on 09/07/2021 23:53:56
One photo-electron per photon (or near offer).If only! The best optical photocathodes have a quantum efficiency of around 30% with a definite wavelength dependence.
Post by: Bored chemist on 10/07/2021 01:27:13
OK, so 30% is the "near offer". Photodiodes do better (but the s/n ratio is worse)One photo-electron per photon (or near offer).If only! The best optical photocathodes have a quantum efficiency of around 30% with a definite wavelength dependence.
However, this
but the output pulse amplitude depends on the energy of the incident photon, which determines how many electrons are emitted from the first interaction.is still wrong- at least for visible light.
If you want to say you get about 0.3 electrons per photon, that's fine. It just means that about 2/3 get absorbed as heat.
But it's pretty much independent of wavelength (and thus, photon energy) in the region where they actually get used.
eg
https://www.sciencedirect.com/science/article/abs/pii/S0168900206009247
You are muddling it with scintillation counters etc. where the output pulse from the PMT is proportional to the gamma photon energy.
Post by: Bored chemist on 10/07/2021 01:28:49
I doubt that the activation of detector means that a photon just hit it.Essentially all the evidence disagrees with you.
Post by: alancalverd on 10/07/2021 09:09:55
Post by: Bored chemist on 10/07/2021 12:08:03
According to Hamamatsu (who know a thing or two about their products) photocathode emission is strongly wavelength-dependentAnd according to you the quantum efficiency is simultaneously less than one
The best optical photocathodes have a quantum efficiency of around 30%And also more than 1
determines how many electrons are emitted from the first interaction.It's never "electrons" i,.e. plural, is it?
Post by: hamdani yusuf on 10/07/2021 14:20:31
According to Hamamatsu (who know a thing or two about their products) photocathode emission is strongly wavelength-dependentThe glass window material contributes significantly to that dependency.
Post by: hamdani yusuf on 10/07/2021 14:29:04
Name one.I doubt that the activation of detector means that a photon just hit it.Essentially all the evidence disagrees with you.
Post by: Bored chemist on 10/07/2021 14:57:51
Einstein's Nobel prize,Name one.I doubt that the activation of detector means that a photon just hit it.Essentially all the evidence disagrees with you.
Post by: hamdani yusuf on 10/07/2021 15:43:30
Einstein's Nobel prize,That's a theoretical work. What's the experimental evidence? Why can't it be interpreted in another way?
Post by: Bored chemist on 10/07/2021 16:06:08
For a start, that's silly.Einstein's Nobel prize,That's a theoretical work. What's the experimental evidence? Why can't it be interpreted in another way?
The experiment on which the work was based is real enough; I did it as a high school experiment.
The evidence for the quantization of light isn't just one experiment.
It's a whole mass of interlocking ideas.
It solves the ultraviolet catastrophe.
It explains the photoelectric effect and, whether you like it ot not, it's the only credible explanation of photomultipliers (and their solid state equivalents- the avalanche photodiodes).
It's also a vital part of photochemistry.
So,
Why can't it be interpreted in another way?Well, go ahead and try.
But make sure that your idea matches all the observations in spectroscopy, electronics, cosmology and so on.
Post by: alancalverd on 10/07/2021 16:23:42
Quote from: alancalverd on Today at 09:09:55The graph I copied is for reflection photocathodes, not for complete photomultipliers or transmission cathodes, where the glass contributes to the overall response.
According to Hamamatsu (who know a thing or two about their products) photocathode emission is strongly wavelength-dependent
The glass window material contributes significantly to that dependency.
Post by: alancalverd on 10/07/2021 16:27:52
Quote from: alancalverd on Yesterday at 23:53:56Suddenly I can't think of an alternative idiomatic phrase that meets the criteria of succinctness and precision. Unless you are prepared to believe that electrons are divisible. "How much of an electron" will only add to HY's confusion!
The best optical photocathodes have a quantum efficiency of around 30%
And also more than 1
Quote from: alancalverd on Yesterday at 14:55:43
determines how many electrons are emitted from the first interaction.
It's never "electrons" i,.e. plural, is it?
Post by: Bored chemist on 10/07/2021 17:09:15
It's not rocket science.Quote from: alancalverd on Yesterday at 23:53:56Suddenly I can't think of an alternative idiomatic phrase that meets the criteria of succinctness and precision. Unless you are prepared to believe that electrons are divisible. "How much of an electron" will only add to HY's confusion!
The best optical photocathodes have a quantum efficiency of around 30%
And also more than 1
Quote from: alancalverd on Yesterday at 14:55:43
determines how many electrons are emitted from the first interaction.
It's never "electrons" i,.e. plural, is it?
The wavelength (among other things) determines the probability of the electron being ejected.
It's about a 1 in 3 chance, but you don't get a third of an electron, (nor do you get 3 electrons).
Post by: hamdani yusuf on 11/07/2021 00:35:33
B is detector showing an activity.
(https://upload.wikimedia.org/wikipedia/commons/thumb/3/39/%D0%92%D0%B5%D0%BD%D0%BE%D0%B2_%D0%B4%D0%B8%D1%98%D0%B0%D0%B3%D1%80%D0%B0%D0%BC.svg/440px-%D0%92%D0%B5%D0%BD%D0%BE%D0%B2_%D0%B4%D0%B8%D1%98%D0%B0%D0%B3%D1%80%D0%B0%D0%BC.svg.png)
We can assume that there are detected photons, depicted by A ∩ B.
We know that the efficiency is less than 100%, so there are undetected photons, depicted by A ∩ ~B.
We also know that temperature can increase detection rate, even when the light source is off. The detections without photon are depicted by ~A ∩ B.
You can define that A=B, but consequently, you must ignore two last cases above.
The electron ejection you described reminds me of Newton's cradle.
Post by: alancalverd on 11/07/2021 12:16:51
temperature can increase detection rate, even when the light source is off.If the light is off there can be no "detection rate" because there is nothing to detect.
The problem of "dark current" is usually resolved by placing a chopper blade in the light beam and using a synchronous detector to subtract the dark current (and other noise) from the signal + noise.
Compensating for the temperature dependence of photon yield is a bit more difficult in many real life applications but you can use a reference beam, temperature stabilisation, or temperature measurement and correction, to get a more consistent measurement.
Post by: Bored chemist on 11/07/2021 13:12:16
You can define that A=B,No, we don't.
If it was then the yield would be exactly 1 and it isn't.
Why did you represent what we say
The electron ejection you described reminds me of Newton's cradle.They both work.
Post by: wolfekeeper on 11/07/2021 16:31:08
Post by: hamdani yusuf on 12/07/2021 02:01:23
Maybe this should be moved to New Theories, it no longer seems to be standard physics. Can classical waves replace the theory of quantum mechanical photons? No. Next question.I agree. If what you mean with classical wave is the one based on Maxwell's equations, it certainly can't explain what we observed. He didn't even know about electron.
I think I'll just create a new thread in new theory section to discuss about this further.
Post by: alancalverd on 12/07/2021 07:56:54
So to answer the question, yes, a simple particle model is an oversimplification, as is a continuous wave model, but we use both to predict (with considerable accuracy) how electromagnetic radiation is generated and what happens when it interacts with stuff.
Post by: hamdani yusuf on 13/07/2021 05:02:16
I don't see any new theory here. We are discussing well-known phenomena that are adequately described by two different mathematical models. The only problem seems to be the recurring misunderstanding that "described by" (physics) doesn't mean "is" (philosophy).You're right. I haven't put any new theory here. I simply pointed out some problems I have with commonly used theory in explaining the behavior of light.
Post by: hamdani yusuf on 13/07/2021 05:54:44
Phonons are quantum mechanical phenomena, just like photons in fact.Do quantum electrodynamics correctly describe how photons interact with macroscopic objects, such as knife edge, single slit and double slit experiment, also their equivalents according to Babinet's principle, i.e. single wire and double wire experiment?
Photons and how they interact with electrons are extremely well described in modern physics by quantum electrodynamics:
https://en.wikipedia.org/wiki/Quantum_electrodynamics
Does it predict the existence of non-diffractive edge and non-diffractive slit?
Post by: alancalverd on 13/07/2021 09:11:45
Post by: hamdani yusuf on 13/07/2021 09:19:51
I'm not even sure you have described any "problems with the theory". We have two mathematical models that describe what actually happens at the macroscopic level, and QED gives us a good insight into photon-electron interactions, leading to probability amplitudes for scattering etc. that do not conflict with observaton.Does the theory describes the criteria, when to use one model, when to use the other?
Do quantum electrodynamics correctly describe how photons interact with macroscopic objects, such as knife edge, single slit and double slit experiment, also their equivalents according to Babinet's principle, i.e. single wire and double wire experiment?
Does it predict the existence of non-diffractive edge and non-diffractive slit?
How does it describe polarization?
If you consider polarisation as a classical electromagnetic phenomenon, there is no "paradox".What makes me wonder is why it was called paradox in the first place if it is clearly explained by classical physics? Have previous generation physicists overlooked it?QuoteAlbert Einstein said of Dirac and polarization,Here is my experiment showing that polarizer does rotate the microwave. You can skip to 3:30.
"Dirac, to whom, in my opinion, we owe the most perfect exposition, logically, of this [quantum] theory, rightly points out that it would probably be difficult, for example, to give a theoretical description of a photon such as would give enough information to enable one to decide whether it will pass a polarizer placed (obliquely) in its way or not." Maxwell's Influence on the Evolution of the Idea of Physical Reality...1931, Ideas and Opinions, p.270
Post by: alancalverd on 13/07/2021 22:51:57
Post by: Bored chemist on 13/07/2021 22:59:04
it is necessarily a mesoscopic, collaborative process that must be modelled by wave analysis.Define mesoscopic in this context.
https://en.wikipedia.org/wiki/Synchrotron_radiation#Polarization_of_synchrotron_radiation
But the important thing is we have mathematical models which work just fine; including for this rather weird case.
So it's not clear what problem the OP is trying to solve.
Post by: alancalverd on 13/07/2021 23:09:27
Quote
Synchrotron radiation is produced when moving particles accelerate, e.g. when electrons move freely in a magnetic field.We aren't dealing with single photon-electron QED interactions but the classical stuff of Maxwell, with the electron movement constrained by the magnetic field. The analysis is similar to a radiofrequency dipole where the alternating current is constrained by a conductor, so that the radiated wave is polarised by the (mesoscopic!) shape of the conductor..
Post by: hamdani yusuf on 14/07/2021 03:55:43
On the other hand, if we think that currently existing theory is not perfect, then there's something we can do to improve it. But first we need to identify its weaknesses/flaws/limitations. We can compare between what the theory predicts and what the observed results are. Are they always identical? If some differences are found, can we identify the causes of those differences?
Post by: hamdani yusuf on 14/07/2021 04:07:09
But the important thing is we have mathematical models which work just fine; including for this rather weird case.Can you demonstrate using your mathematical models that single slit experiment would produce similar pattern as single wire experiment?
So it's not clear what problem the OP is trying to solve.
Can you demonstrate using your mathematical models that double slit experiment would produce similar pattern as double wire experiment?
Can you demonstrate using your mathematical models that shining a laser pointer to a vertically tilted diffraction grating would produce a cone shaped beam?
Does it predict the existence of non-diffractive edge and non-diffractive slit?
What do your mathematical models predict with this proposed experiment?
I have an idea to test the validity of the wave explanation. Let's make the long path of the interfering light beam much longer than the short path, so there's a significant time difference in their arrival at the screen. To balance the intensity, the short path can be reduced further using a filter.
First, the light source is turned on continuously. The single photon production is done using attenuating filter alone. As demonstrated in the video, the screen shows interference pattern.
Second the light source is turned on in short pulses. The pulse width is shorter than time difference between the paths of the light beams. The silence period between the pulses should be randomized to minimize the effect from resonance or echo. If the wave explanation is correct, then the interference pattern should not be observed significantly.
(https://www.thenakedscientists.com/forum/index.php?action=dlattach;topic=68595.0;attach=32168;image)
The half mirrors can be made to reflect more than 50% to compensate longer distance and loss in bottom mirrors.
Post by: alancalverd on 14/07/2021 09:42:14
I have an idea to test the validity of the wave explanation.It's not an explanation! It's a mathematical model of some of the properties of electromagnetic radiation. It (very accurately) predicts the outcome of an experiment.
I can devise an experiment to weigh an elephant, based on the mathematics of levers or hydraulics. It doesn't tell me what an elephant is, or how fast it can run, but it's a damn good model of the effects of gravity on a mammal!
Post by: hamdani yusuf on 14/07/2021 10:20:19
What's the outcome predicted by your mathematical model in following experiments?I have an idea to test the validity of the wave explanation.It's not an explanation! It's a mathematical model of some of the properties of electromagnetic radiation. It (very accurately) predicts the outcome of an experiment.
- Single slit
- double slit
- single wire
- double wire
- vertically tilted diffraction grating?
- non-diffractive edge?
- non-diffractive slit?
Quote
The experiment I proposed was to verify/refute the explanation in the video using wave model.
I can devise an experiment to weigh an elephant, based on the mathematics of levers or hydraulics. It doesn't tell me what an elephant is, or how fast it can run, but it's a damn good model of the effects of gravity on a mammal!
I just found a video explaining this topic pretty convincingly. The best part IMO, is that he made the experiment in real life, not just a simulation.
The result from weighing the elephant can be used to predict if a particular helicopter can be used to lift it.
Post by: Bored chemist on 14/07/2021 10:45:39
What's the outcome predicted by your mathematical model in following experiments?Exactly the same as the observation- or we would have changed the theory.
That's my point; the model works.
Post by: hamdani yusuf on 14/07/2021 12:13:11
What are the results of the observation?What's the outcome predicted by your mathematical model in following experiments?Exactly the same as the observation- or we would have changed the theory.
That's my point; the model works.
What are the results of the calculation?
Are they really the same?
Can you predict the outcomes before performing the experiments, based on the mathematical models alone?
Let's start with the simplest cases, such as single wire diffraction. Let the width of the wire 0.1 mm, wavelength of laser beam 500 nm, distance to a screen 5 m. What the diffraction pattern would look like?
Then tilt the wire 45 degree, top end of the wire is closer to the screen, while bottom end is closer to laser pointer. What the new diffraction pattern would look like?
Science should not be based on blind faith. If we can't proof or demonstrate that something is true, we shouldn't be convinced that it is true. If we can't proof or demonstrate that two things are equal, we shouldn't be convinced that they are equal.
Post by: alancalverd on 14/07/2021 16:13:18
The result from weighing the elephant can be used to predict if a particular helicopter can be used to lift it.But it doesn't tell you what an elephant is, or what it will do next.
Post by: alancalverd on 14/07/2021 16:25:53
Can you predict the outcomes before performing the experiments, based on the mathematical models alone?Yes. You can use Babinet's principle to predict it from the slit experiment, and IIRC from my schooldays, it works.
Let's start with the simplest cases, such as single wire diffraction. Let the width of the wire 0.1 mm, wavelength of laser beam 500 nm, distance to a screen 5 m. What the diffraction pattern would look like?
Post by: Bored chemist on 14/07/2021 18:22:10
Science should not be based on blind faith.Which do you think happened first?
The experiment, or the model?
Post by: hamdani yusuf on 14/07/2021 23:16:52
But it doesn't tell you what an elephant is, or what it will do next.That would need a more complex mathematical model and instrumentation. Although it's not an impossible task.
Nothing in my experiments is as complex as elephant nervous system. Some simple mathematical models should be enough.
Post by: hamdani yusuf on 14/07/2021 23:59:11
Yes. You can use Babinet's principle to predict it from the slit experiment, and IIRC from my schooldays, it works.Is Babinet principle derived from the mathematical models, or is it an additional postulate?
Post by: hamdani yusuf on 15/07/2021 00:19:35
Like many other memes, scientific theories evolve through random mutation and natural selection. Scientists imagine/propose new assumptions or modification of existing theories as hypotheses. Experiments are then used to select compatible hypotheses which will survive to face further tests. Internal selection by the scientists themselves before publications may make the mutations look less random.Science should not be based on blind faith.Which do you think happened first?
The experiment, or the model?
A good mathematical model can be extrapolated to explain new experiments haven't been done before.
Post by: hamdani yusuf on 15/07/2021 07:35:50
Quote
Geoff Pynn (Northern Illinois University) gets you started on the critical thinking journey. He tells you what critical thinking is, what an argument is, and what the difference between a deductive and an ampliative argument is.Critical thinking is essential in doing a good scientific research.
Post by: Bored chemist on 15/07/2021 08:29:23
Critical thinking is essential in doing a good scientific research
It isn't clear that you answered the question.Like many other memes, scientific theories evolve through random mutation and natural selection. Scientists imagine/propose new assumptions or modification of existing theories as hypotheses. Experiments are then used to select compatible hypotheses which will survive to face further tests. Internal selection by the scientists themselves before publications may make the mutations look less random.Science should not be based on blind faith.Which do you think happened first?
The experiment, or the model?
A good mathematical model can be extrapolated to explain new experiments haven't been done before.
Do you understand that science is largely driven by experiments?
Do you realise that anyone who looked through a fine woven silk cloth saw diffraction patterns?
Do you really think that there was a mathematical model of diffraction before there was an observation of it?
Post by: hamdani yusuf on 15/07/2021 10:14:04
Do you understand that science is largely driven by experiments?AFAIK, scientific progress usually started by observation of surprising phenomena, which produced unexpected result when being analyzed using incumbent theory. The scientists then proposed cempeting hypotheses plausible to explain the observation. They then device some experiments to rule out incorrect hypotheses. It's possible that more than one hypotheses can pass the test because they produce similar result for that particular experimental setup. We just need to device other experiments to amplify the differences to finally get the best hypothesis to be selected as accepted theory.
Do you realise that anyone who looked through a fine woven silk cloth saw diffraction patterns?
Do you really think that there was a mathematical model of diffraction before there was an observation of it?
Seeing thing is much easier than explaining it. People has looked at the stars and found planets since ancient time. But the mathematical model for their movement wasn't clearly resolved until Newton's work.
It's not easy to show diffraction pattern using natural light sources. It takes dedicated work to produce clear and unambiguous results. Invention of mass produced laser pointer which are easy to obtain has changed the situation. Now virtually anyone can do the experiment at home.
It's possible to create a mathematical model to describe a phenomena for a limited condition. Galileo used parabolic equation to describe cannon ball trajectory, which is acceptable for most purposes they were dealing with. But for the situation where curvature of the earth is no longer negligible, we need a better mathematical model.
Post by: hamdani yusuf on 15/07/2021 10:33:55
https://en.wikipedia.org/wiki/Diffraction_from_slits#Single_slit
Quote
As an example, an exact equation can now be derived for the intensity of the diffraction pattern as a function of angle in the case of single-slit diffraction.
A mathematical representation of Huygens' principle can be used to start an equation.
https://en.wikipedia.org/wiki/Huygens%E2%80%93Fresnel_principle
Quote
Modern physics interpretationsWhat's unclear for me though, are equations used in photon model derived from Huygen's principle? Or is it the other way around? If we are to be consistent with the meaning of the word principle, the Huygen's principle should be the basis of reasoning, or the starting assumption, and not derived from other assumptions. But I've seen cases where consistency is not always be complied with.
Not all experts agree that the Huygens' principle is an accurate microscopic representation of reality. For instance, Melvin Schwartz argued that "Huygens' principle actually does give the right answer but for the wrong reasons".[1]
This can be reflected in the following facts:
The microscopic mechanics to create photons and of emission, in general, is essentially acceleration of electrons.[1]
The original analysis of Huygens [10] included amplitudes only it does not include neither phases, neither waves propagating at different speeds (due to diffraction within continuous media) and therefore does not take into account interference.
The Huygens analysis also does not include polarization for light which imply a vector potential, where instead sound waves can be described with a scalar potential and there is no unique and natural translation between the two.[11]
In the Huygens description there is no explanation of why we choose only the forward going i.e. Retarded wave or forward envelope of wave fronts, versus the backward propagating advanced wave i.e. backward envelope.[11]
In the Fresnel approximation there is a concept of non-local behavior due to the sum of spherical waves with different phases that comes from the different points of the wave front, and non local theories are subject of many debates (e.g. not being Lorentz covariant) and of active research.
The Fresnel approximation can be interpreted in a quantum probabilistic manner but is unclear how much this sum of states (i.e. wavelets on the wavefront) is a complete list of states that are meaningful physically or represents more of an approximation on a generic basis like in the linear combination of atomic orbitals (LCAO) method.
The Huygens' principle is essentially compatible with quantum field theory in the far field approximation, considering effective fields in the center of scattering, considering small perturbations, and in the same sense that quantum optics is compatible with classical optics, other interpretations are subject of debates and active research.
The Feynman model where every point in an imaginary wave front as large as the room is generating a wavelet, shall also be interpreted in these approximations [12] and in a probabilistic context, in this context remote points can only contribute minimally to the overall probability amplitude.
Quantum field theory does not include any microscopic model for photon creation and the concept of single photon is also put under scrutiny on a theoretical level.
Post by: alancalverd on 15/07/2021 10:52:48
No amount of modelling or instrumentation will tell you what an elephant is. You can use them to extrapolate what an elephant is likely to do next, or interpolate whether it meets the criteria we have assigned to living, dead, grey, mammal, etc., but what it is, is determined by arbitrary human categorisation of things that look and behave similarly. There isn't even a rigid scientific definition of "species" - it's just a convenient box in which to look for large grey animals with prehensile noses. Not that we have a consistent definition of animal....But it doesn't tell you what an elephant is, or what it will do next.That would need a more complex mathematical model and instrumentation. Although it's not an impossible task.
Nothing in my experiments is as complex as elephant nervous system. Some simple mathematical models should be enough.
Post by: alancalverd on 15/07/2021 10:59:43
In most physics textbook I read, explanation for diffraction is derived from Huygens' principle.It's a very good predictive model, not an explanation. That is because it is dependent on a wave and wavelet synthesis of the propagation of light, which is all very well but doesn't consist with the particles you need to model photonuclear phenomena like pair annihilation gamma radiation, or even film photography!
Post by: alancalverd on 15/07/2021 11:02:33
the Huygen's principle should be the basis of reasoning, or the starting assumptionIt's a starting axiom for wavelet analysis and synthesis, and works pretty well for all sorts of waves. But it simply doesn't explain quantum phenomena.
Post by: hamdani yusuf on 15/07/2021 11:09:50
No amount of modelling or instrumentation will tell you what an elephant is. You can use them to extrapolate what an elephant is likely to do next, or interpolate whether it meets the criteria we have assigned to living, dead, grey, mammal, etc., but what it is, is determined by arbitrary human categorisation of things that look and behave similarly. There isn't even a rigid scientific definition of "species" - it's just a convenient box in which to look for large grey animals with prehensile noses. Not that we have a consistent definition of animal....Human minds are a form of model to represent some part of objective reality. Their finiteness inevitably causes imprecision, which leaves some uncertainty. Arbitrary things can be agreed through convention.
Post by: hamdani yusuf on 15/07/2021 11:12:30
Why you need particle model to explain photonuclear phenomena like pair annihilation gamma radiation, or even film photography?In most physics textbook I read, explanation for diffraction is derived from Huygens' principle.It's a very good predictive model, not an explanation. That is because it is dependent on a wave and wavelet synthesis of the propagation of light, which is all very well but doesn't consist with the particles you need to model photonuclear phenomena like pair annihilation gamma radiation, or even film photography!
Post by: hamdani yusuf on 15/07/2021 11:16:28
I've mentioned that classical wave model isn't adequate to explain quantum phenomena. But it doesn't rule out non-classical wave models. And I've also mentioned that I'll discuss them in new theory section, not here.the Huygen's principle should be the basis of reasoning, or the starting assumptionIt's a starting axiom for wavelet analysis and synthesis, and works pretty well for all sorts of waves. But it simply doesn't explain quantum phenomena.
Post by: Bored chemist on 15/07/2021 11:20:24
OK, so you now recognise that the observations came first (thousands of years earlier) and the models came second.Do you understand that science is largely driven by experiments?AFAIK, scientific progress usually started by observation of surprising phenomena, which produced unexpected result when being analyzed using incumbent theory. The scientists then proposed cempeting hypotheses plausible to explain the observation. They then device some experiments to rule out incorrect hypotheses. It's possible that more than one hypotheses can pass the test because they produce similar result for that particular experimental setup. We just need to device other experiments to amplify the differences to finally get the best hypothesis to be selected as accepted theory.
Do you realise that anyone who looked through a fine woven silk cloth saw diffraction patterns?
Do you really think that there was a mathematical model of diffraction before there was an observation of it?
Seeing thing is much easier than explaining it. People has looked at the stars and found planets since ancient time. But the mathematical model for their movement wasn't clearly resolved until Newton's work.
It's not easy to show diffraction pattern using natural light sources. It takes dedicated work to produce clear and unambiguous results. Invention of mass produced laser pointer which are easy to obtain has changed the situation. Now virtually anyone can do the experiment at home.
It's possible to create a mathematical model to describe a phenomena for a limited condition. Galileo used parabolic equation to describe cannon ball trajectory, which is acceptable for most purposes they were dealing with. But for the situation where curvature of the earth is no longer negligible, we need a better mathematical model.
So, the next question is, do you think that when people like Thomas Young, Joseph von Fraunhofer and Christiaan Huygens came up with theories, they didn't actually test them against experimental observation?
Post by: hamdani yusuf on 15/07/2021 12:54:17
OK, so you now recognise that the observations came first (thousands of years earlier) and the models came second.They clearly didn't test their model against all possible experimental setups. They seems to use simplest setups only.
So, the next question is, do you think that when people like Thomas Young, Joseph von Fraunhofer and Christiaan Huygens came up with theories, they didn't actually test them against experimental observation?
Have you seen any literature describing single slit or double slit experiment where the light source is not at normal angle to the aperture?
It's like Galileo's parabolic trajectory which only work when the effect of earth curvature is negligible.
Post by: Bored chemist on 15/07/2021 12:59:15
Have you seen any literature describing single slit or double slit experiment where the light source is not at normal angle to the aperture?Essentially all of them.
The beam is almost always divergent, so the top and bottom of it are not perpendicular to the slit.
It doesn't seem to matter much.
Post by: hamdani yusuf on 15/07/2021 13:07:26
(https://upload.wikimedia.org/wikipedia/commons/thumb/4/4e/Doubleslit.svg/1024px-Doubleslit.svg.png)
If we use particle model, what's the effect of removing the plates above and below the middle plate?
The middle plate alone should be enough to split the trajectory of the photons.
Post by: hamdani yusuf on 15/07/2021 13:10:18
Can you show me one?Have you seen any literature describing single slit or double slit experiment where the light source is not at normal angle to the aperture?Essentially all of them.
The beam is almost always divergent, so the top and bottom of it are not perpendicular to the slit.
It doesn't seem to matter much.
Post by: alancalverd on 15/07/2021 15:07:03
Apparently Newton (Trinity, Cambridge) favored the corpuscular theory whilst his enemy Hooke (Gresham, London) was a wave man. I can't see how you can explain the rings with corpuscles, but Trinity College has produced more Nobel prizewinners than France, so I'm probably out of my depth here.
Post by: Bored chemist on 15/07/2021 15:23:18
Can you show me one where the beam is not divergent?Can you show me one?Have you seen any literature describing single slit or double slit experiment where the light source is not at normal angle to the aperture?Essentially all of them.
The beam is almost always divergent, so the top and bottom of it are not perpendicular to the slit.
It doesn't seem to matter much.
For example the typical specification for a HeNe laser of the type used in physics demos has a divergence of about 1 milliradian.
Since the beam is divergent the top and bottom of the bean can't simultaneously be perpendicular to a straight wire.
Post by: hamdani yusuf on 15/07/2021 22:37:43
Ok. Now we agree that there's no publicized experiment on diffraction where the light beam from the source has deviation from normal by around 45 degrees or more.Can you show me one where the beam is not divergent?Can you show me one?Have you seen any literature describing single slit or double slit experiment where the light source is not at normal angle to the aperture?Essentially all of them.
The beam is almost always divergent, so the top and bottom of it are not perpendicular to the slit.
It doesn't seem to matter much.
For example the typical specification for a HeNe laser of the type used in physics demos has a divergence of about 1 milliradian.
Since the beam is divergent the top and bottom of the bean can't simultaneously be perpendicular to a straight wire.
Post by: Bored chemist on 15/07/2021 22:53:33
For a start, no, we can't agree about that- because nobody looked.Ok. Now we agree that there's no publicized experiment on diffraction where the light beam from the source has deviation from normal by around 45 degrees or more.Can you show me one where the beam is not divergent?Can you show me one?Have you seen any literature describing single slit or double slit experiment where the light source is not at normal angle to the aperture?Essentially all of them.
The beam is almost always divergent, so the top and bottom of it are not perpendicular to the slit.
It doesn't seem to matter much.
For example the typical specification for a HeNe laser of the type used in physics demos has a divergence of about 1 milliradian.
Since the beam is divergent the top and bottom of the bean can't simultaneously be perpendicular to a straight wire.
There may be published work in the field. How would we know?
Define "published"
I tried the experiment earlier and it didn't seem to make any difference.
I am publishing that result here now.
If I get time I might set it up better at the weekend + get pictures.
I'm not claiming to be the first to have done it- as I said, anyone who looked through fine silk...
Post by: hamdani yusuf on 16/07/2021 00:05:26
Post by: hamdani yusuf on 16/07/2021 00:12:59
Define "published"you can put the hyperlink here so everyone can see the result. A picture worths a thousand words, so it clearly helps. Better yet, if it's a video, or moving pictures, so we can see from different perspectives, and information related to passing time is not completely removed, such as the case with still pictures.
I tried the experiment earlier and it didn't seem to make any difference.
I am publishing that result here now.
Post by: hamdani yusuf on 16/07/2021 05:51:36
Here are some experimental observations I uploaded to Youtube. I think you can easily reproduce them to make sure that they are not misleading tricks.
Investigation on Diffraction of light 4 : Non-diffractive obstacle
Investigation on Diffraction of Light 9 : Horizontally Tilted
Investigation on Diffraction of Light 10 : Vertically Tilted Obstacle
Investigation on Diffraction of Light 13 : Non-Diffractive slit; A Challenge to Huygen's Principle
Post by: evan_au on 16/07/2021 09:26:29
Quote from: Hamdani yusuf, Reply #91
How do you split a single photon?There are ways of doing it using non-linear optics. For conservation of energy, the two photons each have lower energy, with a total energy equal to the original photon.
You can also combine two photons into a single photon using non-linear materials, producing a green laser beam from an infra-red laser source.
But the double-slit experiment also works in a vacuum, ie using linear optics.
See: https://en.wikipedia.org/wiki/Nonlinear_optics#Frequency-mixing_processes
Quote from: Kryptid, Reply #61
what would make (light) different from a gravitational wave?One aspect that is very different is in how we detect them.
- Light power from a star follows an inverse square law, so if you double the sensitivity of your telescope (2xArea), you can detect a source 1.4x as far away, ie in a volume of space 2.8 times bigger.
- As I understand it, our current gravitational wave detectors do not detect power, but the amplitude of stress (power being proportional to the square of stress). That means they follow an inverse law (not inverse square law). If you double the sensitivity of your detector, you can detect an object twice as far away, ie in a volume of space 8 times bigger.
The electric-field amplitude of light from a star decreases proportional to distance, but as the power is the square of the amplitude, you get an inverse square law.
Post by: hamdani yusuf on 16/07/2021 11:14:47
See: https://en.wikipedia.org/wiki/Nonlinear_optics#Frequency-mixing_processes
The crystal is supposed to be macroscopic, and consists of many atoms. How can a single photon know where to go after interacting with it, instead of scattered at random direction?
(https://upload.wikimedia.org/wikipedia/commons/thumb/2/2a/Frequency_doubling_with_perfect_pase_matching.gif/400px-Frequency_doubling_with_perfect_pase_matching.gif)
Post by: Bored chemist on 16/07/2021 12:21:09
https://en.wikipedia.org/wiki/Virtual_state
Post by: acsinuk on 16/07/2021 13:07:01
Post by: Bored chemist on 16/07/2021 13:18:38
Photons need a full cubic volume of space to move through. If there is an angle of deflection then real energy will be mixed with virtual/imaginary/quantum magnoflux light.[ Invalid Attachment ]
Post by: Colin2B on 16/07/2021 14:04:34
Photons need a full cubic volume of space to move through. If there is an angle of deflection then real energy will be mixed with virtual/imaginary/quantum magnoflux light.Please refrain from posting new theories in the main section. Continuing to do so may result in limitations on your posting rights.
Post by: Bored chemist on 16/07/2021 16:36:55
Here's the rig. There's a screen off to the right.
[ Invalid Attachment ]
Post by: Bored chemist on 16/07/2021 16:38:37
slit expte 3 resized.jpg (50.96 kB . 746x322 - viewed 2274 times)
Post by: Bored chemist on 16/07/2021 16:41:29
slit expte 2 resized.jpg (44.04 kB . 775x302 - viewed 3612 times)As you can see it curves, but it's essentially the same.
Post by: Bored chemist on 16/07/2021 17:04:37
Again, it's curved as expected.
[ Invalid Attachment ] .
Post by: hamdani yusuf on 17/07/2021 01:12:48
The QM interpretation of things like Raman spectra, frequency mixing and doubling involve these.At some point in the past, this was a new theory. And prior to that, it was merely a hypothesis. It survives because there's no better alternatives to explain known observations. Yet.
https://en.wikipedia.org/wiki/Virtual_state
Post by: evan_au on 17/07/2021 01:28:04
Quote from: hamdani yusuf
The crystal is supposed to be macroscopic, and consists of many atoms. How can a single photon know where to go after interacting with it, instead of scattered at random direction?I expect it would be a diffraction effect, as a result of interacting with many atoms.
- One of the ways to generate entangled photons is to use "spontaneous parametric down-conversion"
- There is a different diffraction pattern for vertically vs horizontally-polarised photons
- Each pattern is a circle
- Where the two circles overlap you are most likely to find the entangled photons.
- As per normal quantum effects, you can't predict where an individual photon will appear before you detect it...
See the diagram here: https://en.wikipedia.org/wiki/Spontaneous_parametric_down-conversion#Applications
Post by: hamdani yusuf on 17/07/2021 01:39:20
This is the comparable image using a diffraction grating (rather than a single slit) at an angle to the beam.What's the expectation based on? Is it derived from Huygen's principle? How?
Again, it's curved as expected.
Post by: hamdani yusuf on 17/07/2021 02:13:53
I expect it would be a diffraction effect, as a result of interacting with many atoms.It would require a photon to interact with many atoms at once. Consequently, we are forced to reject locality. And the distance can be very large for lower frequency light.
Experiment using microwave and radio frequency would require a photon to interact with extremely large number of atoms over large volume of space. Some people don't mind to accept this concepts. Some others find it hard to believe, and try to look for alternatives.
Post by: hamdani yusuf on 17/07/2021 02:36:14
- One of the ways to generate entangled photons is to use "spontaneous parametric down-conversion"What's the mechanism? How the photons know where to go?
Post by: hamdani yusuf on 17/07/2021 02:44:15
And here's the more interesting one.You can get full circle with higher inclination angle, as close as possible to 90 degree.
As you can see it curves, but it's essentially the same.
Post by: alancalverd on 17/07/2021 10:43:04
Experiment using microwave and radio frequency would require a photon to interact with extremely large number of atoms over large volume of space. Some people don't mind to accept this concepts. Some others find it hard to believe, and try to look for alternatives.Which is why we associate a wavelength λ = hc/E with any photon of energy E. The concept has been around since 1900, and where E is very small (as in radio waves) we tend to use wavelength and frequency to describe the phenomena we most often observe.
Post by: hamdani yusuf on 17/07/2021 11:15:52
Is there an experiment demonstrating particle behavior of radio or microwave?Experiment using microwave and radio frequency would require a photon to interact with extremely large number of atoms over large volume of space. Some people don't mind to accept this concepts. Some others find it hard to believe, and try to look for alternatives.Which is why we associate a wavelength λ = hc/E with any photon of energy E. The concept has been around since 1900, and where E is very small (as in radio waves) we tend to use wavelength and frequency to describe the phenomena we most often observe.
Post by: Bored chemist on 17/07/2021 11:42:52
Is there an experiment demonstrating particle behavior of radio or microwave?Countless "experiments" are done with this every day.
https://en.wikipedia.org/wiki/Nuclear_magnetic_resonance
https://en.wikipedia.org/wiki/Electron_paramagnetic_resonance
Post by: Bored chemist on 17/07/2021 11:44:47
It would require a photon to interact with many atoms at once. Consequently, we are forced to reject locality.No.
Nobody ever said how big a photon is.
It turns out that they are soft.
You can squeeze them through small holes, or you can stretch them out across diffraction gratings.
So their size is not well defined.
Post by: Bored chemist on 17/07/2021 11:53:57
To be honest, I don't really know. It seems reasonable to me that you get a curve because, if you imagine looking at a grating in this way, you would see it as foreshortened more at one side than the other. I realise that's not a clear explanation.This is the comparable image using a diffraction grating (rather than a single slit) at an angle to the beam.What's the expectation based on? Is it derived from Huygen's principle? How?
Again, it's curved as expected.
I don't have access to this paper but, based on the abstract, I think it covers the maths you want.
https://aip.scitation.org/doi/pdf/10.1063/1.347388
Personally, I'm prepared to leave it at that.
I know someone else can do the maths, so I accept that the maths works.
You may wish to gain access to the paper to check.
If anyone reading this has (free) access, I'd be grateful if they could check on whether it explains what I think it does.
Post by: Bored chemist on 17/07/2021 11:59:08
https://www.spiedigitallibrary.org/journals/optical-engineering/volume-58/issue-08/087105/Understanding-diffraction-grating-behavior--including-conical-diffraction-and-Rayleigh/10.1117/1.OE.58.8.087105.full?SSO=1&tab=ArticleLinkFigureTable
Post by: hamdani yusuf on 17/07/2021 12:32:07
If anyone reading this has (free) access, I'd be grateful if they could check on whether it explains what I think it does.It's unfortunate that some basic stuff like this are not opened to public. It should be part of physics textbooks on optics.
Thanks anyway for your efforts.
Post by: hamdani yusuf on 17/07/2021 13:04:00
How do they differentiate between the particle behavior of photon and energy transfer among discrete number of resonators?Is there an experiment demonstrating particle behavior of radio or microwave?Countless "experiments" are done with this every day.
https://en.wikipedia.org/wiki/Nuclear_magnetic_resonance
https://en.wikipedia.org/wiki/Electron_paramagnetic_resonance
Post by: Bored chemist on 17/07/2021 15:00:58
I doubt they bother.How do they differentiate between the particle behavior of photon and energy transfer among discrete number of resonators?Is there an experiment demonstrating particle behavior of radio or microwave?Countless "experiments" are done with this every day.
https://en.wikipedia.org/wiki/Nuclear_magnetic_resonance
https://en.wikipedia.org/wiki/Electron_paramagnetic_resonance
But things like the intensity of the absorptions as a function of temperature only make sense if the energy involved is quantised.
Post by: alancalverd on 17/07/2021 15:09:40
Post by: acsinuk on 19/07/2021 15:32:57
Post by: Bored chemist on 19/07/2021 22:10:35
I found that pulsed EMR on Wiki " https://en.wikipedia.org/wiki/Pulsed_electron_paramagnetic_resonance " which shows how a standard microwave cooker worksIt does not show how a microwave over works.
EPR spectra are only obtained for free radical species and food doesn't generally contain many.
On the other hand, a microwave oven uses the absorbance of microwaves (at about 2.4 GHz) by water- there is lots of water in most food.
The page you cited also does not mention "manganflux", because that's bollocks which you made up.
So is
virtual quantum light
Post by: alancalverd on 20/07/2021 00:23:55
Post by: Eternal Student on 20/07/2021 13:31:28
How old are you? Don't look into a microwave oven, especially when it's turned on. Appliances may have been without leaks when they were new but the wire mesh in the glass door can become corroded.
Also, it won't make your dinner cook any faster.
Here's another microwave question that may take a minute to think about: Do most ovens have reflective surfaces and bounce the radiation around inside them, or does the radiation just tend to be absorbed and wasted in the sides of the oven?
Specifically, if 1 pot of noodles can be cooked in 1 minute, then will it take 2 minutes to cook 2 pots simultaneously, or just 1 minute since both pots are being showered by the same intensity of radiation from above (assuming the microwave generator is at the top of the oven)?
In practice I can heat 1 pot in 1 minute and 2 pots in about 1.5 minutes - a sort of hybrid of the two possibilities.
Best wishes.
Post by: hamdani yusuf on 20/07/2021 14:06:36
Hi Alancalverd,You can see in his profile.
How old are you?
Post by: hamdani yusuf on 20/07/2021 16:11:53
Here's another microwave question that may take a minute to think about: Do most ovens have reflective surfaces and bounce the radiation around inside them, or does the radiation just tend to be absorbed and wasted in the sides of the oven?When the microwave is reflected by the metallic surface of the box, only a small part is being absorbed.
I've tried to heat deionized water inside a metal container, and covered with a metal grating. The water got warm very slowly. The inner wall of the microwave was still cool. Eventually the fuse in the power supply blew up. It means that some portion of the reflected microwave energy comes back to the magnetron.
Post by: Eternal Student on 20/07/2021 16:49:24
Can I add another safety warning to Hamdani's latest comment: Please don't try this at home.
Microwave ovens usually generate a high voltage where the microwave radiation is produced (e.g. the top of the microwave) and a maintain a lower voltage (usually a ground) near the bottom. Putting a metallic object into the microwave that reduces the gap between these two regions can result in electical arcing.
Add to this the reflection of the microwaves off the metallic object and the power that is returned to the microwave generator and you will often get the the fuse to blow and/or break the oven in some way.
Post by: hamdani yusuf on 21/07/2021 03:43:15
Hi.
Can I add another safety warning to Hamdani's latest comment: Please don't try this at home.
Microwave ovens usually generate a high voltage where the microwave radiation is produced (e.g. the top of the microwave) and a maintain a lower voltage (usually a ground) near the bottom. Putting a metallic object into the microwave that reduces the gap between these two regions can result in electical arcing.
Add to this the reflection of the microwaves off the metallic object and the power that is returned to the microwave generator and you will often get the the fuse to blow and/or break the oven in some way.
We can see many videos on Youtube showing how microwave ovens work, and various experiments with them. They did it so we don't have to. I only do my own experiment when I can't find the answer from existing videos. I also learned some potential dangers related to operating microwave oven.
AFAIK, the whole inner wall metallic surface of microwave oven is connected to ground. No significant voltage difference between bottom part and top part. The only high voltage supply is the feed of magnetron which comes from a step up transformer. There is a waveguide between magnetron and the cooking chamber, isolating it from high voltage.
(https://th.bing.com/th/id/R.243f464fef6e530b017d915609fc1921?rik=j0BZdnGStKG2Uw&riu=http%3a%2f%2fcdn.preterhuman.net%2ftexts%2fgovernment_information%2fintelligence_and_espionage%2fhomebrew.military.and.espionage.electronics%2fservv89pn0aj.sn.sourcedns.com%2f_gbpprorg%2fmil%2fherf1%2ffig-8.gif&ehk=f8HRW58B5mSp6vDV9rDfUzvFi3H01l3ZyeOy0CSOJN4%3d&risl=&pid=ImgRaw)
(https://qph.fs.quoracdn.net/main-qimg-9551b7f8f4cf8610e2ff2c8a8c6e303b)
Post by: Eternal Student on 21/07/2021 12:18:02
I hope you are well and it sounds like you are being sensible.
My old microwave oven has an interior that is mainly plastic (although there must be some metal or at least a metal mesh underneath most of this). There is a metal spindle that comes through the bottom (floor) and rotates a plate. I haven't tested it but that metal spindle is probably connected to ground. I also don't know for certain where the transformer or microwave generator is located but my oven is supposed to produce the microwaves at the top. The advice about not putting metallic objects into the microwave oven comes from the users guide and the risk of electrical arcing from the top of the oven to the bottom was mentioned. Exact details probably depend on the oven and I have seen some industrial kitchens using metallic bowls inside their microwave ovens.
Post by: Bored chemist on 21/07/2021 12:20:37
Do most ovens have reflective surfaces and bounce the radiation around inside them,Yes.
Post by: Bored chemist on 21/07/2021 12:23:08
You may find it informative to consider how Hertz first detected radio wavesHi.
Can I add another safety warning to Hamdani's latest comment: Please don't try this at home.
Microwave ovens usually generate a high voltage where the microwave radiation is produced (e.g. the top of the microwave) and a maintain a lower voltage (usually a ground) near the bottom. Putting a metallic object into the microwave that reduces the gap between these two regions can result in electical arcing.
Add to this the reflection of the microwaves off the metallic object and the power that is returned to the microwave generator and you will often get the the fuse to blow and/or break the oven in some way.
We can see many videos on Youtube showing how microwave ovens work, and various experiments with them. They did it so we don't have to. I only do my own experiment when I can't find the answer from existing videos. I also learned some potential dangers related to operating microwave oven.
AFAIK, the whole inner wall metallic surface of microwave oven is connected to ground. No significant voltage difference between bottom part and top part. The only high voltage supply is the feed of magnetron which comes from a step up transformer. There is a waveguide between magnetron and the cooking chamber, isolating it from high voltage.
(https://th.bing.com/th/id/R.243f464fef6e530b017d915609fc1921?rik=j0BZdnGStKG2Uw&riu=http%3a%2f%2fcdn.preterhuman.net%2ftexts%2fgovernment_information%2fintelligence_and_espionage%2fhomebrew.military.and.espionage.electronics%2fservv89pn0aj.sn.sourcedns.com%2f_gbpprorg%2fmil%2fherf1%2ffig-8.gif&ehk=f8HRW58B5mSp6vDV9rDfUzvFi3H01l3ZyeOy0CSOJN4%3d&risl=&pid=ImgRaw)
(https://qph.fs.quoracdn.net/main-qimg-9551b7f8f4cf8610e2ff2c8a8c6e303b)
https://en.wikipedia.org/wiki/Heinrich_Hertz#/media/File:Hertz_micrometer_resonator.png
Post by: yor_on on 21/07/2021 13:55:12
All models are an oversimplification, which is why they are called models. A dead mouse is a perfect model of a live mouse, but only for an inifintesimal perod of time.
If you tie a string to its head and one each to the legs you can extend the model. A good name for this would be string theory.
That was evil Jeffrey
Post by: yor_on on 21/07/2021 13:57:08
Einstein's Nobel prize,That's a theoretical work. What's the experimental evidence? Why can't it be interpreted in another way?
awh, ever heard about black body radiation?
Post by: Europa on 21/07/2021 15:27:48
In most theories up to the eighteenth century, light was pictured as being made up of particles. Since particle models cannot easily account for the refraction, polarization, diffraction and birefringence of light.The model is a theory and as such it is neither real nor fake. In order to answer your question more info would be needed.
Photon model gains reputation in early 20th century due to photoelectric effect, especially that low frequency light cannot poduce emitted electron even with increased intensity. But newer development of laser can generate high enough intensity which can poduce emitted electron even with lower frequency than usual threshold.
Is photon model of electromagnetic wave an oversimplification?
Is it possible to explain wave-like phenomena in electromagnetism using photon model?
Is it possible to explain particle-like phenomena in electromagnetism using wave model?
Post by: acsinuk on 21/07/2021 16:03:21
The heating power is held within the spinning magnetic high frequency photons and has nothing to do with electron particles.
Post by: Bored chemist on 21/07/2021 16:13:09
Looking at the wiring diagram we can see that the metal base is bonded to the DC power supply positive anodeYes
and the negative filament is 5,000 volts negative with an isolated/insulated AC filament heaterYes
filament heater inside a vacuum enclosure with a window into the waveguide.Well- you can't see that from the diagram, but it's correct.
which contains the magnetic massless spinning magnofluxAnd there's where you descend into talking bollocks again.
The heating power is held within the spinning magnetic high frequency photons and has nothing to do with electron particles.The likes of Ampere, Faraday and Gauss did the original experiments which show that electricity and magnetism are fundamentally linked.
So your statement is wrong because it requires them to work independently.
Why do you keep doing this?
Do you enjoy getting laughed at?
Post by: hamdani yusuf on 21/07/2021 22:12:50
Einstein's Nobel prize,That's a theoretical work. What's the experimental evidence? Why can't it be interpreted in another way?
awh, ever heard about black body radiation?
https://en.m.wikipedia.org/wiki/Photon
Quote
While trying to explain how matter and electromagnetic radiation could be in thermal equilibrium with one another, Planck proposed that the energy stored within a material object should be regarded as composed of an integer number of discrete, equal-sized parts.So, it's an indirect evidence. In formula E=n.h.f, the only variable required to be an integer is n. E and f can have any real value.
Millikan's experiment shows that electric charge is quantized. Previously, J. J. Thomson have shown the ratio between mass and charge of electrons. Hence the mass of matter is also quantized. So, it's still possible to interpret the quantization of energy generation and absorption as due to quantization of mass and electric charge, instead of photon as a physical entity. When formulating his law, Planck himself didn't think that photon is a real particle.
Post by: alancalverd on 22/07/2021 10:48:48
How the photon model explains the effect of removing the outer edges (which makes it a thin wire diffraction, which according to Babinet’s principle, should produce the same pattern as single slit aperture)?To repeat, for the nth time, it doesn't. Which is why we also have a wave model.
How the photon model explains the result of circular pattern when the diffraction grating is vertically tilted?
Post by: alancalverd on 22/07/2021 10:58:51
Hi Alancalverd,Old enough to never eat pot noodles (life is too short) and a good enough physicist to realise that visible photons pass through the holes in the mesh so you can see what's going on in a microwave oven without frying your eyeballs.
How old are you?
Apropos heating times, if you have a strong absorber (water) in the oven along with a weak one (ice) you can end up boiling the water without significantly melting the ice, which is why "defrost" programs are pulsed - you have to warm the water in your partly-defrosted chicken and let it melt a bit of the ice by conduction, then repeat.
Post by: acsinuk on 22/07/2021 11:20:43
But we know it can be quantized so the formula needs to be Cosine ϴ for real power and Sin ϴ for virtual power.
The maths is then correct but not the physics, as if the photons are attracted forward by a voltage in the z direction then both the current in x direction and flux in y direction must be spinning at right angles to create a volume pulse of energy per second
Post by: alancalverd on 22/07/2021 12:34:16
Post by: Bored chemist on 22/07/2021 12:51:39
Hamdani, Power is energy/second so therefore photon power is P= hf/second.What are you on, and can you get it on prescription?
But we know it can be quantized so the formula needs to be Cosine ϴ for real power and Sin ϴ for virtual power.
The maths is then correct but not the physics, as if the photons are attracted forward by a voltage in the z direction then both the current in x direction and flux in y direction must be spinning at right angles to create a volume pulse of energy per second
Post by: hamdani yusuf on 22/07/2021 14:10:51
Some light phenomena are easier to explain using wave models, some others are easier to explain using particle models. There are many different wave models known in physics. Some of them are compatible with some light phenomena, some of them don't.How the photon model explains the effect of removing the outer edges (which makes it a thin wire diffraction, which according to Babinet’s principle, should produce the same pattern as single slit aperture)?To repeat, for the nth time, it doesn't. Which is why we also have a wave model.
How the photon model explains the result of circular pattern when the diffraction grating is vertically tilted?
If it turns out that a kind of wave model can be used to explain all of observed phenomena, the use of other models, either wave or particle, would be unnecessary.
Post by: Colin2B on 22/07/2021 14:42:38
Hamdani, Power is energy/second so therefore photon power is P= hf/second.You have been warned about posting new theories in the main sections. If you continue your posting will be restricted to new theories only.
But we know it can be quantized so the formula needs to be Cosine ϴ for real power and Sin ϴ for virtual power.
The maths is then correct but not the physics, as if the photons are attracted forward by a voltage in the z direction then both the current in x direction and flux in y direction must be spinning at right angles to create a volume pulse of energy per second
Final warning
Post by: alancalverd on 22/07/2021 22:05:56
If it turns out that a kind of wave model can be used to explain all of observed phenomena, the use of other models, either wave or particle, would be unnecessary.But we already know that you can't describe what happens when an electron-positron pair annihilates, or the photoelectric effect, with a wave model.
Post by: hamdani yusuf on 22/07/2021 22:26:45
What's the real thing being observed? What are the assumptions used in the experiment? How do they correlate?If it turns out that a kind of wave model can be used to explain all of observed phenomena, the use of other models, either wave or particle, would be unnecessary.But we already know that you can't describe what happens when an electron-positron pair annihilates, or the photoelectric effect, with a wave model.
Most of us agree that wave models based on Maxwell's equations can't explain them properly. That's why we needed a newer theory.
Post by: hamdani yusuf on 23/07/2021 08:32:44
We can see many videos on Youtube showing how microwave ovens work, and various experiments with them. They did it so we don't have to. I only do my own experiment when I can't find the answer from existing videos. I also learned some potential dangers related to operating microwave oven.Here is an example
Micron thick steel wool in a microwave and how mirrors work
Quote
Super fine steel wool looks beautiful in a microwave. Tough to film but I'm really happy with the shots that I got! Along the way we get into how mirrors work.
Post by: acsinuk on 23/07/2021 10:55:50
But the wire is on a glass plate insulating it from the metal casing so no DC current is flowing. The wire is surely being over excited by the magnetic fields power =hf/second
Post by: Bored chemist on 23/07/2021 11:05:06
The wire is surely being over excited by the magnetic fields powerIt is still no more possible to separate magnetism and electricity in this context that when I pointed it out here
The likes of Ampere, Faraday and Gauss did the original experiments which show that electricity and magnetism are fundamentally linked.
So your statement is wrong because it requires them to work independently.
Why do you keep doing this?
Do you enjoy getting laughed at?
And this
"hf/second"
is still more or less meaningless.
Post by: alancalverd on 23/07/2021 17:14:49
What's the real thing being observed? What are the assumptions used in the experiment? How do they correlate?No assumptions. The key observation is two 511 keV photons being detected simultaneously at 180 degrees to each other.
Most of us agree that wave models based on Maxwell's equations can't explain them properly. That's why we needed a newer theory.
Post by: hamdani yusuf on 24/07/2021 03:53:35
No assumptions. The key observation is two 511 keV photons being detected simultaneously at 180 degrees to each other.What's the objects being observed? How are they set up?
What's the polarization of the light? Is there only a single frequency? Or is it actually a bunch of similar frequencies, like Gaussian?
IMO, a photon with strictly single frequency is not a feasible mathematical model.
Post by: alancalverd on 24/07/2021 10:36:39
Difficult to measure polarisation of a single photon at 511 keV - if indeed the concept is meaningful.
One point of the photon model is that it can have a single energy defined by the causative interaction. E= mc2 ± 0
The simplest demonstration nowadays is to observe the decomposition of a neutron-deficient nuclide like Ga58 in a PET scanner.
Post by: Bored chemist on 24/07/2021 12:08:57
IMO, a photon with strictly single frequency is not a feasible mathematical model.The model we use includes the uncertainty principle. The energy of the photon is not strictly fixed.
Post by: hamdani yusuf on 24/07/2021 12:28:00
You can measure the photon energy with a proportional counter or a biased detector (we used to use GeLi but I've no doubt there are posher ones available these days). AFAIK it's never been anything other than 511 keV with a bit of wiggle depending on the kinetic energy of the annihilating pair (negligible in most cases).Have you counted how many electrons and positrons before and after the annihilation?
Have you made sure that the detection was not caused by other factors, such as cosmic rays or radioactivity of some materials in the apparatus or instrumental devices?
Post by: hamdani yusuf on 24/07/2021 12:42:38
Difficult to measure polarisation of a single photon at 511 keV - if indeed the concept is meaningful.Of course the concept is meaningful. If a polarizer is placed in the path, perpendicular to the polarization state of the light, it won't pass. The concept can produce observable difference.
Post by: Bored chemist on 24/07/2021 12:56:24
f a polarizer is placed in the path,If no such polariser can be made...
Post by: alancalverd on 24/07/2021 14:34:23
We have a rough idea of the collection efficiency of PET scintillators. It's mostly a geometric calculation as 511 keV penetrates the patient with very little loss. Anyway the numbers don't matter much as the practical dose is small in comparison with the other risks associated wth whatever we are trying to diagnose.
Distinguishing from other factors is easy. No patient - measure the background radiation. Cosmic radiation: not much at 511 keV (pulse height discrimination eliminates other energies) and very unlikely to produce two simultaneous pulses at 180 degrees. Fact is that the signal-noise ratio of PET is pretty good, even though its spatial resolution is poor, which is why it is usually combined with CT or MRI to get a good anatomical image of the hotspot.
BC: I'm not sure how the uncertainty principle applies to E = mc2, which is a scalar equation. Δp.Δx is the scalar product of vectors.
Post by: Bored chemist on 24/07/2021 14:38:21
BC: I'm not sure how the uncertainty principle applies to E = mc2,That's OK, lots of people don't.
http://hyperphysics.phy-astr.gsu.edu/hbase/uncer.html
Post by: hamdani yusuf on 24/07/2021 16:54:54
One point of the photon model is that it can have a single energy defined by the causative interaction. E= mc2 ± 0Planck's law only asserts that energy of a system can only change by discrete multiple of electromagnetic radiation frequency times a constant. It doesn't say anything about quantization of frequency. Thus, radiation energy won't be quantized either.
Post by: hamdani yusuf on 24/07/2021 17:47:23
I've seen some sources said that uncertainty principle is not a problem of measurement. Instead, it's a fundamental principle inherently embedded to the math of wave models through Fourier transform.
Post by: Bored chemist on 24/07/2021 18:34:36
I've seen some sources said that uncertainty principle is not a problem of measurement.They are right.
The uncertainty is intrinsic to the property.
Instead, it's a fundamental principle inherently embedded to the math of wave models through Fourier transform.No.
It's a property of the universe.
The momentum of an electron is not an exact quantity.
If you do the calculations via wave mechanics you get the same answer as you do with the FT approach.
Post by: hamdani yusuf on 24/07/2021 22:19:15
No.How do we know that?
It's a property of the universe.
The momentum of an electron is not an exact quantity.
Post by: Bored chemist on 24/07/2021 22:50:48
How do we know that?It works.
Post by: alancalverd on 24/07/2021 22:53:10
Nothing to do with Planck or a closed system. ε+-ε- annihilation is a single event in which a fixed mass is converted into electromagnetic energy. If we repeat the event with the same masses, we must get the same energy. And we do.
Planck's law only asserts that energy of a system can only change by discrete multiple of electromagnetic radiation frequency times a constant. It doesn't say anything about quantization of frequency. Thus, radiation energy won't be quantized either.
Post by: alancalverd on 24/07/2021 22:56:14
And not a single mention of mass-energy equivalence. Though it is indeed an excellent discussion of indeterminacy and its implications for atomic structure.BC: I'm not sure how the uncertainty principle applies to E = mc2,That's OK, lots of people don't.
http://hyperphysics.phy-astr.gsu.edu/hbase/uncer.html
Post by: Bored chemist on 24/07/2021 23:06:24
And not a single mention of mass-energy equivalence.They probably expect you to work that out for yourself.
If E is uncertain, so is M.
How uncertain depends on the lifetime of the particle. For stable particles it's... not a big effect.
Post by: alancalverd on 24/07/2021 23:09:02
Post by: Bored chemist on 24/07/2021 23:16:27
If that gamma is released from positronium then I think you can calculate the effect of its lifetime(s) on the linewidth,
http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/parlif.html
it's not a big effect.
Post by: hamdani yusuf on 24/07/2021 23:17:39
What works? Assuming that momentum of electron is not an exact quantity?How do we know that?It works.
Why assuming that it has exact quantity, but we can't measure precisely, doesn't work as well?
Post by: hamdani yusuf on 24/07/2021 23:20:06
Nothing to do with Planck or a closed system. ε+-ε- annihilation is a single event in which a fixed mass is converted into electromagnetic energy.How do you count the number of electrons and positrons before and after the annihilation?
Post by: hamdani yusuf on 25/07/2021 10:06:35
Post by: alancalverd on 25/07/2021 10:47:32
We know the mass and charge of each species. We know the energy of the emitted photons. E = mc2 and opposite charges annihilate.Nothing to do with Planck or a closed system. ε+-ε- annihilation is a single event in which a fixed mass is converted into electromagnetic energy.How do you count the number of electrons and positrons before and after the annihilation?
It's clear that we can't capture all the emitted photons, but the simultaneous capture of two 511 keV photons at 180 degrees is excellent evidence that 2me has been annihilated, and the net charge of the patient remains stubbornly at zero.
Post by: Bored chemist on 25/07/2021 10:54:35
How do you count the number of electrons and positrons before and after the annihilation?Broadly the same way chemists count atoms.
Re
Post by: alancalverd on 25/07/2021 10:54:53
I wonder how these experiments would be explained using particle models.It would be foolish to try, which is why we use wave models for low energy electromagnetic radiation. Above about 100 keV the particle model is more useful, and in the diagnostic x-ray range (10 - 100 keV) you need to be aware of both models. Crystallographers, for instance, rely on the particle model to collect their data and the wave model to interpret their results.
Post by: alancalverd on 25/07/2021 10:58:29
By weighing the patient? Neat trick, my friend! ;)How do you count the number of electrons and positrons before and after the annihilation?Broadly the same way chemists count atoms.
Re
Post by: hamdani yusuf on 25/07/2021 12:11:03
We know the mass and charge of each species. We know the energy of the emitted photons. E = mc2 and opposite charges annihilate.I think you know how neutrino was discovered. How do you make sure that no neutrino is produced in the process?
Post by: hamdani yusuf on 25/07/2021 12:25:02
It's clear that we can't capture all the emitted photons, but the simultaneous capture of two 511 keV photons at 180 degrees is excellent evidence that 2me has been annihilated, and the net charge of the patient remains stubbornly at zero.Are they always detected simultaneously in the opposite direction? Detector's efficiency is typically less than 100%. Some of the incoming energy are turned into heat. Do those detectors on opposite sides always fail and succeed concurrently? Is it possible for one detector to fail to capture the photon while its counterpart succeed?
Post by: alancalverd on 25/07/2021 13:22:11
(a) the frequent capture of two simultaneous 511 keV photons at 180 degrees corresponds exactly with the theoretical prediction of pair annihilation
(b) it only happens (at least in a PET scanner) when a positron emitter is present
(c) occasional absence of evidence (failed detector) is not evidence of absence (no positrons).
If you found your fingerprint on a sheet of glass, wouldn't you conclude that you had touched it? The fact that you had touched several other things and several other people had touched the glass, is irrelevant. In nuclear physics, finding the predicted particle at the predicted time and place is considered definitive.
Post by: Bored chemist on 25/07/2021 14:04:58
In nuclear physics, finding the predicted particle at the predicted time and place is considered definitive.Strictly speaking, doing that once is just suggestive.
Once you get a few dozen events, the odds are pretty solid that you have a real effect.
Perhaps you can help us out here. Roughly how many events go into a typical PET scan image?
Post by: alancalverd on 25/07/2021 17:43:37
Post by: acsinuk on 25/07/2021 17:58:16
Photons are massless spinning volumes of magnetic inertia and each unit contains the power = hf Cos ϴ.of energy per second.
Post by: Bored chemist on 25/07/2021 18:09:37
511 keV is the energy of an electron not a photon.Denial of reality is not a good symptom.
Photons are massless
Yes and no.
The photons Alan is talking about have a mass equal to that of an electron (or positron, if you like)
each unit contains the power = hf Cos ϴ.of energy per second.That's gibberish.
Post by: Bored chemist on 25/07/2021 18:11:12
Around 108 to 109. I find that fairly convincing evidence that we are indeed looking at positronium annihilation!Thanks
And roughly how many PET scans are there in the world each day?
And how many have there been in total.
I think we are talking "beyond reasonable doubt" here.
Post by: alancalverd on 25/07/2021 22:57:02
Long before PET-CT imaging, we used the 1.02 MeV threshold for pair production to calibrate small linear accelerators, and photonuclear pair production and subsequent annihilation photons are part of the routine correction factors for all sorts of radiation dosimetry including radiotherapy.
I think the experimental facts are pretty well established. And I can't explain them with a wave model.
Post by: Bored chemist on 25/07/2021 23:11:16
1019 annihilations
That's quite a lot.
About 10 micromoles so something of the order of 30 nanograms of positronium.
If you could get that together in one place you could weigh it of a good balance (apart from the tiny little radiation hazard)
Post by: evan_au on 25/07/2021 23:34:27
Quote from: acsinuk
each unit contains the power = hf Cos ϴ.of energy per second.This sounds like a solar panel.
Quote from: hamdani yusuf
I think you know how neutrino was discovered. How do you make sure that no neutrino is produced in the process?I didn't know how neutrinos were first discovered in practice, so I had to look it up. There was a gap in my knowledge between Pauli and the Homestake experiment:
https://en.wikipedia.org/wiki/Neutrino#Direct_detection
Apparently, the original discovery of neutrinos was adjacent to a nuclear reactor; Uranium fission produces many kinds of radiation, including a broad spectrum of neutron, neutrino and gamma ray energy.
The actual detection of a neutrino absorption (extremely rare) involved not just the emission of a positron (which was detected by its characteristic gamma ray energy), but by emission of a neutron and its subsequent absorption.
https://en.wikipedia.org/wiki/Cowan%E2%80%93Reines_neutrino_experiment#Setup
Typical PET reactions emit a positron only, which then produces a pair of gamma rays with a specific energy, and opposite directions to conserve energy and momentum.
https://en.wikipedia.org/wiki/Isotopes_of_oxygen#Oxygen-15
So the reaction that first detected a neutrino was a more complex reaction than the reactions used in PET scanners.
PET scanners don't try to detect neutrinos - and the energy spectrum and direction of the gamma rays shows that no neutrino is involved.
Post by: alancalverd on 25/07/2021 23:39:32
Quote
If you could get that together in one place you could weigh it of a good balance (apart from the tiny little radiation hazard)...and the fact that it has a halflife of ~100 picoseconds, so there's buggerall left of the stuff we made in 1980, or even yesterday.
Post by: hamdani yusuf on 26/07/2021 05:41:11
Quote
Episode 50. Particles and Waves: Evidence that light can sometimes act like a particle leads to quantum mechanics, the new physics.
“The Mechanical Universe,” is a critically-acclaimed series of 52 thirty-minute videos covering the basic topics of an introductory university physics course.
Each program in the series opens and closes with Caltech Professor David Goodstein providing philosophical, historical and often humorous insight into the subject at hand while lecturing to his freshman physics class. The series contains hundreds of computer animation segments, created by Dr. James F. Blinn, as the primary tool of instruction. Dynamic location footage and historical re-creations are also used to stress the fact that science is a human endeavor.
Post by: hamdani yusuf on 26/07/2021 08:38:43
Quote
No, Changing Electric Fields DON'T Cause Magnetic Fields; The Real Origin of Electromagnetic WavesIs the explanation here correct, according to current understanding of electromagnetic wave?
Doesn't it contradict following statement?
Quote
In addition, E and B are perpendicular to each other and to the direction of wave propagation, and are in phase with each other. A sinusoidal plane wave is one special solution of these equations. Maxwell's equations explain how these waves can physically propagate through space. The changing magnetic field creates a changing electric field through Faraday's law. In turn, that electric field creates a changing magnetic field through Maxwell's addition to Ampère's law. This perpetual cycle allows these waves, now known as electromagnetic radiation, to move through space at velocity c.https://en.wikipedia.org/wiki/Maxwell%27s_equations#Vacuum_equations,_electromagnetic_waves_and_speed_of_light
Post by: hamdani yusuf on 26/07/2021 10:18:56
Quote
In the 1860's and 1870's, a Scottish scientist named James Clerk Maxwell developed a scientific theory to explain electromagnetic waves. He noticed that electrical fields and magnetic fields can couple together to form electromagnetic waves. He summarized this relationship between electricity and magnetism into what are now referred to as "Maxwell's Equations."The diagram from NASA website seems to suggest that there is phase shift between electric and magnetic field in a propagating electromagnetic wave.
(https://science.nasa.gov/files/science-red/s3fs-public/styles/recommended_article_thumb/public/thumbnails/image/anatomy4.jpg?itok=r13T7S7v)
Post by: hamdani yusuf on 26/07/2021 10:40:56
Quote
Summaryhttps://opentextbc.ca/universityphysicsv2openstax/chapter/maxwells-equations-and-electromagnetic-waves/
- Maxwell’s prediction of electromagnetic waves resulted from his formulation of a complete and symmetric theory of electricity and magnetism, known as Maxwell’s equations.
- The four Maxwell’s equations together with the Lorentz force law encompass the major laws of electricity and magnetism. The first of these is Gauss’s law for electricity; the second is Gauss’s law for magnetism; the third is Faraday’s law of induction (including Lenz’s law); and the fourth is Ampère’s law in a symmetric formulation that adds another source of magnetism, namely changing electric fields.
- The symmetry introduced between electric and magnetic fields through Maxwell’s displacement current explains the mechanism of electromagnetic wave propagation, in which changing magnetic fields produce changing electric fields and vice versa.
- Although light was already known to be a wave, the nature of the wave was not understood before Maxwell. Maxwell’s equations also predicted electromagnetic waves with wavelengths and frequencies outside the range of light. These theoretical predictions were first confirmed experimentally by Heinrich Hertz.
And another physics course says:
Quote
Maxwell’s equations encompass the major laws of electricity and magnetism. What is not so apparent is the symmetry that Maxwell introduced in his mathematical framework. Especially important is his addition of the hypothesis that changing electric fields create magnetic fields. This is exactly analogous (and symmetric) to Faraday’s law of induction and had been suspected for some time, but fits beautifully into Maxwell’s equations.https://courses.lumenlearning.com/physics/chapter/24-1-maxwells-equations-electromagnetic-waves-predicted-and-observed/
Since changing electric fields create relatively weak magnetic fields, they could not be easily detected at the time of Maxwell’s hypothesis. Maxwell realized, however, that oscillating charges, like those in AC circuits, produce changing electric fields. He predicted that these changing fields would propagate from the source like waves generated on a lake by a jumping fish.
The waves predicted by Maxwell would consist of oscillating electric and magnetic fields—defined to be an electromagnetic wave (EM wave). Electromagnetic waves would be capable of exerting forces on charges great distances from their source, and they might thus be detectable.
Post by: Bored chemist on 26/07/2021 10:55:51
I hadn't heard about the LW equation.
Wiki shows me how it can be derived from Maxwell's equations so it must be equivalent to those equations.
Re the video: "No, Changing Electric Fields DON'T Cause Magnetic Fields; The Real Origin of Electromagnetic Waves"
At about 06:20 it shows an animation of the magnetic field round a single moving charge.
That's quite interesting in a video that starts by saying that magnetic fields are not created by moving charges.
At 13:44 he says the acceleration dependent part of the equation is usually called the radiation field.
Again, that's a brave statement for someone who kicked off by essentially saying that radiation is not caused by accelerated charges.
This is reiterated at 16:00
At 10 minutes, he says that, if the velocity an acceleration aren't zero then the electric field won't point at where the charge "was".
He then says, when that happens there will be a magnetic field.
Again, that's an interesting claim if you start by saying that a moving charge does not create a magnetic field.
He's right in saying that fields are just bookkeeping and that any "field meter" actually just measures the effects of distant charges on a "test charge".
But that';s pretty much what the definition of a field is.
The gravitational field points "down" because that's the direction in which a test mass will accelerate as a result o the field.
He's restating classical electromagnetics- derived explicitly from Maxwell's equations.
In doing so, he's tying to be controversial, but he doesn't actually seem to say anything new.
And, of course, it's a classical model.
It does not handle QM.
So, while it's fine for a radio transmitter and you can use it to model reflection, refraction and diffraction of light, it's no use at all for interactions with atoms.
And, since it's not relativistic (in the E=MC2 sense) it doesn't deal with the mass or (I think) the momentum of light.
It certainly doesn't cover the creation or annihilation of a positron; and that limits it.
We can observe pair production directly in a cloud chamber.
https://physics.stackexchange.com/questions/127855/is-it-possible-to-produce-images-of-pair-production-in-home-made-cloud-chamber
A model which does not include it is less use than our current model.
Post by: alancalverd on 26/07/2021 11:12:16
So in one dimension we have E = -dB/dt and B = dE/dt (with a few constants that are not relevant at this stage)
Now wiggle one parameter (say E) sinusoidally. E = a sin ωt, so B = dE/dt = a cos ωt, which is the same shape but phase-displaced by π/2.
Post by: alancalverd on 26/07/2021 11:19:14
Quote
Since changing electric fields create relatively weak magnetic fields, they could not be easily detected at the time of Maxwell’s hypothesis.as the deflection of a compass needle by a current-carrying wire was a standard demonstration and the empirical Biot-Savart law that describes it, predates Maxwell.
And whilst I'm riding my pedantic hobby-horse,
Quote
Evidence that light can sometimes act like a particleshould be struck from any syllabus! Light always behaves as light, but we need a particle model to predict that behavior in certain circumstances.
Post by: Bored chemist on 26/07/2021 11:25:10
https://en.wikipedia.org/wiki/Circular_polarization#/media/File:Circular.Polarization.Circularly.Polarized.Light_With.Components_Right.Handed.svg
And we know that photons always carry angular momentum.
The combination of a left and right handed circular polarisation gives a linear one.
For some reason, people seem to think that "ordinary" light is composed of plane polarised light.
In some ways, it works better if you assume it's a 50:50 mix of two sorts of circular polarisation.
Post by: Bored chemist on 26/07/2021 11:28:25
Light always behaves as light,I can see why you were keen on the idea that "Brexit means Brexit".
Both statements are true but meaningless.
Saying that light acts like light is pretty pointless.
Saying that it sometimes seems to act like a particle (and by implication sometimes it doesn't) actually imparts information.
Post by: alancalverd on 26/07/2021 12:19:46
It is absurd to think that light somehow decides how it is going to behave (wave approaching a grating) and then changes its mind (particle detected at the exit). The behavior of light is absolutely consistent (unlike that of a politician) but is not completely modelled by either a particle or a wave formalism. You can't blame the photon! "Seems to " behave is excusable in a public lecture but I wouldn't teach it to an aspiring scientist.
Post by: Bored chemist on 26/07/2021 12:27:16
It is absurd to think that light somehow decides how it is going to behaveAnd nobody said it does "decide".
However, in point of fact, it sometimes behaves like a particle would be expected to and it sometimes behaves like a wave would be expected to.
The fact that it always does what light does is true, but tautology, and thus unhelpful.
The light does not "decide" how to behave.
But we might have to decide which model to use.
Post by: alancalverd on 26/07/2021 12:30:59
Post by: hamdani yusuf on 26/07/2021 13:42:43
Meanwhile, Fourier transform tells us that infinitesimally precise frequency of a wave requires infinitely long sine wave. The most compact waveform in both frequency and time domains is Gaussian. It contains a range of frequency, and has non-zero time duration.Here is an example.
I've seen some sources said that uncertainty principle is not a problem of measurement. Instead, it's a fundamental principle inherently embedded to the math of wave models through Fourier transform.
Post by: hamdani yusuf on 26/07/2021 14:03:47
At about 06:20 it shows an animation of the magnetic field round a single moving charge.What he actually said is
That's quite interesting in a video that starts by saying that magnetic fields are not created by moving charges.
Changing Electric Fields DON'T Cause Magnetic FieldsIn other words, magnetic fields are not created by Changing Electric Fields.
Around 6:00 he does say that charge movement causes magnetic "situation".
Thus, the magnetic field in a spot of space is not generated by changing of electric field in that spot, but instead, by moving charges at some distance away from that spot. Both changing of magnetic field as well as electric field are caused by moving charges. Both field changes are the effect of the same cause.
Post by: Bored chemist on 26/07/2021 14:19:33
However, we can have a charged body. And we can say that, if you put a small test charge near that body there will be a force on it.
And you can plot out the direction of that force as a function o location.
And you can cal that map of "what direction would a test change accelerate if you put it here" an electric field.
You can play the same game (in principle) with a magnet and a hypothetical magnetic monopole.
And that gives you a magnetic field.
If you do something that moves a charged object, then you will change the electric field and if you move a magnet you will change the magnetic field.
So there is no practical difference between moving a charged object and changing the field round the "test charge".
However, if you choose to make this distinction:
Thus, the magnetic field in a spot of space is not generated by changing of electric field in that spot, but instead, by moving charges at some distance away from that spot.you need to explain how the effect on gets from the point where the charge was to the point where you are measuring it.
There must be "something" in between.
Unless you propose to reinvent the luminiferous ether, that "something" is an EM field.
He might say that the field is just "bookkeeping", but there must be something that tells the point A that the charge has moved.
He says the charged thing is sending out it's location velocity and acceleration.
But he skips over how that's sent out.
Is it sent by pigeon?
Given that it travels at C, it's hard to see what it could be apart from EM radiation.
So his model of light conveniently overlooks the fact that it depends on light to make it work.
That's a circular argument.
Post by: alancalverd on 26/07/2021 14:33:23
I've seen some sources said that uncertainty principle is not a problem of measurement. Instead, it's a fundamental principle inherently embedded to the math of wave models through Fourier transform.Neither is true, and the term "uncertainty" is misleading.
It is a principle of indeterminacy: a particle cannot simultaneously have a position and a momentum since position implies stasis and momentum implies continuous motion.
If we arbitrarily assign a limit to the determinacy of both and say Δp.Δx = h, we can determine h by experiment since the indeterminacy defines for example the effective radius of a hydrogen atom.
Post by: hamdani yusuf on 26/07/2021 14:45:02
Neither is true, and the term "uncertainty" is misleading.Did you say that it is a problem of measurement?
Post by: Bored chemist on 26/07/2021 15:23:43
Is that clear enough for you?
Post by: Colin2B on 26/07/2021 15:54:37
The diagram from NASA website seems to suggest that there is phase shift between electric and magnetic field in a propagating electromagnetic wave.I don’t read it that way. I think for graphic effect (3D?) they have displaced the 2 graphs. If you look at the origins of each graph you will see they line up.
Not that it matters, they are only in phase for linear polarisation.
Apologies if someone has already picked this up.
Post by: alancalverd on 26/07/2021 17:37:24
NASA is pretty good at maths and physics.
Apart from converting meters to feet, that is.
Or testing spherical mirrors for aberration.
Or freezing O rings.
Maybe I'll withdraw my Mars application after all.
But Maxwell was a Peterhouse man, so you can trust the phase shift.
Post by: hamdani yusuf on 26/07/2021 22:42:52
He might say that the field is just "bookkeeping", but there must be something that tells the point A that the charge has moved.When Isaac Newton was asked with similar question about gravity, he said hypotheses non fingo.
He says the charged thing is sending out it's location velocity and acceleration.
But he skips over how that's sent out.
Is it sent by pigeon?
Given that it travels at C, it's hard to see what it could be apart from EM radiation.
Post by: hamdani yusuf on 26/07/2021 22:53:42
It looks like a cabinet or cavalier projection. Which means that there is phase difference.The diagram from NASA website seems to suggest that there is phase shift between electric and magnetic field in a propagating electromagnetic wave.I don’t read it that way. I think for graphic effect (3D?) they have displaced the 2 graphs. If you look at the origins of each graph you will see they line up.
Not that it matters, they are only in phase for linear polarisation.
Apologies if someone has already picked this up.
(https://upload.wikimedia.org/wikipedia/commons/thumb/9/93/Comparison_of_graphical_projections.svg/520px-Comparison_of_graphical_projections.svg.png)
Even in circular polarization, they are still in phase. The phase difference here is between electric field in x axis and y axis.
(https://upload.wikimedia.org/wikipedia/commons/d/d1/Circular.Polarization.Circularly.Polarized.Light_Left.Hand.Animation.305x190.255Colors.gif)
Post by: hamdani yusuf on 27/07/2021 07:16:33
I think the experimental facts are pretty well established. And I can't explain them with a wave model.What makes the photons go into opposite direction, instead of random?
Does it also happen in cloud chamber after a positron is detected?
Post by: Bored chemist on 27/07/2021 08:30:10
What makes the photons go into opposite direction, instead of random?
which then produces a pair of gamma rays with a specific energy, and opposite directions to conserve energy and momentum.
Does it also happen in cloud chamber after a positron is detected?Yes.
Post by: hamdani yusuf on 27/07/2021 12:18:21
Does the pair always start with 0 total momentum?What makes the photons go into opposite direction, instead of random?which then produces a pair of gamma rays with a specific energy, and opposite directions to conserve energy and momentum.Does it also happen in cloud chamber after a positron is detected?Yes.
What determines the photons direction?
Post by: Bored chemist on 27/07/2021 12:41:53
Does the pair always start with 0 total momentum?From some point of view, yes.
https://en.wikipedia.org/wiki/Center_of_mass#Barycentric_coordinates
You can pick a different point of view if you wish; it makes the arithmetic harder.
Post by: Bored chemist on 27/07/2021 12:46:22
What determines the photons direction?It seems to be random.
Post by: Colin2B on 27/07/2021 15:06:57
Even in circular polarization, they are still in phase. The phase difference here is between electric field in x axis and y axis.Yes, I agree. My mind was distracted and I’m so used go thinking in E fields only I transposed them.
It looks like a cabinet or cavalier projection. Which means that there is phase difference.I still think it’s an effect of a poor graphic, I would be surprised if the illustrator was thinking cabinet or had been told to put in a phase shift.
If you wanted to show them as out of phase you would either put them both on the same axes or label the z axis to specific points. Have you contacted them to point out the anomaly?
Post by: alancalverd on 27/07/2021 20:57:54
Does the pair always start with 0 total momentum?The initiating positron doesn't have a lot of kinetic energy by the time it is captured by an electron and the pair are in a much greater mass of medium, so the positron's momentum is mostly transferred to everything else by the time the pair annihilates. You may get some forward scatter if you do the experiment in vacuo with very high energy positrons (using an accelerator) but when the radionuclide is embedded in a patient the momentum of the positronium is negligible and random.
What determines the photons direction?
There is thus no preferred direction for the photons.
Post by: evan_au on 28/07/2021 09:14:11
- I don't see anything here that makes a "positronic brain" any less infeasible than it was 70 years ago...
- We have had a fast-paced robot movie based on "I Robot" that pretty much ignored the "3 laws of robotics"
- We have had a slow-moving movie based on "Bicentennial Man" that acknowledged the 3 laws of robotics - but was probably considered too boring for most people
- We have an upcoming series based on "Foundation": I wonder how they will walk this difficult line?
Presumably utilizing a minimal quantity of positrons...
Post by: alancalverd on 28/07/2021 09:46:38
Post by: hamdani yusuf on 28/07/2021 13:30:40
(https://upload.wikimedia.org/wikipedia/commons/thumb/3/35/Parallel_plate_capacitor.svg/220px-Parallel_plate_capacitor.svg.png)
https://en.wikipedia.org/wiki/List_of_PET_radiotracers
Will the gamma ray still generated at random direction? Or they will be more likely directed normal to the plates?
Or the contrary, more gamma ray directed perpendicular to the normal?
Post by: acsinuk on 28/07/2021 17:33:22
Post by: Bored chemist on 28/07/2021 18:04:59
Just found this article about quantum virtual photons whichidentifiesDOESN'T EVEN MENTION the magnoflux spin effect.
FTFY
Post by: hamdani yusuf on 29/07/2021 22:45:09
It's clear that we can't capture all the emitted photons, but the simultaneous capture of two 511 keV photons at 180 degrees is excellent evidence that 2me has been annihilated, and the net charge of the patient remains stubbornly at zero.Getting similar numbers may indicate some correlation. But it can also be coincident.
Quote
A popular Facebook post suggests that the location of the Great Pyramid of Giza is mysteriously connected to the speed of light. It says:https://fullfact.org/online/great-pyramid-speed-of-light/
“Speed of light: 299,792,458 m/s. Coordinates of the Great Pyramid of Giza: 29.9792458°N. Coincidence?”
Human body is not suitable for investigating basic science since it's too complex with many interacting variables. We could start with simpler objects like what I proposed above. It's unfortunate that I have no access to any PET machine. But it should be easy to do for those who has.
Post by: Bored chemist on 29/07/2021 23:14:40
We could start with simpler objectsThey did, and they do.
Did you see the IOP video at about 03:30... the one you posted?
A nice simple object- a vial with a radioactive sample in it.
Post by: Bored chemist on 29/07/2021 23:16:07
I imagine an aluminum plate 100x100 mm, smeared with water containing 15O, and another aluminum plate 100x100 mm on top of it, like a capacitor.You do know that water conducts electricity, don't you?
(https://upload.wikimedia.org/wikipedia/commons/thumb/3/35/Parallel_plate_capacitor.svg/220px-Parallel_plate_capacitor.svg.png)
https://en.wikipedia.org/wiki/List_of_PET_radiotracers
Will the gamma ray still generated at random direction? Or they will be more likely directed normal to the plates?
Or the contrary, more gamma ray directed perpendicular to the normal?
So your idea is impossible.
Why did you propose it?
Post by: alancalverd on 30/07/2021 00:10:57
Getting similar numbers may indicate some correlation. But it can also be coincident.One of James Bond's more memorable statements:
"Once is happenstance, twice is coincidence. Three times is enemy action."
I think 109, exactly as predicted, several times a day in several different laboratories, suggests a law of physics.
But who really knows if the sun will rise tomorrow?
I can't see how an aluminum capacitor can have any effect on the direction of emission of an annihilation photon.
Post by: hamdani yusuf on 30/07/2021 03:07:34
One of James Bond's more memorable statements:If you measure speed of light in vacuum a trillion times a day, you will also get the same number as coordinate of the great pyramid of Giza.
"Once is happenstance, twice is coincidence. Three times is enemy action."
I think 109, exactly as predicted, several times a day in several different laboratories, suggests a law of physics.
But who really knows if the sun will rise tomorrow?
Post by: hamdani yusuf on 30/07/2021 03:41:48
Pure water is not a good electrical conductor.I imagine an aluminum plate 100x100 mm, smeared with water containing 15O, and another aluminum plate 100x100 mm on top of it, like a capacitor.You do know that water conducts electricity, don't you?
(https://upload.wikimedia.org/wikipedia/commons/thumb/3/35/Parallel_plate_capacitor.svg/220px-Parallel_plate_capacitor.svg.png)
https://en.wikipedia.org/wiki/List_of_PET_radiotracers
Will the gamma ray still generated at random direction? Or they will be more likely directed normal to the plates?
Or the contrary, more gamma ray directed perpendicular to the normal?
So your idea is impossible.
Why did you propose it?
I showed the picture of a capacitor to illustrate the shape of the proposed apparatus, not that it will work as a capacitor.
Post by: hamdani yusuf on 30/07/2021 03:49:28
I can't see how an aluminum capacitor can have any effect on the direction of emission of an annihilation photon.The positrons are supposedly produced by 15O in the radioactive water between the plates in random direction. Because the water is shaped as 2 dimensional surface, the effective direction of positrons hitting electrons would be normal to the plates.
At least the experiment will show if the trajectory of the positron before hitting the electron has any effect on the direction of the gamma rays they emmit.
Let's say a positron moves in positive x direction before hitting a stationary electron. Will the gamma ray produced by the collision equally likely to go in x direction as any other direction? Will it be more likely? Or less likely instead?
What's the prediction of existing model?
Post by: Bored chemist on 30/07/2021 08:58:13
Pure water is not a good electrical conductor.Yes it is, in this context.
Saying things that are not true does not make you look good.
Nor does calling me either a liar or a fool.
Without your pointless set-up the positronium decays somewhere in the water.
It's very small; roughly twice the size of a hydrogen atom.
So, it is small enough to "see" the electric field of the water molecule(s) near it.
And that is a very strong field.
So the voltage gradient- the field- you could get across your capacitor would need to be much bigger (or the local field produced by the water molecules would dominate any effect).
But if you put a water molecule in a field which is much bigger then that produced by the water dipole, the molecule will be torn apart- it will no longer be an insulator.
So, don't try to tell me that water is a poor conductor.
In your experiment it would be a plasma with a negative incremental resistance.
Post by: Bored chemist on 30/07/2021 09:10:31
If you measure speed of light in vacuum a trillion times a day, you will also get the same number as coordinate of the great pyramid of Giza.No, actually, you don't.
The number is close, but no more correct than a bunch of other lines of latitude which go through the object.
Also, the point you made is irrelevant.
Measuring the same thing many times does not tell you much.
To be any use, you would need to measure other pyramids and find some connection there as well.
Whereas, as Alan has pointed out, billions of particles in thousands of hospitals in dozens of countries all agree,
The chances of that happening make the national lottery look like a certainty.
Post by: alancalverd on 30/07/2021 10:51:42
If you measure speed of light in vacuum a trillion times a day, you will also get the same number as coordinate of the great pyramid of Giza.This proving that the Egyptians invented SI measurements (it was actually the French revolutionaries), sexagesimal angular measurements (it was actually the Babylonians) and decimal GPS coordinates (actually the US military).
Post by: hamdani yusuf on 30/07/2021 11:23:56
Yes it is, in this context.What context? I wasn't trying to make a capacitor.
Saying things that are not true does not make you look good.
Nor does calling me either a liar or a fool.
It looks like you are being too sensitive.
Post by: hamdani yusuf on 30/07/2021 11:25:43
Whereas, as Alan has pointed out, billions of particles in thousands of hospitals in dozens of countries all agree,What do they agree upon? The gamma ray frequency?
The chances of that happening make the national lottery look like a certainty.
Post by: hamdani yusuf on 30/07/2021 11:26:42
No, it's not. And that's the point.If you measure speed of light in vacuum a trillion times a day, you will also get the same number as coordinate of the great pyramid of Giza.This proving that the Egyptians invented SI measurements (it was actually the French revolutionaries), sexagesimal angular measurements (it was actually the Babylonians) and decimal GPS coordinates (actually the US military).
Post by: alancalverd on 30/07/2021 13:49:07
What do they agree upon? The gamma ray frequency?The gamma energy, 180 degree coincidence, and the fact that it only happens when a positron emitter is present. And something to do with Einstein and Planck, way beyond my pay grade.
Post by: Bored chemist on 30/07/2021 13:58:51
What do they agree upon?That's often the wrong question.
You don't make much progress in science by looking at what agrees.
The interesting question is "What do they disagree on?".
And the answer so far is "nothing".
All the experiments give the expected outcomes.
So this is not a field where you can improve the theories.
Post by: Bored chemist on 30/07/2021 14:02:16
No.Yes it is, in this context.What context? I wasn't trying to make a capacitor.
Saying things that are not true does not make you look good.
Nor does calling me either a liar or a fool.
It looks like you are being too sensitive.
You called it "like a capacitor." If I wanted to consider it as a capacitor I would just quote the time constant; I think it's about 130µseconds- that's how long you have to do your experiment before the charge leaks away.
In the context of the experiment you propose...
I am simply pointing out that in order for the field between your two plates to make a significant difference to the positron behaviour, it would have to be stronger than the field in which the positrons are already present.
But that field is the field produced by the dipoles of the water molecules.
And if you put water in a much stronger field than that, you pull the water apart.
Saying the water isn't a good conductor is simply not true in those conditions is it?.
Post by: Bored chemist on 30/07/2021 14:12:00
Nobody says that coincidences do not happen.No, it's not. And that's the point.If you measure speed of light in vacuum a trillion times a day, you will also get the same number as coordinate of the great pyramid of Giza.This proving that the Egyptians invented SI measurements (it was actually the French revolutionaries), sexagesimal angular measurements (it was actually the Babylonians) and decimal GPS coordinates (actually the US military).
But you are saying that all the PET scan images are a set of about 10^19 coincidences.
That's stupidly lucky, isn't it?
Post by: hamdani yusuf on 30/07/2021 14:33:46
You called it "like a capacitor." If I wanted to consider it as a capacitor I would just quote the time constant; I think it's about 130µseconds- that's how long you have to do your experiment before the charge leaks away.If I called it like a sandwich, perhaps you'll complain that it's not edible.
Post by: hamdani yusuf on 30/07/2021 14:39:14
In the context of the experiment you propose...The purpose of the metal plates are to provide the shape of 2 dimensional surface for positron sources. Nothing more. Because the shape and composition of human tissues are generally too complex for experiments of basic science.
I am simply pointing out that in order for the field between your two plates to make a significant difference to the positron behaviour, it would have to be stronger than the field in which the positrons are already present.
But that field is the field produced by the dipoles of the water molecules.
And if you put water in a much stronger field than that, you pull the water apart.
Post by: hamdani yusuf on 30/07/2021 14:44:03
Nobody says that coincidences do not happen.What actually those PET scan images agree on?
But you are saying that all the PET scan images are a set of about 10^19 coincidences.
That's stupidly lucky, isn't it?
Post by: alancalverd on 30/07/2021 15:02:48
The time sequence of PET activity can also tell you about the metabolic rates of various organs.
The basic science is all sorted out long before anyone injects stuff into a patient!
Post by: Bored chemist on 01/08/2021 18:14:43
Where the stuff is in the body.Nobody says that coincidences do not happen.What actually those PET scan images agree on?
But you are saying that all the PET scan images are a set of about 10^19 coincidences.
That's stupidly lucky, isn't it?
Post by: Bored chemist on 01/08/2021 18:19:31
I apologise; I assumed you were planning to do something useful/ interesting like put a voltage across the plates. (Though , as I said, that would be impossible, if you wanted a strong enough field to make any difference)In the context of the experiment you propose...The purpose of the metal plates are to provide the shape of 2 dimensional surface for positron sources. Nothing more. Because the shape and composition of human tissues are generally too complex for experiments of basic science.
I am simply pointing out that in order for the field between your two plates to make a significant difference to the positron behaviour, it would have to be stronger than the field in which the positrons are already present.
But that field is the field produced by the dipoles of the water molecules.
And if you put water in a much stronger field than that, you pull the water apart.
Without such a voltage, the positrons obviously have no way of knowing that there is an aluminium plate a very long way away from where they are, and so their gamma ray emissions will not be affected by it.
Why did you think a distant metal plate would affect the process?
Post by: TommyJ on 02/08/2021 18:44:14
1. Electromagnetic waves consist of discrete, massless units called photons. A photon travels in vacuum at the speed of light.
2. Each photon has energy where is the frequency of the wave and h is a universal constant called Planck’s constant. The value of Planck’s constant is In other words, the electromagnetic waves come in discrete “chunks” of energy.
3. The superposition of a sufficiently large number of photons has the characteristics of a continuous electromagnetic wave.
Coming just to every-day human detection.
Infrared radiation, with its relatively long wavelength and low photon energy, produces effects in tissue similar to those of microwaves. Infrared is absorbed mostly by the top layer of your skin and simply warms you up, as you know from sitting in the sun or under a heat lamp.
In contrast, ultraviolet photons have enough energy to interact with molecules in entirely different ways, ionizing molecules and breaking molecular bonds. Interactions of ultraviolet radiation with matter are best understood from the photon perspective, with the absorption of each photon being associated with a particular molecular event.
Visible light is at a transition point in the electromagnetic spectrum. The energy of photons of visible light is large enough to cause molecular transitions—which is how the eye detects light. The bending of light by the lens of the eye requires us to think of light as a wave.
Color Vision. The cones, the color-sensitive cells in the retina of the eye, each contain one of three slightly different forms of a light-sensitive photopigment. A single photon of light can trigger a reaction in a photopigment molecule, which ultimately leads to a signal being produced by a cell in the retina. The color vision is a result of the differential response of the three types of cones containing these three different pigments.
Creating the impression of a color Computer monitors and color TVs can create millions of different colors by combining light from pixels of only three colors: red, green, and blue. These are called RGB displays. REASON We’ve seen that there are three different types of cones in the eye. By using differing amounts of three pure colors, we can independently stimulate each of the cone types and thus mimic the response of the eye to light of almost any color. ASSESS The fact that there are three primary colors of light—red, green, and blue— is a function of our physiology, not basic physics.
At the highest energies of the electromagnetic spectrum we find x rays and gamma rays.
Post by: Eternal Student on 02/08/2021 23:55:04
That sounds reasonable, TommyJ.
Fine details may be debatable
A single photon of light can trigger a reaction in a photopigment molecule, which ultimately leads to a signal being produced by a cell in the retina.Some sources of information suggest upto 3 photons are required, one photon on it's own may not be enough to cause a single pigment molecule to trigger. Whatever the precise details, it's fair to say photopigments can be triggered by just a few photons even if it isn't exactly one. Whether your brain can sensibly present that information to you in a way you understand is a different matter. Your brain may process stimulus from the optic nerve in a way similar to "signal averaging" or completely ignore a single stimulation of one pigment molecule if it is not repeated soon or accompanied by signals from near-by photopigment molecules.
Some more discussion was covered by TNS here: https://www.thenakedscientists.com/articles/questions/how-many-photons-does-it-take-see-object.
I think I can see how your comment links with the main topic in this thread. Our eyes work a certain way and this is some evidence for the particle nature of light.
There may be some interesting links with "Olber's paradox". Light intensity will generally fall off as the square of the distance from the source. However, there should eventually be a distance where the photons are spread so thinly that there are whole patches of sky where a detector of finite cross-sectional area can be placed and 0 photons would hit it in a second. When we look up at the night sky we don't see light everywhere, just in a few paces and those are the stars we believe are close. If the Universe was infinite and similar to waht we can see near-by then we should always find a star along any line through space. The "thinning down of photons" as described above is one way to resolve that paradox. The Hubble "deep field" photographs give some indication of what happens if we do collect photons over enough time (we see stars and whole galaxies in regions that superficially appeared black).
Best wishes.
Post by: TommyJ on 03/08/2021 14:38:57
When we look up at the night sky we don't see light everywhere, just in a few paces and those are the stars we believe are close. If the Universe was infinite and similar to waht we can see near-by then we should always find a star along any line through space. The "thinning down of photons" as described above is one way to resolve that paradox. The Hubble "deep field" photographs give some indication of what happens if we do collect photons over enough time (we see stars and whole galaxies in regions that superficially appeared black).Let me put some findings.
The process of detecting light destroys the photon concerned. However for gamma-rays and cosmic rays this is not necessarily true.
For example, when a gamma-ray Comptonscatters on an electron, it will lose energy, but can carry on and do it again. This can produce a series of detectable events which in principle will form a track pointing back in the arrival direction of the photon.
In practice, more indirect information can be obtained. A typical “Compton Telescope” has just two detection layers. In the upper layer a Compton scattering takes place and in the second layer the photon is absorbed. If the incoming photon has energy E1 and scatters through an angle θ then the energy of the scattered photon E2 is given by
(E1 − E2)/ (E1*E2) = (1/ (mc^2) · (1 − cos θ),
where m is the mass of the electron.
After measuring the initial and final energies, and so the scattering angle θ is known, but not the azimuth.
What is possible to get is a ring of possible photon directions. Over time, with the telescope in different orientations, many photons from the same astronomical source may be seen, in which case the intersection of the various rings finally pins down the source position.
This summing of different paths from the same source is used also in telecommunication systems, in order to minimize data burst losses.
Post by: hamdani yusuf on 03/08/2021 16:26:43
Why did you think a distant metal plate would affect the process?Because the positrons are supposed to produce gamma ray only when they hit electrons, which are abundant on metal surface.
After a second thought, I realized that determining the direction of positron-electron collisions can be done better using a small ball of positron source and a small ball of electron source, separated by some distance inside the PET scanner. We can observe if the gamma rays coming out from the target still have random direction.
Post by: hamdani yusuf on 03/08/2021 16:57:56
The photon model of electromagnetic waves consists of three basic postulates:
1. Electromagnetic waves consist of discrete, massless units called photons. A photon travels in vacuum at the speed of light.
2. Each photon has energy where is the frequency of the wave and h is a universal constant called Planck’s constant. The value of Planck’s constant is In other words, the electromagnetic waves come in discrete “chunks” of energy.
3. The superposition of a sufficiently large number of photons has the characteristics of a continuous electromagnetic wave.
Let's compare these postulates to this article I quoted in another thread.
https://en.wikipedia.org/wiki/Planck_constantQuoteThe Planck constant, or Planck's constant, is a fundamental physical constant denoted h, and is of fundamental importance in quantum mechanics. A photon's energy is equal to its frequency multiplied by the Planck constant. Due to mass–energy equivalence, the Planck constant also relates mass to frequency.The unit of Planck constant is J⋅s/cycle. If there are 2 cycles in a chirp of electromagnetic wave, the energy would be twice as much. It's implicitly assumed that the value of cycle must be an integer. Otherwise there would be no observed quantization of energy transfer.
In metrology it is used, together with other constants, to define the kilogram, an SI unit.[1] The SI units are defined in such a way that, when the Planck constant is expressed in SI units, it has the exact value h = 6.62607015×10−34 J⋅Hz−1.[2][3]
At the end of the 19th century, accurate measurements of the spectrum of black body radiation existed, but predictions of the frequency distribution of the radiation by then-existing theories diverged significantly at higher frequencies. In 1900, Max Planck empirically derived a formula for the observed spectrum. He assumed a hypothetical electrically charged oscillator in a cavity that contained black-body radiation could only change its energy in a minimal increment, E, that was proportional to the frequency of its associated electromagnetic wave.[4] He was able to calculate the proportionality constant from the experimental measurements, and that constant is named in his honor. In 1905, Albert Einstein determined a "quantum" or minimal element of the energy of the electromagnetic wave itself. The light quantum behaved in some respects as an electrically neutral particle, and was eventually called a photon. Max Planck received the 1918 Nobel Prize in Physics "in recognition of the services he rendered to the advancement of Physics by his discovery of energy quanta".
Confusion can arise when dealing with frequency or the Planck constant because the units of angular measure (cycle or radian) are omitted in SI.[5][6][7][8][9] In the language of quantity calculus,[10] the expression for the value of the Planck constant, or a frequency, is the product of a numerical value and a unit of measurement. The symbol f (or ν), when used for the value of a frequency, implies cycles per second or hertz as the unit. When the symbol ω is used for the frequency's value it implies radians per second as the unit. The numerical values of these two ways of expressing the frequency have a ratio of 2π. Omitting the units of angular measure "cycle" and "radian" can lead to an error of 2π. A similar state of affairs occurs for the Planck constant. The symbol h is used to express the value of the Planck constant in J⋅s/cycle, and the symbol ħ ("h-bar") is used to express its value in J⋅s/rad. Both represent the value of the Planck constant, but, as discussed below, their numerical values have a ratio of 2π. In this article the word "value" as used in the tables means "numerical value", and the equations involving the Planck constant and/or frequency actually involve their numerical values using the appropriate implied units.QuoteThe word quantum is the neuter singular of the Latin interrogative adjective quantus, meaning "how much". "Quanta", the neuter plural, short for "quanta of electricity" (electrons), was used in a 1902 article on the photoelectric effect by Philipp Lenard, who credited Hermann von Helmholtz for using the word in the area of electricity. However, the word quantum in general was well known before 1900,[2] e.g. quantum was used in E.A. Poe's Loss of Breath. It was often used by physicians, such as in the term quantum satis. Both Helmholtz and Julius von Mayer were physicians as well as physicists. Helmholtz used quantum with reference to heat in his article[3] on Mayer's work, and the word quantum can be found in the formulation of the first law of thermodynamics by Mayer in his letter[4] dated July 24, 1841.https://en.wikipedia.org/wiki/Quantum#Etymology_and_discovery
In 1901, Max Planck used quanta to mean "quanta of matter and electricity",[5] gas, and heat.[6] In 1905, in response to Planck's work and the experimental work of Lenard (who explained his results by using the term quanta of electricity), Albert Einstein suggested that radiation existed in spatially localized packets which he called "quanta of light" ("Lichtquanta").[7]
The concept of quantization of radiation was discovered in 1900 by Max Planck, who had been trying to understand the emission of radiation from heated objects, known as black-body radiation. By assuming that energy can be absorbed or released only in tiny, differential, discrete packets (which he called "bundles", or "energy elements"),[8] Planck accounted for certain objects changing color when heated.[9] On December 14, 1900, Planck reported his findings to the German Physical Society, and introduced the idea of quantization for the first time as a part of his research on black-body radiation.[10] As a result of his experiments, Planck deduced the numerical value of h, known as the Planck constant, and reported more precise values for the unit of electrical charge and the Avogadro–Loschmidt number, the number of real molecules in a mole, to the German Physical Society. After his theory was validated, Planck was awarded the Nobel Prize in Physics for his discovery in 1918.
Do you agree with my underlined statement above?
Post by: Bored chemist on 03/08/2021 17:05:53
Do you agree with my underlined statement above?No.
There are "many" cycles in a photon. There's no reason for it to be an integer.
Post by: hamdani yusuf on 03/08/2021 17:12:44
There may be some interesting links with "Olber's paradox". Light intensity will generally fall off as the square of the distance from the source. However, there should eventually be a distance where the photons are spread so thinly that there are whole patches of sky where a detector of finite cross-sectional area can be placed and 0 photons would hit it in a second. When we look up at the night sky we don't see light everywhere, just in a few paces and those are the stars we believe are close. If the Universe was infinite and similar to waht we can see near-by then we should always find a star along any line through space. The "thinning down of photons" as described above is one way to resolve that paradox. The Hubble "deep field" photographs give some indication of what happens if we do collect photons over enough time (we see stars and whole galaxies in regions that superficially appeared black).Alternatively, the light from extremely distant galaxies might be absorbed by intergalactic media, which convert it into heat which is detected as microwave background.
Post by: hamdani yusuf on 03/08/2021 17:18:49
Can a photon have a fraction number of cycles, such as 1.2345?Do you agree with my underlined statement above?No.
There are "many" cycles in a photon. There's no reason for it to be an integer.
For example, a photon has frequency of 1 THz, and 1.2345 cycle. What's its energy?
Post by: alancalverd on 03/08/2021 17:22:59
After a second thought, I realized that determining the direction of positron-electron collisions can be done better using a small ball of positron source and a small ball of electron source, separated by some distance inside the PET scanner. We can observe if the gamma rays coming out from the target still have random direction.The emitted positrons in a solid or liquid have a range of about 0.5 mm which limits the spatial resolution of PET. Havoing an energy of between 0.5 and 2 MeV (depending on the nuclide) you would need a very strong electric field to determine which way they move before annihilating, and since the p-e pair is electrically neutral, there's no much you can do to orient it in space. What we do know is that the coincident photon pairs are spherically distributed and randomly timed, so the hypothesis that they are randomly distributed seems to be supported.
Post by: alancalverd on 03/08/2021 17:23:34
For example, a photon has frequency of 1 THz, and 1.2345 cycle. What's its energy?E = hf
You are wasting your time trying to apply continuum analytics to a particle model. They are two different models.
Post by: hamdani yusuf on 03/08/2021 17:36:15
Can a positron move further in air, or better yet, vacuum?After a second thought, I realized that determining the direction of positron-electron collisions can be done better using a small ball of positron source and a small ball of electron source, separated by some distance inside the PET scanner. We can observe if the gamma rays coming out from the target still have random direction.The emitted positrons in a solid or liquid have a range of about 0.5 mm which limits the spatial resolution of PET. Havoing an energy of between 0.5 and 2 MeV (depending on the nuclide) you would need a very strong electric field to determine which way they move before annihilating, and since the p-e pair is electrically neutral, there's no much you can do to orient it in space. What we do know is that the coincident photon pairs are spherically distributed and randomly timed, so the hypothesis that they are randomly distributed seems to be supported.
If the source of positron can be concentrated in a small volume, so does the source of electron next to it, then the gamma rays coming from electron source as the product of p-e collision must come from positron moving from positron source to electron source, i.e. the direction is determined. Positron moving to other directions would produce gamma ray somewhere else.
Post by: hamdani yusuf on 03/08/2021 17:38:44
Why Schrodinger developed wave mechanics?For example, a photon has frequency of 1 THz, and 1.2345 cycle. What's its energy?E = hf
You are wasting your time trying to apply continuum analytics to a particle model. They are two different models.
What's the unit of h?
Post by: Bored chemist on 03/08/2021 17:55:17
For example, a photon has frequency of 1 THz, and 1.2345 cycle. What's its energy?E=hf
Post by: hamdani yusuf on 04/08/2021 02:34:35
The symbol h is used to express the value of the Planck constant in J⋅s/cycleLet's do some unit analysis.
The time unit as numerator there, indicates that the value is an accumulated quantity over time, which is second. It's similar to mAh in battery capacity, or kWh in energy consumption.
The cycle as denominator means that the value is the rate of a quantity, which is per cycle.
When a quantity of radiation has value of 6.62607004 x 10-34 J. s/cycle, it means that
6.62607004 x 10-34 Joule of energy is transfered in 1 second interval for each cycle.
If the frequency is 1 Hz, then there is 1 cycle in a second. Thus, in 1 second interval the energy transfered is 6.62607004 x 10-34 Joule.
If the frequency is 1 THz, there are 1 trillion cycles in a second. So the energy transfer accumulated in 1 second is 6.62607004 x 10-22 Joule.
Post by: alancalverd on 04/08/2021 09:43:32
Why Schrodinger developed wave mechanics?to model the probability distribution of a particle.
Quote
What's the unit of h?joule.second So if you multiply by frequency (second -1) you get the energy of the photon, though it's more usually quoted in eV rather than joules. Alternatively if you know the photon energy and use E = hc/λ you can predict its diffraction properties.
Post by: TommyJ on 04/08/2021 09:48:15
Do you agree with my underlined statement above?Agree, in these terms (in case we are on the same page :)
To explain the radiation equation Planck made an assumption that energy in discrete quantities (electron oscillations in atom). Each discrete 'bit' of energy exists as a whole number multiple of minimum energy.
- Energy exists as a discrete quantity
- Exists any minimum quantity of energy for any frequency of oscillation E=hf. E can be greater by a factor of a some positive integer.
If oscillation is greater than the minimum, there exists an integer multiple of minimum quantity. E=nhf, where n is an integer number.
Post by: evan_au on 04/08/2021 10:46:40
Quote from: hamdani yusuf
If there are 2 cycles in a chirp of electromagnetic waveIf you try to generate an electromagnetic wave with a small number of cycles (eg 1 or 2), you generate a signal with an extremely wide bandwidth, so the frequency of the photon(s) is poorly defined, which means that the energy of the photons is also poorly defined. E=hf is still true, but the problem is that you don't know f accurately.
...1.2345 cycle
If you want a precisely measured frequency, you have to measure over many cycles (eg thousands or billions).
This relates back to the Heisenberg uncertainty principle - the more you attempt to constrain one parameter of a quantum variable, the more it becomes uncertain in another parameter.
The 2018 Nobel prize in physics was awarded for research into ultra-short laser pulses. Unlike the more common continuous-wave laser pointers, ultrashort laser pulses have an ultrawide bandwidth.
See: https://en.wikipedia.org/wiki/Ultrashort_pulse
Post by: Bored chemist on 04/08/2021 11:12:10
Let's do some unit analysis.That is not a valid analysis (for a number of reasons).
The time unit as numerator there, indicates that the value is an accumulated quantity over time, which is second. It's similar to mAh in battery capacity, or kWh in energy consumption.
The Joule is a unit of work- i.e. force times distance and a force is a mass times an acceleration.
So Joule is Kg m2 /sec2
So the unit of the Planck constant is
Kg m2 / sec
The "time" part is in reciprocal seconds, not seconds.
And what the hell does the product of an area with a mass mean?
Because, whatever that is, h has units of "the rate of change of whatever".
If the frequency is 1 THz, there are 1 trillion cycles in a second. So the energy transfer accumulated in 1 second is 6.62607004 x 10-22 Joule.No
I can transmit (in principle) any rate of power I like at 1THz.
I am not restricted to sending 6.62607004 x 10-22 Joule per second i.e. 6.62607004 x 10-22 W
It is more correct to say
If the frequency is 1 THz, there are 1 trillion cycles in a second. So the energy transfer accumulated in 1
Photon is 6.62607004 x 10-22 Joule.
It does not take a second to send a photon.
It would be very weird if it did.
What sort of coincidence would it be if the duration of a photon was a sixtieth of a sixtieth of a twenty-fourth of the time it takes the Earth to rotate on its axis?
Post by: hamdani yusuf on 05/08/2021 05:22:19
(https://upload.wikimedia.org/wikipedia/commons/thumb/4/4c/Ultrashort_pulse.svg/330px-Ultrashort_pulse.svg.png)Quote from: hamdani yusufIf there are 2 cycles in a chirp of electromagnetic waveIf you try to generate an electromagnetic wave with a small number of cycles (eg 1 or 2), you generate a signal with an extremely wide bandwidth, so the frequency of the photon(s) is poorly defined, which means that the energy of the photons is also poorly defined. E=hf is still true, but the problem is that you don't know f accurately.
...1.2345 cycle
If you want a precisely measured frequency, you have to measure over many cycles (eg thousands or billions).
This relates back to the Heisenberg uncertainty principle - the more you attempt to constrain one parameter of a quantum variable, the more it becomes uncertain in another parameter.
The 2018 Nobel prize in physics was awarded for research into ultra-short laser pulses. Unlike the more common continuous-wave laser pointers, ultrashort laser pulses have an ultrawide bandwidth.
See: https://en.wikipedia.org/wiki/Ultrashort_pulse
At a glance, the pulse doesn't have a uniform wavelength. It looks longer at beginning and shorter at the end.
But as long as it starts and finish at 0, the number of cycles must be an integer multiple of a half.
Post by: hamdani yusuf on 05/08/2021 08:09:43
So the unit of the Planck constant isYou're missing the cycle part. Although it's dimensionless, it doesn't mean that it's not there. Planck constant has the same unit as angular momentum, when the angle is measured in cycles.
Kg m2 / sec
The "time" part is in reciprocal seconds, not seconds.
And what the hell does the product of an area with a mass mean?
Because, whatever that is, h has units of "the rate of change of whatever".
Quote
The joule-second (symbol J⋅s or J s) is the product of an SI derived unit, the joule (J), and an SI base unit, the second (s).[1] The joule-second is a unit of action or of angular momentum. The joule-second also appears in quantum mechanics within the definition of Planck's constant.[2] Angular momentum is the product of an object’s moment of inertia, in units of kg⋅m2 and its angular velocity in units of rad⋅s−1. This product of moment of inertia and angular velocity yields kg⋅m2⋅s−1 or the joule-second. Planck's constant represents the energy of a wave, in units of joule, divided by the frequency of that wave, in units of s−1. This quotient of energy and frequency also yields the joule-second (J⋅s).https://en.wikipedia.org/wiki/Joule-second
If we try to analyze a derived unit using only its base unit equivalent, we can easily get confused of the meaning.
https://en.wikipedia.org/wiki/SI_derived_unit
How do we get the meaning of this unit?
kg−1⋅m−2⋅s4⋅A2
or this?
kg⋅s−3
Post by: hamdani yusuf on 05/08/2021 08:12:52
I can transmit (in principle) any rate of power I like at 1THz.How would you do that?
I am not restricted to sending 6.62607004 x 10-22 Joule per second i.e. 6.62607004 x 10-22 W
Post by: acsinuk on 05/08/2021 15:50:19
Energy has the equivalent effect of force times distance. So your " kg⋅s−3 " is the equivalent inertia spin energy of the magnon which is at the heart of every photon.
I think it is more accurate to consider the electro-magnetic energy of a photon as Energy= ∑ n h f Cos ϴ per second.
Post by: Bored chemist on 05/08/2021 16:10:33
Start with 1 watt and attenuate it.I can transmit (in principle) any rate of power I like at 1THz.How would you do that?
I am not restricted to sending 6.62607004 x 10-22 Joule per second i.e. 6.62607004 x 10-22 W
Post by: Bored chemist on 05/08/2021 16:11:48
equivalent inertia spin energy of the magnon which is at the heart of every photon.Why do you post meaningless utter bollocks like that?
Post by: Bored chemist on 05/08/2021 16:16:17
You're missing the cycle part. Although it's dimensionless, it doesn't mean that it's not there.It doesn't affect the dimensions and thus does not detract from my point.
If we try to analyze a derived unit using only its base unit equivalent, we can easily get confused of the meaning.Royal "we"?
At a glance, the pulse doesn't have a uniform wavelength. It looks longer at beginning and shorter at the end.So what?
But as long as it starts and finish at 0, the number of cycles must be an integer multiple of a half.
It's not got much to do with a photon, has it?
Post by: Colin2B on 05/08/2021 18:28:16
Hamdani,You have been warned that you could have your posting rights suspended if you continue to post new theories in this section. I’m being more polite that some
Energy has the equivalent effect of force times distance. So your " kg⋅s−3 " is the equivalent inertia spin energy of the magnon which is at the heart of every photon.
I think it is more accurate to consider the electro-magnetic energy of a photon as Energy= ∑ n h f Cos ϴ per second.
Post by: alancalverd on 05/08/2021 21:03:45
I can transmit (in principle) any rate of power I like at 1THz.Just as well you didn't say that in my physics class.
Post by: hamdani yusuf on 06/08/2021 04:25:35
How would you attenuate it?Start with 1 watt and attenuate it.I can transmit (in principle) any rate of power I like at 1THz.How would you do that?
I am not restricted to sending 6.62607004 x 10-22 Joule per second i.e. 6.62607004 x 10-22 W
Post by: hamdani yusuf on 06/08/2021 04:39:27
So what?So,
It's not got much to do with a photon, has it?
How many photons would be needed to produce that particular chirp/pulse?
What's the frequency of each photon?
Post by: hamdani yusuf on 06/08/2021 07:58:47
Energy has the equivalent effect of force times distance. So your " kg⋅s−3 " is the equivalent inertia spin energy of the magnon which is at the heart of every photon.
That unit, kg⋅s−3, is an example I gave to BC that analyzing a derived unit using only its base unit equivalent can be confusing, because we lose the context. You can find this unit in the wikipedia article. https://en.wikipedia.org/wiki/SI_derived_unit#Examples_of_derived_quantities_and_units
watt per square metre W/m2 heat flux density, irradiance kg⋅s−3
Post by: Bored chemist on 06/08/2021 09:20:12
Just as well you didn't say that in my physics class.Indeed, for such an event would imply that you were teaching physics; never a good thing if you don't understand the conservation of angular momentum.
Post by: Bored chemist on 06/08/2021 09:22:04
Many.So what?So,
It's not got much to do with a photon, has it?
How many photons would be needed to produce that particular chirp/pulse?
What's the frequency of each photon?
You can do the Fourier transform if you want a more precise answer.
Post by: Bored chemist on 06/08/2021 09:26:18
How would you attenuate it?Who cares?
There's some data here
https://www.brown.edu/research/labs/mittleman/sites/brown.edu.research.labs.mittleman/files/uploads/normanElecLett.pdf
which suggests that, if I pick the correct angle and polarisation, the fraction of a THz beam reflected by a sheet of glass is about 1/10.
So if I arrange 22 sheets of glass so that the beam is reflected from all of them I will have an attenuation of about 10^22
Why did you think that question was worth asking?
Post by: Bored chemist on 06/08/2021 09:34:38
That unit, kg⋅s−3, is an example I gave to BC that analyzing a derived unit using only its base unit equivalent can be confusing,It seems to have confused you.
You should be aware that you can use different units.
The foot pound second system was in use and worked well for many years.
So you should realise that you don't actually resolve a unit into (for example) SI units, you resolve it into dimensions.
For example the units of energy are mass times distance per time squared.
I can choose pounds or kilograms as the unit of mass, I can pick inches as the distance and weeks as the time.
It will still work- as long as I always use those units.
What you seem to be saying is that changing from cycles per second to radians per second makes a difference to teh dimensional analysis.
It can't.
cycles and radians are both dimensionless.
Numerically, it makes about the same difference as changing from feet to yards- you multiply or divide by about 3.
But the dimensional analysis is the same.
Post by: hamdani yusuf on 06/08/2021 10:59:05
Where does the excess energy go?How would you attenuate it?Who cares?
There's some data here
https://www.brown.edu/research/labs/mittleman/sites/brown.edu.research.labs.mittleman/files/uploads/normanElecLett.pdf
which suggests that, if I pick the correct angle and polarisation, the fraction of a THz beam reflected by a sheet of glass is about 1/10.
So if I arrange 22 sheets of glass so that the beam is reflected from all of them I will have an attenuation of about 10^22
Why did you think that question was worth asking?
Have you considered Planck's law?
Post by: Bored chemist on 06/08/2021 11:06:19
Where does the excess energy go?Who cares?
Have you considered Planck's law?Yes, I forgot to mention that I'm using very cold pieces of glass.
Post by: hamdani yusuf on 06/08/2021 11:10:25
I proposed to use unit analysis instead of dimensional analysis, so we get the correct number.That unit, kg⋅s−3, is an example I gave to BC that analyzing a derived unit using only its base unit equivalent can be confusing,It seems to have confused you.
You should be aware that you can use different units.
The foot pound second system was in use and worked well for many years.
So you should realise that you don't actually resolve a unit into (for example) SI units, you resolve it into dimensions.
For example the units of energy are mass times distance per time squared.
I can choose pounds or kilograms as the unit of mass, I can pick inches as the distance and weeks as the time.
It will still work- as long as I always use those units.
What you seem to be saying is that changing from cycles per second to radians per second makes a difference to teh dimensional analysis.
It can't.
cycles and radians are both dimensionless.
Numerically, it makes about the same difference as changing from feet to yards- you multiply or divide by about 3.
But the dimensional analysis is the same.
The symbol h is used to express the value of the Planck constant in J⋅s/cycleLet's do some unit analysis.
The time unit as numerator there, indicates that the value is an accumulated quantity over time, which is second. It's similar to mAh in battery capacity, or kWh in energy consumption.
The cycle as denominator means that the value is the rate of a quantity, which is per cycle.
When a quantity of radiation has value of 6.62607004 x 10-34 J. s/cycle, it means that
6.62607004 x 10-34 Joule of energy is transfered in 1 second interval for each cycle.
If the frequency is 1 Hz, then there is 1 cycle in a second. Thus, in 1 second interval the energy transfered is 6.62607004 x 10-34 Joule.
If the frequency is 1 THz, there are 1 trillion cycles in a second. So the energy transfer accumulated in 1 second is 6.62607004 x 10-22 Joule.
I'm sorry for misunderstood you. But to me, your statement below makes it look like you are being confused.
The Joule is a unit of work- i.e. force times distance and a force is a mass times an acceleration.
So Joule is Kg m2 /sec2
So the unit of the Planck constant is
Kg m2 / sec
The "time" part is in reciprocal seconds, not seconds.
And what the hell does the product of an area with a mass mean?
Because, whatever that is, h has units of "the rate of change of whatever".
Post by: hamdani yusuf on 06/08/2021 11:27:18
I do care, that's why I started this thread. The rest of the energy must go somehwere else, either being reflected, scattered, or absorbed.Where does the excess energy go?Who cares?Have you considered Planck's law?Yes, I forgot to mention that I'm using very cold pieces of glass.
It seems like you don't consider quantization of radiation energy. Planck didn't either.
Quote
Trying to find a physical explanation of the lawhttps://en.wikipedia.org/wiki/Planck%27s_law#Trying_to_find_a_physical_explanation_of_the_law
See also: Planck–Einstein relation
Once Planck had discovered the empirically fitting function, he constructed a physical derivation of this law. His thinking revolved around entropy rather than being directly about temperature. Planck considered a cavity with perfectly reflective walls; the cavity contained finitely many hypothetical well separated and recognizable but identically constituted, of definite magnitude, resonant oscillatory bodies, several such oscillators at each of finitely many characteristic frequencies. The hypothetical oscillators were for Planck purely imaginary theoretical investigative probes, and he said of them that such oscillators do not need to "really exist somewhere in nature, provided their existence and their properties are consistent with the laws of thermodynamics and electrodynamics.".[89] Planck did not attribute any definite physical significance to his hypothesis of resonant oscillators, but rather proposed it as a mathematical device that enabled him to derive a single expression for the black body spectrum that matched the empirical data at all wavelengths.[90] He tentatively mentioned the possible connection of such oscillators with atoms. In a sense, the oscillators corresponded to Planck's speck of carbon; the size of the speck could be small regardless of the size of the cavity, provided the speck effectively transduced energy between radiative wavelength modes.[82]
Partly following a heuristic method of calculation pioneered by Boltzmann for gas molecules, Planck considered the possible ways of distributing electromagnetic energy over the different modes of his hypothetical charged material oscillators. This acceptance of the probabilistic approach, following Boltzmann, for Planck was a radical change from his former position, which till then had deliberately opposed such thinking proposed by Boltzmann.[91] Heuristically, Boltzmann had distributed the energy in arbitrary merely mathematical quanta ϵ, which he had proceeded to make tend to zero in magnitude, because the finite magnitude ϵ had served only to allow definite counting for the sake of mathematical calculation of probabilities, and had no physical significance. Referring to a new universal constant of nature, h,[92] Planck supposed that, in the several oscillators of each of the finitely many characteristic frequencies, the total energy was distributed to each in an integer multiple of a definite physical unit of energy, ϵ, not arbitrary as in Boltzmann's method, but now for Planck, in a new departure, characteristic of the respective characteristic frequency.[80][93][94][95] His new universal constant of nature, h, is now known as Planck's constant.
Planck explained further[80] that the respective definite unit, ϵ, of energy should be proportional to the respective characteristic oscillation frequency ν of the hypothetical oscillator, and in 1901 he expressed this with the constant of proportionality h:[96][97]
ϵ=hν .
Planck did not propose that light propagating in free space is quantized.[98][99][100] The idea of quantization of the free electromagnetic field was developed later, and eventually incorporated into what we now know as quantum field theory.[101]
Post by: Bored chemist on 06/08/2021 11:28:35
But it's not sensible to then say, as you did
The time unit as numerator there,
Because the actual units are
Mass Distance ^2
Time
Time is clearly the denominator.
Post by: Bored chemist on 06/08/2021 11:32:03
I do care, that's why I started this thread. The rest of the energy must go somewhere else, either being reflected, scattered, or absorbed.You don't seem to care much, or you would have thought for a second about this.
the fraction of a THz beam reflected by a sheet of glass is about 1/10.
I didn't pick glass at random.
The more or less defining property of glass is that it is transparent.
So, if 10% is reflected then nearly 90% is transmitted and a small fraction absorbed.
Were you not able to work that out?
In this particular thought experiment, I catch each of the reflected beams in a mirrored cavity, and gift wrap the cavities so I can give them to my brother for Christmas.
Now, you insisted that it mattered what I did with them.
Please explain why it's important that my brother gets this rather odd Xmas present.
(Alternatively, admit that what happens to the rest of the beam is not actually important)
Post by: Bored chemist on 06/08/2021 11:34:07
ou don't consider quantization of radiation energy. Planck didn't either.Planck is recognised for being among the first to consider the quantisation of radiation.
That led to the law which bears his name.
Are you trying to set some sort of record for how many wrong things you can say in one post?
Post by: hamdani yusuf on 06/08/2021 12:49:42
Although they have the same dimension, they can't represent the same thing.
Post by: hamdani yusuf on 06/08/2021 12:51:01
Are you trying to set some sort of record for how many wrong things you can say in one post?What my record was? What it is now?
Post by: hamdani yusuf on 06/08/2021 12:59:36
I didn't pick glass at random.According to photon model, matter can only emit or absorb light energy in a discrete multiple of hf. That's why I asked how you would reduce the radiation energy to below the minimum nonzero limit.
The more or less defining property of glass is that it is transparent.
So, if 10% is reflected then nearly 90% is transmitted and a small fraction absorbed.
Were you not able to work that out?
In this particular thought experiment, I catch each of the reflected beams in a mirrored cavity, and gift wrap the cavities so I can give them to my brother for Christmas.
Now, you insisted that it mattered what I did with them.
Please explain why it's important that my brother gets this rather odd Xmas present.
(Alternatively, admit that what happens to the rest of the beam is not actually important)
Post by: Bored chemist on 06/08/2021 13:14:20
Post by: hamdani yusuf on 07/08/2021 08:45:37
Do you realise that 1 photon per second is not a limit because you can have one photon per week?How long does a photon with 1 THz frequency last, from beginning to its end?
Post by: hamdani yusuf on 07/08/2021 09:53:05
Post by: hamdani yusuf on 07/08/2021 10:20:16
The video shows that visible light cannot release electron from the metal plate, while UV lamp can.
The follow up questions would naturally occur. What would happen if we use lower frequency radiation, such as infrared, microwave, radio wave, or induction heater?
What if we use higher frequency, such as X ray and gamma ray?
Can strong enough red laser release the electrons from the plate?
Post by: Bored chemist on 07/08/2021 11:31:33
Here is a demonstration of photoelectric effect, which is thought as an evidence that light behaves like particles.Those questions were asked and answered a hundred years ago.
The video shows that visible light cannot release electron from the metal plate, while UV lamp can.
The follow up questions would naturally occur. What would happen if we use lower frequency radiation, such as infrared, microwave, radio wave, or induction heater?
What if we use higher frequency, such as X ray and gamma ray?
Can strong enough red laser release the electrons from the plate?
Post by: Bored chemist on 07/08/2021 11:32:37
Forever.Do you realise that 1 photon per second is not a limit because you can have one photon per week?How long does a photon with 1 THz frequency last, from beginning to its end?
Post by: hamdani yusuf on 07/08/2021 12:41:13
Those questions were asked and answered a hundred years ago.What's the answer?
Post by: Bored chemist on 07/08/2021 12:55:51
Is Google borken?Those questions were asked and answered a hundred years ago.What's the answer?
Post by: hamdani yusuf on 07/08/2021 13:36:52
What if the transmitter is only powered up for 1 second?Forever.Do you realise that 1 photon per second is not a limit because you can have one photon per week?How long does a photon with 1 THz frequency last, from beginning to its end?
Post by: Bored chemist on 07/08/2021 14:59:16
Then you can not tell if the frequency was 999,999,999,999.5 Hz or 1,000,000,000,000.5HzWhat if the transmitter is only powered up for 1 second?Forever.Do you realise that 1 photon per second is not a limit because you can have one photon per week?How long does a photon with 1 THz frequency last, from beginning to its end?
Post by: alancalverd on 08/08/2021 00:26:59
The follow up questions would naturally occur. What would happen if we use lower frequency radiation, such as infrared, microwave, radio wave, or induction heater?no photoelectrons
Quote
What if we use higher frequency, such as X ray and gamma ray?more photoelectrons
Quote
Can strong enough red laser release the electrons from the plate?In some circumstances you can get 2-photon effects but these rely on a nonlinear process which AFAIK does not occur in metals. The effect may have been predicted, but not observed, 100 years ago.
Post by: hamdani yusuf on 08/08/2021 01:30:18
Gamma ray can pass through the metal plate, which means that only small portion of its energy is available to eject the electrons.The follow up questions would naturally occur. What would happen if we use lower frequency radiation, such as infrared, microwave, radio wave, or induction heater?no photoelectronsQuoteWhat if we use higher frequency, such as X ray and gamma ray?more photoelectronsQuoteCan strong enough red laser release the electrons from the plate?In some circumstances you can get 2-photon effects but these rely on a nonlinear process which AFAIK does not occur in metals. The effect may have been predicted, but not observed, 100 years ago.
Induction heater can increase the temperature of the metal plate to produce thermionic effect. So, if the metal plate is kept at low temperature, eg using some cooler, the electron can be kept on the metal plate.
So, if the plate is kept at 0K, is the photoelectric effect by uv light still observed?
Post by: hamdani yusuf on 08/08/2021 04:02:12
Steel wool can be burnt using microwave, so there would be some thermionic effect.
Post by: hamdani yusuf on 08/08/2021 06:33:12
I'm not sure how adding that amount of uncertainty can compensate for the reduction of photon duration from forever to 1 second. Is this really the currently accepted theory?Then you can not tell if the frequency was 999,999,999,999.5 Hz or 1,000,000,000,000.5HzWhat if the transmitter is only powered up for 1 second?Forever.Do you realise that 1 photon per second is not a limit because you can have one photon per week?How long does a photon with 1 THz frequency last, from beginning to its end?
Post by: alancalverd on 08/08/2021 10:23:57
I'm not sure how adding that amount of uncertainty can compensate for the reduction of photon duration from forever to 1 second. Is this really the currently accepted theory?It's the consequence of wave mathematics and fourier analysis. By switching off after 1 second you have terminated the low frequency spectrum at 1 Hz.
Post by: alancalverd on 08/08/2021 10:35:05
There are several possible mechanisms for energetic photon interaction with a substrate. Any gamma rays that have passed through the metal plate without loss of energy obviously have not transferred their energy to any electrons. However several will have scattered, producing photoelectrons (most of which are reabsorbed in the plate) and secondary gammas which themselves can liberate photoelectrons. Some high energy (> 1.02 MeV) gammas will generate e-p pairs which annihilate to produce 511 keV photons, as we have been discussing, and these may also release photoelectrons.
Post by: Bored chemist on 08/08/2021 12:31:04
Microwave can produce observable electric spark on sharp metal objects.You can also do that with high voltage DC.
So what?
It's not got a lot to do with EM radiation, or with photons.
Post by: Bored chemist on 08/08/2021 12:39:53
I'm not sure how adding that amount of uncertainty can compensate for the reduction of photon duration from forever to 1 second. Is this really the currently accepted theory?Then you can not tell if the frequency was 999,999,999,999.5 Hz or 1,000,000,000,000.5HzWhat if the transmitter is only powered up for 1 second?Forever.Do you realise that 1 photon per second is not a limit because you can have one photon per week?How long does a photon with 1 THz frequency last, from beginning to its end?
You can think of it in terms of Fourier analysis, or you can just consider the "counting error".
If my clock ticks exactly once a second and I count the ticks in one minute, I will not always get 60 ticks.
Sometimes, if I start about half a second after the minute, I will get just 59.
If the "tick" has a finite duration, I can get 61 ticks in a minute if I catch the tail endo of one and the start of another
This isn't a matter of "currently accepted theory"; it's just common sense.
So, if the plate is kept at 0K, is the photoelectric effect by uv light still observed?You can not get something to 0K.
Do you understand that the photoelectric effect is not the same as the thermoelectric effect?
Post by: hamdani yusuf on 09/08/2021 00:12:29
We were talking about 1 THz transmitter rated at 1 Watt, attenuated using cascaded filters to produce a single photon. There should be around 1 trillion cycles of electromagnetic waves.I'm not sure how adding that amount of uncertainty can compensate for the reduction of photon duration from forever to 1 second. Is this really the currently accepted theory?It's the consequence of wave mathematics and fourier analysis. By switching off after 1 second you have terminated the low frequency spectrum at 1 Hz.
What do you mean about termination of low frequency spectrum?
Post by: hamdani yusuf on 09/08/2021 04:30:51
(https://www.gaussianwaves.com/gaussianwaves/wp-content/uploads/2014/07/Gaussian-Pulse-Matlab-FFT-how-to-plot-PSD-using-FFT-in-Matlab.png)
https://www.gaussianwaves.com/2014/07/generating-basic-signals-gaussian-pulse-and-power-spectral-density-using-fft/
This kind of pulse can be generated using electronic circuits. The value of σ can be adjusted. The vertical axis of frequency domain is not probability. It's magnitude instead.
(https://slidetodoc.com/presentation_image/4f68a0512a1182de0b7de9b545b1fde3/image-28.jpg)
(https://slidetodoc.com/presentation_image/4f68a0512a1182de0b7de9b545b1fde3/image-29.jpg)
https://slidetodoc.com/fourier-series-the-fourier-transform-what-is-the/
Quote
Shorter light pulses have broader spectra F(w) f(t) t w t w Duration of a light pulse: Width of the spectrum: So: The shorter the pulse, the broader the spectrum! Or inversely, to make a shorter light pulse you need to use light waves with a broader range of frequencies. 29Making the pulse shorter doesn't make the frequency becomes more uncertain. It makes the frequency has a broader range.
Post by: hamdani yusuf on 09/08/2021 08:08:56
(https://image.shutterstock.com/image-vector/fourier-transform-ft-600w-773535130.jpg)
(https://www.thenakedscientists.com/forum/proxy.php?request=http%3A%2F%2Ffaculty.salina.k-state.edu%2Ftim%2FmVision%2F_images%2Fcommon_FT_pairs.png&hash=44c1d607eaa0ed150845de7c608bfedd)
(https://www.thenakedscientists.com/forum/proxy.php?request=http%3A%2F%2Fwww.etc.tuiasi.ro%2Fcin%2FDownloads%2FFourier%2FTable2.gif&hash=953c644f86840785e2cb2bbbbe819cf3)
(https://www.thenakedscientists.com/forum/proxy.php?request=http%3A%2F%2Fwww.dsprelated.com%2Fblogimages%2FSteveSmith%2FSWSmith_Power_Law_Pairs.gif&hash=0515d14c58c6fbc67d286dc0e2156c6d)
Post by: alancalverd on 09/08/2021 09:57:13
Making the pulse shorter doesn't make the frequency becomes more uncertain. It makes the frequency has a broader range.Therefore your attempt to determine the "frequency" of the pulse is subject to greater uncertainty.
"Frequency" only has meaning for a truly repetitive phenomenon. Don't confuse it with the temporal half-width of a single pulse. See, for example, the frequency domain analysis of a delta function, that you have just shown.
Post by: hamdani yusuf on 09/08/2021 11:16:33
If we are forced to represent the curve in frequency domain into a single real number value, then yes, there will be some uncertainty need to be considered. It can be said to have a mean frequency with some standard deviation to represent the uncertainty, hence we need two numeric values. But it loses the information about the shape of the curve.Making the pulse shorter doesn't make the frequency becomes more uncertain. It makes the frequency has a broader range.Therefore your attempt to determine the "frequency" of the pulse is subject to greater uncertainty.
"Frequency" only has meaning for a truly repetitive phenomenon. Don't confuse it with the temporal half-width of a single pulse. See, for example, the frequency domain analysis of a delta function, that you have just shown.
Post by: alancalverd on 09/08/2021 11:32:38
Photon interactions are as close as you can get to delta functions.
Once again: we need two distinct mathematical models to describe the behavior of electromagnetic radiation. There is no reason why they should be interchangeable. Conceptually, particle interactions with a single energy, spatially distributed as described by wave propagation, seem to fit the bill.
Post by: hamdani yusuf on 09/08/2021 12:35:51
The frequency analysis of a delta function has no mean, unless you think that ∞/2 is a number!The curve goes in both positive and negative direction, hence the mean is 0.
Post by: alancalverd on 09/08/2021 12:41:39
More to the point, there is no such thing as a negative frequency. Fact is that it doesn't matter where you put the "center" frequency, the spectrum of a delta function is infinite, but the energy of a photon is absolutely defined.
Post by: hamdani yusuf on 09/08/2021 15:17:03
But E = hf and E > 0 for a photon!You can use online calculators such as wolfram alpha to show that the width of the curve in time domain is proportional to the height of the corresponding curve in frequency domain. So, if the time domain signal is infinitesimally thin, then the frequency domain curve is infinitesimally low. I'd like to elaborate further, but I'm afraid that I'll have to do it in new theory section.
More to the point, there is no such thing as a negative frequency. Fact is that it doesn't matter where you put the "center" frequency, the spectrum of a delta function is infinite, but the energy of a photon is absolutely defined.
Post by: alancalverd on 09/08/2021 17:33:30
0/x = 0 ∀ x, so Limx→0 (0/x) = 0
x/x = 1 ∀ x, so Limx→0 (x/x) = 1
x/0 = ∞ ∀ x, so Limx→0 (x/0) = ∞
Post by: Bored chemist on 09/08/2021 18:16:06
You can think of it in terms of Fourier analysis, or you can just consider the "counting error".which is enough to show that a finite transmission has an imprecisely defined frequency.
Post by: hamdani yusuf on 09/08/2021 23:29:35
Beware of dividing 0 by 0 and coming up with whatever number you like!That's precisely why Alex Kusenko rejected analysis of the Blackbird, which made him agree to take the wager, and eventually cost him 10k.
Remember Wolfram is at best a discrete approximation to a continuous function
For simple enough equations, WolframAlpha can solve them analytically, which means that it leaves no uncertainty. One side of the equation is just a symbolic representation of the other side.
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Algebraic equations consist of two mathematical quantities, such as polynomials, being equated to each other. Solving equations yields a solution for the independent variables, either symbolic or numeric. In addition to finding solutions to equations, Wolfram|Alpha also plots the equations and their solutions.https://www.wolframalpha.com/examples/mathematics/algebra/equation-solving/
Post by: hamdani yusuf on 10/08/2021 08:43:34
-Stephen Hawking.
We can't be sure if a model is accurate just because it works. Humans have been using fire for millenias before they know what it really is. They have been using arrows before they know about aerodynamics and simpler mechanics. They have been using boats long before they know about theory of buoyancy.
Post by: Bored chemist on 10/08/2021 08:51:43
We can't be sure if a model is accurate just because it works.No, but we can be sure a model is wrong if it doesn't work..
So we can rule out lots of possible "new models".
Post by: alancalverd on 10/08/2021 09:49:50
The problem with 0/0 is that the answer depends on the initial condition of an implicit function. You can't guess the initial condition from the endpoint.
You don't need a model to use a product. We have all sorts of useful devices whose physics is not fully understood - cross-country skis are a particularly good example of a clever piece of engineering that is between 2000 and 10000 years old but still not well modelled!
I don't think we have any illusions about the actual behavior of electromagnetic radiation, and most of us are humble enough to recognise that there isn't a useful single model that predicts everything. But we do have two models that work extremely well.
Returning to 0/0, you can see the problem with the downwind Blackbird car. The motive power comes from the difference between wind speed and ground speed, so once the car is moving, you should be able to reduce the wind speed to zero and the car will keep going.....a perpetual motion machine.....Wrong initial condition, wrong model!
Post by: hamdani yusuf on 10/08/2021 10:13:06
Returning to 0/0, you can see the problem with the downwind Blackbird car. The motive power comes from the difference between wind speed and ground speed, so once the car is moving, you should be able to reduce the wind speed to zero and the car will keep going.....a perpetual motion machine.....Wrong initial condition, wrong model!Not really. The force due to speed difference must overcome friction first before it can produce acceleration.
If the speed difference is exactly 0, what's the direction?
Post by: alancalverd on 10/08/2021 11:44:48
Post by: hamdani yusuf on 11/08/2021 06:36:22
More to the point, there is no such thing as a negative frequency.https://www.linkedin.com/pulse/significance-negative-frequency-fourier-transform-samaksh-kr-sharma
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Negative frequency doesn't make much sense for sinusoids, but the Fourier transform doesn't break up a signal into sinusoids, it breaks it up into complex exponentials.
These are actually spirals, spinning around in the complex plane:
(https://www.thenakedscientists.com/forum/index.php?action=dlattach;topic=68595.0;attach=32248)
Spirals can be either left-handed or right-handed (rotating clockwise or counterclockwise), which is where the concept of negative frequency comes from. You can also think of it as the phase angle going forward or backward in time.
In the case of real signals, there are always two equal complex exponentials rotating in opposite directions, so their real parts combine and imaginary parts cancel out, giving you a real sinusoid as the result. Depending on the phase of the two spirals, they could cancel out leaving a purely real sine wave or a real cosine wave or a purely imaginary sine wave, etc.
The negative and positive frequency components are both necessary to produce the real signal, but if you already know that it's a real signal, the other side of the spectrum doesn't provide any extra information, so it's often hand waved and ignored. For the general case of complex signals, you need to know both sides of the frequency spectrum.
The meaning of negative frequencies is just mathematical(not physical) similarly to the imaginary part of a complex signal. In real world, the negative frequency does not exists and the spectral content on negative frequencies must be added to the spectral content at the positive frequencies, to save energy. The concept of negative frequencies is used broad wide namely in this sense, in the connection with complex numbers.
Post by: hamdani yusuf on 11/08/2021 06:47:51
Nothing to do with initial acceleration. I have assumed that the vehicle is moving at ground speed x, and the wind then drops from y to zero. Does the car continue to extract energy from the difference between wind speed (y = 0) and ground speed (x)?The extractable energy is from the speed difference between wind and ground. When it's 0, the wind car can only decelerate.
Post by: TommyJ on 11/08/2021 08:35:45
Negative frequency doesn't make much sense for sinusoids, but the Fourier transform doesn't break up a signal into sinusoids, it breaks it up into complex exponentials.Frequency is the rate at which something happens, so by defintion it must be a positive real number. both negative and imaginary numbers are very useful in the process of calculating the value of real quantities, but they have no physical reality of their own.
There is the angular frequency, which describes the angular velocity of the rotating objects (radian per seconds). Since rotation has two directions then the angular frequency may be positive or negative.
The presentation/formulation of the Fourier spectrum is simplified by using the concept of 'negative frequencies'.
Post by: alancalverd on 11/08/2021 10:22:48
Post by: evan_au on 11/08/2021 11:18:10
Quote from: hamdani yusuf
Negative frequency doesn't make much sense for sinusoids, but the Fourier transform doesn't break up a signal into sinusoids, it breaks it up into complex exponentials.
Euler's formula says rei(ft) = r( cos(ft) + i sin(ft) )
- Where r is the amplitude, f is the frequency and t is time
- So a "complex exponential" can be treated as a sinusoid with an amplitude and phase.
See: https://en.wikipedia.org/wiki/Euler%27s_formula
The point is that the Fourier transform is reversible, with the inverse Fourier transform.
- The Fourier transform of a sine wave produces a positive and negative frequency
- If you throw away the negative frequency component, you have just thrown away half of the energy
- If you then apply the inverse Fourier transform to this corrupted signal, you won't get out what you put in.
See: https://en.wikipedia.org/wiki/Fourier_transform#Definition
The positive and negative frequency components become important for Amplitude-Modulated transmission (which used to be A Thing).
- If you modulate a 1 MHz carrier with a 1kHz sinusoid, you end up with extra frequency components at 1MHz ± 1kHz.
- So the negative frequency components do appear in real applications.
- If you throw away the negative frequency component (and the 1MHz carrier), you end up with "Single Sideband" modulation. It's not very intelligible on an AM receiver...
See: https://en.wikipedia.org/wiki/Amplitude_modulation
https://en.wikipedia.org/wiki/Single-sideband_modulation
Correction: Unmixed the frequency and phase angle
Post by: TommyJ on 11/08/2021 12:04:26
It is proposed that the positive charge propagates in the radial direction and at the same time it rotates clockwise perpendicular to the radial direction. The negative charge rotates counter clockwise resulting that the two charges propagate along in the radial direction.
- Photon, which is a 'pure' energy, spin with no mass, can 'convert' itself in a particle, that has a mass. Electron or a positron can be formed (negative or positive charge). This happens all the time around us with a spontaneous emission or absorption of energy.
Changing colours in a visible spectrum. But not only the colours, but the structure of nature itself is changing. The dynamics of cell life is a fundamental of this process with the organization of positive and negative charge.
'Photon forms the momentary nature of time and evolutionary driving force of everything.'
Photon has momentum = plank constant / frequency = [kg*m/s]
No 'negative frequency' here.
Post by: hamdani yusuf on 11/08/2021 12:32:56
- The present study proposes that the photon consists of a positive elementary charge and a negative elementary charge. The charges are loosely coupled, i.e. they do not form a fixed dipole.Is it a new theory?
It is proposed that the positive charge propagates in the radial direction and at the same time it rotates clockwise perpendicular to the radial direction. The negative charge rotates counter clockwise resulting that the two charges propagate along in the radial direction.
- Photon, which is a 'pure' energy, spin with no mass, can 'convert' itself in a particle, that has a mass. Electron or a positron can be formed (negative or positive charge). This happens all the time around us with a spontaneous emission or absorption of energy.
Changing colours in a visible spectrum. But not only the colours, but the structure of nature itself is changing. The dynamics of cell life is a fundamental of this process with the organization of positive and negative charge.
'Photon forms the momentary nature of time and evolutionary driving force of everything.'
Photon has momentum = plank constant / frequency = [kg*m/s]
No 'negative frequency' here.
Post by: alancalverd on 11/08/2021 13:08:39
- If you modulate a 1 MHz carrier with a 1kHz sinusoid, you end up with extra frequency components at 1MHz ± 1kHz.
- So the negative frequency components do appear in real applications.
I'm being pernickety, but for a good reason: lots of pseudoscience lurks in the wings of this stage, ready to pounce and suck the blood out of loose physics!
999 kHz is not a negative frequency.
Post by: alancalverd on 11/08/2021 13:12:50
- Photon, which is a 'pure' energy, spin with no mass, can 'convert' itself in a particle, that has a mass. Electron or a positron can be formed (negative or positive charge). This happens all the time around us with a spontaneous emission or absorption of energy.But what is observed is that a photon with energy greater than 1.02MeV can be converted into an e-p pair. Nothing else.
If your theory is not matched by observation, it is wrong. Or religion, politics or economics.
Post by: hamdani yusuf on 11/08/2021 13:32:15
Photon interactions are as close as you can get to delta functions.At what frequency of photon?
What's the time resolution of the instruments used for measuring it?
Can the delta function be replaced by a very thin gaussian?
Post by: TommyJ on 11/08/2021 13:34:56
- If you modulate a 1 MHz carrier with a 1kHz sinusoid, you end up with extra frequency components at 1MHz ± 1kHz.Isn't it rather frequency shift than negative frequency.
- So the negative frequency components do appear in real applications.
Post by: TommyJ on 11/08/2021 13:38:28
Is it a new theory?Nothing exceptionally new in theory. Shaw case.
Post by: hamdani yusuf on 11/08/2021 13:45:13
Once again: we need two distinct mathematical models to describe the behavior of electromagnetic radiation. There is no reason why they should be interchangeable.This reminds me of Aristotelian physics.
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Thus, in Aristotelian dynamics, there was a distinction between "natural" downward motion (for example, a rock falling to the ground when dropped) and unnatural violent motion not directed toward the center of Earth (such as that resulting from a catapult).https://aether.lbl.gov/www/classes/p10/aristotle-physics.html
In contrast to earthly motions, in the supralunary regions of the heavens the natural state of motion was circular, because circles were considered to be the perfect geometric figure. Thus the planets would travel forever in circular orbits without the intervention of any force or impetus, because, well, it's the natural thing for planets to do.
I think that someday, someone will come up with a better idea.
Post by: hamdani yusuf on 11/08/2021 13:51:36
It's similar to negative apples or cows for farmers. If you borrow them, negative numbers will come up somewhere in the accounting records.- If you modulate a 1 MHz carrier with a 1kHz sinusoid, you end up with extra frequency components at 1MHz ± 1kHz.Isn't it rather frequency shift than negative frequency.
- So the negative frequency components do appear in real applications.
Post by: alancalverd on 11/08/2021 15:13:51
At what frequency of photon?1. E = hf
What's the time resolution of the instruments used for measuring it?
Can the delta function be replaced by a very thin gaussian?
2. We usually measure wavelength for visible light to soft x-rays, and energy for higher energy photons You can be as accurate as permitted by Heisenberg, and AFAIK we have a good (quantum) explanation for any observed bandwidth.
3. Problem is that however thin you make it, a gaussian still has a curved maximum. The delta function either is or isn't - it's a mathematical concept of an infinitely differentiable function.
There has been some work on attosecond resolution of photon-matter interactions - beyond my current pay grade but some chums at the National Physical Laboratory were investigating them back in the day. Unfortunately most undergraduate texts represent the photon as a smooth chirp, which it isn't!
Post by: evan_au on 11/08/2021 22:59:48
Quote from: evan_au
- If you modulate a 1 MHz carrier with a 1kHz sinusoid, you end up with extra frequency components at 1MHz ± 1kHz.
- So the negative frequency components do appear in real applications.
Quote from: alancalverd
999 kHz is not a negative frequency.True, but it shows the presence of negative frequencies in the Fourier transform.
What I was trying to say (but it was getting late) was the Fourier transform of a product of functions is the convolution of their Fourier transforms:
- Amplitude Modulation multiplies the Carrier wave by the Modulating wave
- This translates to a Convolution of their spectra (ie a convolution of the spectrum of 1kHz and 1 MHz):
Fourier_Transform.png (24.42 kB . 1431x84 - viewed 2263 times)Convolution moves (if I've got my signs right):
-the normally-visible +1kHz to 999kHz
- and the normally-invisible -1kHz to 1001 kHz
- and creating new components at -999 and -1001 kHz (which are still invisible)
- If these negative frequency components did not exist, the product of 1kHz and 1 MHz would produce just one sideband and the carrier. In fact it produces two sidebands.
The reason spectrum analysers don't show you the negative frequencies is that for signals that are real in the time domain, the negative frequencies are a mirror image of the positive frequencies, and don't add any new information.
Quote from: TommyJ
Isn't it rather frequency shift than negative frequency.Yes, convolution can be viewed and calculated as a shift between two functions (in fact, a reversal of one, followed by every possible shift).
Convolution comes up in my work when finding the sum of two probability distributions.
See: https://en.wikipedia.org/wiki/Convolution
Post by: hamdani yusuf on 12/08/2021 01:19:52
The delta function either is or isn't - it's a mathematical concept of an infinitely differentiable function.Is it physically possible?
Post by: alancalverd on 12/08/2021 10:26:33
I guess the Big Bang was pretty close to a delta function: zillions of years later, we can still detect the low frequency components!
Post by: hamdani yusuf on 12/08/2021 12:33:47
No, but it's more useful than a deity.What's the vertical axis? I assume the horizontal axis represents time.
I guess the Big Bang was pretty close to a delta function: zillions of years later, we can still detect the low frequency components!
Afaik, it's billions of years ago.
Post by: hamdani yusuf on 12/08/2021 12:47:18
You can be as accurate as permitted by Heisenberg, and AFAIK we have a good (quantum) explanation for any observed bandwidth.It means that the pulse is not a truly delta function. It looks more like a thin Gaussian instead.
What's the quantum explanation for my experiments with microwave, such as longitudinal propagation in meander sheet, conjoined twin polarizers, and polarization splitter?
Post by: Bored chemist on 12/08/2021 12:59:08
What's the quantum explanation for my experiments with microwave, such as longitudinal propagation in meander sheet, conjoined twin polarizers, and polarization splitter?Photons have a property called polarisation.
Post by: alancalverd on 12/08/2021 13:54:55
What's the quantum explanation for my experiments with microwave, such as longitudinal propagation in meander sheet, conjoined twin polarizers, and polarization splitter?These are cases where the wave model of electromagnetic radiation works really well, and the photon model isn't helpful.
A model is not an explanation. It's a mathematical approximation to observation.
Post by: hamdani yusuf on 13/08/2021 13:16:47
Photons have a property called polarisation.How does the polarization behave?
What determines it when the photon is created?
How can it be changed?
Post by: hamdani yusuf on 13/08/2021 13:24:18
Not all waves have the same behavior. Only in conductors electromagnetic waves can propagate longitudinally.What's the quantum explanation for my experiments with microwave, such as longitudinal propagation in meander sheet, conjoined twin polarizers, and polarization splitter?These are cases where the wave model of electromagnetic radiation works really well, and the photon model isn't helpful.
A model is not an explanation. It's a mathematical approximation to observation.
Post by: Bored chemist on 13/08/2021 13:27:15
How does the polarization behave?Just the way you showed it did in your experiments.
Post by: hamdani yusuf on 13/08/2021 18:23:14
List of random facts doesn't represent scientific knowledge. It should contain general rules governing behaviors of objects in various but related situations. Comprehension is a data compression process.How does the polarization behave?Just the way you showed it did in your experiments.
I designed the filters in my experiments with microwave based on antenna theory as shown in the old training videos by Royal Canadian Air Force.