Post by: Eternal Student on 12/11/2022 20:00:33
Does anyone know of an experiment that shows the oscillation in either the electric or magnetic field (or both) when a light ray travels through some place?
For a radio wave, we can just put a metal rod (an antennae) in the path of the wave and show that an electrical signal is generated in the antenna. We can argue that this is the result of forces from the electric field, F = Ee and also the magnetic field (which is moving relative to the static charges in the rod, giving us some v>0 in the usual formula) F = e (v x B) ). There is plenty of equipment, like a good multimeter, that will directly show we get an electrical signal when it's at these sorts of radio frequencies. Can we do anything like this with e-m radiation in the visible spectrum? Is there an antenna that picks up light (rather than reflects it) and equipment fast enough to respond and show we have a rapidly oscillating electrical signal?
It doesn't have to be an antenna. Is there perhaps some experiment that has been done with a line of charges and then a laser beam was fired along the path of that line of charges? The laser being used to ensure you get a nice ray of light which is all in phase. (Hopefully causing some rapid oscillation in those charges that we could measure).
Anyway, is there any experiment that you know of that directly (or as directly as possible) shows the oscillation in the electric (or magnetic) field for e-m radiation when the frequency is up in the visible band (or higher)?
Is there some ultra-fine probes that we can attach to a multi-meter and just stick them into space and directly measure the oscillating E field as the light ray passes through?
Best Wishes.
Post by: evan_au on 12/11/2022 21:43:28
- When the wall is oriented in one direction, it "short-circuits" the E-field and the microwaves are blocked
- Rotate the wall by 90°, and the microwaves get through, as the wires don't short-circuit the magnetic field.
A similar arrangement (on a smaller scale) is used to produce polarised lenses for visible light - there is a grid of long, parallel crystals in the lense.
Post by: Eternal Student on 13/11/2022 01:18:00
Thanks for your reply, @evan_au . I do quite like the idea of polarising lenses and I am busy thinking about it. It might be enough for my purposes and I'm very grateful for your time.
I'd rather have a more direct demonstration that there are oscillating E and B fields in light, for example, if there was some electrical device you could connect to the crystalline structures in a polarising filter and show that there is an electrical current generated. Otherwise, you just have to accept that polarising lenses do work the way we think and the evidence for that is not at all convincing for the typical person.
Rather than simply blocking light of one polarisation, a polarising lens does sometimes "appear" (I'm not saying they do, only that it may appear) to actively interact with passing light and turn it.
You've probably heard of the experiment that is done to show quantum mechanical effects with polarised light.
Two polarising lenses turned at 90 degrees will block all light passing through. However, if you insert a third lens in between that is offset by only 45 degrees, then you will get some light transmitted through the end of the optical system. This is weird because if the polarising filter is just blocking some light then every filter you add should only remove more light, yet inserting the third filter has somehow boosted the amount of light you get at the end.
See, for example, this reference: https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Quantum_Tutorials_(Rioux)/Quantum_Fundamentals/08%3A_Quantum_Principles_Illuminated_with_Polarized_Light
or this video, if you prefer a more dynamic demonstration: Approx. 1 min 30 secs. "Three polarizing filters: a simple demo of a creepy quantum effect", from YouTube.
You can explain this with QM but a stubborn few will stick with classical physics and suggest that polarising filters just don't work as we thought. If you reduce it down to a simple level, we can fire just two photons (with the same polarisation) at a polarising filter that is off-set by 45 degrees and the filter doesn't seem to be just blocking something. You don't observe something consistent with just stopping (blocking) 0, 1 or 2 of the photons. Most of the time you observe one photon thrown out at the other side and it has been turned so that it is unlike either of the incoming photons.
Anyway, I'm not trying to be awkward. I've just read or heard something presented on a lecture course I was trying to follow and it seems to be wrong, or at least mis-leading. I'm trying to reconcile my existing understanding with what was said. This post is already too long, so I'm not even going to try and discuss the lecture right now. (Besides which, my own difficulties in understanding something aren't all that interesting to others, it's my problem not yours / theirs).
I've always been prepared to accept the oscillatory nature of the E field and B field in light as "a given". It follows from the theory and can be quite directly observed in radio waves. However, I just wondered if it could be stated in a more concrete way: If there is some very direct demonstration of the oscillating E and B fields in visible light.
Best Wishes.
Post by: evan_au on 13/11/2022 06:53:22
Quote from: video
If I ask the same questions in a different order, I get a different answer, which is a little disturbing to meIn my primitive understanding, passing light through a polarising filter is like a test or a measurement of the polarization.
- If a photon passes through the filter, it has "passed" the test, and you have measured its polarisation as being in-line with the filter.
- If you put another polarising filter at 90°, everything passing the first filter will fail the second filter.
- However, if you put a 45° filter in between two 90° polarising filters, you are conducting another test/measurement. The outcome of this measurement is that cos2(45°) = 50%
In a quantum world, taking a measurement changes the thing you are measuring. So it is no longer light polarised vertically, any photons passing the 45° filter are now polarised at 45°. And a fraction of these will pass the third filter.
Corrected, as advised below by Eternal Student...
Post by: paul cotter on 13/11/2022 10:20:58
Post by: Colin2B on 13/11/2022 15:20:49
With regard to polarising filters I can't help wondering is there a rotation taking place?You are still looking at this through classical eyes. The problem is that in subsequent measurements you are changing the bases (similar to coordinates) that you are using to take the measurements.
Post by: Eternal Student on 13/11/2022 15:27:39
Yes that's the right idea @evan_au . Except that you need to take squares, so the probability of a photon passing the filter is 0.5 not 0.71... Under classical field theory (so that is E and B fields) the intensity of light received is proportional to the square of the amplitude of the wave that is transmitted. You resolve the incoming wave into components and get the fraction Cos 45° for the amplitude that is transmitted exactly as you stated. That means Cos2 45° = ½ is the fraction in Intensity. Then you can jump back to the particle and photon interpretation where intensity is just proportional to the number of photons received (per unit time and per unit area of the receiver). Hence, you get the probability ½ for each photon.
Many compounds with chiral asymmetry that exist in right and left handed configurations will rotate the plane of polarisation...I completely agree. It's not clear that polarising lenses based on crystalline macromolecules really are just blockers rather than rotators of polarised light. (I also agree with the other statement but that's why I'm asking if anyone knows of a suitable experiment).
Best Wishes.
LATE EDITING: Overlap with Colin2B who just posted. At a glance, looks OK and this post doesn't need adjustment.
Post by: evan_au on 13/11/2022 20:15:35
Quote from: Paul cotter
I can't help wondering is there a (polarisation) rotation taking place?This is exactly how Liquid Crystal Displays work (or LCD displays, as they are tautologically identified).
- The liquid crystal material is sandwiched between two polarizing filters
- The liquid crystal rotates the polarisation of light under the control of an electric field. This allows light to pass (or not pass), depending on whether a voltage is applied to transparent electrodes.
https://en.wikipedia.org/wiki/Liquid-crystal_display
Post by: alancalverd on 13/11/2022 21:37:27
Post by: Eternal Student on 13/11/2022 22:51:50
I like the idea @alancalverd and thank you for your time.
You do seem to be establishing that light carries energy. I'm not sure if it tells me much about light being an oscillation in the E field. If you fired lumps of blu-tack (that's sticky clay stuff - but other brands are available) at a cup of tea so that they just stick to the cup, then you are transferring kinetic energy to the cup. That's got to give you a warm cup of tea eventually. I think that's how they do it at our local Cafe, it takes years and the cup is sticky. I'm fairly sure they don't use oscillating E field blu-tack, just the regular stuff.
More seriously, I think your suggestion solves half the problem and I'm continuing to think about it, thank you.
Best Wishes.
Post by: Colin2B on 13/11/2022 23:36:30
.Yes that's the right idea @evan_au . Except that you need to take squares, so the probability of a photon passing the filter is 0.5 not 0.71... Under classical field theory (so that is E and B fields) the intensity of light received is proportional to the square of the amplitude of the wave that is transmitted. You resolve the incoming wave into components and get the fraction Cos 45° for the amplitude that is transmitted exactly as you stated. That means Cos2 45° = ½ is the fraction in Intensity. Then you can jump back to the particle and photon interpretation where intensity is just proportional to the number of photons received (per unit time and per unit area of the receiver). Hence, you get the probability ½ for each photon.Yes, you’ve hit on the main reason why a lot of people get confused with the polarisation experiments.
Sorry, I didn’t have time to reply properly before, just unpacking having returned from a 2 week trip to be alongside a relative in hospital. Not a lot of time available.
As you are aware, polarising plastics use long thin chains of molecules such that the electrons will oscillate along the chain and absorb the photons, whereas perpendicular to the chain there there is no absorption and the light passes through. The experiment you are looking for is a tricky one as you are trying to measure specific motion of electrons in the material and to do that means disturbing the very electrons you are looking to measure.
This experiment is takes a different approach and might be similar to what you are looking for https://www.sciencedaily.com/releases/2018/04/180410100941.htm
Post by: Eternal Student on 14/11/2022 00:52:14
That does look like the sort of thing I was after. Thank you very much.
Best Wishes to you and your relative.
Post by: alancalverd on 14/11/2022 09:29:05
Shine white light through an organic filter - say a theatrical gel. Some frequencies have been absorbed (and any theatre lighting technician will tell you how much because the filters get hot!) but the filter pigment has negligible magnetic susceptibility, so the electric field of the incident light must be interacting with the molecules in a frequency-specific manner.
All this presumes, of course, that you assume light to be an electromagnetic wave in the first place. So far, nobody has produced a better model of refraction, diffraction and interference, and the application of Maxwell's equations does indeed give us an exact value for its velocity derived entirely from static measurements.
*There is often a perceptual problem when dealing with high frequency EM radiations. Thanks to transient photoelectrochemistry, our bodies are exquisitely sensitive to the intensity of visible radiation, in the 2 - 5 eV range, but the rest of our senses are very dull in comparison, so we don't associate physiological effects with low incident powers outside that range, hence it is easy to forget that visible light transfers energy.
Post by: Colin2B on 14/11/2022 09:32:25
I like the idea @alancalverd and thank you for your time.Worth having a deep think about Alan’s suggestion. Light doesn’t have mass, so that’s not the source of any energy transfer. When looking at electrons you would expect the charge to be affected by an electric field, which would transfer energy and is certainly the origin of the photoelectric effect. Unless you can think of a better transfer mechanism.
You do seem to be establishing that light carries energy. I'm not sure if it tells me much about light being an oscillation in the E field. If you fired lumps of blu-tack (that's sticky clay stuff - but other brands are available) at a cup of tea so that they just stick to the cup, then you are transferring kinetic energy to the cup. That's got to give you a warm cup of tea eventually. I think that's how they do it at our local Cafe, it takes years and the cup is sticky. I'm fairly sure they don't use oscillating E field blu-tack, just the regular stuff.
I did wonder about experiments using the optical Mossbaur effect with polarised light. There seem to be some experiments out there, but you would have to go through them to decide whether they answer your question.
Note, just got a warning of another post while I’ve been writing. Similar thoughts from Alan
Post by: yor_on on 02/12/2022 12:38:26
This one?
" Another idea I had how to show that light is an electromagnetic wave would be to use a stronger (and well cooled) antenna (a stick in which electrons can be accelerated) than Hertz. The electromagnetic waves in the antenna have the frequency in which the electrons move up and down inside. So if one would increase the frequency more and more (by applying strong alternating voltage with ever increasing frequency), one should eventually see these radiations in form of light (as manifestation of the osciallations of electrons which are electromagnetic waves). "
https://physics.stackexchange.com/questions/145330/what-is-the-experimental-evidence-that-light-is-an-electromagnetic-wave
It's a interesting question, what made you want to ask there?
Post by: Bored chemist on 02/12/2022 12:59:31
https://skullsinthestars.com/2008/05/04/classic-science-paper-otto-wieners-experiment-1890/
Post by: paul cotter on 02/12/2022 20:03:15
Post by: Eternal Student on 02/12/2022 22:39:01
Sory, I haven't been keeping much of an eye on this thread any longer. I'll catch up and follow the links people had suggested.
It's a interesting question, what made you want to ask there?I didn't ask on Physics stack exchange, if that's what you mean. That was someone else.
I asked here for a reason that was releated to something I read or heard from an online lecture course recently. I can try and summarise that but it will take me a day or so to dig up all the details again. I'd like to do that because otherwise I have a tendancy to just "ramble".
Best Wishes.
Post by: Bored chemist on 03/12/2022 00:10:05
it is not possible to run a signal frequency up to that of visible light.That's a matter of definition
https://en.wikipedia.org/wiki/Frequency_comb
Post by: yor_on on 03/12/2022 11:56:40
Post by: Eternal Student on 03/12/2022 23:37:24
...what made you want to ask there?
As I mentioned, my question follows from something I heard in a lecture course and I'm still reluctant to accept. However, I'm open minded and have to give the material some serious consideration. So I'll start by presenting some of that material. (Just to be clear then: Not all of this is my opinion, I'm attempting to present their ideas. I've got some stuff under a spoiler that mentions some of the concerns I would have and I've tried to emphasize *their main claims* wherever I really wouldn't want anyone to think it was my claim). Even though you ( @yor_on ) asked for this, feel free to skip it when you're half way through and bored.
Can we observe light waves?
Superficially you will probably think that you can, of course you can, there are diffraction gratings and all sorts of things about the way light behaves that only makes sense if you treat it as a wave. (That's what I thought)
However, the lecture took an opposing view from the outset. All detectors, for example a detector you have in a lab, just detect energy. A simple detector is really just detecting a heat rise or energy deposit when photons come into it and are absorbed. Some of the more advanced detectors can just about detect a single photon but they are still only detecting an energy deposit and not the oscillation of any wave. Much the same is true for other ways of detecting light such as the human eye. A few individual photons must strike a rod or cone and stuff happens... electrons can be excited, chemical changes could take place etc.. Indeed any atom absorbing light is the textbook example of a quantum effect, a photon must have the right energy. It doesn't matter how many photons you throw at it or how quickly they are thrown one after another, if no photon has the right energy then the electron does not get excited. The key is that the word "energy" is always there, atoms absorbing light, rods and cones in your eye reacting to light or a sophisticated detector in a lab - they all just react to or detect energy. We can observe light, that's for certain, however we don't seem to observe the wave that is supposed to be in a light wave.
Let's refer back to some of the other replies from people on this thread. We're probably all in agreement that light carries energy, however that falls short of establishing that we are detecting the wave aspect of the thing.
It is obvious that light carries energy*........ (and @alancalverd went on to describe experiments where light transfers energy to something and then argued that this could only be due to an electric field)...
Worth having a deep think about Alan’s suggestion. Light doesn’t have mass, so that’s not the source of any energy transfer....
Idea 1: There is a loss of wave characteristics due to time averaging.
A typical plane wave solution to the wave equation can be written as
= E0 . Cos ( k.x - ωt ) 
[Eqns 1]
and 
are unit vectors orthogonal to each other and to the direction of propagation. k = wave vector. x = position vector. ω = angular frequency. t = time.The oscillatory or wave characteristic is clearly provided by the Cosine term.
Now the energy carried in an electromagnetic wave can be determined by using the Poynting vector:
[Eqn 2]
and the intensity, I of a wave is ultimately given by
where < >t denotes an average over time. We're only interested in an average over time because (*it is claimed that*) no detector can ever find an instantaneous intensity from a light wave. Instead, they only detect photons and we must then consider the device to be collecting the energy over a minimum period of time that allows the absorption of one complete photon.Hence,
where we have used [Eqns 1] to substitute for ExB.The basic claim is that T must be one full time period of oscillation. (I'm not happy with that but that's the basic idea presented in the lecture). Anyway, take T as one full time period (T = 2π/ω) and then we can perform the integral easily enough with the substitution θ = ωt to obtain:
[Eqn 3]
Where we have used the relationship c2 = 1/(μ0ε0). The final expression for intensity should be familiar since it holds when we allow T (the time over which we average) to approach infinity and so it's just the usual expression for the (average) intensity of an e-m wave. The key point here is that we have only averaged over one time period, the claim is that this is the minimum amount of time for a detector to absorb a photon. Looking at the RHS of [Eqn 3] we can clearly see that there is no time or space dependence suggesting a wave. There is no Cos ωt term for example. The information about the wave or oscillation is simply lost. The detector responds to the energy delivered by the photon and it really doesn't care about the oscillation in the E field. We could have the e-m wave oscillate in some other way (e.g. a square wave instead of cosine wave, or even just hold constant) and it wouldn't matter provided the energy delivered was the same. I'm not going to repeat all of the lecture so we just need to speed through and consider the possibility that the oscillation of an E field (and B field) doesn't even need to be there. It could just be a convenient model for the energy carried by a photon. There is certainly no information about the wave in what the detector reacts to - which is only the intensity over 1 time period.
Spoiler: show
Idea 2: We only ever observe something related to the square of the modulus of the wave, |E|2 and not E directly.
Consider a generic plane wave solution to the wave equation which we can write in the following form:
E = E0 eiΦ where all the oscillation is due to the complex exponent iΦ. The time dependence resides entirely in the phase term Φ (e.g. Φ = kx-ωt).
Then |E|2 = |E0|2 . | eiΦ |2 = |E0|2 and the phase term is again just lost, there is no term showing an oscillation.
Spoiler: show
- - - - - - - -
That's enough about the lecture content. It went on and said more stuff but overall I finished and thought.... I'm really not sure that was money well spent. None the less, I'm open minded and considered it. Do we have any lab detectors that can genuinely detect or react to the oscillatory nature of light or are all detectors really just picking up the energy of a photon?
Let's take a reply from @Bored chemist while I'm here: Thanks for recommending the article about Wiener's experiments. In that experiment a photographic plate was used to show the wave nature of light. The trouble is the interaction between the light and photosensitive material is very much a 'photon and atoms' interaction much as described earlier. At the point of interaction, the material was simply reacting to a deposit of energy by a photon and not conclusively to the wave or oscillation that may have been in the light. Now don't get me wrong, it's clear that we would have trouble trying to explain the results of that experiment unless we assign some wave properties to light but that's a slightly different thing.
It was agreed, even in the controversial lecture, that it is a very useful model to consider light as an oscillation in the E and B fields. There are many experimental results that can only be explained by assuming light has a wave-like nature. The main issue is only that this wave nature can not be directly observed. We can draw some immediate parallels with Quantum Mechanics and it will probably help to explain what they (the lecturer and his team) were driving at. In QM we assume a wave function, ψ(x,t) , exists and describes the state of a particle but we accept that you cannot directly observe or measure the wave function. It is a model and may not be anything real or tangible that you can measure. There are observables, things that we can measure or observe, you can find the position or momentum of the particle for example - but we can't directly observe or measure the wave function. The best you could hope for is that you measure something that will collapse the wave function to a known state.
Their (the lecturer and team) main argument is that Light may behave like an oscillation in the E field under certain circumstances but when you go to measure the oscillation in the E field you just can't. All you can detect is a photon and the properties of a photon which don't seem to contain any information about the wave or oscillation in the E field that you might assume existed. E.g. the energy of the photon, as outlined above, does not depend on the wave form like E=E0 Cos ωt that you have assigned to the E field.
Anyway, as I've probably indicated I'm not especially convinced that the online lecture was worth the money but I've got to think about it and try to dismiss some of it systematically and categorically. The trouble is, I just do not know of any experiment or piece of equipment that really does detect the oscillation in the E field when some light passes by, hence the original question. There is no shortage of experiments that can only be explained by assuming light has wave-like properties but that just isn't good enough. I'd very much like to establish that the oscillations in the E field are there and can be directly observed and measured. Then there is no doubt that assigning an oscillation in the E field to describe light is NOT just a model to explain wave-like behaviour in light. Compare with the wave function in Quantum Mechanics (i.e. something that could be regarded as a useful model but cannot be directly observed).
Best Wishes.
[Edited to remove the error spotted by @paul cotter in post #51 ]
Post by: Eternal Student on 04/12/2022 01:57:00
I'm sorry to write on the post again when no-one has had any time to respond yet. I'm just a bit troubled and would like to ask another question. Which might help to focus attention on what is happening, or claimed to happen.
In the post just above I mentioned this:
Indeed any atom absorbing light is the textbook example of a quantum effect, a photon must have the right energy. It doesn't matter how many photons you throw at it or how quickly they are thrown one after another, if no photon has the right energy then the electron does not get excited.
Now suppose that making an oscillation in the E field and B field is all you need to do to create some light. You create an appropriate oscillation and what you have is some light there.
Now put an atom (or quite a few atoms like a small block of metal because that will be easier) in the middle of some laser emitters, something like the diagram below:
[ Invalid Attachment ]
The idea is that you surround the atom with lasers at an angle from the atom that is whatever you might need, with the lasers at whatever distance from the atom you might need, have as many of these lasers as you might need, and finally allow the lasers to generate light of a frequency that is whatever you might need EXCEPT the frequency for "turquoise" coloured light. The idea is that the superposition of the E field and B field from all of them combines to give you something that is at least a very good approximation to an e-m wave with the wavelength for "turquoise" at the location of the atom, for at least a short while (the lasers had different frequencies so we can anticipate that there isn't going to be a good standing wave there, the overall superposition of all the E fields and B fields will may change as some lasers slip out of phase with the others BUT for at least a short while, say "long enough" you had some E and B field oscillation that was a good approximation to what we want). Obviously the colour "turquoise" was supposed to be the colour at which an electron will transition between orbitals in that atom (I used "turquoise" because the obvious choices like red, green, blue, yellow were already used on my diagram).
Now the question is: Will the electron make a transition to the new orbital? None of the individual photons from the lasers were of the right frequency for this to happen but will they combine under a superposition of their E and B fields to give you something that will be enough to make the electron jump?
Best Wishes.
Post by: Colin2B on 04/12/2022 08:52:45
Hi again,Hi ES
I'm sorry to write on the post again when no-one has had any time to respond yet.
Now the question is: Will the electron make a transition to the new orbital? None of the individual photons from the lasers were of the right frequency for this to happen but will they combine under a superposition of their E and B fields to give you something that will be enough to make the electron jump?
Best Wishes.
Sorry, still been away with unwell family member so not been following. So excuse if I’m covering something already discussed.
A photon is a measurement phenomenon which is detected when an electron in an atom is given sufficient energy to change state either by changing level or by being ejected (photoelectric effect). So the answer is yes, if the combined energies of the beams hit the sweet spot then the electron will transition and we say we have detected a photon. It’s all a bit circular and incestuous though and misunderstanding of what a photon is can cause a lot of problems, it’s why the phrase travels as a wave and interacts as a photon is often used.
In reality, getting this superposition right is very difficult as the lasers need to be very stable and fixed to large concrete blocks, but there have been some successful experiments on superposition - don’t have references to hand.
Post by: paul cotter on 04/12/2022 10:06:02
Post by: Colin2B on 04/12/2022 10:15:14
One can throw n+1 radio frequencies at a target and they will remain discrete ….How do you know they remain discrete. Describe the detector that shows them to be discrete or non-discrete.
We are now firmly in the realm of rhe measurement problem.
Post by: yor_on on 04/12/2022 10:20:28
Seems like an argument for photons, instead of waves? Don't know if anyone remember but we used to have arguments about that here, and I've seen it elsewhere too, where the most reasonable idea still seemed to be waves. Few accepting the duality as a 'real phenomena'. And now the wind has turned the other way, with new generations of fundamental science. It reminds me a lot of this discussion. Check out juanrga.
https://physics.stackexchange.com/questions/46237/is-the-wave-particle-duality-a-real-duality
=
It's weird though, and I think it says something about how we want the world to be. Because it goes back to Newtonian ideas, both of them. Waves and Particles. One or the other as a simple definition of forces. If we take the example of propagation in a vacuum it doesn't really matter if it is a 'wave' or a 'photon' doing it. Either something propagates a distance in time, or there is something entirely different going on, making us think it's a 'propagation' involved to it.
Post by: yor_on on 04/12/2022 10:29:41
Post by: alancalverd on 04/12/2022 16:09:48
Can we observe light waves?No.
We can observe light (and indeed all electromagnetic radiation) doing things that we can model as a wave
We can predict the speed of light from wave solutions to electromagnetic equations
But when we detect light, it behaves like particles
Which is odd because all the equipment we use for detecting lower energy EM radiation is based on its wavy nature.
Post by: paul cotter on 04/12/2022 16:23:33
Post by: Bored chemist on 04/12/2022 16:30:34
But when we detect light, it behaves like particlesUnless we do it while it's behaving like a wave.
https://skullsinthestars.com/2008/05/04/classic-science-paper-otto-wieners-experiment-1890/
Post by: Eternal Student on 04/12/2022 18:08:09
Thanks for various replies. I'm busy thinking about them while catching up on some housework. I may write some more later.
@Bored chemist , that looks like the same post you made earlier and was addressed. It doesn't matter, you can't read everything. It seems like that experiment only suggests that light has wave-like properties, which we all agree on. However, the photographic paper was catching light as a particle and not responding to the wave nature of it.
Best Wishes.
Post by: Bored chemist on 04/12/2022 19:26:30
The image on the film would be in shades of grey, not black and white.
Post by: evan_au on 04/12/2022 19:54:01
Quote from: Colin2B
Describe the detector that shows (radio frequencies) to be discrete or non-discrete.Adding to Paul Cotter's comment, your mobile phone (or your DSL modem) uses numerous closely-spaced frequencies, which can all be used to transmit signals.
They are separated out by a Fast Fourier transform (and are generated by an Inverse Fourier Transform), usually with some error-correcting code to overcome noise.
Around here, we had problems with VDSL2 broadband and telephone pillars which had "dry" joints or oxidised wire-wrap joints, which can be nonlinear. This would cause intermodulation products across the band, corrupting the signal with the dry joint, and also cross-talking into adjacent copper pairs. Linear crosstalk can be cancelled by (linear) "vectoring", but non-linear crosstalk would take far more processing power to analyse & correct. They resolved it by soldering all of the joints in the noisy pillar.
- This reinforced my view that optical fiber is far superior to copper wires - partly because the fiber (mostly SiO2) is already fully oxidised.
Quote from: alancalverd
all the equipment we use for detecting lower energy EM radiation is based on its wavy nature.The point about lower-frequency EM radiation (eg radio-frequency) is that the individual photons ("radons"?) have ultra-low energy. That means you don't transmit individual photons, but instead a coherent wave consisting of trillions of photons. You can easily modulate or demodulate this wave (eg using FFT/IFFT).
However, as you go to higher frequencies, the energy of individual photons increases, and you are more likely to detect them as individual events, and less likely to detect them as a coherent wave.
- It is possible to produce a coherent wave of light (using a laser), but this gets increasingly difficult at higher frequencies like X-Rays and gamma rays.
- Large astronomical telescopes use photon-counting detectors (cooled to near absolute zero), where you can practically count how many photons struck each pixel of the detector. But the photons are still reflected/refracted as a wave, the very large mirror localising the photon on an individual pixel of the detector, in a way that a smaller mirror would not.
Post by: Bored chemist on 04/12/2022 21:29:17
("radons"?)It's bad enough that radon has several names (thoron, niton, emanation) without using radon to mean two different things.
generated by an Inverse Fourier TransformInteresting thing about the inverse FT is that it's an FT (and then inversion, which is why telescopes typically turn the image upside down.)
Post by: alancalverd on 04/12/2022 23:33:40
Wiener's detector was a photographic film. We now know that the production of an image depends as an absolute minimum on the absorption of at least two visible photons within a fairly short time on a silver halide grain. There is no wave model explanation for reciprocity failure or latent image fade.But when we detect light, it behaves like particlesUnless we do it while it's behaving like a wave.
https://skullsinthestars.com/2008/05/04/classic-science-paper-otto-wieners-experiment-1890/
PS I just noticed that ES has made the same observation!
Post by: Bored chemist on 05/12/2022 08:46:41
And yet.Wiener's detector was a photographic film. We now know that the production of an image depends as an absolute minimum on the absorption of at least two visible photons within a fairly short time on a silver halide grain. There is no wave model explanation for reciprocity failure or latent image fade.But when we detect light, it behaves like particlesUnless we do it while it's behaving like a wave.
https://skullsinthestars.com/2008/05/04/classic-science-paper-otto-wieners-experiment-1890/
PS I just noticed that ES has made the same observation!
It's not that I didn't read it, it's that I don't agree with it.Light isn't waves and it isn't particles.
The image on the film would be in shades of grey, not black and white.
It's light.
Post by: Colin2B on 05/12/2022 09:15:08
Colin2b, if you document the incident frequencies accurately one will not get new frequencies, ie addition and subtraction species unless there is a non linear element present to perform frequency mixing.I understand what you are saying, but additional frequencies do occur in linear systems.
Alan & I regularly use beat frequencies to tune musical instruments. I agree that the individual component frequencies still exist, but it doesn’t change the fact that the beat frequency exists and has a physical effect, you can for example get it to excite a tuned resonator.
ES was very specific in his hypothetical example. If 2 optical waves did superpose to create a beat frequency, is it conceivable that if that frequency coincided with the absorption band of an atom then would an electron in that atom would absorb energy and make the transition.
The set up in ES’s description is difficult to achieve, but straight laser interference of 2 beams has been demonstrated in free space.
Look at a simpler demonstration in the paper below. Two beams are directed along the same path through a beam splitter and although the beat frequency is outside the optical band it’s effect is real and can be measured and monitored electronically. If the frequency happened to be at the right frequency is it not feasible that the atom might also act as a suitable detector.
I think that an atom is too narrow a filter to respond to the components, but in the spirit of ES’s question, if by some fluke the beat frequency were in the optical range would we see the beat as a colour, ES’s turquoise?
Unfortunately, none of this really helps answer ES’s basic question.
[ Invalid Attachment ]
Post by: Bored chemist on 05/12/2022 10:45:16
The photodetectors use in experiments like that depicted, typically measure the power, rather than the field strength of the incident light.
Since the power is proportional to the square of the field, (which is why it's always positive), the detector is inherently non linear.
Post by: alancalverd on 05/12/2022 11:14:22
Quote from: Bored chemist on Yesterday at 19:26:30
It's not that I didn't read it, it's that I don't agree with it.
The image on the film would be in shades of grey, not black and white.
Light isn't waves and it isn't particles.
It's light.
The "shades of grey" actually resolve into "density of dots" at very low intensities.
Photographic film isn't particularly sensitive to x-rays, surprisingly, so Edison invented the fluorescent "intensifying screen" that absorbs incoming x-ray photons and emits large numbers of visible photons which are more efficiently captured by the film. At very low dose rates you can see "quantum mottle" on an x-ray image, each spot corresponding to a single x-ray photon hitting the screen. So at least the quantum behavior of x-radiation is easy to see.
We can also count the random arrival of individual visible photons striking the cathode of a photomultiplier - a phenomemon that can't logically be modelled by continuous wave emission.
But I completely agree (and have often stated) that light isn't waves or particles: the often-stated "duality" is of our models, not the phenomenon itself.
Sadly, pseudoscientists have become increasingly insistent on modelling rather than observation to promote their strange beliefs. I've been engaged in a ridiculous discussion with a local "development partnership" funded at your expense, which insists that a dual carriageway needs a major bypass and bus lane (at your expense) to reduce congestion. I have pointed out that there is no evidence of congestion: the average speed of traffic in both directions throughout the rush hour is 55 mph, but the Powers that Be say "but the computer model says it is congested, so we need to resolve the problem [at your expense]". Thus it is with wave-particle duality - generations of students have been inducted into a mystery with as much validity as the Holy Trinity or Transubstantiation (i.e. none) through the vanity of people who cannot admit that their model(s) are not reality itself.
End of rant (for now).
Post by: paul cotter on 05/12/2022 11:21:14
Post by: Colin2B on 05/12/2022 12:58:32
….. I don't want to derail ES's question.I think he has already done that and I haven’t had time to provide him with additional ideas.
@Bored chemist makes some good points, one for one against:
Most "green lasers" you see are actually the beat frequency of an IR beam with itself i.e frequency doubling.
The photodetectors use in experiments like that depicted, typically measure the power, rather than the field strength of the incident light.
Since the power is proportional to the square of the field, (which is why it's always positive), the detector is inherently non linear.
Not a straightforward subject because what we measure is always affected by the way we take the measurements.
This is probably a separate topic.
Post by: Bored chemist on 05/12/2022 16:21:24
Photographic film isn't particularly sensitive to x-rays, surprisingly, so Edison invented the fluorescent "intensifying screen" that absorbs incoming x-ray photons and emits large numbers of visible photons which are more efficiently captured by the film. At very low dose rates you can see "quantum mottle" on an x-ray image, each spot corresponding to a single x-ray photon hitting the screen. So at least the quantum behavior of x-radiation is easy to seeIf our eyes were about 10 times better, we wouldn't need to explain quantisation of light in textbooks because we would be able to see it.
I wonder if that would leave us more confused about diffraction or less so.
Post by: Bored chemist on 05/12/2022 16:23:37
funded at your expense,
[at your expense]
the vanity of people who cannot admit that their model(s) are not reality itself.You do know that money is just a model, don't you?
Post by: alancalverd on 05/12/2022 17:08:35
If our eyes were about 10 times better, we wouldn't need to explain quantisation of light in textbooks because we would be able to see it.Related topic:
In my youth, all my dogs were interested in dogs that they saw on television. Even a 9 inch 405 line black and white TV made from an old radar unit. More recently, none have shown any interest in life size, full color HD 45 inch damn near real dogs with hi fi stereo sound.
I think the reason is that the old CRT had enough persistence that it presented a fairly complete and slowly changing image, whereas modern LCD units have negligible persistence and rely on the inertia of the human eye to produce an image. But dogs have much faster visual response and just see individual flickering dots.
Post by: Eternal Student on 05/12/2022 17:53:05
...I don't want to derail ES's question....That's fine, discuss anything that's even vaguely relevant, it won't trouble me. I'm still busy doing housework of one type or another, so I'll apologise in advance for not replying but I'll try and keep up where and when I can.
Best Wishes.
Post by: evan_au on 05/12/2022 20:39:22
Quote from:
Alan & I regularly use beat frequencies to tune musical instruments. I agree that the individual component frequencies still exist, but it doesn’t change the fact that the beat frequency exists and has a physical effect, you can for example get it to excite a tuned resonator.I have also used beat frequencies to tune a guitar (in my distant youth)...
- If you pluck an open string, you will get a (decaying) sine wave (with a few harmonics).
- If you shorten the adjacent string, you will also get a (decaying) sine wave of almost the same frequency (also with a few harmonics).
- If you now pluck both together, your ears (a logarithmic sound power detector) will detect increasing and reducing power, as the two waves go into and out of phase (the beat frequency at say, 1Hz).
- If you measured this with an FFT over a complete beat cycle (eg 1 second), you would see just the two string frequencies, but not the beat frequencies. You would not see the difference frequency because the microphone is linear.
- The guitar does not resonate at 1Hz, as it doesn't really work much below 50Hz. What you hear are the two audio frequencies exciting the sounding board more and less, cyclically.
- Your ear does not resonate at 1Hz, as it doesn't really work much below 20-50Hz. What you hear are the two audio frequencies exciting your eardrum more and less, cyclically.
This is very different than if you:
- Wiggled your finger tension on the string while playing a string (vibrato - more commonly used on violins = non-linear), which creates a series of frequency-modulated sidebands, close to the natural string frequency.
- Wiggled the volume control while playing a string (amplitude modulation = nonlinear), which creates a pair of amplitude-modulated sidebands, close to the natural string frequency.
Really, I don't see that beat frequencies in a linear medium can excite events at the beat frequency (unless the pianist gets excited because the piano isn't tuned properly!).
PS: People don't tend to use this method of tuning guitars these days - it is too easy to use a crystal-locked electronic tuner which doesn't just tell you that the frequencies differ by 1 Hz, but tells you whether the string is too high or too low in frequency (to 1% of a semitone...). ...assuming you want an equal-tempered scale.
Post by: Eternal Student on 06/12/2022 06:04:48
Really, I don't see that beat frequencies in a linear medium can excite events at the beat frequency...
Rather than worrying about the medium that the frequencies are travelling through or were generated in behaving in a non-linear way, the object which the beat is incident upon can't always respond linearly, in particluar it can't always "keep up" with the speed of oscillations. An example is an object in the region that is behaving as a damped oscillator while some pair of frequencies that resulted in a beat (an amplitude modulated signal) are incident upon it.
I expect you've seen something like this already but I'll put some rough mathematics down here anyway (I don't know who might be reading this and I need a distraction from housework).
The object (the damped oscillator) has this equation of motion:
[Eqn 1]
with b, k constants (b for the damping and k for the spring constant), x= displacement from equilibrium position and F(t) = driving force which varies with time. There's an m (a mass term) multiplying the d2x/ dt2 usually but I've just divided by m and so my constants k,b and Force F are "per unit mass" if you want to be fussy.
The oscillator is driven by the vibrations in ...whatever... the vibrations in the air for your guitar string examples. So we will take
F(t) = A(t) . Cos Ωt
With A(t) = Cos (ωt+Φ) (with Φ = some constant)
as usual.
A(t) describes the oscillating amplitude envelope for the beat incident on our damped oscillator, while the Cos Ωt describes the carrier waves within it. For a beat, Ω = π (f1 + f2) and ω = π(f1 - f2), with f1 and f2 being the original two frequencies that generated the beat (where we assume they had equal amplitude, or at least equal where they became incident on our object). All that matters is that Ω >> ω. (Minor note: You can set Φ which appeared in the Amplitude wave form to 0 if you like, it doesn't make any difference to what follows).
When Ω >> ω we can solve the equation of motion over a small interval of time t ∈ (t0, t1) such that we can assume A(t) ≈ A0 = constant while the Cos (Ωt) term shows much more variation. Over that small interval the equation of motion then becomes:
A0. Cos Ωt Which is the equation of motion for a damped oscillator with a sinusoidal driving force at an angular frequency Ω (which is assumed to be far above the natural frequency of this oscillator). We have a standard solution for this. The damped oscillator hardly responds to that driving term and we have that x(t) is almost constant throughout that interval, x(t) ≈ x(t0). For convenience we can take t1 - t0 = 2π/Ω (then Ωt moves through 2π over our interval but with ω << Ω we will have that ωt hardly changes giving us A ≈ A0 throughout the interval as we wanted).
If we now consider [Eqn 1] at several discrete values of time t0, t1, t2, .., tn, ....
where the gap between tn and tn+1 is always 2π/Ω then we obtain F(tn) = Cos (ωtn + Φ) . Cos (Ωt0 + n.2π) = c . Cos(ωtn + Φ) with c = constant = Cos (Ωt0).
We can then imagine a new damped oscillator with displacement y(t) and this equation of motion:
c .Cos (ω t + Φ) [Eqn 2]
Confine your attention to just numerical solutions of this O.D.E. Since the RHS of [eqn 2] will be precisely the RHS of [eqn 1] at every value t= tn, we know that numerical solutions of [Eqn 2], y(tn) = f(tn) for some function f, will be precisely the same as numerical solutions of [Eqn 1], x(tn) = f(tn) whenever we take uniform intervals Δt = tn - tn-1= 2π/Ω and start the numerical technique at the same place t=t0. However, [Eqn 2] is just the equation for a damped oscillator being driven at the angular frequency ω << Ω. We have an analytical solution for this. The numerical solution we generate for [Eqn 1] is then well approximated by the anlytical solution y=y(t) for [Eqn 2] evaluated at the points t= tn. Finally we make use the earlier paragraph (about x(t) being reasonably constant over the small intervals because the damped oscillator hardly responds to a driving frequency of Ω) and conclude: The exact solutions of [Eqn 1] take the form x(t) = y(t) + δ(t) where y(t) is the exact solution of [Eqn 2] and δ(t) is a function that is some small correction, i.e. |δ(t)| ≈ 0.Anyway, the general idea or the "bottom line" of the argument, is that the damped oscillator will respond as if it is seeing a driving force of the form Cos (ωt) (as seen in [Eqn 2] ) where ω is the angular frequency of the amplitude modulation.
Best Wishes.
Post by: paul cotter on 06/12/2022 10:01:35
Post by: hamdani yusuf on 07/12/2022 11:13:30
This experiment is takes a different approach and might be similar to what you are looking for https://www.sciencedaily.com/releases/2018/04/
I find this statement in the article inaccurate.
Quote
Polarization occurs when waves, such as electromagnetic or light waves, rotate. Electromagnetic fields called microwaves have a rotating electric field that turns clockwise or counter-clockwise, and most theories predict that microwaves will affect the rotation of electrons. And yet, experimental studies have shown that electrons seem to be unaffected by microwave polarization. These theory-defying results have long perplexed physicists.It describes circular polarization. There's another type of polarization called linear polarization. Elliptical polarization can be seen as combination of both.
Post by: hamdani yusuf on 07/12/2022 11:19:56
I understand that you can show polarisation of a microwave beam by building a wall with row of parallel wires.The microwave gets reflected when it's short-circuited. The metal wires don't absorb much of the energy.
- When the wall is oriented in one direction, it "short-circuits" the E-field and the microwaves are blocked
- Rotate the wall by 90°, and the microwaves get through, as the wires don't short-circuit the magnetic field.
A similar arrangement (on a smaller scale) is used to produce polarised lenses for visible light - there is a grid of long, parallel crystals in the lense.
Thinner wires have higher resistance, and generate heat while absorbing some energy.
Post by: Eternal Student on 07/12/2022 16:15:44
I find this statement in the article inaccurate....The article does seem to be condensed and edited. You're absolutely right, plane polarisation is possible. However, this experiment seemed to require circular polarised microwaves, so they (the editor rather than scientists, I suspect) didn't explain anything else. Later in the article, they do say "circular polarised microwaves" instead of just "polarised microwaves".
The microwave gets reflected when it's short-circuited. The metal wires don't absorb much of the energy.Building a good polarising filter is a problem. The wires can't be too thick and the gap between the wires can't be too large or too small. If the gap between the wires is too small then what you have is a wall and that will often reflect the microwaves. If the gap is too large (imagine a pair of goal posts as used in football) then plenty of microwaves are just going to get straight through the centre without interacting with the wires at all.
Thinner wires have higher resistance, and generate heat while absorbing some energy.
If the wires are too thick then electrons can oscillate (I'll say up and down) significantly even when the wires are running horizontally. So that is saying the electrons can move across the wire instead of along the wire. This tends to block the microwaves regardless of the orientation of the filter. Meanwhile, if the wires are too thin then what tends to happen is that the gap between the wires has got too large compared to the thickness of the wires and/or the resistance of the wires is too high to allow enough oscillation of the electrons.
Throughout the whole thing, when electrons are oscillating they are tending to create or emit radiation of the same frequency as that which was driving them. So the filter itself is generating some microwaves. It's just that the emitted microwaves are thrown out in all directions fairly uniformly, while the horned microwave generators of the type most people use (I've seen some being used by you, @hamdani yusuf , on a different thread) tend to produce a much narrower beam. Hence, even with a filter that is about as perfect as you could build it and orientated so as to block the beam, you will still pick up some microwaves at the other side of it. Hopefully, the microwaves emitted by the electrons in the filter have an intensity that falls off as 1/r2. Meanwhile a good and well focused beam of microwaves shouldn't fall off in intensity with distance travelled. However, in practice nothing is perfect - you probably don't have perfectly focused parallel beams, microwaves get absorbed, re-emitted, scattered and generally disrupted by the filter and everything else like the medium (air) that they travel through.
More generally, I don't suppose classical physics with E and B fields is really going to fully explain what is happening with microwaves and a polarising filter. At the point of interaction with the filter, I would have thought there must be some genuine photon and atom interactions taking place. @alancalverd (and others) have already made some posts on this thread suggesting that since the energy of microwave photons is much lower than light, results like the Ehrenfest theorem and the more generalised correspondence principle will be exhibited. Specifically, that you will get something that is more easily matched to classical physics. (This post is too long, I'm ending now. I'll just make it clear that practical experimentation isn't something that I know much about. Most of my practicals never worked).
Best Wishes.
Post by: paul cotter on 07/12/2022 17:05:18
Post by: Eternal Student on 07/12/2022 19:43:49
It was a typing error,
Where we have used the relationship c = 1/(μ0ε0).
Should have been
Where we have used the relationship c2 = 1/(μ0ε0).
Fortunately that was what was actually done. The original post has now been edited.
Best Wishes.
Post by: DarkKnight on 08/12/2022 05:17:43
Hi, thanks and well spotted @paul cotterI find it interesting that Maxwell explained the speed of light in a vacuum as c when space isn't a vacuum nor empty. I also don't understand why Maxwell ignored that the Suns EM field extends all the way to the Earths surface with enough magnitude to illuminate the surface .
It was a typing error,
Where we have used the relationship c = 1/(μ0ε0).
Should have been
Where we have used the relationship c2 = 1/(μ0ε0).
Fortunately that was what was actually done. The original post has now been edited.
Best Wishes.
Post by: hamdani yusuf on 08/12/2022 06:07:31
I find it interesting that Maxwell explained the speed of light in a vacuum as c when space isn't a vacuum nor empty. I also don't understand why Maxwell ignored that the Suns EM field extends all the way to the Earths surface with enough magnitude to illuminate the surface .What should it be if he didn't ignore it?
Post by: Bored chemist on 08/12/2022 08:41:51
I also don't understand why Maxwell ignored that the Suns EM field extends all the way to the Earths surface with enough magnitude to illuminate the surface .That seems to be one of a long list.
Let's cross one item off it.
https://en.wikipedia.org/wiki/Superposition_principle
Post by: hamdani yusuf on 08/12/2022 10:41:34
Throughout the whole thing, when electrons are oscillating they are tending to create or emit radiation of the same frequency as that which was driving them. So the filter itself is generating some microwaves. It's just that the emitted microwaves are thrown out in all directions fairly uniformly, while the horned microwave generators of the type most people use (I've seen some being used by you, @hamdani yusuf , on a different thread) tend to produce a much narrower beam. Hence, even with a filter that is about as perfect as you could build it and orientated so as to block the beam, you will still pick up some microwaves at the other side of it. Hopefully, the microwaves emitted by the electrons in the filter have an intensity that falls off as 1/r2. Meanwhile a good and well focused beam of microwaves shouldn't fall off in intensity with distance travelled. However, in practice nothing is perfect - you probably don't have perfectly focused parallel beams, microwaves get absorbed, re-emitted, scattered and generally disrupted by the filter and everything else like the medium (air) that they travel through.
The working principle of directional radio antenna is explained clearly in this video by Royal Canadian Air Force.
My model can be thought as an extention to the working principle of antenna, which can be shown clearly here.It uses antenna array as the source of electromagnetic wave. It's different from wave on water surface and Huygen's principle which treat empty space as wave source.
Directional antenna simply shifts the focus of the beam to a point behind the transmitting antenna, which is used as reference to measure r for intensity attenuation.
Post by: hamdani yusuf on 08/12/2022 11:46:31
More generally, I don't suppose classical physics with E and B fields is really going to fully explain what is happening with microwaves and a polarising filter.If you think that classical physics in electromagnetism is well represented by Maxwell's equations as its mathematical model, then we have known its limitations, especially at microscopic scales. Maxwell treated electric charge as continuum, which can be divided infinitesimally. Millikan's oil drop experiment demonstrated quantification of electric charge. This critical false assumption lead this mathematical model to make wrong predictions on microscopic electromagnetic phenomena.
Moreover, Maxwell's equations say nothing about mass of electric charges, nor electromagnetic forces. These dismissals rendered the mathematical model incomplete.
Post by: hamdani yusuf on 08/12/2022 11:55:13
The working principle of directional radio antenna is explained clearly in this video by Royal Canadian Air Force.
Based on this explanation, I've built a model and devised some experiments to test it.
Here's the model I proposed. I'm not really sure if it's new, since it's based on how a dipole antenna work. Can we derive Huygen's principle from equations of antenna? Or can we derive equations of antenna from Huygen's principle?
Investigation on microwave 37 : blocking mechanism
Investigation on microwave 38: blocking mechanism explanation
Investigation on microwave 39: Blocking mechanism evidence
Here's an example how the model can be used to predict experimental results.
Polarization twister design.
Signal splitting.
Asymmetric twister/splitter
Post by: hamdani yusuf on 08/12/2022 13:15:39
Natural resonance frequency of molecular vibration
Many Quality Control Laboratories already use online infrared spectroscopy fed to monitor product composition almost in real time and reduce manual sampling and chemical analyses. Each type of chemical bond has unique "fingerprint/signature absorption spectra" which can be used to calculate the concentration of some chemical compounds in the product/sample.
Post by: alancalverd on 08/12/2022 14:22:01
Maxwell treated electric charge as continuum, which can be divided infinitesimally.No. He merely used the known phenomena of a magnetic field being induced by a moving charge and a potential being induced by a changing magnetic field. He made no assumptions about the nature of either.
Post by: alancalverd on 08/12/2022 14:34:57
I find it interesting that Maxwell explained the speed of light in a vacuum as c when space isn't a vacuum nor empty.
Hardly interesting at all. μ0 and ε0 are arbitrary constants that relate the observed speed of light to our conventional units of time and distance.
Quote
I also don't understand why Maxwell ignored that the Suns EM field extends all the way to the Earths surface with enough magnitude to illuminate the surface .For the same reason that he ignored the Irish Question - it's not relevant to the mathematics. But Maxwell's equations do at least predict that the sun's EM field will extend to infinity in all directions, which is a pretty good approximation considering that Einstein hadn't been born at the time.
Post by: DarkKnight on 08/12/2022 15:08:32
Any linear operator from any position can be viewed as x .I also don't understand why Maxwell ignored that the Suns EM field extends all the way to the Earths surface with enough magnitude to illuminate the surface .That seems to be one of a long list.
Let's cross one item off it.
https://en.wikipedia.org/wiki/Superposition_principle
Unbounded photons can be viewed to travel at
c=
Any bounded EM fields can be viewed as influenced by the x,y,z operator and the speed the EM field travels is dependent to the bounded bodies speed .
Post by: hamdani yusuf on 08/12/2022 15:10:42
What is the equation describing how a magnetic field is induced by a moving charge?Maxwell treated electric charge as continuum, which can be divided infinitesimally.No. He merely used the known phenomena of a magnetic field being induced by a moving charge and a potential being induced by a changing magnetic field. He made no assumptions about the nature of either.
Post by: DarkKnight on 08/12/2022 15:23:51
But Maxwell's equations do at least predict that the sun's EM field will extend to infinity in all directions, which is a pretty good approximation considering that Einstein hadn't been born at the time.
I don't think that is possible because outside of space-time may cause wave-function collapse by the energy being conserved by the empty set [ Invalid Attachment ]
Post by: hamdani yusuf on 08/12/2022 15:26:11
But Maxwell's equations do at least predict that the sun's EM field will extend to infinity in all directions, which is a pretty good approximation considering that Einstein hadn't been born at the time.What did Einstein change?
Post by: Kryptid on 08/12/2022 15:37:26
Any linear operator from any position can be viewed as x .
Unbounded photons can be viewed to travel at
c=
Any bounded EM fields can be viewed as influenced by the x,y,z operator and the speed the EM field travels is dependent to the bounded bodies speed .
Nonsense equations as always, huh?
Post by: paul cotter on 08/12/2022 15:49:26
Post by: DarkKnight on 08/12/2022 15:51:21
Maxwell used μ0 and ε0 for an operator , x works just fine.Any linear operator from any position can be viewed as x .
Unbounded photons can be viewed to travel at
c=
Any bounded EM fields can be viewed as influenced by the x,y,z operator and the speed the EM field travels is dependent to the bounded bodies speed .
Nonsense equations as always, huh?
ex1.jpg (6.07 kB . 365x332 - viewed 768 times)Obviously x = 299,792,458 metres
μ0 and ε0 are not a distance , speed is
so μ0 and ε0 must be equal to x . ex2.jpg (8.24 kB . 365x332 - viewed 828 times)Post by: Origin on 08/12/2022 16:18:29
Maxwell used μ0 and ε0 for an operator , x works just fine.Unfortunately your 'math' is nonsense.
Post by: DarkKnight on 08/12/2022 16:26:31
Maxwells equations readsMaxwell used μ0 and ε0 for an operator , x works just fine.Unfortunately your 'math' is nonsense.
=1 Can you explain where 299,792,458 m/s is derived from that without using x ?
Post by: DarkKnight on 08/12/2022 16:43:28
Wow, what happened? We were discussing physics and suddenly on page #4 we were overtaken by gibberish.In trying to understand any subject , it is firstly of most importance to understand the first principles of a subject. Any branch of knowledge that is taught , should always have strong routes , from a starting point to a conclusion . If this basic principle is not adhered to , then the practitioner becomes ill-informed , having an inadequate awareness of the facts.
Let us now be clear in our understanding of what is a fact compared to interpretation . A fact is something that is known or proved to be true , it is not something that is solely written on paper . A fact has supporting evidence such as observations , a fact can sometimes be an axiom , something that is self evidently true . If we ignore the facts and/or axioms then we are just being subjective as opposed objective. This information is then ill-informed information and can be misleading to a student ,allowing them false ideologies of a subject .
Post by: alancalverd on 08/12/2022 16:48:49
What is the equation describing how a magnetic field is induced by a moving charge?Ampere's Law.
Post by: alancalverd on 08/12/2022 16:51:48
In trying to understand any subject...it is helpful to listen before talking, or read before writing. The essence of science is humility in the face of established facts.
Post by: DarkKnight on 08/12/2022 17:03:50
I have always listened and remain humble . It is helpful if science listened occasionally, Maxwells equation for the speed of light readsIn trying to understand any subject...it is helpful to listen before talking, or read before writing. The essence of science is humility in the face of established facts.
=1 ?
Post by: alancalverd on 08/12/2022 17:16:29
Post by: DarkKnight on 08/12/2022 17:25:28
Not the Maxwell who attended my alma mater, then. Sorry I don't know anything about yours.Then can you explain your input value for the bracketed part of Maxwells equation ?
c = 1/(μ0ε0)
I read 0
Post by: Bored chemist on 08/12/2022 18:08:53
Maxwells equation for the speed of light readsNo, it does not.
1/0=1
c = 1/(μ0ε0)Should be
1/√(μ0ε0)
The subscript zeroes tell you that the permittivity and permeability are the values for a vacuum rather than anything else.
It is helpful if science listened occasionallyIn what way does it help if science listens, when you don't?
Post by: alancalverd on 08/12/2022 18:21:10
Mine used old cgs units but in SI we have
μ0 ≈ 1.26 x 10-6 H/m and
ε0 ≈ 8.85 x 10-12 F/m
It is left as an exercise to the reader to determine the value and dimensions of their product.
Post by: DarkKnight on 08/12/2022 19:10:15
I am well aware of what the values mean but in regards to the measure of the speed of anything ,
Should be
1/√(μ0ε0)
The subscript zeroes tell you that the permittivity and permeability are the values for a vacuum rather than anything else.
is required . Where is the d/t value in the equation ?
Post by: Eternal Student on 08/12/2022 19:13:50
I probably haven't got the time to reply to everything right now but thanks for all replies.
Any linear operator from any position can be viewed as x .I might be missing something here. I'll guess @DarkKnight has written something on a different thread recently and you are talking about something from that?
On the face of it there are many different linear operators but I was going to guess you're talking about Quantum Mechanics and operators acting on the wave function. Assuming x is supposed to be position, the only operator which is given by mutiplication by x is the position operator
. The next statement
Unbounded photons can be viewed to travel atis then not particularly true.
c=1/x
At the very least whatever you have written requires some more explanation. I can't understand what you've written in several of the posts. Before you do explain further, consider if it would be better placed in a new thread. I'm quite happy with even obliquely relevant topics being discussed here but your ideas look quite revolutionary. I'm not a moderator or staff for this forum but as I understand the guidelines, this thread probably shouldn't contain anything that is new to the world of physics. If there is anything new here, then it is just "new to me" or new to someone who contributed to the thread. Hopefully, it is just based on established physics and would have been already known to someone in the world of the physics.
You might be underselling your idea by discussing it here. If it's really new and revolutionary then it deserves a thread of its own.
Best Wishes.
Post by: DarkKnight on 08/12/2022 19:48:15
Hi.I am simply trying to answer your question but to do that we must discuss some issues . x is a vector , it isn't a position .
I probably haven't got the time to reply to everything right now but thanks for all replies.Any linear operator from any position can be viewed as x .I might be missing something here. I'll guess @DarkKnight has written something on a different thread recently and you are talking about something from that?
On the face of it there are many different linear operators but I was going to guess you're talking about Quantum Mechanics and operators acting on the wave function. Assuming x is supposed to be position, the only operator which is given by mutiplication by x is the position operator
The next statementUnbounded photons can be viewed to travel atis then not particularly true.
c=1/x
At the very least whatever you have written requires some more explanation. I can't understand what you've written in several of the posts. Before you do explain further, consider if it would be better placed in a new thread. I'm quite happy with even obliquely relevant topics being discussed here but your ideas look quite revolutionary. I'm not a moderator or staff for this forum but as I understand the guidelines, this thread probably shouldn't contain anything that is new to the world of physics. If there is anything new here, then it is just "new to me" or new to someone who contributed to the thread. Hopefully, it is just based on established physics and would have been already known to someone in the world of the physics.
You might be underselling your idea by discussing it here. If it's really new and revolutionary then it deserves a thread of its own.
Best Wishes.
Yes there is a thread in new theories that presents a model , labelling the operator A(x) .
This is an operator that explains the speed of light , the motion of bodies and gravity .
It is also an operator that can answer your question .
Post by: Bored chemist on 08/12/2022 19:58:57
I am well aware of what the values mean but in regards to the measure of the speed of anything , is required . Where is the d/t value in the equation ?Not really.
It's like saying that the speed of a wave travelling along a string is given by sqrt(t/ (ml)) where t is the tension, m is the mass and l is the length
You just need to do the dimensional analysis and it works out to have the right units.
Post by: DarkKnight on 08/12/2022 20:22:43
I am well aware of what the values mean but in regards to the measure of the speed of anything , is required . Where is the d/t value in the equation ?Not really.
It's like saying that the speed of a wave travelling along a string is given by sqrt(t/ (ml)) where t is the tension, m is the mass and l is the length
You just need to do the dimensional analysis and it works out to have the right units.
You sound more like a politician who completey ignores the questioning . I've done the dimesional analysis and in no way does Maxwell use d/t to measure lights speed . I have asked people what value they would input for e0 and u0 because I read 1/0=1 .
Are you sure that Maxwell wasn't just describing the operator ?
Post by: Kryptid on 08/12/2022 20:54:59
because I read 1/0=1 .
All this time you've been away from the forums, and you still haven't learned something so fundamental as "you can't divide by zero"?
Post by: alancalverd on 08/12/2022 23:47:19
I have asked people what value they would input for e0 and u0If you had bothered to read reply #78 above instead of spouting incoherent drivel about operators, you would know the answer.
Post by: hamdani yusuf on 09/12/2022 02:24:07
Let's say a 1 gram metal ball electrically charged by +1 coulomb is moving to the right at 1 m/s on x axis. Another identical ball is stationary 1 meter above the x axis.What is the equation describing how a magnetic field is induced by a moving charge?Ampere's Law.
What's the magnetic field experienced by the second ball when the first ball is right below it?
What's the magnetic field experienced by the first ball at the same time?
Post by: alancalverd on 09/12/2022 09:07:08
Post by: DarkKnight on 09/12/2022 09:20:07
I have asked people what value they would input for e0 and u0If you had bothered to read reply #78 above instead of spouting incoherent drivel about operators, you would know the answer.
I have read post 78 and what you provided was in my opinion , incoherent drivel . What you provided wasn't a
so the Maxwell equation makes no sense . Additionally all the work done in the universe requires an operator , it isn't drivel .
Post by: DarkKnight on 09/12/2022 09:29:58
because I read 1/0=1 .
All this time you've been away from the forums, and you still haven't learned something so fundamental as "you can't divide by zero"?
I'm not the one claiming c=1/e0u0 which is 1/0
I claim c=Ψ/A(x) where A(x) is the operator , namely an Eigen vector .
I also claim that the B(x) operator , changes the speed of light , this is because B(x) operator represents a vector with a medium .
Ψ/B(x)≠c
Post by: DarkKnight on 09/12/2022 09:48:57
Hi.You seem kinda cool so here is the answer to your question
I probably haven't got the time to reply to everything right now but thanks for all replies.Any linear operator from any position can be viewed as x .I might be missing something here. I'll guess @DarkKnight has written something on a different thread recently and you are talking about something from that?
On the face of it there are many different linear operators but I was going to guess you're talking about Quantum Mechanics and operators acting on the wave function. Assuming x is supposed to be position, the only operator which is given by mutiplication by x is the position operator
The next statementUnbounded photons can be viewed to travel atis then not particularly true.
c=1/x
At the very least whatever you have written requires some more explanation. I can't understand what you've written in several of the posts. Before you do explain further, consider if it would be better placed in a new thread. I'm quite happy with even obliquely relevant topics being discussed here but your ideas look quite revolutionary. I'm not a moderator or staff for this forum but as I understand the guidelines, this thread probably shouldn't contain anything that is new to the world of physics. If there is anything new here, then it is just "new to me" or new to someone who contributed to the thread. Hopefully, it is just based on established physics and would have been already known to someone in the world of the physics.
You might be underselling your idea by discussing it here. If it's really new and revolutionary then it deserves a thread of its own.
Best Wishes.
= -0.5q+0.5q ;)If Einstein were here , he'd say , '' the tip of the rod is forced to have a split end when it interacts with matter'' . Then Coulomb would agree ! :)
Post by: alancalverd on 09/12/2022 10:15:28
I'm not the one claiming c=1/e0u0 which is 1/0Yes you are. Everyone else knows what they are talking about.
Post by: alancalverd on 09/12/2022 10:26:23
At maximum we have a current of 1 coulomb/second , i.e. 1 amp, passing 1 m from the point of interest so the peak field is about 4 microtesla.
Post by: DarkKnight on 09/12/2022 10:26:37
I'm not the one claiming c=1/e0u0 which is 1/0Yes you are. Everyone else knows what they are talking about.
I'm not being awkward here , look at this from my view . Sceince states that speed is derived from
.They then claim the speed of light c = 1/e0u0 which reads 1/0
They then claim that e0u0 doesn't mean 0 it means the numbers you provided which must be imaginary numbers .
In no way does science demonstrate that 1/e0u0 has anything to do with
It is all kinda contradictory when speed is
Please clear up my confusion if you think I am wrong ?
Additionally I have drew a ''diagram'' for this topic that shows the ''split ends'' . (A humourous version) .
[ Invalid Attachment ]
Post by: alancalverd on 09/12/2022 10:31:24
Post by: DarkKnight on 09/12/2022 10:40:52
Your confusion is that you think your incoherent ravings have some meaning to others. You would be better off posting in a psychiatry forum than physics if you want a sympathetic audience.
New information isn't incoherent information , it is information that at this time isn't fully understood by the student .
My math is correct and my explanation of ''split ends'' falls perfectly inline with an atoms components and Coulombs law of charges .
Coulomb would be my citation and additionally Einsteins photo-electrical effect .
No pleasing some people !
Post by: paul cotter on 09/12/2022 10:47:50
Post by: hamdani yusuf on 09/12/2022 12:52:26
Anyway, returning to HY's last question:Let's change the electric charge to 10 Coulomb, and the speed becomes 0.1 m/s. How would it affect the magnetic field?
At maximum we have a current of 1 coulomb/second , i.e. 1 amp, passing 1 m from the point of interest so the peak field is about 4 microtesla.
Post by: alancalverd on 09/12/2022 16:07:24
It's unusual to consider the magnetic properties of a single linearly moving charge but as long as current = charge passing a given plane per unit time, we can estimate the maximum magnetic field from the velocity of the charge at the point of closest approach, multiplied by the value of the charge.
Post by: hamdani yusuf on 09/12/2022 16:39:49
Not at all.1 Ampere means 1 Coulomb per second, or 60 Coulomb per minute, or 3600 Coulomb per hour, or 0.001 Coulomb per millisecond. How did you decide that the ball in my example produced 1 Ampere?
It's unusual to consider the magnetic properties of a single linearly moving charge but as long as current = charge passing a given plane per unit time, we can estimate the maximum magnetic field from the velocity of the charge at the point of closest approach, multiplied by the value of the charge.
Post by: alancalverd on 09/12/2022 17:46:18
Post by: Petrochemicals on 09/12/2022 18:01:27
Post by: Eternal Student on 09/12/2022 20:12:19
I'm sure I'll miss some posts under the number that have appeared. Sorry, post again if it seems important (maybe leave it for a day or so).
@hamdani yusuf posted some stuff about vibrations in molecules caused by IR radiation, way back in post #59.
I like it and I'm still considering it. As you mentioned it really only works for IR radiation not visible light but the frequencies are close and it's hardly worth quibbling over.
Technically it still doesn't really measure the oscillations in the E field directly, it's just that it should be there otherwise it's hard to explain what is happening. Just to be clear: Yes, you'd have a hard time explaining why IR radiation of certain frequencies is absorbed by CO2 without using an oscillation in the E field as a model for the IR radiation - but it could be done. E.g. The explanation does assume that a simple ball-and-spring model was sufficient and this could be challenged. The molecule is better modelled as a Quantum Mechanical structure and there was some sort of "photon and atoms" interaction happening. Additionally I'm not sure how easily you can assert that the molecules were vibrating in some mechanical sense, that's not something you can directly observe. You could measure the temperature of the CO2 and claim that is a measure of these vibrations but that's still a very indirect measure of vibration since temperature is a very complicated thing, all sorts of modes of supporting energy contribute to temperature (I know you've had a thread discussing temperature elsewhere on this forum).
None the less, I like it. It goes high up on the list of experiments that suggest light (or IR radiation in this case) has an oscillating E field.
Best Wishes.
Post by: Eternal Student on 09/12/2022 21:42:43
I don't know whether you have the capabilities, or more to the point whether it would work but you could try slowing light down.Yes we could, we could put the light in a dense medium. However, this affects the wavelength and speed only. The frequency of oscillations would be unchanged.
It's the frequency, or "the speed of oscillation" if you prefer, that makes it so hard to directly measure the oscillation in the E field. We don't seem to have a lot of equipment that can respond to such rapid fluctuations in the E field.
However we could try something similar. In fact, you've made a first class suggestion when I think about. We don't need to change the speed, we can just change the frequency.
Emit some visible light, then get in space rocket and move fast. The light should have a relativistic Doppler shift and then all we have to observe is the oscillation in the E field of some radio waves (which we can more or less do to everyones satisfaction, just stick a pole in the air to act as an radio antenna and directly measure an oscillating current flow).
Best Wishes.
Post by: Eternal Student on 09/12/2022 23:25:12
Probably one of the last replies for a while, I think I've gone through all the other posts.
Posts #21, 22, 23 and some related posts discussing an atom placed in the centre of some lasers.
I would have THOUGHT no, to your question about lasers of different frequenciesI would have thought that a while ago. Now I'm not so sure.
..So the answer is yes, if the combined energies of the beams hit the sweet spot then the electron will transition..That does seem like Colin2B was fairly certain. I didn't know such experiments had already been done.
It's certainly not something that gets a mention in conventional textbooks discussing the photo-electric effect. If you add an asterisked comment like this, then the entire message or implication of the result is eroded:
" ... You can't eject an electron by using light of another frequency, or just by increasing the intensity, there is an absolute minimum frequency of light that must be used**.
** Except when you use a few frequencies all at the same time."
- - - - - - - - - -
Anyway, at about this time I'm thinking that the oscillations in the E field of visible light should be observable. At least as much as we can directly observe the oscillations in radio frequency e-m radiation. There might not be good enough equipment available at the moment but it's at least theoretically an observable and will be directly measured when equipment catches up and improves.
I might not be making it clear what I mean by "theoretically observable", a few people will be thinking - either you have measured it or you haven't. I'll try to re-phrase it this way: There is a quantum mechanical wave function describing visible light (a collection of photons and not just one). However, that wave function is certainly NOT the function (like E=E0 Cos ωt ) that describes oscillation in the E field or B field (there shouldn't be a lot of dispute about that). The oscillation in the E field IS (underline IS ) an observable, the wave function is not. This is the sense in which I would think that the oscillations in the E field are an observable. There is a suitable operator acting on the wave function such that the value of the E field along the path of the light (at a given time) could be obtained as the eigenvalues of that operator.
Compare that with, for example, post #27 from @alancalverd where it was implied that we only model light as an oscillation in the E field, we can't detect the oscillations. I'm thinking that we probably can detect the oscillations in the E field directly. They are "there to be seen" just as much as anything is there to be seen and measured, it may be an observable from a fundamentally quantum mechanical system but that's a separate issue, they are none-the-less observable.
Best Wishes.
Post by: Bored chemist on 10/12/2022 01:19:22
it means the numbers you provided which must be imaginary numbersThey can be measured by high school students.
This is essentially the problem with "Dark knight".
He hasn't a clue.
Post by: hamdani yusuf on 10/12/2022 01:24:48
The oscillation in the E field IS (underline IS ) an observable, the wave function is not. This is the sense in which I would think that the oscillations in the E field are an observable.How would you detect Electric field directly?
In case of radio frequency, we can observe its effects on electronic components at the receiver. But it's not a direct observation either. We rely on the assumptions of how those components work.
At lower/subsonic frequency, we can observe its effect on an electrically charged metal ball, like in the video I posted previously.
At infrared frequency, it can induce vibration on some atoms tied by chemical bonds.
At visible and ultraviolet frequency, it can induce vibration on electrons tied to atomic nuclei.
Post by: Eternal Student on 10/12/2022 02:25:01
How would you detect Electric field directly?That's kind of the main point of starting the thread. I don't know of a good way to do this for light, although some reasonable ideas have been presented and I have a few more ideas now. Thank you very much to everyone who has written something.
So one reasonable option is something like that suggested in post #103, get in a space rocket and reduce the frequency down to something you can measure directly. Another reasonable suggestion is an experiment like the one you ( @hamdani yusuf ) suggested involving IR absorption in CO2 molecules.
In principle, an Electric field should be that which creates a force on a charge, it's that simple - go straight from the definition of what an E field should be. So, to be very specific it should be measured as a force acting on an object that is just due to charge. You could replace the test object with some other material (make it more massive, less permeable to magnetism, a different colour, not having any Hydrogen atoms or whatever...) and the force would be the same provided the charge was the same. To put a numerical quantity on it, the field strength is the force (in Newtons) per unit charge (in Coulombs) and it has a vector property - the direction of the force on a positive charge.
So, working from that definition, the most simple experiment that directly shows an oscillating E field is present would be one where you can measure an oscillating force on a charge and clearly demonstrate it was only due to the charge.
Moving it up one small level of assumption then, an experiment where you could see a small charged blob being moved about and oscillating as if it is being driven by an oscillating E field - that would be a reasonable and fairly direct measurement of an oscillating E field (for most people).
As you state, we tend to increase the number of assumptions when the E field gets small and especially when the frequency of oscillation increases. It's not guaranteed that the effect on an electrical component is really directly measuring or responding to an E field, however there's not that many assumptions being made. E.g. sticking a pole up in the air while radio waves pass across it and measuring an oscillating current being produced is reasonably direct measurement of an oscillating E field being there - let's say it's only another level or two up on the ladder of assumptions you might be asked to make.
By the time you're up to visible light frequencies there's a stack of assumptions that are usually made.
Best Wishes.
Post by: hamdani yusuf on 10/12/2022 05:39:27
Post by: paul cotter on 10/12/2022 09:38:54
Post by: alancalverd on 10/12/2022 10:38:22
How would you detect Electric field directly?Actually not. We detect the current induced in the receiving aerial by the magnetic component.
In case of radio frequency, we can observe its effects on electronic components at the receiver.
Post by: Petrochemicals on 10/12/2022 11:07:15
Hi.You could make your own doppler machine, various moving parts to stretch the frequency. But as you state this would be measuring radio waves rather than light directly.I don't know whether you have the capabilities, or more to the point whether it would work but you could try slowing light down.Yes we could, we could put the light in a dense medium. However, this affects the wavelength and speed only. The frequency of oscillations would be unchanged.
It's the frequency, or "the speed of oscillation" if you prefer, that makes it so hard to directly measure the oscillation in the E field. We don't seem to have a lot of equipment that can respond to such rapid fluctuations in the E field.
However we could try something similar. In fact, you've made a first class suggestion when I think about. We don't need to change the speed, we can just change the frequency.
Emit some visible light, then get in space rocket and move fast. The light should have a relativistic Doppler shift and then all we have to observe is the oscillation in the E field of some radio waves (which we can more or less do to everyones satisfaction, just stick a pole in the air to act as an radio antenna and directly measure an oscillating current flow).
Best Wishes.
Post by: Bored chemist on 10/12/2022 12:18:25
Is that cheating?
Post by: paul cotter on 10/12/2022 12:37:06
Post by: alancalverd on 10/12/2022 14:53:25
Post by: Bored chemist on 10/12/2022 14:56:34
So a charge moving towards the receiver will induce a current in the antenna? And we don't need to rotate our TV antenna to match the polarisation of the transmitter? Hmm. Need to think about that one.That's not what he said, was it?
How would you know if you were lining up the magnetic field or the electric one?
Post by: Eternal Student on 10/12/2022 15:22:47
Basically, you can measure EM wave in any frequency, as long as you have quick enough rectifier.Surely not. The wires which made the dipole aerials are shiny reflective things. If you just shine visible light on an aerial then it just reflects off it, it doesn't interact with the aerial to generate a current in it. Meanwhile, most X-rays or gamma rays can just go straight through without any significant interaction along the way.
The limitation isn't just the diodes or rectifier, it's the interaction with the aerial.
Actually not. We detect the current induced in the receiving aerial by the magnetic component.I'm in general agreement with @paul cotter , it's both Electric and magnetic field.
Is that cheating?No more so than getting in a space rocket and using a Doppler shift to bring the frequency down. So yes, it's cheating, you aren't really measuring an oscillation in e-m radiation of visible light frequency but it is suggesting that the oscillations would have been there in the visible light even if the equipment you have wasn't fast enough to find it. So it's OK for the bigger picture that an oscillation is there and is theoretically observable.
In the early days of investigations on the photoelectric effect I am sure a strictly monochromatic light source was not being used....Probably true. However, the multiple sources and frequencies of light probably weren't positioned in the right places, they wouldn't have been kept in phase anything like as well as some lasers, they wouldn't have been polarised in the right way etc. By random chance a few rays of light might combine so as to make a photon of the right frequency appear at an atom of the metal and there could have been a few electrons released. The equipment may not have been good enough to identify that from general noise. You would not have been able to eliminate the possibility that a cosmic ray from space, or a high energy photon from a decaying piece of granite inside the building, had just come in and hit the metal. That sort of thing is going to happen quite often and is going to be much of the "noise" that would have present in the experiment. I mean, on a hot day when a thunderstorm is due the air itself will be quite ionised anyway even before you start liberating electrons from the surface of the metal.
As I implied in an earlier post, I'm not entirely sure what will happen with the arrangement of lasers around an atom. However, @Colin2B was fairly sure and seemed to be suggesting that experiments have already been done like this. In my limited experience on this forum, Colin2B is rarely so wrong as to imagine experiments that were never actually done.
Let's think about this in a different way: A positron and an electron can annihilate to produce a pair of photons, that's everyday physics now with the PET scanners they use in hospital. We think that during the early moments of the universe the opposite happened, pairs of photons combined to form some matter, this was the earliest synthesis of matter. Now, if photons can combine to make particles of ordinary matter, then it's really not asking a lot for photons to combine and just make a new photon with a higher amount of energy. I'm just an ordinary person and I like a nice simple model with photons as particles just as much as the next guy. However, according to an entirely quantum mechanical model for light like QED or even QFT, there are no particles, it's all just waves in an underlying field. It shouldn't be too surprising that those waves can combine under certain circumstances. We just choose to interpret that as something that was identifiable as individual photons emitted by the laser combining to form a new higher energy photon at the atom.
As I mentioned before, the wave function describing the oscillations in an E and B field for light is NOT the quantum mechanical wave function for light. However, I find it quite interesting that it is a step closer to the QM wave function. For the atom surrounded by lasers, if we can arrange it so that the E and B fields oscillate correctly at the atom, then we seem to have also arranged the QM wave oscillations so that a superposition will happen and we can get something that looks like a photon of a higher energy. That, I think, is what @Colin2B was alluding to in his comments about the photon being detected at the atom.
Best Wishes.
Postscript: New posts have come in.
So a charge moving towards the receiver will induce a current in the antenna?Yes it will. However, it would be a one-off or DC current you could observe. The free charges in the aerial will move so as to oppose the E field that is created by the approaching charge and there will have to be a significant difference in the number density of free charges on the conductor (the aerial) at different places. It's especially evident when comparing the surface closest to the moving charge with the surface furthest away from the moving charge. You would get an AC current provided you move the charge to and fro. You know this @alancalverd, you've answered questions about conductors in an E field elsewhere and and at other times on this forum.
Post by: paul cotter on 10/12/2022 16:32:03
Post by: Colin2B on 10/12/2022 23:37:13
If you have a really bright light source, like an exploding universe, and you wait until the expansion has stretched out the EM radiation to the microwave region of the spectrum, then you can measure that frequency directly using a big enough antenna, an amplifier and a frequency meter.I was going to suggest the same, but can’t find an example of optical to microwave redshift. Haven’t had a lot of time, will give so e thought next week.
Is that cheating?
@Eternal Student
I’ve been away with limited wifi. Because threads like this shift focus frequently assumptions are made and the direction changes which means original brief comments can be misunderstood, but will expand after tomorrow.
Interesting topic which has resulted in a weaving of different threads, some of which I’ll remove when I have time ie DarkKnight
Post by: hamdani yusuf on 11/12/2022 01:41:00
Actually not. We detect the current induced in the receiving aerial by the magnetic component.The components used in this video aren't magnetic.
Basically, you can measure EM wave in any frequency, as long as you have quick enough rectifier.
Post by: hamdani yusuf on 11/12/2022 05:30:41
Surely not. The wires which made the dipole aerials are shiny reflective things. If you just shine visible light on an aerial then it just reflects off it, it doesn't interact with the aerial to generate a current in it.The antenna in radio frequency is also reflective, which is how the electric current is generated there. The current is then converted to other type of energy in the receiving unit.
The problem with higher frequency is the antenna becomes impractically short, and the electric charge, i. e. electron's inertia makes it harder to move, and dampen the vibration.
Quote
Meanwhile, most X-rays or gamma rays can just go straight through without any significant interaction along the way.That's why even at high frequency, particle model for light is still not accurate. Ordinary particles never pass through other particles regardless their energy magnitude.
Post by: paul cotter on 11/12/2022 09:45:22
Post by: Bored chemist on 11/12/2022 09:47:48
I was going to suggest the same, but can’t find an example of optical to microwave redshift. ...The em radiation emitted by the recombination of electrons and protons in the early universe was a mix of UV and visible. (Up to about 13.6 ev)
So the low energy end of the CMBR spectrum started out as visible light.
Post by: hamdani yusuf on 11/12/2022 11:10:30
Your proposed rectifier also becomes impractically small as the frequency goes up. As I said before, low microwave frequencies are the highest standard electronics can go.That's true, for now. Perhaps advancements in nanotechnology can push the limits further.
Post by: alancalverd on 11/12/2022 14:56:38
Ordinary particles never pass through other particles regardless their energy magnitude.I've just been sieving some flour. I wish I had read your post before wasting my time. Or am I making biscuits with waves and butter?
Post by: paul cotter on 11/12/2022 15:56:47
Post by: alancalverd on 11/12/2022 18:06:29
Post by: hamdani yusuf on 11/12/2022 21:35:53
Did your flour particles pass through the particles of your sieve?Ordinary particles never pass through other particles regardless their energy magnitude.I've just been sieving some flour. I wish I had read your post before wasting my time. Or am I making biscuits with waves and butter?
Or did they pass through the holes?
Have you tried to look closer?
Post by: hamdani yusuf on 11/12/2022 22:09:42
https://youtube.com/shorts/Sndz8Mm52u8?feature=share
Blowing through glass. I guess everyone here knows how it works.
Post by: Bored chemist on 11/12/2022 22:22:42
Somehow this video just popped up in my YouTube recommendations.https://en.wikipedia.org/wiki/Coand%C4%83_effect
https://youtube.com/shorts/Sndz8Mm52u8?feature=share
Blowing through glass. I guess everyone here knows how it works.
Why did you think it was relevant?
Post by: alancalverd on 11/12/2022 23:00:33
Did your flour particles pass through the particles of your sieve?Solid state physics has progressed since the days of Democritus. We now know, thanks to some classic experiments just down the road at the Cavendish laboratory, that practically all of every atom is empty space.
Or did they pass through the holes?
Have you tried to look closer?
X-ray photons do occasionally interact with electrons and very rarely with nuclei, but in diagnostic radiology (where we want maximum interaction with, say, bones) we rely on around 5 - 10% passing unhindered through the animal to form an image. And we do detect them as individual photons, as I mentioned some way back.
If 90% of the flour remained in the sieve, I'd probably complain, but I'm sure you get the picture.
Post by: evan_au on 12/12/2022 08:20:11
Quote from: paul cotter
Your proposed rectifier also becomes impractically small as the frequency goes up. As I said before, low microwave frequencies are the highest standard electronics can go.How does this apply to the several m2 of rectifier on my roof?
It seems to interact quite well with visible and near IR frequencies.
Post by: paul cotter on 12/12/2022 08:43:50
Post by: Bored chemist on 12/12/2022 08:48:05
Your roof generates DC, rather than AC at multi THz frequencies.Quote from: paul cotterYour proposed rectifier also becomes impractically small as the frequency goes up. As I said before, low microwave frequencies are the highest standard electronics can go.How does this apply to the several m2 of rectifier on my roof?
It seems to interact quite well with visible and near IR frequencies.
Post by: Bored chemist on 12/12/2022 08:51:07
We now know, thanks to some classic experiments just down the road at the Cavendish laboratory, that practically all of every atom is empty space.And yet we know we can't walk through walls.
Particles like those from which we are made, generally do not go through particles such as those from which walls are made.
It's to do with the exclusion principle or electrostatic repulsion- it seems to depend on who you ask.
Post by: alancalverd on 12/12/2022 10:04:37
And yet we know we can't walk through walls.Sir Arthur Eddington observed that "The student of physics must become accustomed to having his common sense violated five times before breakfast. If he were to fall through the floor and materialise in the basement, he should not consider it magic, but merely a highly improbable coincidence."
Einstein was more succinct in saying that quantum mechanics is weirder than you can imagine.
For those of a quantitative bent, the deBroglie wavelength of a 70 kg mass is very small indeed, which makes the probability of your being somewhere else of no practical value. Not that the concept is entirely devoid of engineering applications, as demonstrated by the electrons tunnelling through Evan's pn junction when the sun shines.
Post by: hamdani yusuf on 12/12/2022 10:19:40
Someone here thought that particles can go pas through other particles.Somehow this video just popped up in my YouTube recommendations.https://en.wikipedia.org/wiki/Coand%C4%83_effect
https://youtube.com/shorts/Sndz8Mm52u8?feature=share
Blowing through glass. I guess everyone here knows how it works.
Why did you think it was relevant?
Post by: Bored chemist on 12/12/2022 12:56:54
A "particle" that is everywhere all the time isn't very particulate, is it?And yet we know we can't walk through walls.Sir Arthur Eddington observed that "The student of physics must become accustomed to having his common sense violated five times before breakfast. If he were to fall through the floor and materialise in the basement, he should not consider it magic, but merely a highly improbable coincidence."
Einstein was more succinct in saying that quantum mechanics is weirder than you can imagine.
For those of a quantitative bent, the deBroglie wavelength of a 70 kg mass is very small indeed, which makes the probability of your being somewhere else of no practical value. Not that the concept is entirely devoid of engineering applications, as demonstrated by the electrons tunnelling through Evan's pn junction when the sun shines.
Post by: hamdani yusuf on 12/12/2022 13:00:25
Sir Arthur Eddington observed that "The student of physics must become accustomed to having his common sense violated five times before breakfast. If he were to fall through the floor and materialise in the basement, he should not consider it magic, but merely a highly improbable coincidence."It would be more probable to get stuck half way through the floor, or even pass through to somewhere under the surface of the earth.
Post by: alancalverd on 12/12/2022 14:07:46
(b) no
The Beer-Lambert law applies when n is large, and there is a plausible "quantum theory of ghosts" where the object particles are light and the barrier N>>n and very dense.
Post by: alancalverd on 12/12/2022 14:10:06
A "particle" that is everywhere all the time isn't very particulate, is it?But a particle that could be anywhere at any time but only interacts at one point and one time, is neatly described by Schrodinger and Planck, properly understood.
Post by: Petrochemicals on 12/12/2022 14:23:25
Sounds like some sort of field that becomes activated somehow, producing a particle.A "particle" that is everywhere all the time isn't very particulate, is it?But a particle that could be anywhere at any time but only interacts at one point and one time, is neatly described by Schrodinger and Planck, properly understood.
Post by: alancalverd on 12/12/2022 17:51:39
It's much easier to stick with the experimental observations: we generally need to model propagation with wave equations, and note that detection is better modelled at high photon energies (optical and above) with discrete quanta.
Anything else is philosophy, not physics.
I've never understood why people find this difficult, and have to introduce anthropocentric observer effects and other mysticisms. You can write down the probability function of the score from throwing n dice. There is a very low probability (1/n6) of scoring n or 6n and a very high probability of scoring around 3.5n. That is a continuous Schrodinger-type predictive wave function, and gets more "continuous" as n increases. When you throw the dice you will always and only get a single quantised integer score - a Planck result. If there was such a thing as an observer effect, you could make a fortune by asking everyone else to shut their eyes when the croupier rolls the dice.
Post by: evan_au on 12/12/2022 19:45:42
Quote from: alancalverd
When you throw the dice you will always and only get a single quantised integer scoreMy high school geology teacher was preparing us for a field trip, and wanted to make a point about the way shells tend to lie when they die (as I recall).
He grabbed the curved plastic lens cap off his slide projector (imagining it like one side of a bivalve) and flipped it in the air. It bounced on the floor, and came to a stop, neatly balanced on the narrow rim. This was so spectacularly unlikely that I think the class entirely missed the point of the illustration...
Post by: Eternal Student on 12/12/2022 21:49:09
There is a very low probability (1/n6) of scoring ...You did that deliberately didn't you? I'm just to pretend I didn't see it and hope it reads (1/6)n when I get back here.
Best Wishes.
Post by: alancalverd on 12/12/2022 23:57:17
Post by: Petrochemicals on 14/12/2022 20:15:56
Lasers can theoretical turn microwave to photon, there are filters that can change infra red to blue, as seen in confinement fusion.
I was going to suggest the same, but can’t find an example of optical to microwave redshift. Haven’t had a lot of time, will give so e thought next week.
Post by: Bored chemist on 14/12/2022 20:35:28
there are filters that can change infra red to blue, as seen in confinement fusion.Not really.
Post by: hamdani yusuf on 17/12/2022 04:49:59
https://en.wikipedia.org/wiki/Amp%C3%A8re%27s_circuital_lawWhat is the equation describing how a magnetic field is induced by a moving charge?Ampere's Law.
Quote
In classical electromagnetism, Ampère's circuital law (not to be confused with Ampère's force law)[1] relates the integrated magnetic field around a closed loop to the electric current passing through the loop. James Clerk Maxwell (not Ampère) derived it using hydrodynamics in his 1861 published paper "On Physical Lines of Force"[2] In 1865 he generalized the equation to apply to time-varying currents by adding the displacement current term, resulting in the modern form of the law, sometimes called the Ampère–Maxwell law,[3][4][5] which is one of Maxwell's equations which form the basis of classical electromagnetism.Where do you find electric charge or velocity in the article above?
Post by: paul cotter on 17/12/2022 09:52:14
Post by: alancalverd on 17/12/2022 10:14:44
Where do you find electric charge or velocity in the article above?Current = charge per unit time passing a given point. I = dQ/dt
Post by: Bored chemist on 17/12/2022 11:18:48
... the interaction between the light and photosensitive material is very much a 'photon and atoms' interaction much as described earlier. At the point of interaction, the material was simply reacting to a deposit of energy by a photon ...That energy is transferred when a force moves through a distance.
The force is applied to an electron and promotes it to a higher energy state.
The nature of the force is electrostatic in nature.
The electron reacts to that electromagnetic force.
Or do you think it's one of the other 3 fundamental forces that does it?
Post by: hamdani yusuf on 17/12/2022 13:39:55
In my example, Q is constant over time.Where do you find electric charge or velocity in the article above?Current = charge per unit time passing a given point. I = dQ/dt
Post by: Eternal Student on 17/12/2022 17:04:15
That energy is transferred when a force moves through a distance.That's a result from mechanics, especially Newtonian mechanics. Not all energy is transferred in a way that can be identified as some force moved through some physical distance.
For the transfer of heat between two bodies, there doesn't need to be some force identified and some distance over which it was applied. For example, a hot body does not push a colder body away, it just transfers heat. (You can try to look microscopically and consider particles being agitated or accelerated by some force but if you look again with different glasses on then there are no particles, just waves. Alternatively you just need to recognise something you ( @Bored chemist ) said in a different thread about temperature - temperature can be a measure of all sorts of internal energy in a substance and not just translational, rotational or vibrational motion of particles).
The transfer of energy by waves is, of course, another example. A water wave is a wave in something, you could imagine that a superposition of two waves into a bigger wave (which is then a bigger lump of energy) happens because the water is being pushed up by some force from the other wave. For an e-m wave, it does not have to be a wave in any material like "the aether". Somehow the two waves just do combine and there is a big wave BUT there may not be any material or any force acting on that material that can be identified. You obtain a bigger amount of energy in the final e-m wave but there was no material where mechanical forces had been applied over some physical distance.
Getting directly to the situation being discussed: For an atom and photon interaction, the energy is transferred in some way that is not like some sort of mechanical force applied over some physical distance. For example, you can't have two small forces that would sum up to the sufficient force (such as two low energy photons striking the electron). You must have one photon of the right energy all in one go. There is also no way you could use a smaller force but allow it to act over a larger physical distance (I don't even know what that could mean or look like - the photon just interacts and there was no "distance" over which that force was applied).
Best Wishes.
Post by: Bored chemist on 17/12/2022 17:44:25
Hi.The latent image in a photograph is composed of "out of place" electrons.That energy is transferred when a force moves through a distance.That's a result from mechanics, especially Newtonian mechanics. Not all energy is transferred in a way that can be identified as some force moved through some physical distance.
For the transfer of heat between two bodies, there doesn't need to be some force identified and some distance over which it was applied. For example, a hot body does not push a colder body away, it just transfers heat. (You can try to look microscopically and consider particles being agitated or accelerated by some force but if you look again with different glasses on then there are no particles, just waves. Alternatively you just need to recognise something you ( @Bored chemist ) said in a different thread about temperature - temperature can be a measure of all sorts of internal energy in a substance and not just translational, rotational or vibrational motion of particles).
The transfer of energy by waves is, of course, another example. A water wave is a wave in something, you could imagine that a superposition of two waves into a bigger wave (which is then a bigger lump of energy) happens because the water is being pushed up by some force from the other wave. For an e-m wave, it does not have to be a wave in any material like "the aether". Somehow the two waves just do combine and there is a big wave BUT there may not be any material or any force acting on that material that can be identified. You obtain a bigger amount of energy in the final e-m wave but there was no material where mechanical forces had been applied over some physical distance.
Getting directly to the situation being discussed: For an atom and photon interaction, the energy is transferred in some way that is not like some sort of mechanical force applied over some physical distance. For example, you can't have two small forces that would sum up to the sufficient force (such as two low energy photons striking the electron). You must have one photon of the right energy all in one go. There is also no way you could use a smaller force but allow it to act over a larger physical distance (I don't even know what that could mean or look like - the photon just interacts and there was no "distance" over which that force was applied).
Best Wishes.
There's not many explanations for that which don't involve electromagnetic forces.
Post by: alancalverd on 17/12/2022 18:04:56
In my example, Q is constant over time.and it is moving
Post by: Petrochemicals on 17/12/2022 18:29:34
Petro, there are no filters that can change the frequency of incident radiation. What I believe you are thinking about is certain crystals that perform frequency doubling operations such as infrared to visible, similar to the frequency doublers in common use in electronics. A lot of visible lasers generate at infrared and then double up to visible.
Petro, there are no filters that can change the frequency of incident radiation. What I believe you are thinking about is certain crystals that perform frequency doubling operations such as infrared to visible, similar to the frequency doublers in common use in electronics. A lot of visible lasers generate at infrared and then double up to visible.Yep something like that, a crystal I think they used in the fusion lab recently. What is incident radiation.
Post by: Eternal Student on 17/12/2022 20:51:36
The latent image in a photograph is composed of "out of place" electrons.? I'm not sure what that was about.
In a simple photographic material, like some silver halide held in a gelatin emulsion, there is something called a "latent image centre" that forms when silver ions absorb an electron. Those latent image centres are only stable when there's a few silver atoms all located close together, the text I was just looking at suggested 3 or 4 silver atoms must be located close to each other to remain stable. An earlier post from @alancalverd suggested you needed two or more, I'm not going to quibble over precisely how many you need, it's a small number. Anyway, the latent image centre has a collection of whole silver atoms and not just some out of place electrons.
The free delocalised electrons were there, available for the silver ions to absorb, only because some photon had ejected an electron from the valence band of one halide particle and into the conduction band of the combined silver-halide lattice. That happened somewhere in the structure. Some of the Silver ions are free to move around through the crystal lattice and the electron is also quite free to move around. So, usually the electron doesn't need to travel too far before it encounters some Silver ion that can absorb it, forming atomic silver. Hence the latent image centre does usually form very close to where the original photon struck the photographic material (and that's why the photograph works, we have a latent image centre within nano-metres of where the photon struck the photographic material). Anyway, the ejection of that electron to the conduction band was a quantum mechanical effect, it happened due to some interaction between a photon and an atom (well, a bromide ion rather than a whole atom). That interaction is what I might have described as a 'photon-and-atom' interaction.
There's not many explanations for that which don't involve electromagnetic forces.There is no attempt to explain what mechanical forces applied when an atom modelled with Quantum Mechanics has an electron excited to another orbit by a photon. Mechanical forces and solid particles on which they can be applied are not there or required to be there in a QM model of an atom.
(There are, as you suggest, some explanations involving forces like an electrostatic force when the atom and its electrons are modelled just as some ordinary mechanical system).
Best Wishes.
Post by: evan_au on 17/12/2022 21:34:50
Quote from: Paul Cotter
change the frequency of incident radiation...certain crystals that perform frequency doubling operations such as infrared to visible
Quote from: Petrochemicals
What is incident radiation?This description refers to the familiar hand-held green laser pointer, which has an infra-red laser at 808nm, and a non-linear neodymium-doped frequency-doubling crystal which produces the emitted green beam at 404nm.
- The incoming, or "incident" radiation is infra-red from the pump laser
- The outgoing, or emitted radiation is green
https://en.wikipedia.org/wiki/Laser_pointer#Colors_and_wavelengths (see subsection "Green")
Quote
a crystal I think they used in the fusion lab recentlyThis description refers to the US National Ignition Facility.
In this case, Xenon flash lamps are used to pump electrons in neodymium-doped glass into a high-energy state.
The incident radiation from the Xenon flash lamps is white light over a wide range of frequencies; only a tiny fraction of the light is at exactly the right frequency to produce the desired 1053nm laser pulse. This results in low efficiency for this type of laser, and is why NIF needs to cool down for 12 hours before attempting another pulse.
- This is unlike the hand-held laser pointer, where the incident infra-red energy is at just the right wavelength to produce the desired outgoing green beam, achieving quite high efficiency and allowing continuous operation (until the battery goes flat...).
https://en.wikipedia.org/wiki/National_Ignition_Facility#Laser
Post by: Bored chemist on 17/12/2022 23:29:56
the text I was just looking at suggested 3 or 4 silver atomsAnd you get those atoms by pulling an electron off a halide ion and sticking it onto a silver ion.
Compared to the silver halide, those electrons are in the wrong places.
Post by: hamdani yusuf on 18/12/2022 02:41:53
A 1 Coulomb charged particle moves at 1 m/s speed. What's the current?In my example, Q is constant over time.and it is moving
In cgs system, the same experiment would read:
A 1 Coulomb charged particle moves at 100 cm/s speed. What's the current?
Post by: Bored chemist on 18/12/2022 10:42:23
A 1 Coulomb charged particle moves at 1 m/s speed. What's the current?It depends
Imagine I put that coulomb into a 1 metre cube box. At 1 m/s the whole coulomb goes past me in 1 second and that's a current of 1 amp.
Now imaging I put the same charge in a box 10 metres long.
It now takes 10 seconds to go past me.
So that's 1 C in 10 S or 0.1 C/S so that's 0.1 amps.
You really need to study science a bit more in order to avoid asking meaningless question.
Post by: hamdani yusuf on 18/12/2022 12:04:05
It's meaningless to you because you haven't understood the problem yet. It shows that Maxwell's equations are not adequate to describe electrodynamics systems.A 1 Coulomb charged particle moves at 1 m/s speed. What's the current?It depends
Imagine I put that coulomb into a 1 metre cube box. At 1 m/s the whole coulomb goes past me in 1 second and that's a current of 1 amp.
Now imaging I put the same charge in a box 10 metres long.
It now takes 10 seconds to go past me.
So that's 1 C in 10 S or 0.1 C/S so that's 0.1 amps.
You really need to study science a bit more in order to avoid asking meaningless question.
Let's distribute the electric charge into a thin metal disc with 10 m diameter and 0.1 mm thick. The disc moves axially at 1 m/s. What's the electric current?
Here's another example.
Electrons move in a CRT at approximately 0.1 c. What's the current generated by each electron? What's the expected B field at a point 1 mm from the trajectory of the electron?
Post by: Bored chemist on 18/12/2022 13:31:57
It's meaningless to you because you haven't understood the problem yet.Here's is the problem you set.
A 1 Coulomb charged particle moves at 1 m/s speed. What's the current?What part of it do you think I didn't understand.
More importantly, what do you think the answer is?
Post by: Bored chemist on 18/12/2022 13:33:39
What's the current generated by each electron?Can I ask you to do something that will help a lot.
Before you post anything here, can you just check that the units of your question make sense.
Your question is as pointless as if you asked "How fast is a mile?"
Post by: alancalverd on 18/12/2022 13:48:52
Post by: Bored chemist on 18/12/2022 15:54:05
Current is dQ/dt,Good point.
What's the current generated by each electron?Charge at start of experiment about 10^-19 Coulombs
Charge at end of experiment about 10^-19 Coulombs
Change in charge
Zero
Rate of change of charge
Zero
Current
Zero
Usefulness of question
Pretty close to zero.
Post by: alancalverd on 18/12/2022 16:21:18
Usefulness? not a lot, but the inverse phenomenon was used to deflect the electron beam in CRT televisions, radar displays and scanning electron microscopes, though magnetic drive is generally too slow for use in oscilloscopes.
Post by: Bored chemist on 18/12/2022 16:45:56
e.g "the galvanometer sensitivity is 2 divisions per microcoulomb."
From
https://gulpmatrix.com/the-working-principle-and-uses-of-a-ballistic-galvanometer/#gsc.tab=0
And one thing that makes then useful is that sometimes, the current vs time plot is complicated, but you only need to know the total charge (e.g testing magnets).
So, yes, you can in principle measure a charge of a single electron but, like the mA. Sec meter, it's measuring charge not current.
You can actually measure currents so small that, on average, there's less than 1 electron per second flowing through a circuit.
https://download.tek.com/document/2648%20Counting%20Electrons1.pdf
But the "per second" bit is important.
And... it's a bit fiddly.
Post by: Eternal Student on 18/12/2022 23:35:01
In regards to @hamdani yusuf asking questions about a magnetic field due to a single moving charge.
You were deceived a little and that's all. @alancalverd provided a simplified calculation of the B field as if the moving charge was an ordinary current in a wire (I think, not sure, it was a lot of posts ago).
For a more complete answer you would need to consider all of Maxwell's laws and then the Biot-Savart law will emerge. Note that there are TWO different things that are often called the Biot-Savart law. We want the Biot-Savart law for a single moving charge and not for a steady current flowing in a wire.
See https://en.wikipedia.org/wiki/Biot%E2%80%93Savart_law#Point_charge_at_constant_velocity.
At low velocities the non-relativistic version is often used:
[Equation 1]
where vectors and scalars appear as usual:
is the velocity of the charged particle. r is the scalar magnitude of r which is almost the position vector of the moving charge. r is the vector to the moving charged particle from the point where you are trying to determine the B field. So if you are determining B at position r' then r = R - r' with R = position of the charged particle (from the origin). is the unit vector in the direction orthogonal to both v and r (it's positive in the direction v x r ).As you can see in [Equation 1] the current, I, does not appear, only the charge and the velocity of that charge.
- - - - - - - -
As for the multiple questions about the current due to some moving charges with constant velocity. For a genuine point charge (with charge Q), the current becomes a Dirac delta function but Maxwell's equations do hold quite well using these functions. Specifically I = current at position x and time t is defined as:
I(x , t) = 0 if the charge is never at position x.
Q. δ(t-t0) if the moving charge is at position x at time t0.
More usually we'd be interested in a current density vector, J, rather than just a current. You hardly need to make use of such a peculiar definition anyway. When applying Ampere's law (to the situation with just the one moving point charge), just choose a surface which does not have the moving charge on it at the time t. Then J = 0 everywhere on that surface and the contribution to the magnetic field is then entirely due to the changing E field over that surface. If you're interested, this is almost precisely the opposite conditions that apply for a steady current flowing in a wire and that version of the Biot-Savart law. For a steady current in a wire, the E field would be unchanging and only the flow of charges matter and contribute to the magnetic field. In the situation with just one charge, the E field is definitely changing (as the source moves) but the current density vector is 0 almost everywhere. The fact that the Biot-Savart law for a steady current in a wire does look like the Biot-Savart law for a single moving charge is almost a happy co-incidence, indeed you can quickly prove one from the other (if you assume the magnetic field from a current in a wire is just the sum of the magnetic fields due to individual charges moving in that wire). It's not a pure co-incidience, of course, but discussing that probably needs a different thread.
Best Wishes.
Post by: hamdani yusuf on 19/12/2022 08:38:04
See https://en.wikipedia.org/wiki/Biot%E2%80%93Savart_law#Point_charge_at_constant_velocity.The equations in the link are:
(https://wikimedia.org/api/rest_v1/media/math/render/svg/89e2eeb70fb6766d16db953b66c16a16482bb363)
For electron in CRT, the equations below don't apply.
(https://wikimedia.org/api/rest_v1/media/math/render/svg/9b4530d6cb2b24d71b8c43a2f390b4a6b95ea981)
and
(https://wikimedia.org/api/rest_v1/media/math/render/svg/d912ca72799f3c6e362e68bd7ca3e7ee7b0a65d1)
The most obvious limitation of Maxwell's equations is lacking of explanation for permittivity and permeability of various media. How can different distribution of electric charges change the permittivity and permeability of a point in space?
Post by: Bored chemist on 19/12/2022 12:57:27
The most obvious limitation of Maxwell's equations is lacking of explanation for permittivity and permeability of various media.That's not their job.
Post by: Eternal Student on 19/12/2022 17:11:45
For electron in CRT, the equations below don't apply.Why not?
They can be quick, up to 1/10 c according to one piece of text. Have you tried the relativistic versions?
Also usually a CRT device like an oscilloscope doesn't try to determine or measure the magnetic and electric field generated by the ray.
Best Wishes.
Post by: paul cotter on 19/12/2022 18:44:41
Post by: hamdani yusuf on 19/12/2022 22:20:58
That's why they don't work well at microscopic scale.The most obvious limitation of Maxwell's equations is lacking of explanation for permittivity and permeability of various media.That's not their job.
Post by: hamdani yusuf on 19/12/2022 22:22:23
Hi.For high speed charged particles, retardation needs to be accounted. That's why the article mentioned Jefimenko.For electron in CRT, the equations below don't apply.Why not?
They can be quick, up to 1/10 c according to one piece of text. Have you tried the relativistic versions?
Also usually a CRT device like an oscilloscope doesn't try to determine or measure the magnetic and electric field generated by the ray.
Best Wishes.
Post by: hamdani yusuf on 19/12/2022 22:27:41
Permittivity and permeability are properties of various materials. Maxwell's equations deal with the behaviour of fields, not with material properties, though these properties do influence the maths.Materials are made of electrically charged particles. Permeability and permittivity emerge from their distribution in space.
Post by: Eternal Student on 20/12/2022 04:54:13
For high speed charged particles, retardation needs to be accounted. That's why the article mentioned Jefimenko.Yes, Jefimenko's equations are better for very high speed charged particles.
However, you weren't originally asking questions about high speed particles (posts #148 through to #174 started with particles moving at ~ 1m/s and even the upgrade to Cathode rays had velocities ~ 0.1c). I'm not sure what the problem was or where you were going next. I had guessed it was about charge being infinitely divisible but perhaps it wasn't. Anyway, that's perfectly fine (a forum should have the discussion move and change direction for all sorts of reasons - otherwise what's the point of having a forum?)
Best Wishes.
Post by: paul cotter on 20/12/2022 08:35:19
Post by: alancalverd on 20/12/2022 17:22:05
That's why they don't work well at microscopic scale.citation needed
AFAIK any moving charge creates a magnetic field, and any changing magnetic field can induce a current in a conductor. I've only worked with atomic nuclei (in MRI systems) but BC may well have played with electrons (chemists like ESR measurements). How microscopic did you have in mind?
Back in the mists of my youth I was involved in the control and measurement of 400 keV electron beams. The relativistic effects were clearly demonstrable but I can't remember what fraction of c - we set it up eventually as an undergraduate teaching experiment.
Post by: Bored chemist on 20/12/2022 18:42:48
I understand what Hamdani is alluding to and it is something I have often thought about-permeability and permittivity are macroscopic properties derived fundamentally from the presence of charges in said material. Do they have a meaning at the atomic level?
You can calculate a refractive index from the relative permeability and permittivity.
That's a measure of how much light slows down as it passes through the material.
But light doesn't actually slow down, simplistically, it gets "stuck" to the particles
So, on the microscopic level, a refractive index isn't well defined and thus nor is the permeability or permittivity.
On the other hand, I can say the plastic they make thin spectacle lenses from has a high refractive index because it's full of sulphur compounds and the sulphur is big and polarisable- the nuclei don't keep a very tight grip on the electrons so, a passing ray of light interacts strongly with the electrons and thus with the plastic.
That's a good enough model down to the atomic level or thereabouts
It's not a clear cut issue.
Post by: hamdani yusuf on 21/12/2022 03:17:59
I understand what Hamdani is alluding to and it is something I have often thought about-permeability and permittivity are macroscopic properties derived fundamentally from the presence of charges in said material. Do they have a meaning at the atomic level? Have to run now, working today, mv switchroom in a stinking meat plant-yuck.
Water has the same chemical composition, but different pressure and temperature can change its structure, which results in different electromagnetic properties, as shown in this article.
Quote
https://pubs.acs.org/doi/10.1021/jp105975c#Note also that the permittivity shown here is the bulk value. It's not clear if local electric permittivity or magnetic permeability of a point in space closer to the Oxygen atom differ from another point closer to the Hydrogen atom, or another point between two water molecules.
Dielectric Constant of Ices and Water: A Lesson about Water Interactions
J. L. Aragones, L. G. MacDowell, and C. Vega
Abstract
(https://pubs.acs.org/cms/10.1021/jp105975c/asset/images/medium/jp-2010-05975c_0009.gif)
In this paper, the dielectric constant has been evaluated for ices Ih, III, V, VI, and VII for several water models using two different methodologies. Using Monte Carlo simulations, with special moves to sample proton-disordered configurations, the dielectric constant has been rigorously evaluated. We also used an approximate route in which proton-disordered configurations satisfying the Bernal−Fowler rules were generated following the algorithm proposed by Buch et al. (Buch, V.; Sandler, P.; Sadlej, J. J. Phys. Chem. B1998, 102, 8641), and the dielectric constant was estimated assuming that all configurations have the same statistical weight (as Pauling did when estimating the residual entropy of ice). The predictions of the Pauling model for the dielectric constant differ in general from those obtained rigorously by computer simulations because proton-disordered configurations satisfying the Bernal−Fowler rules can differ in their energies by as much as 0.10−0.30 NkT (at 243 K). These differences in energy significantly affect properties that vary from one configuration to another such as polarization, leading to different values of the dielectric constant. The Pauling predictions differ from the simulation results, especially for SPC/E and TIP5P, but yield reasonable results for TIP4P-like models. We suggest that for three charge models the polarization factor (G) in condensed phases depends on the ratio of the dipole to the quadrupole moment. The SPC/E, TIP5P, TIP4P, TIP4P/2005, TIP4P/ice models of water are unable to describe simultaneously both the experimental dielectric constants of water and ice Ih. Nonpolarizable models cannot describe the dielectric constants of the different condensed phases of water because their dipole moments (about 2.3 D) are much smaller that those estimated from first principles (of the order of 3 D). However, the predictions of TIP4P models provide an overall qualititatively correct description of the dielectric constant of the condensed phases of water, when the dipole moment of the model is scaled to the estimated value obtained from first principle calculations. Such scaling fails completely for SPC/E, TIP3P, and TIP5P as these models predict a completely different dielectric constant for ice Ih and water at the melting point, in complete disagreement with experiment. The dielectric constant of ices, as the phase diagram predictions, seems to contain interesting information about the orientational dependence of water interactions.
Post by: hamdani yusuf on 21/12/2022 03:23:05
I think any introduction to quantum mechanics mention some limitations of Maxwell's equations, that's why quantum mechanics was developed in the first place. Different sources may emphasize different limitations.That's why they don't work well at microscopic scale.citation needed
AFAIK any moving charge creates a magnetic field, and any changing magnetic field can induce a current in a conductor. I've only worked with atomic nuclei (in MRI systems) but BC may well have played with electrons (chemists like ESR measurements).
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How microscopic did you have in mind?What about a point of space between hydrogen and oxygen atom in a water molecule?
Post by: hamdani yusuf on 21/12/2022 03:40:33
However, you weren't originally asking questions about high speed particles (posts #148 through to #174 started with particles moving at ~ 1m/s and even the upgrade to Cathode rays had velocities ~ 0.1c).My original question is about limitation of Maxwell's equations to describe point to point interactions between two electrically charged particles, similar to Newton's mechanics and universal gravitation. Coulomb's law is only good for non-moving charges. How their movements affects the interacting forces is not well defined yet.
For my example about slowly moving electrically charged particle is to show that B field and E field are not constant over time when measured at a specific point in space, and they affect each other, which may complicate the calculation further.
Post by: Bored chemist on 21/12/2022 08:40:35
Water has the same chemical composition, but different pressure and temperature can change its structure, which results in different electromagnetic properties, as shown in this article.We know that snow looks different from rain, even without that article.
What about a point of space between hydrogen and oxygen atom in a water molecule?We can certainly measure the electron density there and, from that , we can get a fair idea of the permeability and permittivity.
Post by: alancalverd on 21/12/2022 10:38:04
The wavelength of visible light is orders of magnitude larger than an atom or molecule. When you walk on sand your speed depends on its bulk properties (wet, dry, compacted...) not the interaction of your foot with each individual grain, though the inter-grain mechanics (sharp, soft....) actually determine the bulk property.
The fact that for instance μiron is sometimes enormously greater than most other materials is clearly a function of group (domain) behavior rather than that of a single atom.
As the wavelength of EMR approaches atomic dimensions so we need to model the interaction by quantum rather than wave mechanics, as I stated several posts ago. At the other end of the scale we can measure μ and ε statically for any substance, and as we can see with the dispersion of white light in glass, these parameters vary with wavelength and with the nature of the transmitting medium, since bulk properties ultimately depend on atomic properties.
So to address HY's problem: Maxwell's equations describe wave propagation. The wave model works well when considering propagation at wavelengths greater than an atom or molecule diameter and also describes diffraction from a crystal lattice (a bulk property), but does not describe the interaction of EMR with individual atoms, for which we have a particle model.
Maxwell's equations don't "break down" any more than a train timetable "breaks down" when you want to catch a bus - they predict only and exactly what they say they predict.
Post by: hamdani yusuf on 23/12/2022 06:57:43
There is an underlying misunderstanding here.OK.
In Maxwell's theory, speed of light in a medium depends on its ε and μ. But Maxwell didn't say how distribution of electrically charged particles in the medium affects the values of ε and μ. The change of speed when light propagate from one medium to another is commonly attributed to the cause of refraction. Observations show that change of direction in refraction depends on frequency, which in turn implies that ε and μ of the medium also depend on frequency.
I have uploaded three more videos investigating behavior of microwave. This time I use meta-material.These are videos showing experiments on refraction of microwave using metamaterials.
The first is constructing meta-material to demonstrate interference by partial reflector
Second, we emulate refraction in microwave using meta-material, which is a multilayer metal grating
Lastly, reconstructing prism for microwave using meta-material to demonstrate refraction and internal reflection.
NB: This is not an April fool
The air between the aluminum tubes is the same in composition as the air outside of the meta-prism. Yet, it demonstrates significant difference in ε and μ in microwave frequency. It shows that Maxwell's equations are incomplete. They haven't described the relationships between distribution of particles and electromagnetic characteristics of space around them.
Post by: Bored chemist on 23/12/2022 08:45:06
But Maxwell didn't say how distribution of electrically charged particles in the medium affects the values of ε and μ.It isn't just their distribution that matters.
How tightly held they are also maters.
But, while Maxwell didn't go into this aspect, others did.
https://en.wikipedia.org/wiki/Clausius%E2%80%93Mossotti_relation
And, once again, it looks like you didn't study before asking.
Post by: hamdani yusuf on 23/12/2022 08:53:05
Maxwell's equations don't "break down" any more than a train timetable "breaks down" when you want to catch a bus - they predict only and exactly what they say they predict.Did Maxwell mention anything about the limitations of his model, or in what conditions was his model expected to fail in explaining observations?
Post by: hamdani yusuf on 23/12/2022 09:02:33
How tightly held they are also depends on the distribution of the particles, i.e. protons, electrons, and neutrons.But Maxwell didn't say how distribution of electrically charged particles in the medium affects the values of ε and μ.It isn't just their distribution that matters.
How tightly held they are also maters.
But, while Maxwell didn't go into this aspect, others did.
https://en.wikipedia.org/wiki/Clausius%E2%80%93Mossotti_relation
And, once again, it looks like you didn't study before asking.
The article you cited doesn't work for metamaterials, as shown in my experiments.
Which question did you try to answer?
Post by: Bored chemist on 23/12/2022 09:14:19
He probably thought that they were too obvious to mention.Maxwell's equations don't "break down" any more than a train timetable "breaks down" when you want to catch a bus - they predict only and exactly what they say they predict.Did Maxwell mention anything about the limitations of his model, or in what conditions was his model expected to fail in explaining observations?
Which question did you try to answer?I was answering your misunderstanding.
Post by: alancalverd on 23/12/2022 10:11:59
They do not purport to describe attenuation, diffraction, interference, or any other interaction with anything, any more than a train timetable purports to predict the arrival of buses. (Toronto residents may disagree on that point, but few other cities are as efficiently coordinated).
Post by: paul cotter on 23/12/2022 10:13:26
Post by: hamdani yusuf on 23/12/2022 14:44:00
Maxwell's equations have no known limitations. AFAIK they describe the propagation of EM radiation at all frequencies and in all materials.Do they describe photoelectric effect?
Refraction of X-ray in glass?
Post by: hamdani yusuf on 23/12/2022 14:45:52
He probably thought that they were too obvious to mention.Or he didn't know their limitations, and seemingly, you don't either.
Post by: hamdani yusuf on 23/12/2022 14:46:38
I was answering your misunderstanding.So, you were talking to yourself.
Post by: hamdani yusuf on 23/12/2022 14:51:40
They do not purport to describe attenuation, diffraction, interference, or any other interaction with anything, any more than a train timetable purports to predict the arrival of buses. (Toronto residents may disagree on that point, but few other cities are as efficiently coordinated).Those exclusions would make Maxwell's equations not very useful.
So why did Kelvin confidently say that physics was almost complete back then?
Post by: hamdani yusuf on 23/12/2022 14:54:45
Ferrimagnetic materials often display a variable μ dependent on the frequency and intensity of an applied MAGNETIC field but these materials are opaque to em radiation.Gamma ray is also em radiation, and it will likely pass through ferromagnetic materials to a significant depth.
Post by: paul cotter on 23/12/2022 15:36:42
Post by: alancalverd on 23/12/2022 16:31:05
Those exclusions would make Maxwell's equations not very useful.Like a train timetable, eh? Or the periodic table, which doesn't predict the winner of a horse race.
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So why did Kelvin confidently say that physics was almost complete back then?Because he was wrong. Kapitza said the same thing in 1964 (I was there). As they say in aviation
after 100 hours you know everything
after 1000 hours you know you don't know everything
after 10,000 hours you know you can't know everything.
Post by: evan_au on 23/12/2022 23:26:34
Quote from: Paul cotter
ε is a material bulk property and does not vary with frequency, to the best of my knowledgeWater (H2O) and Glass (SiO2) have a μr close to 1, but they are still dispersive (produce rainbows) - is this is due to variation in ε or μ with wavelength?
- There are particular wavelengths of infra-red light where optical fibers have near-zero dispersion (ie over that part of the spectrum, refractive index does not vary very much). These wavelengths are of great interest for high-speed telecommunications, since dispersion distorts optical signals - different wavelengths arrive at different times.
- The other trick used for high-speed telecommunications is to have a long length of "normal" dispersive fiber (eg 30-70km), with a short length of dispersion-compensating fiber (eg 30-70m), which has the opposite dispersion characteristic to "normal" fiber, ensuring that all wavelengths arrive simultaneously.
https://www.fiberopticsonline.com/doc/understanding-and-measuring-chromatic-dispers-0002
Post by: hamdani yusuf on 24/12/2022 10:49:05
Then why did you write this statement?Those exclusions would make Maxwell's equations not very useful.Like a train timetable, eh? Or the periodic table, which doesn't predict the winner of a horse race.QuoteSo why did Kelvin confidently say that physics was almost complete back then?Because he was wrong. Kapitza said the same thing in 1964 (I was there). As they say in aviation
after 100 hours you know everything
after 1000 hours you know you don't know everything
after 10,000 hours you know you can't know everything.
Maxwell's equations have no known limitations. AFAIK they describe the propagation of EM radiation at all frequencies and in all materials.
Post by: alancalverd on 24/12/2022 17:17:22
Post by: hamdani yusuf on 24/12/2022 23:58:45
Because it is true.When two statements are contradicting each other, then at least one of them must be false.
Maxwell's equations have no known limitations.Followed by their limitations.
They do not purport to describe attenuation, diffraction, interference, or any other interaction with anything
Post by: alancalverd on 25/12/2022 00:35:34
Maxwell's equations are intended to describe the propagation of electromagnetic radiation in a medium. As far as we know they do so for all EMR in all media. If that is what interests you, they have no limitation. If however you are interested in the color of Manchester United's 2023 away strip, they are admittedly of no use whatever.
Post by: Bored chemist on 28/12/2022 15:40:02
When two statements are contradicting each other, then at least one of them must be false.And if they are both accurate answers to a question, (which they are) it must be a very poorly constructed question.
Post by: hamdani yusuf on 29/12/2022 03:52:00
Maxwell's equations are intended to describe the propagation of electromagnetic radiation in a medium. As far as we know they do so for all EMR in all media. If that is what interests you, they have no limitation. If however you are interested in the color of Manchester United's 2023 away strip, they are admittedly of no use whatever.Michelson & Morley's, Fizeau's, and Sagnac's experiments showed the limitations of Maxwell's equations. There's also Faraday's paradox. They should still be within the scope of Maxwell's model.
Post by: hamdani yusuf on 29/12/2022 03:55:02
Water (H2O) and Glass (SiO2) have a μr close to 1, but they are still dispersive (produce rainbows) - is this is due to variation in ε or μ with wavelength?Then εr is the varying factor.
Post by: hamdani yusuf on 29/12/2022 03:57:30
Which question?When two statements are contradicting each other, then at least one of them must be false.And if they are both accurate answers to a question, (which they are) it must be a very poorly constructed question.
Is there something you don't understand?
Post by: alancalverd on 29/12/2022 10:45:12
Michelson & Morley's, Fizeau's, and Sagnac's experiments showed the limitations of Maxwell's equations. There's also Faraday's paradox. They should still be within the scope of Maxwell's model.M&M and Fizeau are entirely consistent with Maxwell. Faraday's paradox has nothing to do with the propagation of light.
Post by: Bored chemist on 29/12/2022 17:19:20
How did you come to that conclusion?Water (H2O) and Glass (SiO2) have a μr close to 1, but they are still dispersive (produce rainbows) - is this is due to variation in ε or μ with wavelength?Then εr is the varying factor.
Post by: Bored chemist on 29/12/2022 17:23:36
Which question?Have you forgotten what you asked?
You asked what current an electron was.
Post by: hamdani yusuf on 31/12/2022 06:33:08
M&M and Fizeau are entirely consistent with Maxwell.If you modify Newton's framework, just like Lorentz did.
Post by: hamdani yusuf on 31/12/2022 06:34:33
Faraday's paradox has nothing to do with the propagation of light.It has to do with generating electromagnetic field, which is what light is, according to Maxwell.
Post by: hamdani yusuf on 31/12/2022 06:36:57
Because c = 1/√εμHow did you come to that conclusion?Water (H2O) and Glass (SiO2) have a μr close to 1, but they are still dispersive (produce rainbows) - is this is due to variation in ε or μ with wavelength?Then εr is the varying factor.
If c changes while μ doesn't, then ε must change.
Post by: hamdani yusuf on 31/12/2022 07:22:20
It looks like you came late to join the party.Which question?Have you forgotten what you asked?
You asked what current an electron was.
Here's my original question.
What is the equation describing how a magnetic field is induced by a moving charge?Maxwell treated electric charge as continuum, which can be divided infinitesimally.No. He merely used the known phenomena of a magnetic field being induced by a moving charge and a potential being induced by a changing magnetic field. He made no assumptions about the nature of either.
Post by: alancalverd on 31/12/2022 10:06:33
Quote from: alancalverd on 29/12/2022 10:45:12No, it's a misconception of Gaussian induction.
Faraday's paradox has nothing to do with the propagation of light.
It has to do with generating electromagnetic field, which is what light is, according to Maxwell.
Maxwell combines two observed phenomena: an electric current generates a magnetic field, and a changing magnetic field can induce a current.
The Faraday "paradox" only involves the second phenomenon.
Post by: hamdani yusuf on 31/12/2022 22:00:51
Maxwell combines two observed phenomena: an electric current generates a magnetic field, and a changing magnetic field can induce a current.Check again the 4th equation. Magnetic field is generated by electric current PLUS changing electric field.
Post by: alancalverd on 01/01/2023 00:18:38
Post by: hamdani yusuf on 01/01/2023 06:09:11
His addition was the critical point which led Maxwell to conclude the existence of electromagnetic waves.