If time is a variable, which we know it is, then assigning a wave function as time related is far less outlandish than assigning an unobserved and mechanically unaccounted for energy to propel everything in the universe outwards at an ever accelerating rate.
I daresay the world of physics would be quite happy to consider a quantised dark energy though, aye?
But you reckon that subtracting time from the length of a standard second for a linear data curve is a no, no...? Kg
Not in the slightest, and it was Planck who was quoted as saying that he had fudged the maths, I'm certainly not qualified to make such an assertion.
I'm simply suggesting a possibility that I've seen to do what Planck did not manage, despite his best efforts, and linearise the maths.
Can Planck's law curve be matched to Rayleigh-Jean's law curve like this?The Rayleigh curve does not peak so could not be matched.
…..
Do the curves now match?
This concept ties in with the measurement of joules per 'standard' second and should eliminate the quantised nature of the data.Could you explain your thinking on this?
But time is not standard, it is variable,
Colin - I am suggesting that by calculating joules added as per a second that is becoming shorter as energy is added, that the quantised nature of calculating joules added as per standard second will be eliminated, and that the Planck data curve will no longer peak, but follow the Rayleigh-Jean law classical curve.
There is evidence of time varying within the same reference frame. Food put in a fridge will last longer than food that is not. Cryogenics is a consideration also.
I have been following your posts on these speculations of yours for a considerable length of time now timey. And frankly, I've been considering your ideas and been inclined to give you every benefit of doubt taking into account your sincerity.
There is evidence of time varying within the same reference frame. Food put in a fridge will last longer than food that is not. Cryogenics is a consideration also.
Colin - I am suggesting that by calculating joules added as per a second that is becoming shorter as energy is added, that the quantised nature of calculating joules added as per standard second will be eliminated, and that the Planck data curve will no longer peak, but follow the Rayleigh-Jean law classical curve.But the Rayleigh curve goes off the scale at higher frequencies, are you suggesting that energies in the uv zone are tending towards infinite?
There is evidence of time varying within the same reference frame. Food put in a fridge will last longer than food that is not. Cryogenics is a consideration also.Please show the evidence.
Jeff - What does it matter about Planck saying that he'd fudged the maths. That's what he apparently said according to Manjit Kumar and I've no doubt it is cited. But so what? ... And the link you posted - are you saying that the text suggests that my synopsis is incorrect?
Jeff - What does it matter about Planck saying that he'd fudged the maths. That's what he apparently said according to Manjit Kumar and I've no doubt it is cited. But so what? ... And the link you posted - are you saying that the text suggests that my synopsis is incorrect?
You made an issue of it or does your memory conveniently fail you?
Forget it
the cesium atomic clock must be cooled radically to a constant temperature. Any increase in temperature will increase its rate of time.The cesium atoms in a cesium fountain are cooled to a low temperature, but not for the reason you suggest.
Can Planck's law curve be matched to Rayleigh-Jean's law curve like this?
https://en.m.wikipedia.org/wiki/Ultraviolet_catastrophe
Frequency is a count of how many wave cycles complete within the time scale of a standard second.
E=hf and Wavelength=hv
What I'm going to do is decrease the length of a second as energy increases.
So for this calculation, by replacing Wavelength=hv with Wavelength=hf, where frequency is the velocity of a second - we are now keeping the 'distance' of a wavelength constant and instead decreasing the length of a second that the cycles of the wave complete within...
Do the curves now match?
Thank you for your input. Please forgive my impatience, but I am tempted to get my son to make a video of me trying to make posts on this broken phone. The screen constantly freezes up, or directs me to wrong page, and I'm uncertain about the future of my internet connection.OK, but just bear in mind that I almost didnt come back.
The question was concerning if decreasing the length of a standard second proportionally to increased energy as the Planck data curve diverges from the classical curve in the higher frequencies - or indeed adding to the length of a standard second proportionally to decreased energy where the Planck curve diverges from the classical curve in the lower frequencies - and if measuring joules per second added to the blackbody via these variable seconds as energy is added, would the Planck data be rendered linear?This is actually 2 separate questions:
The question is not concerning if doing so is appropriate, just if Planck's data could then be linear instead of quantised.
2.Would this linearise the data
The thing that is missing from E=hv in most quotes is n, more correctly E=nhv where n is an integer. Altering v doesn't change n, it is still there. Again you need to go well back to Planks work on oscillators and try to get rid of it, he tried and boy did he try hard!
Colin was saying that time dilation/contraction can only happen when viewing another reference frame than your own. (please see NIST 2010 ground level cesium clock experiments where 2 clocks are running at different rates 1 metre apart).
Although time varies between reference frames and places of differing gravity, there is no indication that it varies within an inertial frame.
We can also see that the notion of a waves length is dimensionless. Although the length of the light wave is proportional to that frequency of lights energy, the length of a wave in terms of a distance is completely meaningless as light of any frequency travels the same distance in the same time period.Can you please explain how a wavelength is dimensionless and meaningless, I don't follow your reasoning.
Alas there are no quantum discontinuities in the spectrum of a black body, so there's something wrong with the argument here.
these motions cause electronic transitions in atoms, and photons are subsequently emitted.
Jeff - yes, you were referring to frequency. So am I correct in my interpretation that the count of frequency is dimensionless?The count is dimensionless (remember Sesame Street), frequency and wavelength are not.
I do not know why you keep referring to me in relation to Thebox.Not a reference to you at all, but aimed at the box - perhaps unfairly. I have a great deal of respect for the box, he does try to think, but gets bogged down by misunderstanding what he is observing. I'll try not to comment on him.
Yes the count is dimensionless, but the duration of time the count is held relative to is not dimensionless. True or false?Yes, that's what I said. The count of cycles is dimensionless, the frequency (cycles per unit time) is not.
Alan - yes indeed, but the energy input to radiation frequency ratio is quantised as a result of the blackbody experiment data.
OK Colin - Thanks. Though I too like Thebox and I have no wish to hurt his feelings.By the way, he does understand that light takes the 8mins, he is confused by an interesting variation of Zeno's paradox.
The frequency of wavelength doesn't really take on a meaningful dimension until velocities are attributed. Am I correct?I'm having difficulty deciphering your sentence because you are still misusing wavelength, I think? What do you mean by frequency of wavelength? Frequency or cycles?
Would a blackbody conducted in a sphere give different data?No, the shape of the body has no effect on the radiation curve. You need to reread Alan's post which explains it all.
Think flute (clean sinusoid) compared with a rectangular organ pipe.I like Alan's eg.
But Hubble adds the aspect of velocity to the scenario.Obviously, if the source is moving wirh respect to the observer, you will see both Doppler and relativistic changes in the received frequency.
And yes, that's right, because a sphere is the most efficient shape. But the change in the frequency of the light is occurring within the changes in the molecular and atomic structure of the blackbody not the cavity. Correct?What change? Blackbody radiation is a continuous spectrum. If you aise the temperature the peak and upper limit shift towards a higher energy and the total photon flux increases. Atomic structure is not changed by temperatures below several zillion degrees. There are no molecules in a metallic structure.
... by attributing lesser (or extra) length of seconds to the process proportionally as energy is added (or decreased), a bit like wavelength being inversely proportional to frequency, except it would be a variable second, the length of which being inversely proportional to frequency, and using these variable seconds to calculate as the applied energy is increased, ...... I can see that the logic might work, but it's the maths that are the defining factor. Does what I suggest actually do what I suggest it does mathematically?You can get the same effect in many different ways
The reason E = hv is that h is joule second and v is cycles divided by seconds. Since cycles have no dimension you can simply cancel out the unit of seconds in the numerator and denominator leaving the dimension of the answer as joules. Note that cancelling a unit has no effect on the values in the equation. This is what we use dimensional analysis for. I have explained this in words rather than just using maths since that is how you requested answers.The unit for h is actually Joule second per cycle. Reduced Planck's constant, called ħ (h bar) has a unit of Joule second per radian. They have the same dimension, but different in numerical value.
The unit for h is actually Joule second per cycle.
The SI units are defined in such a way that, when the Planck constant is expressed in SI units, it has the exact value h = 6.62607015?10−34 J⋅Hz−1
The unit for h is actually Joule second per cycle.
Er, no.QuoteThe SI units are defined in such a way that, when the Planck constant is expressed in SI units, it has the exact value h = 6.62607015?10−34 J⋅Hz−1
Since 1 Hz = 1/second, the term Hz-1 has the dimension of time, so h is simply 6.62607015?10−34 joule.sec.
h/2π = ħ is still joule.sec but 2π turns up in so many calculations that it is simply a more compact way of writing equations.
The hertz (symbol: Hz) is the unit of frequency in the International System of Units (SI), equivalent to one event (or cycle) per second.If a wheel is turning at 1/2π cycle per second, then it's turning at 1 radian/cycle. Because 1 cycle = 2π radian.
https://en.wikipedia.org/wiki/Hertz
Photons are not rotating, so radians are not involved. How many radians in a square wave? Or a piano string?You seem to forget about Fourier transform.
I never forget Fourier, but photons and pianos have never heard of him.Do you think photons have square waves?
Hammond organs and power stations use a rotating shaft to generate various frequencies.
None of which changes the dimensions of h or ħ : joule.second. 2π is a dimensionless constant.Radian is indeed a dimensionless unit.
The unit was formerly an SI supplementary unit and is currently a dimensionless SI derived unit
radian/sec is not a frequency but a rate of change of angle or heading. A "standard rate 1 turn" in an airplane is 0.0524 radian/second.Perhaps you should stick to commonly used definition of any concepts, unless you have a strong reason not to do so.
The radian per second (symbol: rad⋅s−1 or rad/s) is the unit of angular velocity in the International System of Units (SI). The radian per second is also the SI unit of angular frequency (symbol ω, omega). The radian per second is defined as the angular frequency that results in the angular displacement increasing by one radian every second.
https://en.wikipedia.org/wiki/Radian_per_second
h/2π = ħ is still joule.sec but 2π turns up in so many calculations that it is simply a more compact way of writing equations.What do you think is the unit of ħ?
An antenna is oscillating at 1 billion radian per second. What's the energy of its photon?Seriously? What do you think the energy of a photon is from an antenna oscillating at 57.3 billion degrees per second? Does the question even make sense to you?
If it wasn't obvious to you yet, I referred to the electric field in the antenna.An antenna is oscillating at 1 billion radian per second. What's the energy of its photon?Seriously? What do you think the energy of a photon is from an antenna oscillating at 57.3 billion degrees per second? Does the question even make sense to you?
109 rad/sec is a rate of rotation,Not necessarily.
The unit hertz (Hz) is dimensionally equivalent, but by convention it is only used for frequency f, never for angular frequency ω. This convention is used to help avoid the confusion that arises when dealing with quantities such as frequency and angular quantities because the units of measure (such as cycle or radian) are considered to be one and hence may be omitted when expressing quantities in SI units.
The question doesn't make sense. 109 rad/sec is a rate of rotation, so presumably we are looking at a betatron, not an antenna. The electron energy, and hence the energy of any photons emitted, depends on the radius of the torus.Photonic interpretation for Planck's law states that hf is the energy of one photon. Thus radius of the
I may have time to do the calculation later, but the dog needs a walk!
For those who are not familiar with electronics in radio communications, I can give a more visually accessible example. A bar magnet is rotated with axis perpendicular to the magnetic field inside the magnet. The angular speed is 3000 rpm (rotation per minute) , which is typical for industrial motors. The question is the same, what is the photon energy radiated by the rotating magnet?If it wasn't obvious to you yet, I referred to the electric field in the antenna.An antenna is oscillating at 1 billion radian per second. What's the energy of its photon?Seriously? What do you think the energy of a photon is from an antenna oscillating at 57.3 billion degrees per second? Does the question even make sense to you?
Photonic interpretation for Planck's law states that hf is the energy of one photon. Thus radius of theBut it does affect the energy of the electron that generates the photon. It all gets a bit complicated as it's quite easy to get an electron up to 0.9c in a betatron, at which point the relativistic corrections become very significant.
torus or antenna doesn't affect photon energy, as long as the frequency can be kept the same.
It supposed to affect the number of photons radiated by the antenna.Photonic interpretation for Planck's law states that hf is the energy of one photon. Thus radius of theBut it does affect the energy of the electron that generates the photon. It all gets a bit complicated as it's quite easy to get an electron up to 0.9c in a betatron, at which point the relativistic corrections become very significant.
torus or antenna doesn't affect photon energy, as long as the frequency can be kept the same.
The question is the same, what is the photon energy radiated by the rotating magnet?Small.
Earth calling Alan.Photonic interpretation for Planck's law states that hf is the energy of one photon. Thus radius of theBut it does affect the energy of the electron that generates the photon. It all gets a bit complicated as it's quite easy to get an electron up to 0.9c in a betatron, at which point the relativistic corrections become very significant.
torus or antenna doesn't affect photon energy, as long as the frequency can be kept the same.
It supposed to affect the number of photons radiated by the antenna.Nonsense. It affects the energy of the photons, not the number. More electrons -> more photons, electrons oscillating faster -> higher photon frequency and E = hf
How does relativistic corrections affect the photon frequency, and its energy?You need to apply relativistic corrections to the electron velocity as you increase its kinetic energy. If energy E gives it a speed of 0.9c, 2E won't make it travel at 1.8c,
He's not talking about a betatron.
frequency of 159 MHz. That's the top end of the radio frequencies, maybe early microwave.No big deal - it's between the VHF (civilian) and UHF (military) aircraft comms bands.
Have your read my first post here carefully?The question is the same, what is the photon energy radiated by the rotating magnet?Small.
3000 RPM is 50Hz so the photon energy is 50 times Planck's constant.
That's high school maths. Why have you reopened a long-dead thread to ask about it?
I posted it because I think the statement in bold above needs clarification. Just because a unit has no dimension, it doesn't mean that it can just be ignored. You will get different value if you use different units, such as radian, degree, grad, brad, etc.The reason E = hv is that h is joule second and v is cycles divided by seconds. Since cycles have no dimension you can simply cancel out the unit of seconds in the numerator and denominator leaving the dimension of the answer as joules. Note that cancelling a unit has no effect on the values in the equation. This is what we use dimensional analysis for. I have explained this in words rather than just using maths since that is how you requested answers.The unit for h is actually Joule second per cycle. Reduced Planck's constant, called ħ (h bar) has a unit of Joule second per radian. They have the same dimension, but different in numerical value.
Using ħ can reduce the number of symbols used in equations of quantum mechanics, where we need to write down 2π if h were used instead.
I posted it because I think the statement in bold above needs clarification. Just because a unit has no dimension, it doesn't mean that it can just be ignored. You will get different value if you use different units, such as radian, degree, grad, brad, etc.Jeffrey didn't cancel "cycles" but "seconds", because [E] = ML2T-2 and [ν]= T-1 so [h] = ML2T-1.
Is a rotating magnet really emitting energy in free space? It certainly induces a current in a conductor, ......Inducing current in a conductor is a separate issue. It creates e-m waves in the region of space around itself, these will propogate. There is no reason it would stop emitting e-m radiation in free space.
In open space, this rotating magnet is emitting radiation and thus losing energy. Where is that energy coming from? If the rotating magnet reduces its (rotational) k.e. - where is the torque coming from that makes that happen?It's not a question I can easily answer and I would like to get some opinions on it. There are a few solutions already existing in some Q&A websites like Quora but their quality.... is... variable.
By Newtonian mechanics, some torque would seem to operate on the rotating object.This simply does not make sense. Consider a magnet rotating in an infinite vacuum. Where does its angular momentum go? Newtonian mechanics says it is conserved!
Consider a magnet rotating in an infinite vacuum. Where does its angular momentum go? Newtonian mechanics says it is conserved!That is indeed an interesting problem to resolve. Just to be clear, I didn't make up the original situation and the questions that follow from it.
Starting from the beginning, a rotating magnetic dipole (or electric dipole) should generate e-m radiation.I think not. You need to accelerate a charge to generate a photon. So it is entirely possible that a spinning neutron star could excite radiation from any gas cloud, plasma or passing ion, but a magnet in an infinite vacuum won't.
You need to accelerate a charge to generate a photon.It is a way but do really think it's the only way to generate e-m radiation?
He didn't cancel cycle, but ignored it as if it wasn't there. You can't say always if there are exceptions.I posted it because I think the statement in bold above needs clarification. Just because a unit has no dimension, it doesn't mean that it can just be ignored. You will get different value if you use different units, such as radian, degree, grad, brad, etc.Jeffrey didn't cancel "cycles" but "seconds", because [E] = ML2T-2 and [ν]= T-1 so [h] = ML2T-1.
[X] means "dimensions of X "
Conventionally, the frequency of radiation is always stated in Hz, to avoid the confusion you have made for yourself.
Frequency (symbol f), most often measured in hertz (symbol: Hz), is the number of occurrences of a repeating event per unit of time.[1] It is also occasionally referred to as temporal frequency for clarity and to distinguish it from spatial frequency. Ordinary frequency is related to angular frequency (symbol ω, with SI unit radian per second) by a factor of 2π. The period (symbol T) is the interval of time between events, so the period is the reciprocal of the frequency: f = 1/T.
https://en.wikipedia.org/wiki/Frequency
Necromancy?, BC. I always thought necromancy was the divination of future events by dissecting a dead creature and examining it's entrails. This method has not as yet been ascertained to be effective!And someone tried to divine the future from the entrails of a long-dead thread.
You seem to have invented a requirement for an orbit.He's not talking about a betatron.
He's talking about the angular speed of an electron. What else would you call a device that makes electrons orbit at a constant 109radians per second?
I also am fairly sure a rotating magnet would radiate em. It would have to rotate at quite a high speed to generate easily detectable radiation. I have been trying for several hours to mentally model the maths involved but I have failed so far. We normally look at charge acceleration to produce em as it is the most convenient method but one has to remember that the magnetic field is simply the electric field as seen from a different frame of reference.I never could do the maths, but I don't care.
..one has to remember that the magnetic field is simply the electric field as seen from a different frame of reference...Well said, I very nearly added that to the list of things @alancalverd has probably said to other people.
I never could do the maths, but I don't care.You've made a good argument without the Mathematics.
I have looked up the derivation of the wave equation via the Laplacian of the E or H field in phasor form and it is completely symmetric: a time variation in either produces an em wave.Thank you.
If the magnet falls over in a non-existent forest with no observers, does it matter?Hmm... I've often wondered, if I'm alone in the woods and my wife and daughters cannot hear me, am I still wrong?
A monopole would throw a spanner in the maths where divH is required to be =0.You've already given the smart answer. Maxwells equations will not permit a magnetic monopole, so we can't use them and expect to get a sensible answer. As far as I know, quantum theories that do predict magnetic monopoles will have them behave much like an electric charge.
The force acting on the needle is electromagnetic.It's certainly magnetic, but where is the E field?
Pushing electrons to and fro in an antenna is cyclical (Particularly if the antenna is resonant.)but doesn't involve any rotation.
And had anyone said it should?Pushing electrons to and fro in an antenna is cyclical (Particularly if the antenna is resonant.)but doesn't involve any rotation.
What do you mean 'to accelerate a charge to generate a photon'?Starting from the beginning, a rotating magnetic dipole (or electric dipole) should generate e-m radiation.I think not. You need to accelerate a charge to generate a photon. So it is entirely possible that a spinning neutron star could excite radiation from any gas cloud, plasma or passing ion, but a magnet in an infinite vacuum won't.
So let's have a 1 tesla bar magnet spinning at 1000Hz in a vacuum. Can anyone please describe its electromagnetic emission spectrum? I just can't visualise it!I suspect the spectrum will be frame dependent.
So let's have a 1 tesla bar magnet spinning at 1000Hz in a vacuum. Can anyone please describe its electromagnetic emission spectrum? I just can't visualise it!Can you visualise the emission if you had an electric dipole rotating at that rate?
...Apparently there are virtual photons that enable fields, wave propagation, potential gradients, ...
The carrier of the EM force is the photon.
So, in this case, I know that a rotating magnet emits EM radiation. That force is periodic and thus there must be a component of the EM radiation with the same period.So it must have photons with energies corresponding to the frequency of rotation of the record player.
Re. "but a magnet in an infinite vacuum won't. "
If the magnet falls over in a non-existent forest with no observers, does it matter?
I contend that it will still emit EM radiation.
Imagine that we somehow have a magnet in an infinite empty universe.
We set it spinning (We know it is spinning, because an ant who happens to be standing on it notices the centrifugal effect).
After a while, and at some distance from the magnet, we magically call a compass needle into being.
Does the needle have to "wait" for EM radiation from the magnet to reach it, or is that changing field already there?
I can't see how it would so I think the magnet must have been emitting EM radiation all along.
But... if there's EM radiation, then there are photons.
And, if there are photons, the universe isn't empty.
So the solution may be that you can't have a rotating magnet in an empty universe.
All seems a bit esoteric..
But let's ignore nearly the whole of the universe and consider some hydrogen atoms.
Some of them have the magnetic dipoles of the electron and the proton aligned parallel, and in others it's antiparallel.
And if one happens to flip from the first state to the second, it emits a photon of about 21 cm wavelength.
That photon crosses space and is picked up many years later by a detector here on earth.
But we have only been constructing such detectors for about 100 years.
So, for a source more than 100 light years away, the detector had not been built when the photon was emitted.
I think that's close enough to " we magically call a compass needle into being." for the analogy to work.
An electron- with a magnetic diploe moment- was flipped and sent out EM radiation. It did so in a universe in which the detector did not exist.
You may say that, without the proton of the hydrogen atom, the energy of the photon would be undefined. Which is a fair point
But, at that point, I think you need to be able to do Laplace transforms and, as I said, I can't do the maths.
What do you mean 'to accelerate a charge to generate a photon'?I'm not alancalverd.... but one way you could do this is with a device called a cyclotron. Put a charged particle like a proton in to a cyclotron with some initial velocity (and remember to switch the magnetic field generator on). Then e-m radiation will be emitted.
How does it work?
So let's have a 1 tesla bar magnet spinning at 1000Hz in a vacuum. Can anyone please describe its electromagnetic emission spectrum? I just can't visualise it!I guess you couldn't find Griffith's introduction to electrodynamics then. I don't have a copy of it either, which is a bit shameful, I may get one for Christmas if I'm lucky.
Quote from: Jaaanosik on Yesterday at 20:37:01Adding energy to a charged particle, acceleration in a straight line, does not generate a photon.QuoteWhat do you mean 'to accelerate a charge to generate a photon'?
How does it work?
I'm not alancalverd.... but one way you could do this is with a device called a cyclotron. Put a charged particle like a proton in to a cyclotron with some initial velocity (and remember to switch the magnetic field generator on). Then e-m radiation will be emitted.
Adding energy to a charged particle, acceleration in a straight line, does not generate a photon.I'm fairly sure it would generate some photons.
Hi.Adding energy to a charged particle, acceleration in a straight line, does not generate a photon.I'm fairly sure it would generate some photons.
The link that you provided gives a formula for the power radiated when acceleration is parallel to the velocity of the charged particle.
When any charged particle (such as an electron, a proton, or an ion) accelerates, energy is radiated in the form of electromagnetic waves.
- Taken from a slightly different Wikipedia page. https://en.wikipedia.org/wiki/Larmor_formula
The only minor issue that will arise is whether all of this oscillation in E and B fields would always be recognisable as "a photon", a discrete packet of energy. Classically, it's an oscillation in the E and B fields but it should be a continous emission rather than something that occurrs in bursts at discrete places and times.
Best Wishes.
This comes back to 'who is looking'.Interesting.
you said to use LaTex. I had no clue what you were talking about and I should have asked-It's a markup language especially intended for mathematical symbols.
To [u]underline[/u] you enclose the start and end tags [u] and [/u] around the thing you want to underline.
That displays like this:the [tex] and [/tex] tags mark the start and end of the LaTeX code block.
[tex] \sum \limits_{i=1}^n {3i} [/tex]
Inside of the LaTeX code block you can use code that signifies or "marks up" the symbols you want.
The bit of code above would have produced a Sigma symbol for a sum over an index i from 1 to n of terms 3i.
This should be the sum from i=1 to 2n of terms 3i.
[tex] \sum \limits_{i=1}^{2n} {3i} [/tex]
the end-result is curved trajectory accelerations in K'1, and K'2 frames.That is an interesting situation but I'm not sure what differences you are trying to imply would be seen.
Is a stationary charged particle release photons when seen by an inertially moving observer?No. If the charged particle is at rest in some inertial frame, then an observer at rest in any other inertial frame records the same acceleration for the charged particle (that will be 0 since it was at rest in some inertial frame). So the Larmor formula shows 0 e-m radiation from it.
Hi.Do you know that magnetic field generated by moving charged particle is velocity dependent?the end-result is curved trajectory accelerations in K'1, and K'2 frames.That is an interesting situation but I'm not sure what differences you are trying to imply would be seen.
Take some care to distinguish between a curved trajectory and an acceleration which is altered or "curved" as you phrased it.
The acceleration experienced by the charged particle is the same in all of those inertial reference frames.
The acceleration vector is a in frame K'1 and also the same vector in K'2 and the original frame which I'll call K0. For convenience let's say the acceleration is in in the z-axis direction, it will be in the z-axis direction for every frame and assuming the gap between plates and the potential across them remains constant, this acceleration is just a constant vector a independant of time for every experiment.
The acceleration is not curved or altered in any way.
Now we can imagine the experiment is run theree times, where the plates are at rest in the three different frames (and the charged particle is initially at rest on the blue plate) and you also maintain observers at rest in the frames K'1, k'2 and K0 for every experiment. Then the observers will each see two experiments where the charged particle takes a curved trajectory and only one experiment where it was straight.
The Larmor formula ( https://en.wikipedia.org/wiki/Larmor_formula ) depends only on the acceleration and not on whether the path was straight or curved. Using the simple Larmor formula which ignores relativistic effects (so the particle never reaches a high speed ) we see that every observer records exactly the same total amount of power radiated in the e-m field in every experiment.
There will be some differences in the e-m radiation. We have a Doppler effect, so if the apparatus and particle are moving toward an observer for half the experiment and then away for the other half, the radiation incident on them changes frequency and energy content. However, if they just collect all the radiation coming off the particle (for example construct a sphere around the apparatus that moves with the apparatus, and just measure the total power incident on that sphere at all times) the Larmor formula applies and everyone sees the same total radiated power.
...
Best Wishes.
That is an interesting situation but I'm not sure what differences you are trying to imply would be seen.This is a huge difference...
Do you know that magnetic field generated by moving charged particle is velocity dependent?Yes (but thanks for the information).
The magnetic field generated by an accelerated charged particle is changing in magnitude and direction because it is velocity/trajectory dependent.
K0 does not predict this directional change but K'1 and K'2 do predict the change.
The only minor issue that will arise is whether all of this oscillation in E and B fields would always be recognisable as "a photon", a discrete packet of energy. Classically, it's an oscillation in the E and B fields but it should be a continous emission rather than something that occurrs in bursts at discrete places and times.Someone may argue that the issue is fundamental rather than minor.
The unit for h is actually Joule second per cycle.
Er, no.QuoteThe SI units are defined in such a way that, when the Planck constant is expressed in SI units, it has the exact value h = 6.62607015?10−34 J⋅Hz−1
Since 1 Hz = 1/second, the term Hz-1 has the dimension of time, so h is simply 6.62607015?10−34 joule.sec.
h/2π = ħ is still joule.sec but 2π turns up in so many calculations that it is simply a more compact way of writing equations.
In many applications, the Planck constant
ℎ naturally appears in combination with 2π as ℎ/(2π), which can be traced to the fact that in these applications it is natural to use the angular frequency (in radians per second) rather than plain frequency (in cycles per second or hertz). For this reason, it is often useful to absorb that factor of 2π into the Planck constant by introducing the reduced Planck constant[38][39]: 482 (or reduced Planck's constant[40]: 5 [41]: 788 ), equal to the Planck constant divided by
2π[38] and denoted by ħ (pronounced h-bar[42]: 336 ).
https://en.wikipedia.org/wiki/Planck_constant#Reduced_Planck_constant_%E2%84%8F
QuoteQuote from: hamdani yusuf on Yesterday at 13:09:12No. If the charged particle is at rest in some inertial frame, then an observer at rest in any other inertial frame records the same acceleration for the charged particle (that will be 0 since it was at rest in some inertial frame). So the Larmor formula shows 0 e-m radiation from it.
Is a stationary charged particle release photons when seen by an inertially moving observer?
What do you think about this Wikipedia entry?
Not exactly. See the panel on top of what you quoted.What do you think about this Wikipedia entry?
Here's what Wikipedia thinks about it (panel at the top right of the article)
Reduced Planck constant
Common symbols
ℏ
SI unit joule-seconds
exactly the same units as h
Does the unit of the radiation frequency have no effect on the numeric value of their quantum of energy?The number of joules of energy which a photon carries doesn't know or care if you measure frequency in 1/seconds or 1/weeks
Does the unit of the radiation frequency have no effect on the numeric value of their quantum of energy?No. The unit of energy is joules or electron volts and the number is (by definition) fixed for any quantum.
doesn't know or care if you measure frequency in 1/seconds or 1/weeksSchlumberger used to boast that they could supply customised oscilloscopes "calibrated in millifurlongs per microfortnight", if you wanted.
Among the statements above, which one(s) do you think is wrong?The following one:
Changing of electromagnetic field is often called radiation.
The value of h is 6.62607015x10−34Does the unit of the radiation frequency have no effect on the numeric value of their quantum of energy?No. The unit of energy is joules or electron volts and the number is (by definition) fixed for any quantum.
becauseThe value of h is 6.62607015x10−34Does the unit of the radiation frequency have no effect on the numeric value of their quantum of energy?No. The unit of energy is joules or electron volts and the number is (by definition) fixed for any quantum.
energy of a single photon is h.f.
If the photon frequency is 1 Hz, then its energy is 6.62607015x10−34 Joule.
If the photon frequency is 1 GHz, then its energy is not 6.62607015x10−34 Joule.
If the photon frequency is 1 rpm, then its energy is not 6.62607015x10−34 Joule.
If the photon frequency is 1 rad/s, then its energy is not 6.62607015x10−34 Joule.
The value of Planck's constant (h) would change.
Correct.And a grand total of zero people said otherwise.
The familiar numerical value of h is in the unit of Joule/Hz, or Joule second/cycle. In different unit, the value would be different.
Are you sure?Correct.And a grand total of zero people said otherwise.
The familiar numerical value of h is in the unit of Joule/Hz, or Joule second/cycle. In different unit, the value would be different.
he value of h is 6.62607015x10−34No, it is 6.62607015x10−34 joule.second. The number would be quite different if it was expressed in BThU.hours
energy of a single photon is h.f.Hardly a stunning observation, but true nonetheless.
If the photon frequency is 1 Hz, then its energy is 6.62607015x10−34 Joule.
If the photon frequency is 1 GHz, then its energy is not 6.62607015x10−34 Joule.
If the photon frequency is 1 rpm, then its energy is not 6.62607015x10−34 Joule.1 rpm or 1 rad/second is not a frequency, but a speed of rotation.
If the photon frequency is 1 rad/s, then its energy is not 6.62607015x10−34 Joule.
No, it is 6.62607015x10−34 joule.second. The number would be quite different if it was expressed in BThU.hoursJoule second per cycle.
1 rpm or 1 rad/second is not a frequency, but a speed of rotation.They are angular frequency, but frequency nonetheless.
I haven't seen any rotating photons.Haven't you learned something about circular polarization?
Quotefrom: alancalverd on Yesterday at 18:10:22Joule second per cycle.
No, it is 6.62607015x10−34 joule.second. The number would be quite different if it was expressed in BThU.hours
I give up.Your frustration is caused by your misunderstanding of frequency. Don't blame your own mistake to someone else.
Clearly the International Bureau of Weights and Measures (and everyone else) have no idea what they are talking about, our concept of dimensions, and the entire SI system of units, is faulty.
I suggest that, for the benefit of all mankind, you should dispel their ignorance with a direct communication to BIPM, Sevres, France (Quantum Mechanics Division).
1 rpm or 1 rad/second is not a frequency, but a speed of rotation.Or it's a rate of change of phase angle.
Then why does the SI unit not mention it?Feel free to ask them.
Then why does the SI unit not mention it?SI organization consists of people, who view the world based on their contemporary science community. They are not infallible, and they have publicly changed the standards. They may have their blind spots, just like us human beings. That's why it's important to let different ideas to compete and be discussed to reveal any residual mistakes.
1 rpm or 1 rad/second is not a frequency, but a speed of rotation.1 kilocycle per second, 1 Gigacycle per second, are also frequency. They also have 1/second dimension.
I haven't seen any rotating photons.
I haven't seen any rotating photons.Now I come to think of it, all the photons I have heard of carry angular momentum.