# Looking for more accurate energy-momentum transformations for photons

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## Main Question or Discussion Point

Photons deviate from the above energy-momentum transformations under certain circumstances while still in flat space-time, I'm wondering what set of transformations would more accurately describe them over as wide a range of circumstances as possible, still in flat space-time, I've searched and I'm hoping I can get some useful info.

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Orodruin
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Photons deviate from the above energy-momentum transformations under certain circumstances while still in flat space-time
No they don't.

Photons deviate from the above energy-momentum transformations under certain circumstances while still in flat space-time
No they don't.
Hmm, ok, I was looking at the transformation of frequency of a monochromatic point source:

$f'=γ(1-βcosθ)f$

It is equivalent to the energy transformation above. But it is only a far-field approximation.

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Orodruin
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Hmm, ok, I was looking at the transformation of frequency of a monochromatic point source:

$f'=γ(1-βcosθ)f$

It is equivalent to the energy transformation above. But it is only a far-field approximation.
What makes you think that that is any different from the transformation you gave above?

What makes you think that that is any different from the transformation you gave above?
hmmm? I said it is equivalent mathematically. But it is a far-field approximation, so there should ostensibly be another more accurate transformation.

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Ibix
I think you need to lay out what you think is a far-field approximation to what.

I think you need to lay out what you think is a far-field approximation to what.
As mentioned above, for a monochromatic point source. The formula gets closer to the value of f' the further away the "photon-event" is from the source.

Ibix
So far you have given nine formulae. One of them you think is an approximation to something. I have no idea which formula you think is an approximation or why you think this.

You need to explain what you think is an approximation and why you think it is an approximation. Simply saying "the formula is a far field approximation" is leaving me with no idea what it is you are asking about.

So far you have given nine formulae. One of them you think is an approximation to something. I have no idea which formula you think is an approximation or why you think this.

You need to explain what you think is an approximation and why you think it is an approximation. Simply saying "the formula is a far field approximation" is leaving me with no idea what it is you are asking about.
Ok, the formula in #3 is equivalent to the energy transformation, which is the top right one of the eight formulae.

The left side are the space-time transformations, but they just came with the image, they aren't part of the discussion.

The formula in #3 is derived from far-field approximation considerations, so it ostensibly necessitates another more accurate transformation.

Ibix
The formula in #3 is derived from far-field approximation considerations
Is it? As far as I'm aware it can be derived from applying the Lorentz transforms to to a distance $\lambda$ making an angle $\theta$ to the x axis in order to get $\lambda'$ and then using $\nu'=c/\lambda'$. So I still don't see the problem.

Is it?
It appears to be the case as Einstein mentions the approximate application in section 7 of "On the Electrodynamics of Moving Bodies".

Ibix
He only does that so he has a plane wave. The argument applies equally to any other shape.

It's easy enough to repeat the derivation with a spherical wavefront, or you can just observe that a plane wave tangent to a spherical wave looks locally like the spherical wave and just skip the re-derivation.

He only does that so he has a plane wave. The argument applies equally to any other shape.

It's easy enough to repeat the derivation with a spherical wavefront, or you can just observe that a plane wave tangent to a spherical wave looks locally like the spherical wave and just skip the re-derivation.

I see, how do you show that its equal for a spherical wavefront, or any shape for that matter

Ibix
See post #10.

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Nugatory
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From the English version, Einstein writes, "relatively to an infinitely distant source", does this mean that he should've removed "infinitely distant"?
That's just how he gets to the plane wave approximation.

This might be a good time to mention that it is seldom effective to learn the physics directly from what the pioneers wrote (the history of science is of course a different matter). First there are wrong turns that are only apparent in hindsight, so are unmarked in Einstein's writing but safely bypassed in modern textbooks. Second, and often a greater challenge for the student, is the pioneering papers are written for other specialists. The audience is expected fill in missing steps without prompting and to recognize unstated assumptions from the context.

vela
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I think your confusion is arising because you're equating a classical electromagnetic wave, which is what Einstein was talking about, with a photon. Einstein uses the far-field approximation because that's where only the wavelike solution survives. Near the source, you'd have other significant contributions to the changing electric and magnetic fields, which would make the analysis harder, and which would obscure the points he was trying to make.

That's just how he gets to the plane wave approximation.

This might be a good time to mention that it is seldom effective to learn the physics directly from what the pioneers wrote (the history of science is of course a different matter). First there are wrong turns that are only apparent in hindsight, so are unmarked in Einstein's writing but safely bypassed in modern textbooks. Second, and often a greater challenge for the student, is the pioneering papers are written for other specialists. The audience is expected fill in missing steps without prompting and to recognize unstated assumptions from the context.
To ensure no misunderstanding, the line I quoted comes from the paragraph immediately before he introduces the frequency transform.

pervect
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For a photon, the (semi)classical relationship is that $E = |\vec{p}|c$. This comes from the fact that a photon is massless, and it's an exact relationship.

Einstein however, was writing special relativity as a purely classical theory. I do not believe that Einstein would have used the word "photon" at all. A search of the paper you cite didn't find the word anywhere in the paper.

So your question is about something that Einstein didn't write about. Thus you won't find the answer to your question as you write it in Einstein's paper, because he was writing about somethign else. As others have remarked, you may find your answers couched in terms of "photons", which Einstein didn't use in his 1905 paper, in more modern sources.

Let's look at the section heading of the section in which your quote is taken from.

§6. Transformation of the Maxwell-Hertz Equations for Empty Space.
In this context, Einstein's remarks make perfect sense. If you pick up a book on Maxwell's equations, a section on antenna theory specifically should talk about the near and far fields of antennas, and provide you with the missing context of Einsein's remarks.

THis might be an interesting exercise, but it won't really help you much to learn special relativity, if that's your goal. I'm guessing that's what your goal is, but I don't really know, so if you have some other goal, perhaps you do want to read more about the Maxwell-Hertz equations so that you can understand why Einstein made remarks about the near and far fields.

Let's look at the section heading of the section in which your quote is taken from.

§6. Transformation of the Maxwell-Hertz Equations for Empty Space.
Sorry, my quote in #15 comes from §7.

Einstein uses the far-field approximation because that's where only the wavelike solution survives. Near the source, you'd have other significant contributions to the changing electric and magnetic fields, which would make the analysis harder, and which would obscure the points he was trying to make.
I'm not sure that's his intention, because, at the beginning of §7, he uses the far-field approximation to approximate a plane wave from the circular waves produced by a point source, and he continues his derivations from there, with no mention of near-field significant contributions.

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As far as I'm aware it can be derived from applying the Lorentz transforms to to a distance $\lambda$ making an angle $\theta$ to the x axis in order to get $\lambda'$ and then using $\nu'=c/\lambda'$.
thanks, i'm a little confused about how to begin. does it involve the distance between two events? because we also have to take note of the relativity of simultaneity, I'm guessing the wavelengths are all calculated at a simultaneous moment.

thanks, i'm a little confused about how to begin. does it involve the distance between two events? because we also have to take note of the relativity of simultaneity, I'm guessing the wavelengths are all calculated at a simultaneous moment.
Here's one possible setup, involving "standard configuration" of a primed and unprimed frame:

In the primed frame, a spectrometer remains at rest at the origin while a light-source moving parallel to the $x^\prime$-axis [edit: in the negative $x^\prime$-direction, as per standard configuration] briefly sends a monochromatic light wave directly toward the spectrometer. The unprimed frame is the light-source's rest frame.

The goal is to find an expression for the wavelength (and ultimately the frequency) of the light as measured in the primed frame. Steps:
• Use time dilation to express the wave's primed period in terms of its unprimed wavelength.
• Express the (primed) distance the light travels during the primed period in terms of the unprimed wavelength.
• Express the (primed) radial component of the light-source's displacement during the primed period in terms of the unprimed wavelength. (This will involve the cosine of the primed angle between the light's velocity vector and the positive $x^\prime$-direction.)
• Sum the previous two results—that's your primed wavelength in terms of the unprimed wavelength and the aforementioned primed angle. You may also Lorentz-transform the primed angle to the equivalent corresponding [edited] unprimed angle (aberration).

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Nugatory
Mentor
It's worth pointing out that this thread has segued from the question in the title (energy-momentum transformations for photons) to the transformations for light. This is a good thing.

It's worth pointing out that this thread has segued from the question in the title (energy-momentum transformations for photons) to the transformations for light. This is a good thing.
True, but if we're not going to invoke the Planck–Einstein relation then aren't we barking up the wrong tree by transforming frequency?