Energy of Single Photon

1. Aug 12, 2008

JayKo

is hv as we all know, but how do with derive it from Einstein's special theory of relativity? any pointer to help me kickstart off? thanks ;)

2. Aug 12, 2008

Beto Pimentel

We don't, I am afraid. I suppose E=hv is an experimental result from Einstein's ananlysis of the photoelectric effect through Planck's idea of quantized energy in a blackbody radiator.

3. Aug 12, 2008

Avodyne

$E=h\nu$ comes from quantum mechanics, not special relativity, and is true for all particles, not just photons. (Photons are special because they can be built up into a coherent electromagnetic wave whose frequency is directly measurable.)

4. Aug 12, 2008

nicksauce

If you REALLY want to use special relativity, you can start with p=h/lambda, and then use E = sqrt(p^2c^2 - m^2c^4), and then use m=0.

5. Aug 12, 2008

lightarrow

It doesn't come from anything. When you will have found its derivation (as long with the value of h) you will won the Nobel prize.

6. Aug 12, 2008

nicksauce

Err should be E = sqrt((pc)^2 + (mc^2)^2)

7. Aug 12, 2008

cesiumfrog

For any particle (e.g. photon) that you wish to express relativistically as a wave packet:

The phase of a plane wave at different points in space-time is a scalar tensor and it can obviously be expressed as the contraction of the position tensor with what we shall call the wave four-vector. Therefore that vector is a tensor, its scalar magnitude is frame invariant. The magnitude equation constitutes a dispersion relation, for the class of relativistic plane waves that are "the same as this one" in another frame, and from this you can trivially calculate the velocity of packets of such waves. This will show that the four-momentum tensor of the "particle" wave-packet is co-linear with the mean wave tensor, from which we can read off the fact that in every frame (and for all same particles) the energy is proportional to frequency. QED. https://www.physicsforums.com/showthread.php?t=243135"

Last edited by a moderator: Apr 23, 2017
8. Aug 13, 2008

lightarrow

This seems to show that if the proportionality holds in a frame, then it must hold in every frame, but how do you show that it holds in the first frame?

Last edited by a moderator: Apr 23, 2017
9. Aug 13, 2008

cesiumfrog

It shows the four-velocity of a wave-packet is always parallel to the four wave-vector. That implies E=hf and p=h/$\lambda$ (where h is just some constant).

Plus, if one particle can be represented relativistically by a wave-packet then the same proportionality constant must apply, not only for the same particle in another frame (as you acknowledged), but also for identical particles (of different velocities) in the first frame. The key concept is that if a law obeyed by one plane wave in some frame is relativistic, that means that there exists a whole class of different plane waves (and hence packets) obeying the same law (because in some other frame they would be equivalent). The argument doesn't prove that the *same* proportionality constant will apply to photons as does for electrons.

As for that Nobel prize you mentioned, this derivation is basically from de Broglie's PhD thesis.

Last edited: Aug 13, 2008
10. Aug 13, 2008

Staff: Mentor

But this is also true for a classical electromagnetic wave, isn't it?

In a classical em wave, the energy and momentum don't depend on the frequency and wavenumber; they depend instead on the square of the amplitude of the wave.

In the QM wave function, it's the other way around: the energy and momentum are proportional to the frequency and wavenumber, and they don't depend on the amplitude, which is fixed by the normalization condition.

11. Aug 13, 2008

cesiumfrog

Of course (in fact, the line you quoted is even true for sound waves and such, it was the following stage of the argument that specifically pertains to EM and de Broglie matter waves).

Do you dispute that E=nhw and p=nhk for any classical EM wave-packet?

12. Aug 13, 2008

Beto Pimentel

Point acknowledged. I wouldn't have been able to even think of this line of reasoning, and would vote for your Noble Prize, were it not for you sharing with us that it was De Broglie's reasoning in the first place. I would vote for him anyway, but he's got one Nobel already, so...
Anyway, this does not invalidate the previous affirmatives (that E=hv cannot be deduced from Relativity alone). The fact that the formalism states that E IS PROPORTIONAL to the frequency does not mean the proportionality constant is h, which still needs to be determined experimentally or through assumptions based on Planck's hypothesis, right?

Last edited by a moderator: Apr 23, 2017
13. Aug 13, 2008

peter0302

First, everyone is of course right that E=hv is experimentally derived and cannot simply be derived from special relativity alone.

What is derivable from special relativity alone is the idea that massless particles still have momentum and energy: E=pc. That comes from the larger E^2 = (mc^2)^2 + (pc)^2 where m is equal to zero.

It can also be shown in SR, though, that potential energy is inversely proportional to distance, so E proportional to 1/lambda and therefore proportional to frequency is inferable from SR. But the exact relation - involving h - must be derived experimentally.

14. Aug 13, 2008

cesiumfrog

Historically, I understand that at the time of de Broglie's work, the electron diffraction experimental results already existed to determine the value of h for electron wave-packets, but nobody had thought to interpret it in that way yet (this being only... 7? years after Planck and Einstein attributed a particle nature to light waves). Instead, de Broglie noted that fitting the same h for electrons as for photons would reproduce (and justify) the Hydrogen model of the day.

People forget how much QM was influenced by SR. This relativity argument explained E=ihd/dt and p=ihd/dx, which became the starting point for Schroedinger (about a decade later), and if BECs had been familiar at the time then we would have had relativistic QM from the very start.

Last edited: Aug 13, 2008
15. Aug 15, 2008

lightarrow

Cesiumfrog, could you please elaborate that in detail? I have no books on which to study these computations, it would be very much appreciated!
Thanks
lightarrow.

16. Aug 15, 2008

cesiumfrog

Did you follow the link? Let me know if there's a specific detail you don't follow. I found the argument in the RQM texts by Crewther or Dyson.