# NON-QED derivation of E=pc

1. Jul 5, 2006

### clj4

Can anybody point out if there is a NON-QED derivation of E=pc ?
I would need a reference if such a derivation existed. Thank you in advance.

PS: I am not looking for the definition of 4 vector invariants that leads to $$E^2 = (pc)^2 + (mc^2)^2$$.
In the same vein, I am not looking for setting m=0 in the above equation because it brings up the question of deriving $$E^2 = (pc)^2 + (mc^2)^2$$ from first principles without resorting to QED.

Last edited: Jul 6, 2006
2. Jul 5, 2006

### masudr

Derived from which principles?

3. Jul 5, 2006

### James R

E=pc can be derived from special relativity. The general relationship between energy and momentum is:

$$E^2 = (pc)^2 + (mc^2)^2$$

For a photon with rest mass m=0, the result follows.

p=hf is incorrect. The correct de Broglie wavelength formula is $p=h/\lambda$.

This can be derived from the formula E=pc for a photon, by using the photon energy E=hf.

4. Jul 6, 2006

### clj4

From base principles, in the style of Einstein's derivation of $$E=mc^2$$ (not from a definition and nor from the trivial exercise of setting m=0 in $$E^2=(pc)^2+(mc^2)^2$$. The latter derivation is of no use.
This appears to be a tough problem, I do not know if it has a solution, this is why I asked.

Last edited: Jul 6, 2006
5. Jul 6, 2006

### clj4

Thank you, I appreciate the effort. I should have mentioned that I am not looking for setting m=0 in
$$E^2 = (pc)^2 + (mc^2)^2$$
either, since that brings about deriving this from first principles.
Since I am on this subject, I am not looking for the definition of 4 vector invariants that leads to $$E^2 = (pc)^2 + (mc^2)^2$$

6. Jul 6, 2006

### Haelfix

It most likely follows from stat mech, if you *assume* the various distributions are true apriori and run it in reverse, though I have never seen it done before.

The field theory proof is sort of the nontrivial one that kinda jumps out at you, but I guess that doesnt interest you.

7. Jul 6, 2006

### pervect

Staff Emeritus
I imagine you could find the energy and momentum of a plane wave, and point out that they are related by E=pc? (At least in a vacuum).

I suppose you'd also need a proof that the far field can always be written as a sum of plane waves. (I think this is true, but actually I'm not 100% sure).

8. Jul 6, 2006

### clj4

I tried that , didn't manage. I tried with a spherical wave , for some reason I thought it would be easier due to symmetry. Thiis is why I a asking for a reference :-)

9. Jul 6, 2006

### HallsofIvy

You have said you wanted to know a "non-QED" method to derive E=pc (which is peculiar in itself since E= pc doesn't involve QED to begin with) and you don't want to use "one method" and you don't want to use "another method". Exactly what do you want? Your question is very unclear.

10. Jul 6, 2006

### masudr

Shankar derives it by quantizing the electromagnetic field, but it's not really fully-fledged QED, in my opinion. See Shankar's text on QM, chapter 18, pg. 516 (2nd ed.), equation 18.5.69.

11. Jul 6, 2006

### clj4

From base principles, in the style of Einstein's derivation of $$E=mc^2$$ (not from a definition and nor from the trivial exercise of setting m=0 in $$E^2=(pc)^2+(mc^2)^2$$. The latter derivation is of no use.
This appears to be a tough problem, I do not know if it has a solution, this is why I asked.

Last edited: Jul 6, 2006
12. Jul 6, 2006

### clj4

Thank you, I will have a look though I would want to avoid QM as well as QED.

I am looking for a derivation in the style of Einstein's derivation of $$E=mc^2$$ (not from a definition and nor from the trivial exercise of setting m=0 in $$E^2=(pc)^2+(mc^2)^2$$. The latter derivation is of no use.
This appears to be a tough problem, I do not know if it has a solution, this is why I asked.

13. Jul 6, 2006

### masudr

Well, the photon, which is the notion of a massless boson which are the excitations of the EM field, can only reasonably be described by QED, since it is in this framework that it arises.

And from that, one can show that E = pc must be satisfied by those particles. It's kind of unfair asking to derive a property of photons without using the model/theory that predicts its existence, don't you think?

14. Jul 6, 2006

### clj4

I don't know . The whole idea is to avoid the quantization (the notion of photon) altogether and to use wave theory only. After all , Einstein did this in deriving radiation pressure. In general, it is much more difficult to prove that something cannot be done. In effect, I am interested in how Einstein would have established E=pc with what he had at hand in 1905. Is it possible? has it been done? can we prove that it is impossible?

One of the posters informed me that the issue might be covered in:

A.P.French, Special Relativity Nelson Chapter 1

I do not have the book, anyone can help with scanning the text and posting it ?
Thank you

Last edited: Jul 6, 2006
15. Jul 6, 2006

### masudr

Sorry to go on about it, but am I right in thinking that what you want is to show that the momentum of a classical EM wave is equal to it's energy divided by c?

16. Jul 6, 2006

### clj4

Yes, this is correct.

17. Jul 7, 2006

### Ich

I did a quick search and found http://www.tphys.physik.uni-tuebingen.de/faessler/Physik3/optik4_2.pdf" [Broken] (in German).
Eq. 4.43-4.46, figure 4.4
The Lorentz Force always pushes a charged particle in the same direction. By comparing work and force you get E=pc.

Last edited by a moderator: May 2, 2017
18. Jul 7, 2006

### clj4

Thank you!

Following the same search principle I found another derivation:

http://www.mathpages.com/home/kmath601/kmath601.htm

Last edited by a moderator: May 2, 2017
19. Jun 21, 2008

### qucquc

E=pv

You can refer to http://www.geocentricity.com/ba1/no82/byl.html [Broken]
dU=Fx=(dp/dt)x=dp(x/dt)=dpv, after integral, you can get E=pv

Last edited by a moderator: May 3, 2017