# I Do the Energy-Momentum Transformations apply to Photons?

1. Mar 27, 2016

I apologize in advance for this dumb question, I think I know the answer but I just want to be sure.

A photon has energy E = pc = hf

Do the Energy-Momentum transformations:

apply exactly to photons?

Or must we introduce certain corrective terms? Let's say all this takes place in free space.

2. Mar 27, 2016

### Andrew Mason

The speed of the photon, of course is c in all frames but the wavelength (which determine its p and E) will depend on motion of the observer relative to the source. For that you need to apply the relativistic. doppler factor.

AM

3. Mar 27, 2016

Can we also apply the Energy-Momentum transformations without introducing corrective terms?

4. Mar 27, 2016

### Ibix

The energy and momentum transformations come from the requirement that the energy-momentum four-vector has an invariant length in all inertial frames. So yes, they must apply to all things including photons.

5. Mar 27, 2016

I just want to be sure that we can apply them exactly to photons like they are applied to massive particles, without the need for any corrective terms.

6. Mar 27, 2016

### PeroK

Yes, in fact the energy-momentum of a photon is an eigenvector of the Lorentz Transformation; and, the Doppler red and blue-shift factors are the eigenvalues!

7. Mar 27, 2016

Do these apply exactly to photons without the need for any additional corrective terms?

8. Mar 27, 2016

### PeroK

I've no idea what corrective terms you may be talking about. The Lorentz Transformation applies to both particles and light rays travelling at $c$. One of the first things you should do with the LT is check that it transforms a light ray travelling at $c$ into a light ray travelling at $c$ in the new frame; likewise, that it transforms the energy-momentum for a photon into the energy-momentum for a photon.

9. Mar 27, 2016

### Staff: Mentor

Here's a way to check: in the original frame (which is the primed frame the way you have the equations written), you have $E' = p' c$. Plug that into the two transformation equations and verify that you obtain $E = p c$.

10. Mar 27, 2016

### Andrew Mason

If you are using the Lorentz transformations for the photon, where v=c, γ would be undefined. In any event, the wavelength does not depend on the speed of the photon. It depends on thd speed of the source. So where does the speed of the source appear in the Lorentz transformations?

AM

11. Mar 27, 2016

### PeroK

In the Lorentz Transformation $v$ and $\gamma$ relate to the relative velocity between the reference frames.

12. Mar 27, 2016

### Andrew Mason

Exactly! v is the relative speed of two different inertial reference frames. But, since a photon does not have or define an inertial reference frame the Lorentz transformations, and the energy-momentum transformations derived from them, do not apply to a photon.

AM

13. Mar 27, 2016

### PeroK

That's an unorthodox point of view!

Neither reference frame needs to coincide with the rest frame of any particle under consideration.

Last edited: Mar 27, 2016
14. Mar 27, 2016

### Andrew Mason

I am not sure what you mean. How do the transformations apply to a photon? Can you show us?

AM

15. Mar 27, 2016

### PeroK

I'll let Michael Fowler from the University of Virginia do the work. The last section on this page "Photon Energies in Different Frames" shows how to derive the relativistic Doppler shift formula from the E-M transformation for a photon.

http://galileo.phys.virginia.edu/classes/252/energy_p_reln.html

16. Mar 27, 2016

### Staff: Mentor

He's not. He's just showing that the energy and momentum for a photon transform appropriately when you change frames, using an ordinary Lorentz transformation where $v < c$. See below.

Sure they do; they preserve the property $E = pc$ for a photon in every frame. What's more, by plugging in $E = pc$ in the transformation equations, you can see the Doppler factor pop out.

17. Mar 27, 2016

### Staff: Mentor

Yes. Note that $v$ is the velocity between two inertial frames so $v< c$
Yes.
Yes.
Yes.

Note, pulses of light do not have a rest frame, but the energy-momentum of a pulse of light in one inertial frame is related to the energy-momentum in another inertial frame through the transformation equations posted.

Last edited: Mar 27, 2016
18. Mar 27, 2016

### Staff: Mentor

Consider a person throwing a ball on a train. You can express the momentum of the ball in the frame of the train or in the frame of the embankment. It is moving in both frames, but its energy-momentum is perfectly valid. There is no need to consider the ball's frame.

Similarly with light. If a person is shining a light on a train then you can express the energy-momentum of the light in the train frame or in the embankment frame. The energy-momentum transformation is the same for the ball and the light. The fact that the light does not have a frame where it is at rest does not mean that the energy-momentum transformation equation is invalid or inapplicable between other valid inertial frames.

19. Mar 27, 2016

### Andrew Mason

I must apologize to everyone for the confusion that I may have caused by my last couple of posts. You are right - for some reason I was thinking about a transformation to the rest frame of the moving particle. Of course, and as you point out, this would not have much use since in the rest frame p is always 0 and E is just the rest energy, $m_0c^2$

The Lorentz and Energy-momentum transformations can be used to compare the frequencies/wavelengths of the photon as measured in two different inertial frames moving at speed v relative to each other. While the transformations may be used to calculate the energy and momentum of a particle with mass measured by an observer travelling at speed v relative to the rest frame of the particle, they cannot be used that way with a photon (i.e. by letting v = c), since a photon does not have or define an inertial reference frame.

AM

corrected and added to - see post below.

Last edited: Mar 28, 2016
20. Mar 27, 2016

### Staff: Mentor

If you mean a photon, there is no such thing as the rest frame of a photon. A photon always moves at $c$; there is no frame in which it is at rest.

If you mean a particle with nonzero rest mass $m$, then in its rest frame, $p = 0$ but $E = m$; the energy is not zero.