Are there any energy change of photon in different frame?

Click For Summary
Photons experience changes in energy and momentum when observed from different inertial frames, similar to the Doppler effect, which is crucial for astronomical calculations. The energy and momentum of photons transform according to the Lorentz transformation, affecting their frequency based on the relative motion of the source and observer. In the context of the Compton process, the energy-momentum four-vector illustrates how these properties shift between frames. Specifically, for photons, the transformation results in a modified energy expression that accounts for the relative velocity. Understanding these transformations is essential in both theoretical physics and practical applications in astrophysics.
ndung200790
Messages
519
Reaction score
0
Please teach me about this:
Are there any change in energy and momentum of photon in different frame?Are there any analogies with Doppler effect?
Thank you very much in advance.
 
Physics news on Phys.org
Yes! Indeed photons can be red or blueshifted, this doppler effect is how we do many calculations in astronomy (e.g. finding the speed of objects moving towards or away from us).
 
Except for Doppler effect,are there any changing in energy-momentum of photon in different frame.Example:Consider Compton process in lab frame and center of mass frame.Because the frequence of photon in Doppler effect depends on the moving towards each other or far away of receiver and sourse
 
Energy and momentum transform between different inertial reference frames exactly the same way as time and position, via the Lorentz transformation.

The time-position four-vector: (ct, x, y, z)

One way to write the energy-momentum four-vector: (E, p_x c, p_y c, p_z c)

If we need to deal only with x-components, the Lorentz transformation for time and position looks like this:

ct^\prime = \gamma (ct - \beta x)

x^\prime = \gamma (x - \beta ct)

For energy and momentum:

E^\prime = \gamma (E - \beta p_x c)

p^\prime_x c = \gamma (p_x c - \beta E)

where as usual \beta = v / c and

\gamma = \frac{1}{\sqrt{1 - v^2 / c^2}} = \frac{1}{\sqrt{1 - \beta^2}}

and v is the relative velocity of the two frames.

For a photon, E = pc, so the Lorentz transformation for the one-dimensional case becomes

E^\prime = \gamma (E - \beta E)

E^\prime = \gamma (1 - \beta) E

E^\prime = \sqrt {\frac {1 - \beta}{1 + \beta}} E
 
Last edited:
In Birkhoff’s theorem, doesn’t assuming we can use r (defined as circumference divided by ## 2 \pi ## for any given sphere) as a coordinate across the spacetime implicitly assume that the spheres must always be getting bigger in some specific direction? Is there a version of the proof that doesn’t have this limitation? I’m thinking about if we made a similar move on 2-dimensional manifolds that ought to exhibit infinite order rotational symmetry. A cylinder would clearly fit, but if we...

Similar threads

  • · Replies 42 ·
2
Replies
42
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 26 ·
Replies
26
Views
2K
  • · Replies 17 ·
Replies
17
Views
3K
  • · Replies 87 ·
3
Replies
87
Views
5K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 19 ·
Replies
19
Views
1K
  • · Replies 35 ·
2
Replies
35
Views
5K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 17 ·
Replies
17
Views
2K