Are there any energy change of photon in different frame?

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SUMMARY

Photons experience energy and momentum changes when observed from different inertial reference frames, a phenomenon explained by the Lorentz transformation. The discussion highlights the Doppler effect as a primary example, where photons can be redshifted or blueshifted based on the relative motion of the source and observer. Additionally, the Compton process is mentioned to illustrate energy-momentum transformations in both the lab frame and the center of mass frame. The energy-momentum four-vector is defined as (E, p_x c, p_y c, p_z c), and the transformations for energy and momentum are provided, emphasizing the relationship between energy and momentum for photons.

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  • Understanding of Lorentz transformations
  • Familiarity with the Doppler effect
  • Knowledge of energy-momentum four-vectors
  • Basic principles of special relativity
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  • Explore the Compton effect in detail
  • Learn about the mathematical derivation of the Doppler effect for light
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Physicists, students of relativity, and anyone interested in the behavior of light and energy in different inertial frames.

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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.
 
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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
 
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