Are there any energy change of photon in different frame?

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.
 

Matterwave

Science Advisor
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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
 

jtbell

Mentor
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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: [itex](E, p_x c, p_y c, p_z c)[/itex]

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

[tex]ct^\prime = \gamma (ct - \beta x)[/tex]

[tex]x^\prime = \gamma (x - \beta ct)[/tex]

For energy and momentum:

[tex]E^\prime = \gamma (E - \beta p_x c)[/tex]

[tex]p^\prime_x c = \gamma (p_x c - \beta E)[/tex]

where as usual [itex]\beta = v / c[/itex] and

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

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

[tex]E^\prime = \gamma (E - \beta E)[/tex]

[tex]E^\prime = \gamma (1 - \beta) E[/tex]

[tex]E^\prime = \sqrt {\frac {1 - \beta}{1 + \beta}} E[/tex]
 
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