does the gravitational Pull a photon exerts depend on the observer? Since a photons energy changes due to length contraction depending on the observers speed, shouldn't it exert a different gravitational pull depending on the observer? This seems counterintuitive to me. Is there something wrong with my reasoning?
2 Answers
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The existing answer is a bit technical, so I thought I would say a few words.
In Newtonian physics "gravitational pull" is the gravitational force exerted by one body on another. In Newtonian physics you would get the same force (between two given bodies) no matter which observer you pick, as long as you only consider observers whose motion is at constant velocity.
In General Relativity, the more accurate description of gravity, the "gravitational pull" is a more complicated sort of thing, not just a force. However some of its effects are somewhat like a force. The force (on any given body) will be found to be different for different observers moving at constant velocity with respect to one another, but the change will only be significant when the velocities are high, a significant fraction of the speed of light.
To find the change in the frequency (and thus the energy) of a light wave you can use the Doppler effect formula $$ f' = f \sqrt{ \frac{1 + v/c}{1-v/c} } $$ where $f'$ is the received frequency if a source moving towards the receiver at speed $v$ has frequency $f$ in its rest frame. The gravitational effects are related not just to the energy of the light but also how it is arranged spatially (the shape of the beam). Once you have found the gravitational force on some nearby object (a rock or something) then the force on that same rock, as found for observers moving relative to the rock, can be obtained by ordinary methods in special relativity, involving the Lorentz transformation.
answered Oct 24 at 18:20
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Yes:
$$ p_{\mu} = (E/c, \vec p) = \hbar(\omega /c, \vec k) $$
is frame dependent.
The only problem with your reasoning is that you forgot the spacetime metric $g_{\mu\nu}$ also transforms covariantly, as does the resulting curvature tensor $G_{\mu\nu}$ such that it is $\kappa T_{\mu\nu}$, where the stress energy tensor goes something like:
$$ T_{\mu\nu} \propto p_{\mu}p_{\nu} $$
