1
$\begingroup$

Two similar looking photons (locally) are emitted, one from Earth and the other from the Moon, and they are observed at some point X out in space.

On Earth, time is slightly more dilated due to gravity than it is on the Moon, because gravity is stronger. So will the energy measured at point X be similarly reduced?

$\endgroup$
3
$\begingroup$

Yes, the gravitational red shift will be greater for a photon emitted from the Earth than for a similar photon emitted from the Moon.

The energy is given by $E=hf$, and the frequency $f$ is decreased due to the time dilation i.e. the frequency observed far from the Earth or Moon is given by:

$$ f = f_0 \sqrt{1 - \frac{2GM}{c^2 R}} \tag{1} $$

where $M$ is the mass of the Earth/Moon and $R$ is the radius of the surface$^1$. When the time dilation is very small, as it is for both the Earth and the Moon, we can approximate equation (1) using a binomial expansion. In this case we get the expression:

$$ f \approx f_0 \left(1 - \frac{GM}{c^2 R}\right) \tag{2} $$

Note that the Newtonian gravitational potential energy is given by:

$$ \Phi = -\frac{GM}{R} $$

So the frequency and therefore the energy of the photon is simply related to the gravitational potential energy at the surface:

$$ f \approx f_0 \left(1 + \frac{\Phi}{c^2}\right) \tag{3} $$


$^1$ Strictly speaking the $R$ in equation (1) is not the radius of the Earth but rather the circumference of the Earth divided by $2\pi$. In flat spacetime these are of course the same thing, but in a curved spacetime they are different. However for weak gravitational fields like that of the Earth the difference is negligible.

$\endgroup$
1
  • $\begingroup$ I agree with your answer, which is why I accepted it. It does imply though, since the rate of time is the same for both measurements, that the speed of light must be different for the two photons at point X. However, instead of making the natural and straightforward assumption that the light accelerated at different rates through the different gravitational fields, we must believe that they went through a mathematically complex warping of space-time on their way. How strange is that? $\endgroup$
    – Alan Gee
    Feb 24 '19 at 14:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.