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I am a beginner in general relativity. I read the relevant section 9.1.5 in Relativity make relatively easy by Andrew Steane. After that, I thought a problem in gravitational red shift and gravitational time dilation.

Suppose there are two observers A, B, at two heights above Earth and a light source at A emits a signal to B. Then there is blue shift at B. The frequencies relation at two heights is:
enter image description here

$$ \frac{f_B}{f_A}=\exp{(-\frac{(\Phi_B-\Phi_A)}{c^2})} $$

$f_B$ is the frequency measured by observer B and $f_A$ is the frequency measured at A. $\Phi_B,\ \Phi_A$ are the potential energy at B and A.

Now suppose there is another observer C far away from A and B, and there are two oscillators at A and B, with frequencies $f_A$ and $f_B$ which are measured by A and B.

enter image description here

I set at C such that someone at C observes the same frequencies from A and B. Then by the above relation:

$$ \begin{cases} \frac{f_A}{f'_A}=\exp{(-\frac{(\Phi_A-\Phi_C)}{c^2})}\\ \frac{f_B}{f'_B}=\exp{(-\frac{(\Phi_B-\Phi_C)}{c^2})} \end{cases} $$

$f'_A=f'_B$ by my setting. So I got a relation between $f_A$ and $f_B$.

$$ \frac{f_A}{f_B}=\exp{(-\frac{(\Phi_A-\Phi_B)}{c^2})} $$

Because $\Phi_A>\Phi_B$, $f_A<f_B.$ This does not satisfy with the statement of gravitational time dilation on the Wikipedia(https://en.wikipedia.org/wiki/Gravitational_redshift), which is higher oscillation frequency at larger gravitational potential.

What did I do it in a wrong way? How can I get gravitational time dilation at the radiation source?

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  • $\begingroup$ You "set C such that A and B frequencies are the same." How can this happen if A and B are at different potentials? By your equation it seems like the frequency ratio can only be 1 if the potentials are the same. $\endgroup$
    – RC_23
    Commented Apr 19, 2022 at 23:36
  • $\begingroup$ I suppose that $f_A$ and $f_B$ are different and the observer at C measures two same frequencies from A and B. $\endgroup$
    – Hsu Bill
    Commented Apr 19, 2022 at 23:51

1 Answer 1

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You have arranged it so that C observes the same frequency from two sources at different depths in the potential. Both are redshifted, but B is redshifted by more than A because it is at a (numerically) smaller potential. Since it is redshifted by more than the signal from A, then in order to be observed at the same frequency at C, $f_B > f_A$.

The statement on the Wikipedia page is surely correct if observing two sources from C where $f_A = f_B$. In that case, using your notation, $f'_A> f'_B$ and the source higher in the potential well would be observed to have a higher frequency.

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  • $\begingroup$ Thank you for your kind reply! But how can I know there is gravitational time dilation at the radiation source? $\endgroup$
    – Hsu Bill
    Commented Apr 20, 2022 at 21:31

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