I am unable to get the maths and neither the physics behind how gravity affects clock rate. Note- As far as possible I was able to express and clear I have understood.

"Suppose if we keep spatial components to be zero, the clock runs at different rates in gravity." My sir showed me the proof by sending a radiation beam from ground to a height and said frequency changes, then it comes back down creating a electron positron pair. He then wrote the energy for down path as: $E_{down}=E_{up}+mgL$, where $E_{up}$ is m$c^{2}$ per particle. He then's find the relation of frequency as

$\nu_{down}=\nu_{up}(1+\frac{gL}{c^2})$. Then he uses this theory to convert how clock operates in gravitational field for N-cycles and finds the relation as:

$\delta t_{x}=\delta t_{fudicial}(1+\frac{2\phi(x)}{c^2})$. He then compares it with $g_{00}(t,x)dt^2$ and proves it and says that one can show for length contraction gets affected by gravity (which i have no clue).

  • 1
    $\begingroup$ Have you learned about how the metric tensor determines the proper time measured by a clock? $\endgroup$
    – G. Smith
    Jun 24, 2020 at 17:11
  • 1
    $\begingroup$ Well, if you carry an atomic clock up a mountain and set it down, it will tick at a different rate than a clock back in the lab. $\endgroup$
    – Jon Custer
    Jun 24, 2020 at 17:19
  • $\begingroup$ @G.Smith I have not reached till there? $\endgroup$ Jun 25, 2020 at 6:11

3 Answers 3


One can assume that as a photon goes up in a gravitational field, it gains potential energy and looses kinetic energy (and frequency). Then a distant observer would say that the system which emitted the photon was vibrating slower than it would in his position high in the field. There will also be a corresponding change in the wavelength. This approach closely approximates the effects predicted by general relativity for the earth's field, and must be considered in the operation of the GPS system.

Energy of a photon, $E = hf = mc^2$, and acceleration of gravity, $g = GM/(R^2)$. (In general relativity the curvature of space-time is determined by distribution of mass and energy, and photons respond to that curvature. As far as gravity is concerned, energy acts like mass.) For a small increase in height: $dE = hdf = -mgdr = -(hf/(c^2))(GM/(r^2))dr$. Separating variables and integrating gives:

$\int_{f_1}^{f_2}\,\frac{df}f= -B\int_{r_1}^{r_2}\frac{dr}{r^2}$ where $B = GM/c^2$. Then $\ln f_2 - \ln f_1= \ln\frac{f_2}{f_1} = B(\frac{1}{r_2}-\frac{1}{r_1})$ and

$$\frac{f_2}{f_1} = e^{B\frac{1}{r_2}-B\frac{1}{r_1}} =\frac{e^{\frac{B}{r_2}}}{e^{\frac{B}{r_1}}}= \frac{T_1}{T_2}$$

These calculations can be approximated using power series expansions.

  • $\begingroup$ Can you give me a reference site or book where these detail mathematics and proof is given? $\endgroup$ Jun 25, 2020 at 6:10
  • $\begingroup$ I have tried to edit the equations the best I could. You are using way to many unneccessary braces in your equations. And you can wrap one equation in '$' signs. You don't need to break random parts of the equation between such signs. The formatting is still not ideal, but at least I've gotten you a step in the right direction. I would suggest being consistent in your formatting, and using the \frac command rather than division with '/' and many parentheses. $\endgroup$ Jun 27, 2020 at 21:35
  • $\begingroup$ Also, I think you meant "moderators" rather than "monitors". Note that is not the moderators' job (or really any user's job) to heavily edit your posts and teach you how to use MathJax $\endgroup$ Jun 27, 2020 at 21:43
  • $\begingroup$ BioPhysicist: Your interpretation is exactly what I had in mind. Thanks for your time and effort. $\endgroup$
    – R.W. Bird
    Jun 28, 2020 at 13:34

The phenomenon of time dilatation or contraction can be explained and calculated with the Schwarzschild metrics which leads to the formula: $$ \Delta\tau_2=\sqrt{\frac{1-\frac{2GM}{c^2r_2}}{1-\frac{2GM}{c^2r_1}}}\Delta\tau_1 $$ where $ \Delta\tau_1 $ and $ \Delta\tau_2 $ are the proper time intervals measured between two same events respectively by a static observer $ 1 $ at radial coordinate $ r_1 $ and by a static observer $ 2 $ at radial coordinate $ r_2 $.

This means that if the observer $ 2 $ is at a distance $ r_2 $ greater than the distance $ r_1 $ of the observer $ 1 $, his "time" will pass more quickly than the "time" of the observer $ 1 $.

If the observer $ 2 $ is at a very great distance from the massive object of mass $ M $, and if the observer $ 1 $ is at a radial coordinate $ r $, you will have: $$ \Delta\tau_2\simeq\frac{\Delta\tau_1}{\sqrt{1-\frac{2GM}{c^2r}}} $$ At last, if the massive object is a black hole with the observer $ 1 $ close enough (low value of $ r $), a proper time interval of $ 1 $ hour for him will correspond to a proper time interval of several years for the observer $ 2 $ (refer to Interstellar movie for instance).

Hoping to have answered your question,

Best regards.


If we fire a bullet upwards, it gains potential energy as it rises in the magnetic field, but due to conservation of energy it loses kinetic energy and eventually comes to a stop and falls.

For a photon, it also gains gravitational potential energy as it rises, but it does not slow down because the speed of light is constant locally. The energy of a photon is given by $\hbar f$ where $\hbar$ is a constant and f is the frequency. To lose energy the photon has to reduce its frequency. Reducing frequency is basically the same thing as slowing the tick rate of a clock.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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