At higher altitudes above a body, clocks tick more slowly, and gravitational field is weaker. But what is the relationship? It is tempting for a GR newbie such as myself to think that anywhere that the gravity is equal to that of the earth's surface that time-dilation would be the same, i.e. dilation as a simple function of local gravitational acceleration. I have been assured that this is not the case. I wonder if someone can calculate the following situation, which strikes me as the simplest 2nd case to compare with earth's surface: At the surface of the Sun, gravity is roughly 28 times that at the Earth's surface. If we move out to a suitable distance, we can find a place where the gravitational acceleration is the same as the surface of the Earth - 1g - and we can hold a clock there with a rocket. What is the time dilation there with regard to a "Far away" observer? I understand to the dilation on the Earth's surface to be 0.0219 seconds per year compared with the distant observer.

  • $\begingroup$ I gave you a +1 because somebody gave you a -1 without a comment. To the downvoter, you should really post. Anyway, time dilation has to do with your speed. At that distance from the sun, you will be able to calculate an orbital speed. Try doing that and seeing the results :) $\endgroup$ – markovchain Feb 4 '13 at 5:09
  • $\begingroup$ @markovchain, time dilation in special relativity is to do with speed. In general relativity, gravitational fields can cause time dilation with respect to an observer even when stationary with respect to the same observer. $\endgroup$ – Kenshin Feb 4 '13 at 7:17
  • $\begingroup$ Dudes, this is a general relativity question. Speed is to be neglected here. $\endgroup$ – Jonathan Dunn Feb 4 '13 at 14:05

For a Schwarzschild observer the gravitational time dilation means the local time at a distance $r$ from a mass $M$, appears to be running slow by a factor of:

$$ \sqrt {1 - \frac{2GM}{r c^2}} $$

You get this straight from the Schwarzschild metric:

$$ ds^2 = -\left(1-\frac{2GM}{r c^2}\right)dt^2 + \left(1-\frac{2GM}{r c^2}\right)^{-1}dr^2 + r^2 d\Omega^2 $$

because hovering above the Sun, both $dr$ and $d\Omega$ are zero.

Anyhow, I reckon the acceleration due to the Sun's gravity is 9.81m/s$^2$ at a distance of about $3.68 \times 10^9$m (using the Newtonian expression for $a$), and at this distance I work out the time dilation to be about 12.6 seconds per year. So the time dilation is not simply proportional to the gravitational acceleration.

You can get an approximate, but very good, relationship between the acceleration and the time dilation by using the Newtonian expression for the acceleration to get $r$ and substituting in the expression for the time dilation. Doing this gives:

$$ \sqrt{1 - \frac{2\sqrt{aGM}}{c^2}} $$

which I don't think is terribly illuminating!

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    $\begingroup$ If you take the limit of your last expression for small $a$ and $GM$, and take the difference between the values of the expression for Earth and the sun you get that $\delta\sqrt{\cdots}\approx\sqrt{\frac{M_{sun}}{M_{Earth}}}\sqrt{\text{Earth}}$ (I hope the notation is obvious): the difference is proportional to the square root of the mass ratio. So if I made no mistakes the time dilation at the sun should be 577 times bigger than the Earth's, which agrees with your numbers. $\endgroup$ – Michael Brown Feb 4 '13 at 14:05

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