# Gravitational Time Dilation and Schwarzschild Coordinates

So I'm trying to use this equation for the time dilation of an object, but I don't know how to get the distance that I have (in meters) to a radial coordinate in terms of schwarzschild coordinates. How can I do this?

$$t_0 = t_f \sqrt{1 - \frac{2GM}{rc^2}} = t_f \sqrt{1 - \frac{r_s}{r}}$$

where

• $t_0$ is the proper time between events A and B for a slow-ticking observer within the gravitational field

• $t_f$ is the coordinate time between events A and B for a fast-ticking observer at an arbitrarily large distance from the massive object (this assumes the fast-ticking observer is using Schwarzschild coordinates, a coordinate system where a clock at infinite distance from the massive sphere would tick at one second per second of coordinate time, while closer clocks would tick at less than that rate),

• $G$ is the gravitational constant,

• $M$ is the mass of the object creating the gravitational field,

• $r$ is the radial coordinate of the observer (which is analogous to the classical distance from the center of the object, but is actually a Schwarzschild coordinate),

• $c$ is the speed of light, and

• $r_s = 2GM/c^2$ is the Schwarzschild radius of M.

• It isn't obvious what you are asking. Are you asking about exactly how the Schwarzschild radial coordinate $r$ is defined? Oct 22, 2015 at 16:31
• The Schwarzschild r-coordinate is also given in metres (or whatever other distance unit you choose). But careful, for almost all observers the r-coordinate does not correspond to ruler measurements. Does this answer the question?? May 8, 2018 at 23:59

You need the full metric.

If you have two events and a curve from one to another then you can use the metric to find the analog of the distance along the curve. If the curve is timelike you can use

$$d\tau=\sqrt{\left(1-\frac{r_s}{r}\right)dt^2-\left(1-\frac{r_s}{r}\right)^{-1}dr^2-r^2d\theta^2-r^2\sin^2\theta d\phi^2},$$

and if the curve is spacelike you can use

$$ds=\sqrt{-\left(1-\frac{r_s}{r}\right)dt^2+\left(1-\frac{r_s}{r}\right)^{-1}dr^2+r^2d\theta^2+r^2\sin^2\theta d\phi^2}.$$

To way you use it is to break a curve into small curves so that the coordinates don't change much in that small curve then take the actual coordinate changes $\Delta t,$ $\Delta r,$ $\Delta \theta$ and $\Delta \phi$ in that small curve and put $(\Delta t)^2$ where $dt^2$ was and compute the square root and add up the results for each small curve.

This is like approximating the length of a curve by breaking it into small pieces and then finding the length of the straight line version of each small piece. You get the actual length by doing an integral, which means breaking into smaller and small pieces and taking the limit of the sum ad the pieces gets small.

So you do the same for your curve. Break into into pieces compute e.g. $\sqrt{\left(1-\frac{r_s}{r}\right)\Delta t^2-\left(1-\frac{r_s}{r}\right)^{-1}\Delta r^2-r^2\Delta\theta^2-r^2\sin^2\theta \Delta\phi^2}$ for each piece, add them up and then take the limit of the result of those sums in the limit the pieces get smaller. The actual limit is then the length of the curve.

If the curve was timelike, the limit is the amount a clock ticks if it travels on the curve. If the curve is spacelike the limit is the amount a ruler measures when placed along that curve.