# Symmetry in gravitational time dilation

In special relativity, when two observers move in respect to each other, each observes the other's clock tick slower.

An observer floating in space far from gravitational fields, observes earth clock ticking slower; at what rate is the clock of the floating observer ticking as observed from earth?

## 1 Answer

Assuming that the Schwarzschild metric is a good representation of the spacetime geometry round the Earth, the ratio of Earth time to space time is:

$$\frac{t_E}{t_S} = \sqrt{1 - \frac{2GM}{c^2r_E}}$$

where $r_E$ is the radius of the Earth. So the clock on Earth runs slow as observed from space.

The ratio of space time to Earth time is simply the inverse of this:

$$\frac{t_S}{t_E} = \frac{1}{\sqrt{1 - \frac{2GM}{c^2r_E}}}$$

so the clock in space runs fast as observed from Earth.