# Eddington Finkelstein Coordinates + Black Holes

1. Jun 4, 2006

### discjockey

When deriving the Eddinton-Finkelstein Coordinates from the Schwarzschild metric, we start to examine light rays. However, in my relativity book, it states that ds^2=0: why do we assume that?

2. Jun 4, 2006

### Garth

Hi, and welcome to these Forums discjockey!

Not knowing exactly what text you are using makes a response a little risky, however the most obvious answer is that light rays travel on null-geodesics with ds2 = 0.

In flat space-time:
ds2 = dx2 + dy2 + dx2 - c2dt2.

As a light ray travels at the speed of light c then
dx2 + dy2 + dx2 = c2dt2.

Hence ds2 = 0.

The result also holds in curved space-time, it is a basic property of light to travel always at c as measured by local rulers and clocks.

Garth

Last edited: Jun 4, 2006
3. Jun 4, 2006

### discjockey

Thanks a lot!

4. Aug 4, 2010

### mersecske

if we use ingoing null coordinate (v) and the radius (r) for basis
then we have a null-coordinate, and a space-coordinate above the horizon,
and a null-coordinate, and a time-coordinate below the horizon.
Usually we use time ad space coordinates together,
and we evolve a system in the direction increasing time.
In the case of EF-coordinates we can evolve the system
using the null-coordinate (v)?
But under the horizon r is a time coordinate,
we can evolve the system in r also?

5. Aug 4, 2010

### George Jones

Staff Emeritus
6. Aug 28, 2010

Thanks