# The Schwarzschild metric

1. Dec 16, 2007

### Orion1

The solution for the Schwarzschild metric is stated from reference 1 as:

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

The solution for the Schwarzschild metric is stated from references 2 as:

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

The solution for the Schwarzschild metric is stated from references 3 as:
$$ds^2 = \left(1 - \frac{r_s}{r} \right) c^2 dt^2 - \left(1 - \frac{r_s}{r} \right)^{-1} dr^2 - r^2 \left(d \theta^2 - \sin^2 \theta d \phi^2 \right)$$

$$r_s$$ - Schwarzschild radius

There is a difference in the sign of the elements between the stated solutions.

Which is the correct solution?

Reference:
http://en.wikipedia.org/wiki/Deriving_the_Schwarzschild_solution
http://en.wikipedia.org/wiki/Tolman-Oppenheimer-Volkoff_equation#_ref-ov_3
http://en.wikipedia.org/wiki/Schwarzschild_metric

Last edited: Dec 16, 2007
2. Dec 16, 2007

### cesiumfrog

Did you check any references outside of wikipedia?

By the looks, the only difference is the sign convention: is it spacelike or timelike intervals that are represented by real lengths? For what purpose do you want one to be "correct"?

3. Dec 16, 2007

### SpaceTiger

Staff Emeritus
Might want to double-check your signs on this one. The third reference looks like it gives the same as the second reference (but distributing the last parenthetical). The differences between the first two are probably just convention, as cesiumfrog said.

4. Dec 17, 2007

### CompuChip

Why don't you read chapter 7 of Carroll's notes and check for yourself (for one thing, I know that the arbitrariness of the sign is mentioned)