# The Schwarzschild metric

1. Sep 9, 2012

1. The problem statement, all variables and given/known data

The problem is I am wanting to know if the expression on the right hand side is dimensionless.

2. Relevant equations

$$ds^2 = (1 - \frac{2GM}{c^2 r})c^2 dt^2$$

3. The attempt at a solution

Since the Schwarzschild radius is $$r = \frac{2GM}{c^2}$$ would I be right in saying that

$$\frac{2GM}{c^2 r}$$

is dimensionless?

2. Sep 9, 2012

Yes, it is.

3. Sep 9, 2012

### vela

Staff Emeritus
Note that it would have to be if the formula is valid since you're subtracting that quantity from 1, which is dimensionless.

4. Sep 10, 2012

### karlzr

Of course it is. Note that sometime we write metric in this form:$$1-\frac{2GM}{r}$$
just a matter of unit conventions.

5. Sep 11, 2012