Is the Schwarzschild metric dimensionless?

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
4 replies · 2K views
help1please
Messages
167
Reaction score
0

Homework Statement



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

Homework Equations



[tex]ds^2 = (1 - \frac{2GM}{c^2 r})c^2 dt^2[/tex]

The Attempt at a Solution



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

[tex]\frac{2GM}{c^2 r}[/tex]

is dimensionless?
 
Physics news on Phys.org
help1please said:

The Attempt at a Solution



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

[tex]\frac{2GM}{c^2 r}[/tex]

is dimensionless?

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