# B Help with units here

1. Jul 27, 2016

### needved

Hi people :)

I'm learning some of General Relativity topics but still im a beginer, uh! i use the Schutz "A first course of general relativity", but i a little confused about the units, the author say it use c = G = 1 all around the book.
Just right now i reading chapter 11: Schwarchild Geometry and Black Holes". especifically "Conserved quantities" unit, and there, there is a couple of graphs V(r) vs r and in both graphs the root is "r = 2M" where M is the mass of the Schwarchild Black Hole, so my question is how i recover the original lenghts units? i mean if M =1000M(Sun) then i cant say the distance r is 2000M(Sun).

(Sorry if a silly question )

2. Jul 27, 2016

### Staff: Mentor

The mass $M$ in length units can be found by taking the mass $M_{\text{conv}}$ in conventional units and applying the formula:

$$M = \frac{G M_{\text{conv}}}{c^2}$$

where $G$ is Newton's gravitational constant and $c$ is the speed of light. Basically, the mass $M$ in "length units" is the mass in a system of units in which $G = c = 1$; in this system mass and length have the same units.

3. Jul 28, 2016

### Ibix

More generally, dimensional analysis. You want a length (dimension $L$), you've got a mass (dimension $M$), and you only have G (dimension $M^{-1}L^3T^{-2}$) and c (dimension $LT^{-1}$) to play with. You need to multiply by $G^ac^b$ ($a$ and $b$ are powers, not tensor indices) such that the dimensions match.

In this case, you have $r=G^ac^bM$, the dimensions of which are $L=M^{-a}L^{3a}T^{-2a}L^bT^{-b}M$. Comparing powers of M, L, T gives you

M: $0=-a+1$
L: $1=3a+b$
T: $0=-2a-b$

any pair of which solves to give you Peter's answer.

Informally, stick the SI units in to what you have and multiply/divide by powers of G and c until they match.

Last edited: Jul 28, 2016