B Help With Units in General Relativity

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Hi people :)

I'm learning some of General Relativity topics but still I am 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 can't say the distance r is 2000M(Sun).

Thanks in advance

(Sorry if a silly question :frown: )
 
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needved said:
how i recover the original lenghts units?

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.
 
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 into what you have and multiply/divide by powers of G and c until they match.
 
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