1. Jul 13, 2008

### snoopies622

I am looking at some Christoffel symbols that are expressed using geometrized units, and the only variables to appear are m,r and theta. If I want to convert these to non-geometrized units, how do I know where to place the G's and c's?

Is there a consistent way of doing this sort of thing? Do I simply change the m's to mG/(c^2)'s and the t's to ct's? I understand that if I know in advance the dimensionalities that I am looking for then it is not a problem, but that is not always the case.

2. Jul 13, 2008

### Staff: Mentor

I don't know of any way to do that correctly in all cases. I usually just take exactly the approach you mentioned, and it often works.

3. Jul 13, 2008

### Mentz114

You can see from the geodesic equation that the Christoffel symbols have dimension L-1, if that helps. It also follows from the definition of the CS's. To get back to non-geometrical units you should multiply terms containing g00 by c2, and terms like g0k by c ( k = 1,2,3, time is x0). You have the correct substitution for m.

4. Jul 13, 2008

### snoopies622

Thank you both.

Now I have a quick follow-up: If the Christoffel symbols have dimension L-1 (which is what I thought) then what about the case of

$$\Gamma ^{r} _{\theta \theta} = -r$$

in plane polar coordinates? That doesn't look like L-1. There is an analogous counterexample in spherical coordinates, too.

5. Jul 14, 2008

### Mentz114

Yes, it is only L-1 for the Cartesian coords. However, you can easily work out the dimensions by looking at the definition, and the metric. For the Schwarzschild metric

$$\Gamma ^{r} _{\theta \theta} = 2m - r$$ which clearly has dimension L, which it must in order that

$$\frac{d^2r}{d\lambda^2}$$ has the same dimensions as

$$\Gamma ^{r} _{\theta \theta}\left(\frac{d\theta}{d\lambda}\right)^2$$

Sorry for my hasty first post, it looks like you can read off the dimensions of the connections from the geodesic equation.

6. Jul 14, 2008

### snoopies622

It looks like the dimensions of $$\Gamma ^{a}_{bc}$$ always = $$\frac {(dimensions\ of\ a)}{(dimensions\ of\ b)(dimensions\ of\ c)}$$.

Well that's easy to remember. Thanks again, Mentz.