# Deriving the Schwarzschild Metric

1. Aug 23, 2011

### JDoolin

I've worked through a common-sense argumenthttp://www.mathpages.com/rr/s8-09/8-09.htm" [Broken] showing the time-time component of the Schwarzschild metric

$$g_{tt} = \left (\frac{\partial \tau}{\partial r} \right )^2\approx 1-\frac{2 G M}{r c^2}$$

On the other hand, I've not worked through any common-sense argument for the grr component of the Schwarzschild metric:
$$\left (\frac{\partial s}{\partial r} \right )^2\approx \frac{1}{1-\frac{2 G M}{r c^2}}$$

I can see there is a derivation in the http://www.blatword.co.uk/space-time/Carrol_GR_lectures.pdf" [Broken] on pages 168-172. That remains a goal for me, to work through that derivation as well, but I'm not comfortable with most of the concepts involved here.

I am interested in the reasons behind these steps.

• assuming $$g_{11}=-1/g_{00}$$
• finding the connection coefficients
• finding the nonvanishing components of the Reimann tensor
• taking the contraction (as usual?) to find the Ricci tensor
• setting all the components of the Ricci tensor to zero
• discovering that the g00 and g11 had to be functions of r, only,
• Setting R00=R11=0
• Doing some differential equations with boundary conditions, and deriving the metric

My trouble is that I don't have a common-sense understanding of the Reimann tensor or the Ricci tensor. I'm also having trouble distinguishing the relevant equations, like definitions of these tensors, as I find myself, as I read through the Carroll Lectures, filling up page after page of undefined components, but never really getting to the heart of the matter.

I'm only beginning to have some common-sense understanding of the connection (Christoffel) coefficient, based on the Cartesian to polar connection coefficients, diagrammed on page 6, http://mysite.verizon.net/vze11jx21/GR2c-Derivatives.pdf".

Where should I begin? Perhaps with step 1. Why do we start with the assumption that gtt is the negative reciprocal of grr?

Last edited by a moderator: May 5, 2017
2. Aug 23, 2011

### Bill_K

JDoolin, My advice is to forget about the common-sense approach and learn the mathematics. Common sense will quickly lead you astray.

3. Aug 23, 2011

### K^2

Once you make use of spherical symmetry, it's a diff eq with two variables. (Three at the most, if you don't make assumptions about r-t cross term). It might not be obvious from equations themselves, but once you know the form you're looking for, it should be easy enough to work out.

4. Aug 23, 2011

### JDoolin

Common sense, or mathematics, either way, why does Carroll start with the assumption that gtt is the negative reciprocal of grr?

5. Aug 23, 2011

### Samshorn

It isn't an assumption. See, for example, the explanation following equation (4) of this link:

http://www.mathpages.com/rr/s5-05/5-05.htm

Last edited by a moderator: May 5, 2017
6. Aug 23, 2011

### WannabeNewton

It isn't an assumption. The metric you start with (after the spherical symmetry conditions and orthogonality conditions) is $ds^{2} = -e^{2\alpha (r)}dt^{2} + e^{2\beta (r)}dr^{2} + r^{2}d\Omega ^{2}$. You know that $R_{tt} = 0$ and $R_{rr} = 0$ because you are solving the vacuum field equations and if you calculate the components for each and add the two components together you find that $\partial _{r}\alpha = - \partial _{r}\beta$ so, after using the freedom to rescale constants, $\alpha = -\beta$ which explains the reciprocal. This is also explained pretty clearly in Caroll's text.

7. Aug 23, 2011

### JDoolin

Actually that's not what I meant, anyway. When I ask "Why do you assume?" I mean "How do you derive?" I meant no disrespect to Carroll.

Thank you. That looks very promising.

8. Aug 23, 2011

### pervect

Staff Emeritus
As far as I know, g_00 is determined to low order (second order, I think), by requiring that GR match up with Newtonian physics in the Newtonian limit. Which is (I hope) what your common sense argument boils down to.

The spatial components of the metric require you to actually use Einstein's field equations to find. And the full metric makes new non-Newtonian predictions, such as spatial curvature (spatial, using the Schwarzschild time-slice) and the extra bending of light. So I think you'll need to actually understand the heart of the theory, where Einstein's field equations, came from to get this part, and not just look for correspondence with Newton's theory.

Given the field equations, though, it's not terribly hard to show what the metric must be.