# Christoffel Symbols for Schwarzschild Metric (?)

1. Jul 24, 2009

### Widdekind

ROUGH DRAFT

I have a beginner's basic question:

1. Schwarzschild Metric components

Let $$\epsilon$$ = rs / r, where rs is the Schwarzschild Radius. Then, as is is well-known:

$$g_{00} = 1 - \epsilon$$
$$g_{11} = - \left( 1 - \epsilon \right)^{-1}$$
$$g_{22} = - r^{2}$$
$$g_{33} = - r^{2} \; sin^{2}(\theta)$$​

B/c this Schwarzschild Metric Tensor gij is Diagonal, its Inverse gij is also Diagonal, w/ components equal to "one over" those above.

2. Christoffel Symbol components

As is well-known:

$$\Gamma^{i}_{k\ell} = {1 \over 2} g^{im} (g_{mk,\ell} + g_{m\ell,k} - g_{k\ell,m})$$​

But, since the Schwarzschild Metric Tensor is diagonal, $$g^{im} = \delta^{im} \; g^{ii}$$. So:

$$\Gamma^{i}_{k\ell} = {1 \over 2} g^{ii} (g_{ik,\ell} + g_{i\ell,k} - g_{k\ell,i}) + 0$$​

Thus, in this Schwarzschild Polar Coordinate System, w.h.t.:

$$\Gamma^{0}_{k\ell} = {1 \over 2} g^{00} $\left( \begin{array}{cccc} 0 & g_{00,1} & 0 & 0 \\ g_{00,1} & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right)$$$

$$\Gamma^{1}_{k\ell} = {1 \over 2} g^{11} $\left( \begin{array}{cccc} -g_{00,1} & 0 & 0 & 0 \\ 0 & g_{11,1} & 0 & 0 \\ 0 & 0 & -g_{22,1} & 0 \\ 0 & 0 & 0 & -g_{33,1} \end{array} \right)$$$

$$\Gamma^{2}_{k\ell} = {1 \over 2} g^{22} $\left( \begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & 0 & g_{22,1} & 0 \\ 0 & g_{22,1} & 0 & 0 \\ 0 & 0 & 0 & -g_{33,2} \end{array} \right)$$$

$$\Gamma^{3}_{k\ell} = {1 \over 2} g^{33} $\left( \begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & g_{33,1} \\ 0 & 0 & 0 & g_{33,2} \\ 0 & g_{33,1} & g_{33,2} & 0 \end{array} \right)$$$​

Or, noting that $$\partial \epsilon / \partial r = - \epsilon / r$$, w.h.t.:

$$\Gamma^{0}_{k\ell} = {1 \over 2} (1 - \epsilon)^{-1} $\left( \begin{array}{cccc} 0 & \epsilon / r & 0 & 0 \\ \epsilon / r & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right)$$$

$$\Gamma^{1}_{k\ell} = -{1 \over 2} \left( 1 - \epsilon \right) $\left( \begin{array}{cccc} -\epsilon / r & 0 & 0 & 0 \\ 0 & \left( 1 - \epsilon \right)^{-2}(\epsilon / r) & 0 & 0 \\ 0 & 0 & 2 r & 0 \\ 0 & 0 & 0 & 2 r \; sin^{2}(\theta) \end{array} \right)$$$

$$\Gamma^{2}_{k\ell} = -{1 \over 2} r^{-2} $\left( \begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & 0 & -2 r & 0 \\ 0 & -2 r & 0 & 0 \\ 0 & 0 & 0 & r^{2} \; sin(2 \theta) \end{array} \right)$$$

$$\Gamma^{3}_{k\ell} = -{1 \over 2} r^{-2} \; sin^{-2}(\theta) $\left( \begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -2 r \; sin^{2}(\theta) \\ 0 & 0 & 0 & -r^{2} \; sin(2 \theta) \\ 0 & -2 r \; sin^{2}(\theta) & -r^{2} \; sin(2 \theta) & 0 \end{array} \right)$$$​

3. Geodesic Equation (?!)

$${\partial^{2} \over \partial s^{2}} \left( \begin{array}{c} c t \\ r \\ \theta \\ \phi \end{array} \right) + \left( \begin{array}{c} (1 - \epsilon)^{-1} {\epsilon \over r} {\partial (ct) \over \partial s} {\partial r \over \partial s} \\ {1 \over 2} \left( 1 - \epsilon \right) \left( {\epsilon \over r}{\partial (ct) \over \partial s}^{2} - \left( 1 - \epsilon \right)^{-2}{\epsilon \over r}{\partial r \over \partial s}^{2} - 2 r {\partial \theta \over \partial s}^{2} - 2 r \; sin^{2}(\theta) {\partial \phi \over \partial s}^{2} \right) \\ {2 \over r}{\partial r \over \partial s} {\partial \theta \over \partial s} - {sin(2 \theta) \over 2}{\partial (\phi) \over \partial s}^{2} \\ {2 \over r}{\partial r \over \partial s} {\partial \phi \over \partial s} + 2 cot(\theta){\partial \theta \over \partial s} {\partial \phi \over \partial s} \end{array} \right) = 0$$

4. Zero-Gravity limit (??)

If $$\epsilon = 0$$, w.h.t.:

$${\partial^{2} \over \partial s^{2}} \left( \begin{array}{c} c t \\ r \\ \theta \\ \phi \end{array} \right) + \left( \begin{array}{c} 0 \\ - r {\partial (\theta) \over \partial s}^{2} - r \; sin^{2}(\theta) {\partial (\phi) \over \partial s}^{2} \\ {2 \over r}{\partial r \over \partial s} {\partial \theta \over \partial s} + {sin(2 \theta) \over 2}{\partial (\phi) \over \partial s}^{2} \\ {2 \over r}{\partial r \over \partial s} {\partial \phi \over \partial s} + 2 cot(\theta){\partial \theta \over \partial s} {\partial \phi \over \partial s} \end{array} \right) = 0$$

Further restricting $$\theta = {\pi \over 2}$$, w.h.t.:

$${\partial^{2} \over \partial s^{2}} \left( \begin{array}{c} c t \\ r \\ \theta \\ \phi \end{array} \right) + \left( \begin{array}{c} 0 \\ - r {\partial (\phi) \over \partial s}^{2} \\ 0 \\ {2 \over r}{\partial r \over \partial s} {\partial \phi \over \partial s} \end{array} \right) = 0$$

Is this the equation of a straight line in Polar Coordinates ?

5. Weak-Gravity limit (??)

If $$\epsilon << 1$$, w.h.t.:

$${\partial^{2} \over \partial s^{2}} \left( \begin{array}{c} c t \\ r \\ \theta \\ \phi \end{array} \right) + \left( \begin{array}{c} {\epsilon \over r} {\partial (ct) \over \partial s} {\partial r \over \partial s} \\ {1 \over 2} \left( {\epsilon \over r}{\partial (ct) \over \partial s}^{2} + {\epsilon \over r}{\partial (r) \over \partial s}^{2} - \left( 1 - \epsilon \right) 2 r {\partial (\theta) \over \partial s}^{2} - \left( 1 - \epsilon \right) 2 r \; sin^{2}(\theta) {\partial (\phi) \over \partial s}^{2} \right) \\ {2 \over r}{\partial r \over \partial s} {\partial \theta \over \partial s} + {sin(2 \theta) \over 2}{\partial (\phi) \over \partial s}^{2} \\ {2 \over r}{\partial r \over \partial s} {\partial \phi \over \partial s} + 2 cot(\theta){\partial \theta \over \partial s} {\partial \phi \over \partial s} \end{array} \right) = 0$$

Further restricting $$\theta = {\pi \over 2}$$, w.h.t.:

$${\partial^{2} \over \partial s^{2}} \left( \begin{array}{c} c t \\ r \\ \theta \\ \phi \end{array} \right) + \left( \begin{array}{c} {\epsilon \over r} {\partial (ct) \over \partial s} {\partial r \over \partial s} \\ {1 \over 2} \left( {\epsilon \over r}{\partial (ct) \over \partial s}^{2} + {\epsilon \over r}{\partial (r) \over \partial s}^{2} - \left( 1 - \epsilon \right) 2 r {\partial (\phi) \over \partial s}^{2} \right) \\ 0 \\ {2 \over r}{\partial r \over \partial s} {\partial \phi \over \partial s} \end{array} \right) = 0$$

Does this reduce to Newton's equations ?

Last edited: Jul 24, 2009
2. Jul 24, 2009

### Widdekind

3. Jul 24, 2009

### Widdekind

Verifying Geodesic Equation by applying Euler-Lagrange Equation to Scwarzschild Metric-derived Lagrangian

We apply the Euler-Lagrange Equation to the Scwarzschild Metric-derived Lagrangian:

$$L \equiv g_{\mu \nu} {d x^{\mu} \over ds} {d x^{\nu} \over ds} = 1$$​

Explicitly, w.h.t.:

$$L \equiv (1 - \epsilon) c^{2} {dt \over ds}^{2} - (1 - \epsilon)^{-1} {dr \over ds}^{2} - r^{2} {d \theta \over ds}^{2} - r^{2} sin^{2}(\theta) {d \phi \over ds}^{2}$$​

Applying the Euler-Lagrange Equation:

$${\partial L \over \partial x^{\mu}} - {d \over ds}{\partial L \over \partial ({\partial x^{\mu} \over \partial s})} = 0$$​

w.h.t.:

$$\left( \begin{array}{c} 0 \\ c^{2} {\epsilon \over r}{\partial t \over \partial s}^{2} + (1 - \epsilon)^{-2} {\epsilon \over r}{\partial r \over \partial s}^{2} - 2 r {\partial \theta \over \partial s}^{2} - 2 r sin^{2}(\theta) {\partial \phi \over \partial s}^{2}\\ - r^{2} sin(2 \theta) {\partial \phi \over \partial s}^{2} \\ 0 \end{array} \right) - \left( \begin{array}{c} 2 c^{2} {\epsilon \over r} {\partial r \over \partial s}{\partial t \over \partial s} + 2 c^{2} (1 - \epsilon) {\partial^{2} t \over \partial s^{2}} \\ 2 (1 - \epsilon)^{-2} {\epsilon \over r}{\partial r \over \partial s}^{2} - 2 (1 - \epsilon)^{-1}{\partial^{2} r \over \partial s^{2}} \\ -4 r {\partial r \over \partial s}{\partial \theta \over \partial s} - 2 r^{2}{\partial^{2} \theta \over \partial s^{2}} \\ -2 r^{2} sin^{2}(\theta){\partial^{2} \phi \over \partial s^{2}} - 2 r^{2} sin(2 \theta) {\partial \theta \over \partial s}{\partial \phi \over \partial s} - 4 r {\partial r \over \partial s}{\partial \phi \over \partial s} \end{array} \right) = 0$$​

This seems to be substantially similar to the above-derived Geodesic Equation (so far).

Last edited: Jul 24, 2009
4. Jul 25, 2009

### Widdekind

UPDATE

4. Zero-Gravity limit (??)

If $$\epsilon = 0$$, w.h.t.:

$${\partial^{2} \over \partial s^{2}} \left( \begin{array}{c} c t \\ r \\ \theta \\ \phi \end{array} \right) + \left( \begin{array}{c} 0\\ - r {\partial \theta \over \partial s}^{2} - r \; sin^{2}(\theta) {\partial \phi \over \partial s}^{2} \\ {2 \over r}{\partial r \over \partial s} {\partial \theta \over \partial s} - {sin(2 \theta) \over 2} {\partial (\phi) \over \partial s}^{2} \\ {2 \over r}{\partial r \over \partial s} {\partial \phi \over \partial s} + 2 cot(\theta){\partial \theta \over \partial s} {\partial \phi \over \partial s} \end{array} \right) = 0$$

Further restricting $$\theta = {\pi \over 2}$$, w.h.t.:

$${\partial^{2} \over \partial s^{2}} \left( \begin{array}{c} c t \\ r \\ \theta \\ \phi \end{array} \right) + \left( \begin{array}{c} 0 \\ - r {\partial (\phi) \over \partial s}^{2} \\ 0 \\ {2 \over r}{\partial r \over \partial s} {\partial \phi \over \partial s} \end{array} \right) = 0$$

Is this the equation of a straight line in Polar Coordinates ?

5. Weak-Gravity limit (??)

If $$\epsilon << 1$$, w.h.t.:

$${\partial^{2} \over \partial s^{2}} \left( \begin{array}{c} c t \\ r \\ \theta \\ \phi \end{array} \right) + \left( \begin{array}{c} {\epsilon \over r} {\partial (ct) \over \partial s} {\partial r \over \partial s} \\ {1 \over 2} \left( {\epsilon \over r}{\partial (ct) \over \partial s}^{2} - {\epsilon \over r}{\partial r \over \partial s}^{2} - \left( 1 - \epsilon \right) 2 r {\partial \theta \over \partial s}^{2} - \left( 1 - \epsilon \right) 2 r \; sin^{2}(\theta) {\partial \phi \over \partial s}^{2} \right) \\ {2 \over r}{\partial r \over \partial s} {\partial \theta \over \partial s} + {sin(2 \theta) \over 2}{\partial \phi \over \partial s}^{2} \\ {2 \over r}{\partial r \over \partial s} {\partial \phi \over \partial s} + 2 cot(\theta){\partial \theta \over \partial s} {\partial \phi \over \partial s} \end{array} \right) = 0$$

Further restricting $$\theta = {\pi \over 2}$$, w.h.t.:

$${\partial^{2} \over \partial s^{2}} \left( \begin{array}{c} c t \\ r \\ \theta \\ \phi \end{array} \right) + \left( \begin{array}{c} {\epsilon \over r} {\partial (ct) \over \partial s} {\partial r \over \partial s} \\ {1 \over 2} \left( {\epsilon \over r}{\partial (ct) \over \partial s}^{2} - {\epsilon \over r}{\partial r \over \partial s}^{2} - \left( 1 - \epsilon \right) 2 r {\partial \phi \over \partial s}^{2} \right) \\ 0 \\ {2 \over r}{\partial r \over \partial s} {\partial \phi \over \partial s} \end{array} \right) = 0$$

Does this reduce to Newton's equations ?

5. Jul 26, 2009

### cristo

Staff Emeritus
There doesn't really appear to be a question here, which is probably why you haven't received any replies. If I were you, I would re-check your first calculation, and compare to well known results. (See, for example, http://arxiv.org/abs/0904.4184 for a useful catalogue of spacetimes).