# Finding the correct Christoffel-symbols in a 2+1D space-time

I'm trying to calculate the Christoffel Symbols in a 2+1D space-time with the following metric:

$$ds^2 = N^2(\vec r)c^2dt^2-\phi(\vec r)(dx^1)^2-\phi(\vec r)(dx^2)^2$$

To find the Christoffel ymbols I need to invert the metric tensor $g_{\mu\nu}$ to $g^{\mu\nu}$.
Am I correct in assuming that this last tensor only has non-zero elements on the diagonal and that these are the corresponding elements from the covariant metric tensor inverted? (i.e. $g^{00} = \frac{1}{g_{00}}$ etc.)

Because if that's right then I don't know how I'm supposed to find the correct Christoffel symbols.

For example, when calculating $\Gamma^{2}_{\hphantom{2}12}$ I get $\frac{1}{2(-\phi)}\frac{\partial (-\phi)}{\partial x^1}$ which apparently has 1 minus too much.

• Why do you think there is a minus too much? In case it's bothering you, Christoffel symbols of diagonal metrics can still have "non diagonal" entries (i.e. it's possible to have $\Gamma^\sigma_{\hphantom{2}\mu\nu}\neq 0$ for $\mu\neq\nu$). Commented Aug 12, 2016 at 13:59
• I get he same for $\Gamma^2_{12}$. Your inverse metric is correct. Are you sure that your source states a minus for $\Gamma^2_{12}$? $\Gamma^1_{22}$ is $-\frac{1}{2\phi}\frac{ \partial \phi}{ \partial x^1}$ if I am not mistaken. Are they using the standart convention for the indices?
– N0va
Commented Aug 12, 2016 at 14:51
• The Christoffel symbol of the first kind $\Gamma_{212}=-\frac{1}{2}\frac{\partial \phi}{\partial x^1}$ has a minus. But this one does not have the metric factor $g^{22}$ in it.
– N0va
Commented Aug 12, 2016 at 15:04
• Your metric does not have off-diagonal elements. Still, it is a nice exercise to assume the most general inverse metric, say, {{A,B},{C,D}} multiply it with your metric and insist that the result gives the unit matrix. Commented Aug 12, 2016 at 16:58

Taking the coordinate system to be $\{t,x,y\}$, with metric,
$$ds^2 = N(\vec r)^2 dt^2 - \phi(\vec r)(dx^2+ dy^2)$$
I presume $N(\vec r)$ to mean that $N =N(x,y)$, i.e. there is only a spatial dependence. Computing the Christoffel symbols is straightforward and simply requires applying the formula:
$\Gamma^t_{tx} = N^{-1}\partial_x N, \Gamma^t_{ty} = N^{-1}\partial_y N$. $\Gamma^x_{tt} = \frac{N}{\phi}\partial_x N, \Gamma^x_{xx} = - \Gamma^x_{yy} = \frac{1}{2\phi}\partial_x\phi.$ $\Gamma^{x}_{xy} = \frac{1}{2\phi}\partial_y \phi$. Then for $\Gamma^y_{ab}$ the matrix is the same, but with all derivatives w.r.t. $x \leftrightarrow y$. The scalar curvature of the space is,
$$R = \frac{1}{N\phi} \left( 2\nabla^2 N\right) - N\left[(\partial_i\phi)(\partial^i \phi) - \phi\nabla^2 \phi\right]$$
where the Laplacian is on $\mathbb R^2$ and $i=x,y$. All other curvature tensors are quite complicated in terms of $N$ and $\phi$.