1
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ 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 $\mu\neq\nu$). $\endgroup$ Commented Aug 12, 2016 at 13:59
  • $\begingroup$ 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? $\endgroup$
    – N0va
    Commented Aug 12, 2016 at 14:51
  • $\begingroup$ 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. $\endgroup$
    – N0va
    Commented Aug 12, 2016 at 15:04
  • $\begingroup$ 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. $\endgroup$
    – Marion
    Commented Aug 12, 2016 at 16:58

1 Answer 1

0
$\begingroup$

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 $\phi$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.