## Tuesday 6 August 2019

### Question

 A diagonal metric in 4-space
Imagine we had a diagonal metric $g_{\mu\nu}$. Show that the Christoffel symbols are given by
\begin{align}
\Gamma_{\mu\nu}^\lambda&=0&\phantom {10000}(1)\nonumber\\
\Gamma_{\mu\mu}^\lambda&=-\frac{1}{2}\left(g_{\lambda\lambda}\right)^{-1}\partial_\lambda g_{\mu\mu}&\phantom {10000}(2)\nonumber\\
\Gamma_{\mu\lambda}^\lambda&=\partial_\mu\left(\ln{\sqrt{\left|g_{\lambda\lambda}\right|}}\right)&\phantom {10000}(3)\nonumber\\
\Gamma_{\lambda\lambda}^\lambda&=\partial_\lambda\left(\ln{\sqrt{\left|g_{\lambda\lambda}\right|}}\right)&\phantom {10000}(4)\nonumber
\end{align}In these expressions, $\mu\neq\nu\neq\lambda$ and repeated indices are not summed over.

### Answer

At last I have recovered from the grind of that geodesic on a sphere and completed this question. It involved a bit of index manipulation, some knowledge about the inverse of a diagonal matrix and the chain rule (taken slowly).

All three pages in Ex 3.03 Diagonal metric.pdf.

#### 2 comments:

1. There is an easy way to do this. Simply solve the Euler-Lagrange equations for a Lagrangian which takes a very simple form with a diagonal metric L = g_ii (dx^i/dT)^2 (it's a function of the lagrangian used to derive the geodesic equation, so we can do this).

For coordinate j this is: d/dT[dL/d(dx^j/dT)] = dL/dx^j

Then compare to the geodesic equation d^2x^j/dT^2 + Christoffel^j_kl = 0. Applying Leibniz rule to LHS you will get a second order derivative with a coefficient + some derivative of a metric times derivative of one coordinate, and on the right derivatives of the metric times square of a derivative of coordinates. Divide by the coefficient near the second order derivative, group on one side, and you get the first equation required from what was on the RHS. Then apply the chain rule on the remainder of what was on the LHS and you'll get the other two.

1. Done this way, it takes about 5 lines to get the answer.