Christoffel symbols of semilog metric

Question

Problem 17.3a in Moore's "A general relativity workbook" requires to determine the only non-zero Christoffel symbol for the 2D semilog $(p,q)$ coordinate system defined as $p=x$ and $q=e^{by}$.

What puzzles me is that I found more than one non-zero Christoffel symbol, and I wanted to ask you what am I doing wrong.

First of all, I calculated the metric $g_{\mu\nu}'$ of the coordinate system as $$ g_{\mu\nu}' =\frac{\partial x^\alpha}{\partial {x^\mu}'}\frac{\partial x^\beta}{\partial {x^\nu}'}g_{\alpha\beta} =\frac{\partial x}{\partial {x^\mu}'}\frac{\partial x}{\partial {x^\nu}'} + \frac{\partial y}{\partial {x^\mu}'}\frac{\partial y}{\partial {x^\nu}'} $$ given that $g_{\alpha\beta}$ for the 2D Euclidean plane is the identity matrix. Calculating this, I got $$ g_{\mu\nu}'= \begin{bmatrix} 1 & 0\\ 0 & \frac{1}{p^2 b^2} \end{bmatrix} $$ Is this correct?

Once this is done, I calculated the Christoffel symbols by comparing the geodesic equation expressed in terms of the metric elements $0=\frac{d}{d\tau}\left(g_{\mu\nu}\frac{dx^\nu}{d\tau}\right)-\frac{1}{2}\partial_\mu g_{\alpha\beta}\frac{dx^\alpha}{d\tau}\frac{dx^\beta}{d\tau}$ with the same equation expressed via the Christoffel symbols $0=\frac{d^2x^\mu}{d\tau^2}+\Gamma^\mu_{\alpha\beta}\frac{dx^\alpha}{d\tau}\frac{dx^\beta}{d\tau}$. If we set $\mu=p$ in the first equation we get $$ 0=\frac{d}{d\tau}\left(g_{p\nu}\frac{dx^\nu}{d\tau}\right)-\frac{1}{2}\frac{\partial g_{\alpha\beta}}{\partial p}\frac{dx^\alpha}{d\tau}\frac{dx^\beta}{d\tau}=\\ =\frac{d^2p}{d\tau^2} - \frac{1}{2}\frac{\partial g_{qq}}{\partial p}\left(\frac{dq}{d\tau}\right)^2=\\ =\frac{d^2p}{d\tau^2} + \frac{1}{b^2p^3}\left(\frac{dq}{d\tau}\right)^2 $$ Setting $\mu=p$ in the Christoffel version of the geodesic equation, we then can conclude that $\Gamma^p_{pp}=\Gamma^p_{pq}=\Gamma^p_{qp}=0$, while $\Gamma^p_{qq}=\frac{1}{b^2p^3}$. By doing the same procedure with $\mu=q$, I got that $\Gamma^q_{pp}=\Gamma^q_{qq}=0$ and that $\Gamma^q_{pq}=\Gamma^q_{qp}=-\frac{1}{p}$.

This means that three Christoffel symbols are non-zero, in opposition with the text of the exercise. Am I missing something, or is the text wrong?

Answer 1

As pointed out in the comments to your question, there is a mistake in the metric expression. I'll work it out in a slightly different manner just to show how it should look like at the end. Noticing that $x = p$ and $y = \frac{1}{b}\log(q)$, we see that $$\begin{align} \textrm{d}s^2 &= \textrm{d}x^2 + \textrm{d}y^2, \\ &= \textrm{d}p^2 + \frac{1}{b^2}\textrm{d}(\log{q})^2, \\ &= \textrm{d}p^2 + \frac{1}{b^2 q^2}\textrm{d}q^2. \end{align}$$ The chain rule manipulations I used to make this computation end up being exactly the terms you computed explicitly, so this is essentially the same calculation. Notice that we get a $\frac{1}{b^2 q^2}$ term, not a $\frac{1}{b^2 p^2}$ term.

From here, you should be able to obtain the correct expression by whichever method you prefer to compute the Christoffel symbols with. I'll sketch it in my favorite, which consists in using the fact that the geodesic equation is equivalent to the Euler–Lagrange equations given by the Lagrangian $$L = \frac{1}{2} g_{\mu\nu} \frac{\textrm{d} x^\mu}{\textrm{d} \tau}\frac{\textrm{d} x^\nu}{\textrm{d} \tau},$$ and this can be exploited to obtain the Christoffel symbols, as discussed in this answer and references therein.

The Lagrangian will be $$L = \frac{\dot{p}^2}{2} + \frac{\dot{q}^2}{b^2 q^2},$$ with the dots standing for derivatives with respect to your favorite affine parameter. The Euler–Lagrange equations are $$\ddot{p} = 0$$ and $$\ddot{q} - \frac{\dot{q}^2}{q} = 0,$$ from which we read that the only non-vanishing Christoffel symbol should be $$\Gamma^{q}{}_{qq} = - \frac{1}{2q}.$$