Dodelson - Problem 2.2

Chapter 2 - The Smooth, Expanding Universe

Exercise 2
Show that the geodesic equation gets the correct equations of motion for a particle traveling freely in two dimensions using polar coordinates. You can get the Christoffel symbols one of two ways (or both!) and then proceed to (b).

(a) Get the Christoffel symbol either directly from the term in brackets in Eq. (2.17) or from the 2D metric $$g_{ij} = \left( \begin{array} {cc} 1&0 \\ 0&r^2\\ \end{array} \right)$$ using Eq. (2.19). Show that the only nonzero Christoffel symbols are $$\Gamma_{12}^{2} = \Gamma_{21}^{1} = \frac{1}{r} ; \Gamma_{22}^{1} = -r$$ with 1,2 corresponding to r,$\theta$.

(b) Write down the two components of the geodesic equation using these Christoffel symbols. Show that these give the proper equations of motion for a particle traveling in a plane.

Solution
(a) First note that $$g^{ij} = g_{ij}^{-1} = \left( \begin{array} {cc} 1&0 \\ 0&\frac{1}{r^2}\\ \end{array} \right)$$ Now we make use of Eq. (2.19): $$ \Gamma_{\alpha\beta}^{\mu} = \frac{g^{\mu\nu}}{2} \left[ \frac{\delta g_{\alpha\nu}}{\delta x^{\beta}} + \frac{\delta g_{\beta\nu}}{\delta x^{\alpha}} - \frac{\delta g_{\alpha\beta}}{\delta x^{\nu}} \right] $$ We have only two variables, r and $\theta$, labelled as 1 and 2 respectively so $\mu$ can only take on values of 1 and 2. First we consider the case $\mu = 1$. Then $$ \Gamma_{\alpha\beta}^{1} = \frac{g^{1\nu}}{2} \left[ \frac{\delta g_{\alpha\nu}}{\delta x^{\beta}} + \frac{\delta g_{\beta\nu}}{\delta x^{\alpha}} - \frac{\delta g_{\alpha\beta}}{\delta x^{\nu}} \right] $$ $$ = \frac{g^{11}}{2} \left[ \frac{\delta g_{\alpha 1}}{\delta x^{\beta}} + \frac{\delta g_{\beta 1}}{\delta x^{\alpha}} - \frac{\delta g_{\alpha\beta}}{\delta x^{1}} \right] $$ since $g^{12} = 0$. Also notice that the only non zero derivative is $\frac{\delta g_{22}}{\delta x^{1}}$ and so the above reduces to: $$ \Gamma_{22}^{1} = \frac{g^{11}}{2} \left[ - \frac{\delta g_{22}}{\delta x^{1}} \right] $$ $$ = - \frac{1}{2} 2r = -r$$ as required.

Following the same procedure as above except letting $\mu = 2$ we obtain $$ \Gamma_{\alpha\beta}^{2} = \frac{g^{22}}{2} \left[ \frac{\delta g_{\alpha 2}}{\delta x^{\beta}} + \frac{\delta g_{\beta 2}}{\delta x^{\alpha}} - \frac{\delta g_{\alpha\beta}}{\delta x^{2}} \right] $$ The third term in the bracket is always zero and we can only get non-zero values if either $\alpha = 1$ and $\beta = 2$ or $\alpha = 2$ and $\beta = 1$. In either case we will get $$ \Gamma_{12}^{2} = \Gamma_{21}^{2} = \frac{g^{22}}{2} \left[\frac{\delta g_{22}}{\delta x^{1}} \right] = \frac{1}{r}$$ again as we expected.

(b) Recall the geodesic equation is given by $$ \frac{d^2x^l}{dt^2} + \Gamma^{l}_{jk}\frac{dx^k}{dt}\frac{dx^j}{dt} = 0 $$ This time we have two choices for l, again because we have two independent variables. The first equation of motion is going to be for $l = 1$ and $x^1 = r$. $$ \frac{d^2r}{dt^2} + \Gamma^{1}_{jk}\frac{dx^k}{dt}\frac{dx^j}{dt} = 0$$ $$ \frac{d^2r}{dt^2} + \Gamma^{1}_{22}\frac{d\theta}{dt}\frac{d\theta}{dt} = 0$$ $$ \frac{d^2r}{dt^2} - r \left( \frac{d\theta}{dt} \right)^2 = 0$$ where we made use of the fact that $\Gamma^{1}_{22} = -r$ and $\Gamma^{1}_{12} = \Gamma^{1}_{21} = \Gamma^{1}_{11} = 0$. Thus we obtain the equation of motion for r: $$\ddot{r} - r\dot{\theta}^2 = 0$$ To obtain the second equation of motion we choose $l = 2$ from which $x^2 = \theta$ follows. Then we proceed as before to obtain: $$ \frac{d^2\theta}{dt^2} + \Gamma^{2}_{12}\frac{d\theta}{dt}\frac{dr}{dt} + \Gamma^{2}_{21}\frac{d\theta}{dt}\frac{dr}{dt} = 0$$ $$ \frac{d^2\theta}{dt^2} + \frac{2}{r}\frac{d\theta}{dt}\frac{dr}{dt} = 0$$ Thus we obtain the equation of motion for $\theta$: $$\ddot{\theta} + \frac{2}{r}\dot{\theta}\dot{r} = 0$$ The obtained equations are the equations of motion for a particle in a plane and our job is done.