Category: Martin’s General Relativity: A First Course For Physicists

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201903030526 Problem 6.1

The covariant derivative of v^\mu is defined:

v^\mu_{;\alpha}=\partial_\alpha v^\mu +\Gamma^\mu_{\alpha\beta}v^\beta.

The Christoffel symbols are all zeros for Cartesian coordinates in the plane. For the metric tensor of Cartesian coordinates being the 2\times 2 identity matrix, and all its elements constants, the Christoffel symbols of the first kind as well as the second should be zero:

\begin{aligned} V^x_{;x}&=\partial_xV^x\\ V^y_{;x}&=\partial_xV^y\\ V^x_{;y}&=\partial_yV^x\\ V^y_{;y}&=\partial_yV^y \end{aligned}

The metric tensor for polar coordinates in the plane is derived below:

\mathbf{r}=(r\sin\theta ,r\cos\theta)

From

\begin{aligned} \hat{\mathbf{e}}_r&=\frac{\partial}{\partial r}(\mathbf{r})=(\sin\theta ,\cos\theta) \\ \hat{\mathbf{e}}_\theta &=\frac{\partial}{\partial \theta}(\mathbf{r})=(r\cos\theta ,-r\sin\theta ) \end{aligned},

and hence

\begin{aligned} \hat{\mathbf{e}}_r\cdot \hat{\mathbf{e}}_r&=\sin^2\theta +\cos^2\theta =1\\ \hat{\mathbf{e}}_\theta\cdot \hat{\mathbf{e}}_\theta &=r^2\cos^2\theta +(-r)^2\sin^2\theta =r^2 \\ \hat{\mathbf{e}}_r\cdot \hat{\mathbf{e}}_\theta &=\hat{\mathbf{e}}_\theta\cdot \hat{\mathbf{e}}_r=r\sin\theta\cos\theta -r\sin\theta\cos\theta =0 \end{aligned},

it follows that the metric tensor \mathbf{G}=(g_{ij}) is

\begin{bmatrix} 1 & 0 \\ 0 & r^2 \end{bmatrix}.

Since only the (2,2)^{\textrm{th}} element is a non-constant dependent on r, I expect that some of the Christoffel symbols will be zeros.

Two formulae for Christoffel symbols of the second kind:

\begin{aligned} \Gamma^\alpha_{\alpha\beta}&(=\Gamma^\alpha_{\beta\alpha})=\displaystyle{\frac{\partial}{\partial x^\beta}\bigg( \frac{1}{2}\ln |g_{\alpha\alpha}| \bigg)}\\ \Gamma^\alpha_{\beta\beta}&=\displaystyle{-\frac{1}{2g_{\alpha\alpha}}\frac{\partial}{\partial x^\alpha}(g_{\beta\beta})}\qquad\quad (\alpha\neq\beta) \end{aligned}

Because the (1,1)^\textrm{th}, (1,2)^\textrm{th}, and (2,1)^\textrm{th} elements are constants, referring to the preceding formulae I can preclude these terms 0= \Gamma^1_{11}=\Gamma^1_{12}= \Gamma^1_{21}=\Gamma^2_{11}=\Gamma^2_{22} and consider only the terms \Gamma^1_{22}, \Gamma^2_{12}, and \Gamma^2_{21}.

Neatly written,

\begin{aligned} \Gamma^1_{22}&(=\Gamma^r_{\theta\theta})=\displaystyle{-\frac{1}{2(1)}\frac{\partial }{\partial r}(r^2)}=-r\\ \Gamma^2_{12}&(=\Gamma^\theta_{r\theta})=\displaystyle{\frac{\partial}{\partial r}}\bigg( \displaystyle{\frac{1}{2}\ln r^2}\bigg) =\displaystyle{\frac{1}{r}} \\ \Gamma^2_{21}&(=\Gamma^\theta_{\theta r}) =\displaystyle{\frac{1}{r}} \end{aligned}

By the formula for covariant differentiation

v^\mu_{;\alpha}=\partial_\alpha v^\mu +\Gamma^\mu_{\alpha\beta}v^\beta

the following can be obtained:

\begin{aligned} V^r_{;r}&=\partial_rV^r,\\ V^r_{;\theta}&=\partial_\theta V^r-rV^\theta ,\\ V^\theta_{;r}&=\partial_rV^\theta +\frac{1}{r}V^\theta ,\\ V^\theta_{;\theta}&=\partial_\theta V^\theta +\frac{1}{r}V^r \end{aligned}.