3

A metric on $\mathbb{R}^2$ is given by the form $dr^2+ f(r,\theta)d\theta ^{2}$ in polar coordinates. Let $\gamma(t)$ be a curve in $\mathbb{R}^2$ given by $\gamma(t) = (t,\theta_0)$ in polar coordinates where $\theta_0$ is a constant.

Find $\nabla_{\gamma'(t)}\gamma'(t)$.

I know that the given metric is not an induced metric, so you must find out the Christoffel symbols to calculate the covariant derivative. Also, the Euclidean metric on $\mathbb{R}^2$ is given by the form $dr^2+ r^2d\theta^2$ in polar coordinates.

Any hints or solutions are greatly appreciated.

1233dfv
  • 5,625
  • What a bunch of douchebags. Thank for closing this, you're all so helpful and full of knowledge that you want to spread. Pricks. – 1233dfv Apr 28 '15 at 18:08

1 Answers1

3

Now you have $g_{rr} = 1$ and $g_{\theta \theta} = f(r,\theta)$, otherwise zero, so you can start computing all Christoffel symbol, using the formula $$ \Gamma_{ij}^k = \frac{1}{2} \sum_l g^{kl}(g_{il,j} + g_{lj,i} - g_{ij,l}). $$ Here i,j,k are either "r" or "$\theta$".

In this coordinates system $(r,\theta)$, you have $ \gamma(t) = (r,\theta) = (a(t), b(t))$. By chain rule, we get $$ \gamma'(t) = a'(t) \partial_r + b'(t) \partial_\theta. $$

Therefore, $ \nabla_{\gamma'(t)} \gamma'(t) = \nabla_{\gamma'(t)} (a'(t) \partial_r + b'(t) \partial_\theta) $.

$ = a''(t) \partial_r + a'(t)^2 \nabla_{\partial_r} \partial_r + a'(t)b'(t) \nabla_{\partial_\theta} \partial_r + b''(t) \partial_\theta +a'(t)b'(t) \nabla_{\partial_r} \partial_\theta +b'(t)^2 \nabla_{\partial_\theta} \partial_\theta$

It is just an expansion of geodesic equations. You can put Christoffel symbols back here.

Tony Low
  • 201
  • This is very helpful, thank you. – 1233dfv Apr 28 '15 at 00:37
  • I'm not sure how to do the computation. Is there any ways you could show me how to compute this for for one of the Christoffel symbols? – 1233dfv Apr 28 '15 at 00:38
  • 1
    $$\Gamma_{rr}^\theta = \frac{1}{2} g^{\theta r} (g_{rr,r} + g_{rr,r} - g_{rr,r}) + \frac{1}{2} g^{\theta \theta} (g_{r\theta ,r} + g_{\theta r,r} - g_{rr, \theta}) . $$

    So it becomes $$0 + \frac{1}{2} (\frac{1}{f}) (0+ 0 -0) = 0. $$

     – Tony Low Apr 28 '15 at 21:51