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Metric of polar coordinates

  1. May 27, 2015 #1
    Just started self teaching myself differential geometry and tried to find the christoffel symbols of the second kind for 2d polar coordinates. I am checking to see if I did everything correctly.

    With a line element of:
    mathtex.cgi?ds^2%20=%20dr^2%20+%20r^2%20d\theta^2.gif
    therefore the metric should be:
    nu}%20=%20\[%20\left(%20\begin{array}{cc}%201%20&%200%20\\%200%20&%20r^2%20\end{array}%20\right).gif
    The christoffel symbols of the second kind can be found by:
    tex.cgi?\Gamma^{m}_{ij}%20=%20\frac{1}{2}%20g^{mk}%20(%20g_{ki,j}%20+%20g_{kj,i}%20-%20g_{ij,k}).gif
    And the non-zero christoffel symbols I found:

    mathtex.cgi?\Gamma^{r}_{\theta\theta}%20=%20r.gif
    mathtex.cgi?\Gamma^{\theta}_{r\theta}%20=%20\Gamma^{\theta}_{\theta%20r}%20=%20\frac{1}{r}.gif

    I noticed since mathtex.cgi?g^{mk}.gif is symmetric it is non-zero when m=k so summing over k is not needed, i do not know if I missed anything by doing this.
     
    Last edited by a moderator: Apr 19, 2017
  2. jcsd
  3. May 27, 2015 #2

    PeterDonis

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    You didn't miss any non-zero christoffel symbols, but you need to check the sign of ##\Gamma^r_{\theta \theta}##.
     
  4. May 27, 2015 #3
    Ahh, I see was just looking at christoffel equation while doing derivatives in head and forgot to carry the negative. Thank you.
     
  5. May 27, 2015 #4

    Orodruin

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    Just a quick note, there are easier ways of getting the Christoffel symbols than inserting the metric into the expression from the Levi-Civita connection.
     
  6. May 27, 2015 #5

    PAllen

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    I'm curious - whenever I do one by hand, I do as the OP did. What are these easier ways?
     
  7. May 27, 2015 #6

    Orodruin

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    I would start by deriving the geodesic equations for the metric and identify the Christoffel symbols from there. Perhaps you do not find that simpler?

    Edit: In the case of polar coordinates, minimise
    $$
    S = \int (r^2 \dot \phi^2 + \dot r^2) d\tau
    $$
    Euler-Lagrange equations give:
    \begin{eqnarray}
    \ddot r - r \dot\phi^2 &=& 0 \\
    \ddot \phi + 2\frac{1}{r} \dot r \dot \phi &=& 0
    \end{eqnarray}
    Identification with ##\ddot x^\mu + \Gamma^\mu_{\nu\rho} \dot x^\nu \dot x^\rho = 0## along with the zero torsion of the Levi-Civita connection gives
    $$
    \Gamma^r_{\phi\phi} = - r, \quad \Gamma^\phi_{r\phi} = \Gamma^\phi_{\phi r} = \frac 1r.
    $$
     
    Last edited: May 27, 2015
  8. May 27, 2015 #7

    HallsofIvy

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    Good job! Now, go on to the sphere!

    Actually, a bit harder would be for an ellipse or the surface of an ellipsoid.
     
    Last edited by a moderator: Apr 28, 2017
  9. May 27, 2015 #8

    PAllen

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    No, I just never thought of doing it that way. Since Euler-Lagrange are used to derive the general geodesic equation with Christoffel symbols, it really never occurred to me that for simple metrics their direct application would be simpler than the general definition.
     
  10. May 27, 2015 #9

    Orodruin

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    To be honest I think it is even more simplifying for worse metrics (in particular off-diagonal ones), where otherwise you would have a large number of terms and partial derivatives of the metric to compute and keep track of. I have not done the counting, but I do believe that this approach significantly reduces the amount of necessary book keeping. Although you will still need to compute the same derivatives, it somehow feels lighter that you do not have to write them down more than once and the book keeping of knowing which derivative goes where is handled automagically.
     
  11. May 27, 2015 #10

    PAllen

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    That makes a lot of sense. For transforming the metric from one set of coordinates to another, I basically never use the Jacobian definition; I substitute in the physicists metric short hand. It just never occurred to me that using Euler-Lagrange would avoid lots of bookkeeping for computation of Christoffel symbols. A years PF contribution is now paid for in saved computation.
     
  12. May 27, 2015 #11
    For spherical coordinates this is what I found:
    ##ds^2 = dr^2 + r^2 d\theta^2 + r^2 sin\theta^2 d\phi^2##

    ##\Gamma^r_{\theta \theta} = -r \\
    \Gamma^r_{\phi \phi} = -rsin^2 \theta \\
    \Gamma^\theta_{r \theta} = \Gamma^\theta_{\theta r} = \frac{1}{r} \\
    \Gamma^\theta_{\phi \phi} = -sin\theta cos\theta \\
    \Gamma^\phi_{r \phi} = \Gamma^\phi_{\phi r} = \frac{1}{r} \\
    \Gamma^\phi_{\theta \phi} = \Gamma^\phi_{\phi \theta} = cot\theta ##
     
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