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Covariant derivative in polar coordinates

  1. Aug 16, 2009 #1
    I calculated the christoffel symbols and know that I have them right. I want to take the covariant derivative of the basis vector field [tex]e_{r}[/tex] on the curve s(t) = (a, t/a). I differentiate it and get s' = (0, 1/a) and according to the metric, this is a unit vector because a will always be equal to r. Every way that I take the covariant derivative, [tex]\nabla_{s'(t)}e_{r}[/tex], yields [tex]\frac{1}{r^{2}}[/tex] for the [tex]e_{\theta}[/tex] component and 0 for the [tex]e_{r}[/tex] component.

    If this is correct, then I don't know how I'm supposed to interpret it. the [tex]e_{r}[/tex] component being 0 makes perfect sense to me but based on the formula for the circumfrence of a circle, I was expecting [tex]\frac{1}{r}[/tex] for the [tex]e_{\theta}[/tex] component.

    Am I making a mistake in my calculations somewhere? If i'm calculating the covariant derivative correctly, whats the geometric interpretation of the square in the denominator?

    Thanks for any replies.
    Last edited: Aug 16, 2009
  2. jcsd
  3. Aug 16, 2009 #2
    I forgot to give the christoffel symbols so you don't have to calculate them yourself. they are

    [tex]\Gamma_{r}_{\theta}^{\theta} = \Gamma_{\theta}_{r}^{\theta} = r^{-1} [/tex]

    [tex]\Gamma_{\theta}_{\theta}^{r} = -r[/tex]

    The other five are all 0.
  4. Aug 18, 2009 #3
    You appear to have mistaken a factor of [tex] a [/tex] for a factor of [tex] r [/tex]. Your Christoffels are all correct, so the correct calculation is as follows (let [tex] T = s'(t) [/tex]):

    \nabla_T \hat{\mathbf{e}}_{(r)}^a &= T^a \nabla_a \hat{\mathbf{e}}_{(r)}^a\\
    &= \frac{1}{a} \nabla_{\theta} \hat{\mathbf{e}}_{(r)}^a\\
    &= \frac{1}{a} \Gamma^a_{r \theta} \textrm{,}
    which gives [tex] 0 [/tex] for the [tex] r [/tex] component and [tex] \displaystyle \frac{1}{ar} [/tex] for the [tex] \theta [/tex] component.

    By the way, if you think you've made a mistake in a calculation like this, you can always revert to Cartesian coordinates to check your work, using the formulas

    \hat{\mathbf{e}}_{(r)} &= \frac{1}{\sqrt{x^2 + y^2}} \langle x, y \rangle\\
    \hat{\mathbf{e}}_{(\theta)} &= \frac{1}{\sqrt{x^2 + y^2}} \langle -y, x \rangle
    and remembering that covariant derivatives in Cartesian coordinates are the same as partials. In fact, you can (in principle) use this method of checking for any surface that admits an embedding into 3-space, or into any convenient 3-manifold of your choosing; in that case, the covariant derivative of the embedded surface is the projection of the covariant derivative associated with the ambient manifold onto the tangent space of the embedded surface.
  5. Aug 18, 2009 #4
    Thank you for your reply. I the curve in question was just a circle centered at the origin so [tex]a = r[/tex] and I was conflating the two once all the calculations were done. Sorry if that was confusing.

    I think that I figured out the geometric interpretation problem that I was having. I wanted it to be [tex]\frac{1}{r}[/tex] representing the fact that moving along a given length of the circle changes the angle in a way that is inversely proportional to the radius. To get that, I have to do

    [tex]\langle\nabla_{T}\hat{e}_{(r)}, T\rangle = g_{i}_{j}(\nabla_{T}\hat{e}_{(r)})^{i}T^{j} = r^{2}\frac{1}{ra}\frac{1}{a} = \frac{r}{a^{2}}[/tex]

    which just reduces to [tex]r^{-1}[/tex] like I wanted. That's what I integrate over [tex]S[/tex] to figure out how much [tex]\hat{e}_{(r)}[/tex] changes in going around some part of the circle and it's giving me all the answers that I want. I wish that I could prove it but I have no idea what I would prove. Does this make sense to anyone at least? Is [tex]\langle\nabla_{T}\hat{e}_{(r)}, T\rangle[/tex] important for anything other than as an integrand? I'm just trying to calibrate my geometric understanding of all of these numbers.

    EDIT: he he. That last question about what [tex]\langle\rangle[/tex] is good for sounds kinda stupid in retrospect. I guess it means the same thing as in linear algebra since it's in the tangent space.
    Last edited: Aug 19, 2009
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