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Lie derivation of a metric

  1. Dec 13, 2012 #1
    1. The problem statement, all variables and given/known data

    I've searched everywhere, and I cannot find an example of calculation of Lie derivation of a metric.

    If I have some vector field [itex]\alpha[/itex], and a metric [itex]g[/itex], a lie derivative is (by definition, if I understood it):

    [itex]\mathcal{L}_\alpha g=\nabla_\mu \alpha_\nu+\nabla_\nu \alpha_\mu[/itex]

    So if my vector field is given in this form (polar coordinates for instance):

    [itex]\alpha=r\sin^2\theta \partial_t+r\partial_\varphi[/itex]

    (this is something I made up btw), so if I got this right (and I'm not sure, that's why I'm asking) I should find the Christoffel symbols from my metric, and use the definition of covariant derivative, and just calculate term by term (for [itex]\mu,\ \nu=t,\ r,\ \theta,\ \varphi[/itex])?

    In that case, is [itex]\alpha_t=r\sin^2\theta[/itex] ? And so on? Or did I missed the point entirely, because I'm at loss :\
     
  2. jcsd
  3. Dec 14, 2012 #2
    I think you more or less got it, although I agree it's pretty confusing to think about Lie derivatives of the metric. The definition of Lie derivative is I guess usually something like
    [tex] \mathcal{L}_\alpha(g_{\mu \nu}) = V^\lambda \nabla_\lambda g_{\mu \nu} + g_{\nu \lambda} \nabla_\mu \alpha^\lambda + g_{\mu \lambda} \nabla_\nu \alpha^\lambda [/tex] and using the fact that covariant derivative of the metric vanishes, you get the form [itex] \mathcal{L}_\alpha(g_{\mu \nu}) = \nabla_\mu \alpha_\nu + \nabla_\nu \alpha_\mu [/itex]
     
  4. Dec 14, 2012 #3
    The reason that is confusing me is that I have a vector field given in the form I've mentioned, and I have a nondiagonal metric, and I'm trying to get the same result that I found in one article, and I can't.

    Here: http://arxiv.org/abs/0908.0184 he makes a Lie derivative along a diffeomorphism [itex]\xi[/itex] of the metric, and he gets a 4x4 matrix. I tried my 'reasoning' and I got a scalar, so I must be doing something wrong :\
     
  5. Dec 14, 2012 #4
    I think the indices are the confusing part here -- you forgot them on the left side on your expression, but it's clear when you write it out, that the Lie derivative of a rank n tensor is also a rank n tensor. Not sure what you mean by your reasoning giving you a scalar. [itex] \nabla_\mu \alpha_\nu + \nabla_\nu \alpha_\mu [/itex] is very manifestly a tensor.
     
  6. Dec 15, 2012 #5
    Hmmm then I definitely did something wrong :\

    I found an example of Lie derivative of a metric of 3 sphere, along some vector... I'll try to follow that, and see that I'll get...
     
    Last edited: Dec 15, 2012
  7. Dec 15, 2012 #6
    Ok, I am in a bit of a problem here. I found one example of Lie derivative of a vector, along another vector on a sphere. And there was given the formula for each of the component. Here I would have (if I'm correct):

    [itex]\mathcal{L}_\xi g_{\mu \nu}=\nabla_\mu \xi_\nu+\nabla_nu\xi_\mu=\frac{\partial \xi_\nu}{\partial x^\mu}-\Gamma^\rho_{\mu\nu}\xi^\rho+\frac{\partial \xi_\mu}{\partial x^\nu}-\Gamma^\sigma_{\nu\mu}\xi^\sigma[/itex]

    right?

    So I say, that that new tensor I get by doing a Lie derivative, let's call it a, with components [itex]a_{\mu\nu}[/itex] is given by above formula, by putting all the possible combinations for [itex]\mu\nu[/itex] ([itex]\tau,\ r,\ \theta,\ \varphi[/itex])?

    EDIT:

    I tried with one component, and I didn't get the answer like in article :\ so I must be doing something wrong... again :(
     
    Last edited: Dec 15, 2012
  8. Dec 15, 2012 #7
    Found it!!! Finally! I did some digging and found a great book called Geometrical methods of
    mathematical physics by Bernard Shutz, and in it a proper way to calculate the Lie derivative of a metric, and I am getting good results ^^

    I'm so happy right now! Thanks for all the help :)
     
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