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Homework Help: Metric variation of the covariant derivative

  1. Oct 30, 2015 #1
    1. The problem statement, all variables and given/known data
    Hi all, I currently have a modified Einstein-Hilbert action, with extra terms coming from some vector field [itex]A_\mu = (A_0(t),0,0,0)[/itex], given by

    [itex]\mathcal{L}_A = -\frac{1}{2} \nabla _\mu A_\nu \nabla ^\mu A ^\nu +\frac{1}{2} R_{\mu \nu} A^\mu A^\nu [/itex].

    The resulting field equation has been given as [itex] R_{\mu \nu} - \frac{1}{2} R g_{\mu \nu} = 8 \pi G (T_{\mu \nu} + T^A _{\mu \nu}) [/itex] i.e. the same as Einstein's equation but with an extra energy-momentum term coming from [itex]A_\mu[/itex]. I'm trying to find the form of [itex]T^A _{\mu \nu}[/itex] and I only know that the density associated with it (I assume [itex]T^A _{00}[/itex]) is [itex]\rho_A = \frac{3}{2}H^2A_0^2+3HA_0 \dot{A}_0 - \frac{1}{2} \dot{A}^2 _0[/itex]. (The metric used is FRW [itex]ds^2 = dt^2 - a^2(t) \delta_{ij}dx^idx^j[/itex]).

    2. Relevant equations

    Searching around has found these (I think) useful relations: [itex]\frac{\delta g_{\mu \nu}}{\delta g^{\rho \sigma}} = -g_{\mu \rho}g_{\nu \sigma}[/itex], [itex]\frac{A^\mu}{\delta g^{\rho \sigma}} = 0[/itex] and [itex]\frac{A_\mu}{\delta g^{\rho \sigma}} = -g_{\mu \rho}A_\sigma[/itex]

    3. The attempt at a solution

    So my problem is getting a analytical equation for variations in the covariant derivative. I can rewrite [itex]R_{\mu \nu} A^\mu A^\nu [/itex] as [itex]\nabla _\mu A^\mu \nabla _\nu A^\nu - \nabla _\mu A^\nu \nabla _\nu A^\mu [/itex], so everything is in terms of covariant derivatives. The problem arises with variations in the Christoffel symbols. This gives me terms that are derivatives of [itex]\delta g_{\mu \nu}[/itex], which I'm not sure how to deal with (ultimately I'd want some expression all multiplied by [itex]\delta g^{\rho \sigma}[/itex]. In the original EH action for example the variations in the connections are in a total derivative so they aren't needed to be evaluated. I tried to see what would happen if I ignored variations in the connection and with [itex]T^A _{\mu \nu} = - 2 \frac{\delta \mathcal{L}}{\delta g^{\rho \sigma}} + g_{\rho \sigma}\mathcal{L} [/itex], which gave me some of the terms in the density, but ultimately I believe I'm missing extra terms. Ignoring the variations in the connection, I get

    [itex]T^A _{\rho \sigma} = \nabla_\rho A_\nu \nabla_\sigma A^\nu - g^{\mu \beta}\nabla_\mu A_\rho \nabla_\beta A_\sigma -\frac{1}{2} g_{\rho \sigma} g^{\alpha \mu} \nabla _\mu A_\nu \nabla _\alpha A ^\nu +\frac{1}{2} g_{\rho \sigma} R_{\mu \nu} A^\mu A^\nu [/itex].

    The [itex]R_{\mu \nu} A^\mu A^\nu [/itex] term gives me (finding the (00) term in T) a [itex]-\frac{3}{2}(\dot{H} + H^2)A_0^2[/itex] term. I also get that the first 2 terms in [itex]T_{00}[/itex], after using the substitution of connections in FRW, all cancel out or go to zero. While the third term gives me the [itex]-\frac{1}{2} \dot{A}^2 _0[/itex] and [itex]+\frac{3}{2}H^2A_0^2[/itex]. The problem is then I'm missing a few terms which I think come from me ignoring variations in the connection.

    Thanks for any help you might be able to give me.
  2. jcsd
  3. Nov 4, 2015 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
  4. Sep 23, 2016 #3
    If I am understanding you, you arrive at terms like ∂(δgμν) and you don't know how to factor it in terms of δgμν, is that right?

    If so, I recently arrived to the same problem through a different action. Sadly I haven't found an answer yet, so I'll be watching this space... Also, if I find an answer I'll post it here.

    By the way, would you mind letting me know where you found the relation

    Thanks in advance
  5. Sep 27, 2016 #4
    Please forget this last point, I got confused with δ and ∂ again...
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