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Clarification on Robertson-Walker Metric

  1. Jun 30, 2015 #1
    Here is the Robertson Walker metric:

    ds2= (cdt)2 - R2(t)[dr2/(1- kr2) + r2(dθ2 + sin2(θ)dΦ2)]

    This metric is seen and discussed in this link: http://burro.cwru.edu/Academics/Astr328/Notes/Metrics/metrics.html

    Now I am in the process of deriving the general relativistic mathematical objects for this metric such as the Christoffel symbols, Ricci tensor, etc... However, one thing is bothering me.

    As you can see both in the link and at the top of this post, they did not omit the c term using the c=1 convention in the first term of the metric. However, that scale factor R(t) only has t in it and not ct.

    This bothers me because I am on the fence about whether I should treat R(t) as a constant when deriving my Christoffel symbols or if I should treat it as a function of x0 and differentiate accordingly when deriving my Christoffel symbols. Note that x0 = ct , x1= r , x2=θ , x3 = Φ

    It is possible that they may be assuming that c=1 inside of the R(t) function and that is why they omit the c there, or it could just simply be that R(t) is not a function of x0 and I should just treat it as a constant when differentiating terms of my metric tensors.

    Which option is the correct choice?

    For those who need clarification on what I am asking, here is a numerical example:

    The metric tensor element g11 = -R2(t)/(1- kr2)

    While deriving the Christoffel symbols, one of the derivatives I will have to take is:
    ∂g11 /∂x0

    If I treat the term -R2(t) as a function of x0, then the above derivative would evaluate to be:

    -2R(t)R'(t)/(1- kr2) where R'(t) is simply the derivative of R(t) with respect to t.

    However, if I treat the term -R2(t) as a constant, then the derivative is 0.

    Which case is the correct case?
  2. jcsd
  3. Jun 30, 2015 #2


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    Gold Member

    I think your difficulty is in ##x^0=ct##. Just drop the ##c## there. ##R(t)## is a function of ##x^0##. If not you get no curvature as you say.
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