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Find the Riemannian curvature tensor component

  1. Mar 29, 2017 #1
    Given the metric of the gravitational field of a central gravitational body:

    ds2 = -ev(r)dt2 + eμ(r)dr2 + r2 (dθ2 + sin2θdΦ2)

    And the Chritofell connection components:

    361c882fb4714b209f951c04bd9e9f1c.png

    Find the Riemannian curvature tensor component R0110 (which is non-zero).

    I believe the answer uses the Ricci tensor:

    Rμv = Rλμvλ = Γλμλ,v - Γλμv,λ + ΓρμλΓλ - ΓλμvΓρλρ

    This is far as I've got:

    R0110 = Γ010,1 - Γ011,0 + Γρ10Γ0 - Γ011Γρ

    R0110 = 0.5v'' + Γρ10Γ0 - Γ011Γρ

    I'm not entirely sure on the meaning of ρ, at first I thought it would cycle through 0, 1, 2 and 3 to represent the 4 dimensions but after working that through I found it didn't work. Any help is much appreciated!
     
  2. jcsd
  3. Mar 29, 2017 #2

    TSny

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    Hello.

    The summation convention implies that the right-hand side is summed over ##\lambda = 0, 1, 2, 3##. You cannot just set ##\lambda = 0## on the right-hand side (and ##\mu = 1##, ##\nu = 1##) in order to claim that ##R_{11} = R^0_{\; 110}##.
     
  4. Mar 29, 2017 #3
    Brilliant thank you! Am I right at least in using the Ricci tensor? If so, what's the significance of ρ?
     
  5. Mar 29, 2017 #4

    TSny

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    In the Ricci tensor, ##\rho## is a repeated index. So it plays the role of a summation index.

    I don't see how the Ricci tensor is going to be helpful for this problem. Instead, use the fundamental expression for ##R^\mu_{\; \nu \alpha \beta}## in terms of the Christoffel symbols.
     
  6. Mar 30, 2017 #5
    Rρλμν = - Γρλμ,ν + Γρλν,μ - ΓσλμΓρσν + ΓρλνΓρσμ

    R0110 = - Γ011,0 + Γ010,μ1 - Γσ11Γ0σ0 + Γσ10Γ0σ1

    R0110 = - (Γ011Γ000 + Γ111Γ010 + Γ211Γ020 + Γ311Γ030) + (Γ010Γ001110Γ011 + Γ210Γ021 + Γ310Γ031)

    R0110 = - (0 + 0.5v'*0.5μ' + 0 + 0) + ((0.5v')2 + 0 + 0 + 0)

    R0110 = (0.5v')2 - 0.25 ν' μ'

    I'm still unsure of what σ is, again, thank you for your help.
     
  7. Mar 30, 2017 #6

    TSny

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    You have a typographical error in the last term. Can you spot it?

    OK, except for the ##\mu## subscript in the second term on the right. Did you mean for that to be there? When going to the next line (3rd line in your post), you have dropped the first two terms of the second line. Although one of these terms is zero, the other one is not zero. Every thing else looks OK to me. So, I think your final result is correct except for the term that you dropped that is not zero.

    ##\sigma## is a summation index because is appears twice in a term; once as a superscript and once as a subscript. It looks to me that you handled ##\sigma## correctly.
     
  8. Mar 30, 2017 #7
    Great thank you! Your help is greatly appreciated!
     
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