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Riemann Tensor

  1. Feb 12, 2016 #1
    Using Ray D'Inverno's Introducing Einstein's Relativity. Ex 6.31 Pg 90.

    I am trying to calculate the purely covariant Riemann Tensor, Rabcd, for the metric

    gab=diag(ev,-eλ,-r2,-r2sin2θ)

    where v=v(t,r) and λ=λ(t,r).

    I have calculated the Christoffel Symbols and I am now attempting the solution using

    Rabcd=gae(∂cΓedb-∂dΓecbecfΓfdbedfΓfcb)

    With e and f as summation indices I have assumed that where the e and f occur in the same Christoffel symbol a summation of all Christoffel symbols, of all combinations of the two variables, should be summed.

    For R0101 I have the equation to be

    R0101=g00(∂0011111)-∂1001101)+(Γ000001100101)(Γ011111)-(Γ010110011111)(Γ001101)

    Yielding the solution

    R0101=(1/2)eλ(∂2λ/∂t2) - (1/4)eλ(∂λ/∂t)(∂v/∂t)-(1/4)eλ(∂λ/dt)2-(1/2)evv''-(1/4)ev(v')^2-(1/4)evv'λ'-(1/4)ev(λ'(∂v/∂t)-(∂λ/∂t)v')

    where ' represents ∂/∂r.

    The only term that is not in the solution given in the textbook is -(1/4)ev(λ'(∂v/∂t)-(∂λ/∂t)v') .

    The relevant Christoffel symbols for R0101 are

    Γ000=1/2 (∂v/dt) , Γ001=1/2 v'

    Γ011=(1/2)e(λ-v)(∂λ/∂t) , Γ100e(v-λ)v'

    Γ101=(1/2)(∂λ/∂t) , Γ111=(1/2)λ'


    I feel like I'm missing something rather simple as I have yet to come across a thread or example where the covariant Riemann Tensor has been calculated and the workings have been displayed. I may of course be using the wrong search terms for finding such a thing.
     
  2. jcsd
  3. Feb 12, 2016 #2

    andrewkirk

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    Can you fill in the steps that are covered by the words 'yielding the solution'?
    Your presentation of the formula for the Riemann tensor in terms of Christoffel symbols is correct, but to check the process by which you reach your solution, without your providing any steps, would require any helper to do the entire problem themself.
    If you can write down your steps, it shouldn't be too hard for somebody to find where it goes wrong - if indeed it does (textbook answers sometimes contain errors).
     
  4. Feb 13, 2016 #3
    Thank you Andrew.

    To expand on the 'Yielding the solution' part the steps in between are

    substituting the relevant Christoffel Symbols into

    R0101=g00(∂0(Γ011+Γ111)-∂1(Γ001+Γ101)+(Γ000+Γ001+Γ100+Γ101)(Γ011+Γ111)-(Γ010+Γ110+Γ011+Γ111)(Γ001+Γ101)

    =ev(∂t(½eλ-v(∂λ/∂t) + ½λ') - ∂r(½v'+½(∂λ/∂t) +(¼eλ-v(∂λ/∂t) + ½λ')(½(∂v/∂t) + ½(∂λ/∂t)) - (½v' + ½(∂λ/∂t)(½v' +½λ'))

    Multiplying out bracket

    R0101==ev(½eλ-v(∂2λ/∂t2) - ½eλ-v(∂λ/∂t)(∂v/∂t) + ½eλ-v(∂λ/∂t)2 + ½(∂λ'/∂t) -½v'' - ½(∂λ'/∂t) + ¼eλ-v(∂λ/∂t)(∂v/∂t) + ¼eλ-v(∂λ/∂t)2 + ¼(∂v/∂t)λ' + ¼λ'(∂λ/∂t) - ¼v'2 - ¼v'λ' - ¼(∂λ/∂t)v' - ¼(∂λ/∂t)λ')

    Then I collected like terms to get my final solution stated in the post.
     
  5. Feb 13, 2016 #4

    andrewkirk

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    It looks like the signs of the two components of the non-matching term in the OP are opposite from the signs those components have in post 3.

    It's a bit hard to be sure because the post is very hard to read since there is no latex used and no differentiation of bracket sizes to clarify where expressions for terms and factors begin and end.
    If you know latex, it would help greatly if you could post your formulas using that. The only difference between this forum's latex and the standard is that in-line formulas use two consecutive # symbols as delimiters for the beginning and end of code, rather than the usual single $.
     
  6. Feb 13, 2016 #5

    andrewkirk

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    Also, you have a ##\tfrac{1}{4}## coefficient in the following line in post 3:
    All numeric coefficients should be ½ prior to multiplying out brackets.
     
  7. Feb 14, 2016 #6
    I will try using latex once I can get to my laptop. :)
    Thank you for your help so far.
     
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