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How to get components of Riemann by measuring geodesic deviation?

  1. Jul 4, 2012 #1
    Hi all, I'm now reading Chap 11 of Gravitation by Wheeler, etc.

    In exercise 11.7, by introducing Jacobi curvature tensor, which contains exactly the same information content as Riemann curvature tensor, we are asked to show that we can actually measure ALL components of Jacobi curvature tensor by measuring the geodesic deviation and thus we can measure all components of Riemann tensor because Riemann tensor can be expressed in terms of Jacobi tensor.

    The component of Jacobi tensor: [tex]J^\mu_{\nu\alpha\beta}=J^\mu_{\nu\beta\alpha}=1/2(R^\mu_{\alpha\nu\beta}+R^\mu_{\beta\nu\alpha})[/tex], where R is Riemann tensor
    The geodesic deviation equation in terms of Jacobi tensor is:

    [tex](n^\mu;_\alpha u^\alpha);_\beta u^\beta+J^\mu_{\alpha\beta\gamma}u^\alpha u^\beta n^\gamma =0[/tex]

    We are allowed to choose arbitrary vectors u and n, where u is the tangent vector of geodesic and n is the separation vector between fiducial geodesic and neighboring geodesic.

    In other words, by choosing u and n, we can directly measure the 'relative acceleration' of geodesics using say, test particles, which is the first term in the geodesic equation above. If we choose u and n to be some basis vectors, we surely can obtain corresponding components of Jacobi tensor directly. But these components are all of form of [tex] J^\mu_{\alpha\alpha\gamma}[/tex], i.e. having the same value on the second and the third index(this is because we do derivatives to separation vector n twice in the same direction, which is by definition the deviation of geodesic with tangent vector u ). For instance, if we choose n = (1,0,0,0) and u = (0,1,0,0), we get [tex]n^\mu;_0;_0+J^\mu_{001}=0[/tex], thus obtain [tex]J^\mu_{001=-n^\mu;_0;_0}[/tex] . If we want to calculate other components, we have to choose u to be the form like (1,1,0,0). But in this way, while [tex] J^\mu_{01\gamma}[/tex] appears in the equation, [tex] J^\mu_{10\gamma}[/tex] also shows up(this is because we do derivatives to separation vector n twice in the same direction, which is by definition the deviation of geodesic with tangent vector u ). Then we have two unknowns in one equation. How to solve this problem?

    (The rest 2 terms if of the form [tex] J^\mu_{00\gamma}[/tex] and [tex] J^\mu_{11\gamma}[/tex] which we already obtain directly by setting u to be basis vectors in 0 and 1 directions as indicated above)

    Hope I clearly state the problem.

    Thanks very much for any help.
     
    Last edited: Jul 4, 2012
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  3. Jul 13, 2012 #2
    no.

    You gone wrong in the part

    J[itex]^{\mu}_{\alpha\alpha\gamma}[/itex] part.If you do it like that you will have same index 4 times in an expression.

    (nμ;αuα);βuβ+Jμαβγuαuβnγ=0 this is correct. just think every u component =1 but keep u indices different. so you will can find every component of J by changing components of basis vectors.
     
  4. Jul 13, 2012 #3

    TSny

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    I think this is a way to do it (if I didn't make a mistake).

    Use J[itex]^{\mu}_{\alpha\beta\gamma}[/itex] = J[itex]^{\mu}_{\alpha\gamma\beta}[/itex] along with J[itex]^{\mu}_{(\alpha\beta\gamma)}[/itex] = 0 to show J[itex]^{\mu}_{\alpha\beta\gamma}[/itex] + J[itex]^{\mu}_{\beta\alpha\gamma}[/itex] = -J[itex]^{\mu}_{\gamma\alpha\beta}[/itex]

    Then check the following. Suppose you want the specific component J[itex]^{\mu}_{\nu\rho\tau}[/itex] where the indices are considered fixed chosen values. Choose the [itex]\nu[/itex] component of n to be 1 and the other three components of n zero. Choose the [itex]\rho[/itex] and [itex]\tau[/itex] components of u to be 1 and the other two components zero. I think you will find that

    J[itex]^{\mu}_{\alpha\beta\gamma}[/itex] u[itex]\alpha[/itex] u[itex]\beta[/itex] n[itex]\gamma[/itex] will reduce to J[itex]^{\mu}_{\rho\rho\nu}[/itex] + J[itex]^{\mu}_{\tau\tau\nu}[/itex] - J[itex]^{\mu}_{\nu\rho\tau}[/itex]

    Since you already know how to get the first two terms, you will be able to find J[itex]^{\mu}_{\nu\rho\tau}[/itex] from the geodesic deviation equation.
     
    Last edited: Jul 13, 2012
  5. Jul 13, 2012 #4
    the thing is that , in question it asks you to use the geodesic equation to find the components.then you do not need to use the identity of symmetric part of jacobian curvature.
    Using basis vectors in the geodesic equation doing just fine. he just made a mistake in indices.the second equation in his post was correct , just write for u[itex]^{\alpha}[/itex]=1 and u[itex]^{\beta}[/itex]=1 and keep indices of J as same. Then he is done.
     
  6. Jul 13, 2012 #5

    TSny

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    Hi, crimsonidol. I don't understand your comment. In the exprssion J[itex]^{\mu}_{\alpha\beta\gamma}[/itex]u[itex]\alpha[/itex]u[itex]\beta[/itex]n[itex]\gamma[/itex] you are required to sum over all values of [itex]\alpha[/itex] from 0 to 3 and similarly for [itex]\beta[/itex] and [itex]\gamma[/itex]. If you were to let all the components of u and all the components of n be equal to one, for example, then you would get a sum of 64 terms J[itex]^{\mu}_{\alpha\beta\gamma}[/itex] (where [itex]\alpha[/itex], [itex]\beta[/itex], and [itex]\gamma[/itex] would take on all values 0, 1, 2, 3).

    Maybe I'm misunderstanding your comment.
     
    Last edited: Jul 13, 2012
  7. Jul 14, 2012 #6
    Hi TSny, thanks very much, I think that's the kind of answer I'm looking for.
     
  8. Jul 14, 2012 #7
    Yeah Tsny you are right. I think i got confused.Sorry for the confusion. in Jacobian always two indices same if we choose u's as basis vectors in some direction.I was thinking something else probably , but when i read again I got it.

    But there is one thing confusing me. Should not we use a one form to get components of jacobian tensor.
    I suppose our equation should look like
    <w[itex]^{\mu}[/itex],[itex]\nabla_{u}[/itex][itex]\nabla_{u}[/itex]n>+J[itex]^{\mu}_{\alpha\alpha\gamma}[/itex]=0
    where u and n chosen as u=e[itex]_{\alpha}[/itex] n=e[itex]_{\gamma}[/itex]
     
  9. Jul 14, 2012 #8

    TSny

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    Yes, that looks correct to me. So, if you choose u and n as coordinate basis vectors, then you only get an expression for a component of J which has two subscripts that are identical. This is what was bothering shichao116. He was wondering how to get other components of J, such as those in which all three subscripts are different.

    Actually, I just now noticed that MTW define J such that the order of the subscripts is different than what shichao116 wrote. MTW ties the first subscript with n and the last two subscripts with u whereas shichao116 is tying the first and second subscripts with u and the last subscript with n. So, in your example, MTW notation would be J[itex]^{\mu}_{\gamma\alpha\alpha}[/itex]. But, that doesn't change the general thrust of shichao116's question.
     
  10. Jul 14, 2012 #9

    TSny

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    shichao116,

    As I noted in a reply to crimsonidol, MTW appear to be using a different order of subscripts for J than you did. If you take a look at equation (11.35) you can see that the vector n is contracted with the first subscript of J rather than the last subscript. But that's not a big deal.

    I believe I have another way to get any of the components of J. You already know how to get components of the form J[itex]^{\mu}_{\gamma\alpha\alpha}[/itex] (MTW's notation).

    Now let u and v be two independent vectors and let w = u + v. Consider the geodesic deviation equation for a geodesic in the direction of w: ∇wwn + J(w,w)n = 0. Replace w by u + v in J(w,w) and use linearity of J along with J(u,v)n = J(v,u)n to get ∇wwn + J(u,u)n + J(v,v)n + 2J(u,v)n= 0. We assume we have knowledge of the value of the geodesic deviation term ∇wwn, so we can use the equation to solve for J(u,v)n in terms of things we know. Since we can choose u and v arbitrarily, we can get all the components of J.
     
  11. Jul 14, 2012 #10
    Hi TSny, thanks for sharing. This seems to be a more elegant way, I like it :)
     
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