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Understanding curvature tensor equation

  1. Feb 6, 2012 #1
    hi,


    I am trying to understand the meaning of the following equation in the simplest way possible



    thanks in advance
     
    Last edited: Feb 6, 2012
  2. jcsd
  3. Feb 6, 2012 #2
    Your equation is incorrect because the free indices don't match. Anyway, the simplest way to think about it would probably be a matrix multiplication of the 1x4 row vector A with the 4x4 matrix g. I'm also not really sure what this has to do with the curvature tensor.
     
    Last edited: Feb 6, 2012
  4. Feb 6, 2012 #3

    HallsofIvy

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    Staff Emeritus
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    As elfmotat suggests, what you want is
    [tex]A^\sigma= g^{\sigma\mu}A_{\mu}[/tex]

    And, as he says, this is not directly related to the curvature tensor, it is simply the relation between the covariant and contravariant components of a tensor. In fact, it can be considered the definition of "contravariant components".
     
  5. Feb 6, 2012 #4
    Or it can be understood as: metric tensor gives us a particular linear map between cotangent and tangent spaces, the spaces of of forms and the spaces of tangent vectors. In this particular case the contravariant metric can be even degenerate, like it is the case for Newton-Galilei space-times. If it is nondegenerate, then we have an isomorphism, if it is degenerate, we gave only morphism (general linear map).
     
  6. Feb 6, 2012 #5
    thank you Elfmotat and HallsofIvy and arkajad

    from what I understand its a matrix that transform a 1xn row vector
    if Aμ is a (1x2 row vector) in the two dimensional surface of the sphere.
    how can I understand the following equation with a simple numerical example

    (gμμ) has a fixed valued? or it depends on Aμ

    thanks in advance
     
    Last edited: Feb 6, 2012
  7. Feb 6, 2012 #6
    One possible meaning can be: suppose the metric is diagonal. Then you look at the general formula:

    [tex]A^\mu= g^{\mu\sigma}A_{\sigma}[/tex]

    where you have to sum over sigma. But only one term is nonzero, where sigma=mu. This is your equation.

    Or, if you prefer: if the matrix is diagonal, then each component of the transformed (row) vector is proportional to the corresponding component of the column vector.
     
    Last edited: Feb 6, 2012
  8. Feb 6, 2012 #7
    Don't overload your indices. An index shouldn't appear more than twice in any term. Valid ways of writing your original equation would be:

    [itex]A^\mu =g^{\sigma \mu}A_{\sigma }[/itex]
    [itex]A^\rho =g^{\mu \rho}A_{\mu}[/itex]
    [itex]A^\nu =g^{\lambda \nu }A_{\lambda }[/itex]

    etc.

    Taking t=constant and r=constant, the metric of a two-sphere is:

    [itex]ds^2=r^2(d\theta ^2+sin^2\theta d\phi^2)[/itex]

    Written as a matrix:

    [tex]g_{\mu \sigma}=\begin{bmatrix}
    r^2 & 0\\
    0 & r^2sin^2\theta
    \end{bmatrix}[/tex]

    where μ,σ range from θ to ϕ.

    The inverse metric is therefore:

    [tex]g^{\mu \sigma}=\begin{bmatrix}
    r^{-2} & 0\\
    0 & r^{-2}csc^2\theta
    \end{bmatrix}[/tex]


    If the components of the one-form A are [itex]A_\mu=\begin{bmatrix}
    a & b
    \end{bmatrix}[/itex], then the components of the vector A are given by:

    [tex]A^{\sigma}=g^{\mu \sigma}A_{\mu}=
    \begin{bmatrix}
    a & b
    \end{bmatrix}
    \begin{bmatrix}
    r^{-2} & 0\\
    0 & r^{-2}csc^2\theta
    \end{bmatrix} =
    \begin{bmatrix}
    ar^{-2} & br^{-2}csc^2\theta
    \end{bmatrix}[/tex]

    The components of the (inverse) metric tensor are independent of the components of any particular one-form.
     
    Last edited by a moderator: May 5, 2017
  9. Feb 6, 2012 #8
    Well, here it appears. That means it is no longer free! It is arbitrary, but fixed.
     
  10. Feb 6, 2012 #9
    (gμμ) has a fixed value. It is mu-th diagonal element of the metric. It should not depend Aμ, thou mu in (gμμ) and in Aμ should be the same.
     
  11. Feb 6, 2012 #10

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    I feel compelled to point out that a tensor can be represented by a matrix, in a given coordinate system, in the same sense that a velocity vector can be represented as, say [itex]<v_x, v_y, v_z>[/itex]. But neither the tensor nor the vector is that representation.
     
    Last edited by a moderator: May 5, 2017
  12. Feb 6, 2012 #11
    It doesn't matter. First of all, gμμ doesn't really make sense. It's also not clear which index is being summed over (first or second) which would matter if the metric wasn't symmetric.
     
  13. Feb 8, 2012 #12
    thank you all for your help
     
  14. Feb 8, 2012 #13
    Except for the case of [itex]\mathbf{R}^n[/itex], where we often define vector as an n-tuple of numbers [itex](x_1,...,x_n)[/itex].
     
  15. Feb 9, 2012 #14
    hi elfmotat
    is this transformation needed in "Riemann curvature tensor" ?
     
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