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Energy-momentum pseudotensor example problem

  1. Jul 7, 2012 #1
    1. The problem statement, all variables and given/known data
    I'm following the derivation of finding the energy flux of a gravitational wave propagating along the z-axis where they use the energy-momentum pseudotensor to achieve this, but I can't seem to get an answer that matches theirs.



    2. Relevant equations
    We are given a general gravitational plane wave where its wave vector k is given by [tex]k_{a}=(\omega/c,0,0,-\omega/c).[/tex] We then have a stress-energy pseudotensor given by
    [tex]t_{ab}=\frac{c^{4}}{64\pi G}k_{a}k_{b}A_{ij}^{TT}A^{ij\, TT}.[/tex] where the amplitude polarization tensor is [tex]A_{ij}^{TT}=\begin{bmatrix}A_{+} & A_{\times} & 0\\
    A_{\times} & -A_{+} & 0\\
    0 & 0 & 0
    \end{bmatrix}.[/tex]


    3. The attempt at a solution
    It seems to be just some basic algebraic manipulation that is needed but for some reason I can't get the answer of [tex]t_{ab}=\frac{c^{2}\omega^{2}}{32\pi G}(A_{+}^{2}+A_{\times}^{2})\begin{bmatrix}1 & 0 & 0 & -1\\
    0 & 0 & 0 & 0\\
    0 & 0 & 0 & 0\\
    -1 & 0 & 0 & 1
    \end{bmatrix}[/tex] which was given in the textbook. I tried some manipulation, but then could see nowhere to go next as all of my matrices ended up being zero. I started out by putting everything in where [tex]t_{ab}=\frac{c^{4}}{64\pi G}(\omega/c,0,0,-\omega/c)(\omega/c,0,0,-\omega/c)\begin{bmatrix}0 & 0 & 0 & 0\\
    0 & A_{+} & A_{\times} & 0\\
    0 & A_{\times} & -A_{+} & 0\\
    0 & 0 & 0 & 0
    \end{bmatrix}\begin{bmatrix}0 & 0 & 0 & 0\\
    0 & A_{+} & A_{\times} & 0\\
    0 & A_{\times} & -A_{+} & 0\\
    0 & 0 & 0 & 0
    \end{bmatrix}.[/tex] I then simplified to get [tex]t_{ab}=\frac{c^{2}\omega^{2}}{64\pi G}(1,0,0,-1)(1,0,0,-1)\begin{bmatrix}0 & 0 & 0 & 0\\
    0 & A_{+} & A_{\times} & 0\\
    0 & A_{\times} & -A_{+} & 0\\
    0 & 0 & 0 & 0
    \end{bmatrix}\begin{bmatrix}0 & 0 & 0 & 0\\
    0 & A_{+} & A_{\times} & 0\\
    0 & A_{\times} & -A_{+} & 0\\
    0 & 0 & 0 & 0
    \end{bmatrix}.[/tex] Once this was done though I found out that multiplying [tex](1,0,0,-1)\begin{bmatrix}0 & 0 & 0 & 0\\
    0 & A_{+} & A_{\times} & 0\\
    0 & A_{\times} & -A_{+} & 0\\
    0 & 0 & 0 & 0
    \end{bmatrix}[/tex] gave zero and that's essentially where I stopped. I know I must be making a really stupid mistake but I just can't see it. Any help would be appreciated, thanks guys.
     
  2. jcsd
  3. Jul 7, 2012 #2

    gabbagabbahey

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    The RHS of this equation looks more like the averaged stress-energy pseudotensor... is that what you are asked to find?
     
  4. Jul 8, 2012 #3
    Yup that's the one. :) We were already given a plane wave of the form [tex]h_{ij}^{TT}=A_{ij}^{TT}\cos\left(k_{a}x^{a}\right)[/tex] which when averaged in the energy-momentum pseudotensor came out to the form that you quoted.
     
  5. Jul 8, 2012 #4

    gabbagabbahey

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    Okay. Well what kind of quantity is [itex]A_{ij}A^{ij}[/itex]; tensor, vector or scalar? What kind of quantity is [itex]k_bA_{ij}[/itex]? Does it make sense to compute it as a vector times a matrix (which results in a vector)?

    Aside: Typically one uses Greek indices when they are to range over spatial and temporal dimensions, and Latin for spatial only.
     
  6. Jul 11, 2012 #5
    Well, [tex]A_{ij}^{TT}A^{ij\, TT}[/tex] are just two tensors in the transverse-traceless gauge (where i,j=1,2,3) and [tex]k_{a}[/tex] is just a vector which can be generalized to a (0,1) tensor (where a=0,1,2,3,4) which can then be generalized to a 1x4 matrix. I think it would be possible to multiply the amplitude matrices with the wave vectors because the final answer in the text book seems to look like that root was taken somewhere.
     
  7. Jul 11, 2012 #6

    gabbagabbahey

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    No, [itex]A_{ij}[/itex] and [itex]A^{ij}[/itex] are 2 tensors. The quantity [itex]A_{ij}A^{ij}[/itex] is, according to the Einstein summation convention, a scalar:

    [tex]A_{ij}A^{ij}\equiv \sum_{i,j}A_{ij}A^{ij} = A_{11}A^{11} + A_{12}A^{12} + \ldots + A_{33}A^{33}[/tex]

    Do you get a scalar if you naively multiply the matrix representations of [itex]A_{ij}[/itex] and [itex]A^{ij}[/itex] together using the usual matrix multiplication rules?

    Sure, [itex]k_a[/itex] is a vector, or a rank 1 tensor, but look at how many different indices you have in the quantity [itex]k_aA_{ij}[/itex]: 3 indices means this quantity is a rank 3 tensor (or pseudo-tensor, depending on how it transforms), which is not what you get when you naively multiply the matrix representation of [itex]k_a[/itex] with the matrix representation of [itex]A_{ij}[/itex].

    The point I was hoping that you would see from my previous questions is that when you are given something like

    [tex]A_{ij} =\begin{pmatrix} A_{+} & A_{\times} & 0 \\ A_{\times} & -A_{+} & 0 \\ 0 & 0 & 0 \end{pmatrix}[/tex]

    it does not mean that [itex]A_{ij}[/itex] is that matrix, only that [itex]A_{ij}[/itex] is the component in the [itex]i[/itex]th row and [itex]j[/itex]th column of that matrix. The same thing holds true for the expression [itex]k_a=(-\frac{\omega}{c}, 0 , 0, \frac{\omega}{c})[/itex]; that is [itex]k_a[/itex] is not that 1x4 matrix, rather it is the [itex]a[/itex]th component of it.

    You cannot just naively multiply the matrix representation of tensors together and expect to get a result that makes sense. You need to pay attention to the indices and what the equation is actually telling you to do

    Instead, try starting by calculating [itex]A_{ij}A^{ij}=A_{ij}g^{di}g^{ej}A_{de}[/itex]. You should end up with a nice scalar value of [itex]2(A_{+}^{(2)}+A_{\times}^{(2)})[/itex].
     
  8. Jul 11, 2012 #7
    Sometimes I just forget about the simplest of rules! I'll give that a try and see what I end up with, thanks very much.
     
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