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Another question from Schutz

  1. Oct 4, 2005 #1
    I got so much excellent help from my last question that I have decided to take advantage as much as I can. Here is another question.

    On page 202, I combine equations 8.18 and 8.19 and change bars for primes (to match eqn. 8.22)

    [tex]g_{\alpha' \beta'} = \eta_{\alpha' \beta'} + \Lambda^{\mu}_{\alpha'}\Lambda^{\nu}_{\beta'}h_{\mu \nu}[/tex]

    where [itex]\Lambda[/itex] is a boost. Here is eqn 8.21 with the non-linear terms deleted

    [tex]\Lambda^{\alpha}_{\beta'} = \delta^{\alpha}_{\beta} - \xi^{\alpha}_{,\beta}[/tex]

    Combining these I get:

    [tex]g_{\alpha' \beta'} = \eta_{\alpha' \beta'} + (\delta^{\mu}_{\alpha} - \xi^{\mu}_{,\alpha})(\delta^{\nu}_{\beta} - \xi^{\nu}_{,\beta})h_{\mu \nu}[/tex]

    using [itex]\eta_{\alpha' \beta'} = \eta_{\alpha \beta}[/itex], expanding the factors, and dropping the term quadratic in [itex]\xi[/itex] I get:

    [tex]g_{\alpha' \beta'} = \eta_{\alpha \beta} + \delta^{\mu}_{\alpha}\delta^{\nu}_{\beta}h_{\mu \nu} - \xi^{\mu}_{,\alpha}\delta^{\nu}_{\beta}h_{\mu \nu} - \delta^{\mu}_{\alpha}\xi^{\nu}_{,\beta}h_{\mu \nu}[/tex]

    [tex]g_{\alpha' \beta'} = \eta_{\alpha \beta} + h_{\alpha \beta} - \xi^{\mu}_{,\alpha}h_{\mu \beta} - \xi^{\nu}_{,\beta}h_{\alpha \nu}[/tex]

    Now, finally, comes my question:

    How can I use eqn 8.23 to simplify this to 8.22? It looks like Schutz is using h as if it were [itex]\eta[/itex]

    Here is 8.23

    [tex]\xi_{\alpha} = \eta_{\alpha \beta}\xi^{\beta}[/tex]

    Here is 8.22

    [tex]g_{\alpha' \beta'} = \eta_{\alpha \beta} + h_{\alpha \beta} - \xi_{\alpha,\beta} - \xi_{\beta,\alpha}[/tex]
    Last edited: Oct 5, 2005
  2. jcsd
  3. Oct 4, 2005 #2

    George Jones

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    I think you need to transform [itex]\eta[/itex] too, i.e.,

    [tex]g_{\alpha' \beta'} = (\delta^{\mu}_{\alpha} - \xi^{\mu}_{,\alpha})(\delta^{\nu}_{\beta} - \xi^{\nu}_{,\beta})(\eta_{\mu \nu} + h_{\mu \nu})

    After multiplying everything out and dropping terms that are "second order small", I get (8.22).

    Because of (8.13),what I call "second order small" includes terms like

    \xi^{\nu}_{,\beta}h_{\mu \nu}.

  4. Oct 4, 2005 #3


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    A specific example I recently worked might be of some interest
    Starting with the metric

    \left[ \begin {array}{cccc} 1-2\,{\it \Phi0} \left( {\it x1},{\it y1},{\it z1} \right) &0&0&0\\\noalign{\medskip}0&-1-2\,{\it \Phi0} \left( {
    \it x1},{\it y1},{\it z1} \right) &0&0\\\noalign{\medskip}0&0&-1-2\,{
    \it \Phi0} \left( {\it x1},{\it y1},{\it z1} \right) &0
    \\\noalign{\medskip}0&0&0&-1-2\,{\it \Phi0} \left( {\it x1},{\it y1},{
    \it z1} \right) \end {array} \right]

    when boosted in the 'x' direction by the substitutions for a Lorentz boost
    x = (x1+beta*t1)/sqrt(1-beta^2), t=(t1+beta*x1)/sqrt(1-beta^2)


    \left[ \begin {array}{cccc} 1+{\frac { \left( 2+2\,{\beta}^{2} \right) {\it \Phi0} \left( {\it x1},{\it y1},{\it z1} \right) }{-1+{
    \beta}^{2}}}&-4\,{\frac {\beta\,{\it \Phi0} \left( {\it x1},{\it y1},{
    \it z1} \right) }{-1+{\beta}^{2}}}&0&0\\\noalign{\medskip}-4\,{\frac {
    \beta\,{\it \Phi0} \left( {\it x1},{\it y1},{\it z1} \right) }{-1+{
    \beta}^{2}}}&{-1+\frac { \left( 2+2\,{\beta}^{2} \right) {\it '\Phi0}
    \left( {\it x1},{\it y1},{\it z1} \right) }{-1+{\beta}^{2}}}}&0&0\\\noalign{\medskip}0&0&-1-2\,{\it
    \Phi0} \left( {\it x1},{\it y1},{\it z1} \right) &0\\\noalign{\medskip}0
    &0&0&-1-2\,{\it \Phi0} \left( {\it x1},{\it y1},{\it z1} \right)
    \end {array} \right]

    of course, for a general transformation [itex]\Lambda[/itex], [itex]\eta_{uv}[/itex] might not turn out to be the same after the transform as it did for this example, you have to assume that the transform preserves the lengths of space-time intervals to make this happen.

    ps a minor Latex point - note that Lambda^a{}_b gives [tex]\Lambda^a{}_b[/tex], properly aligned, as opposed to Lambda^a_b which gives [tex]\Lambda^a_b[/tex]
  5. Oct 5, 2005 #4


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    What sort of transformation is [itex]\Lambda[/itex] supposed to be? It looks like my boost example, for instance, does not have [itex]\Lambda[/itex] in this form for large beta, because the diagonal coefficients are 1/sqrt(1-beta^2), not unity.
  6. Oct 5, 2005 #5
    A boost. I will edit my original post to indicate this.
  7. Oct 5, 2005 #6
    Thanks. As long as you guys are so willing to help, I intend to continue to take advantage. I have no access to a teacher or tutor.
  8. Oct 5, 2005 #7


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    OK, that's exactly what I did in my specific example. But note that for large values of beta, the term that maps x into x' is

    x' = x/sqrt(1-beta^2)

    which has a magnitude of 1/sqrt(1-beta^2), which can be very large, approaching infinity as beta->1.

    While your relation 8.21 requires

    [tex]\Lambda^{\alpha}{}_{\beta'} = \delta^{\alpha}{}_{\beta} - \xi^{\alpha}{}_{,\beta}[/tex]

    which means that you are assuming that the mapping from x to x' is unity, or close to it.

    Perhaps this is a Lorentz boost restricted to small beta, and that only the linear terms in beta are being kept (because beta is assumed to be <<1)?
  9. Oct 6, 2005 #8
    This is my fault. I left out that information (I assumed the reader had a copy of Schutz even though I knew that you don't). There is an extra condition:

    [tex]|\xi^{\alpha}{}_{,\beta}| \ll 1[/tex]

    That is why George called the product of this term with h, second order small.
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