Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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


    User Avatar
    Staff Emeritus
    Science Advisor

    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


    User Avatar
    Staff Emeritus
    Science Advisor

    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


    User Avatar
    Staff Emeritus
    Science Advisor

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook