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Question from Schutz, A First Course in GR

  1. Sep 27, 2005 #1
    Can someone help me understand something on page 220 of the book 'A First Course in General Relativity' by Bernard Schutz?
    Near the middle of the page, the line element is given as

    [itex]ds^2 = -dudv + f^2(u)dx^2 + h^2(u)dy^2[/itex]

    (I changed g to h so I can talk about the metric tensor) which I think is:

    [tex]
    \Large
    \[
    \mathbf{g}_{\alpha\beta} =
    \left( \begin{array}{cccc}
    0 & -\frac{1}{2} & 0 & 0 \\
    -\frac{1}{2} & 0 & 0 & 0 \\
    0 & 0 & f^2(u) & 0 \\
    0 & 0 & 0 & h^2(u)
    \end{array} \right) \]
    [/tex]

    Is this correct? If so, then:

    [tex]
    \Large
    \[
    \mathbf{g}^{\alpha\beta} =
    \left( \begin{array}{cccc}
    0 & -2 & 0 & 0 \\
    -2 & 0 & 0 & 0 \\
    0 & 0 & \frac{1}{f^2(u)} & 0 \\
    0 & 0 & 0 & \frac{1}{h^2(u)}
    \end{array} \right) \]
    [/tex]

    How am I doing so far?

    Then using equation 5.75 on page 143

    [itex]\Large \Gamma^{\gamma}_{\beta\mu} = \frac{1}{2}\mathbf{g}^{\alpha\gamma}(g_{\alpha\beta,\mu} + g_{\alpha\mu,\beta} - g_{\beta\mu,\alpha})[/itex]

    we have

    [itex]\Large \Gamma^{v}_{xx} = \frac{1}{2}\mathbf{g}^{\alpha v}(g_{\alpha x,x} + g_{\alpha x, x} - g_{xx,\alpha})[/itex]

    but the only value of [itex]\alpha[/itex] for which [itex]\mathbf{g}^{\alpha v}[/itex] is not zero is u. so

    [itex]\Large \Gamma^{v}_{xx} = \frac{1}{2}\mathbf{g}^{uv}(g_{ux,x} + g_{ux, x} - g_{xx,u}) = \frac{1}{2}(-2)(-2f\dot{f}) = 2f\dot{f}[/itex]

    but the book has:

    [itex]\Large \Gamma^{v}_{xx} = 2\dot{f}/f[/itex]

    Is the book wrong or am I?
     
  2. jcsd
  3. Sep 27, 2005 #2

    pervect

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    I don't have the book, so I can't tell if they might be doing something different than what you describe, but everything you typed here looks OK to me.
     
  4. Sep 28, 2005 #3
    Thank you for these encouraging words. However, I wouldn't have posted if I didn't think what I typed looked OK. I find it hard to believe that Schutz would have made such a mistake. I hope someone who has a copy will take a look and see if I haven't made some error.
     
  5. Sep 28, 2005 #4

    pervect

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    Sorry I can't be of more help. If it helps your confidence any, I did put your line element into GRTensorII and had it verify the value of [itex]\Gamma^v{}_{xx}[/itex].

    The only issue that I'm aware of is that the equations you are using do require you to be using a coordinate basis. Unless Schutz was talking about a basis of one forms somewhere in the text (possibly an orthonormal basis) in addition to giving you the line element, it seems unlikely that you would be using other than a coordinate basis.
     
    Last edited: Sep 28, 2005
  6. Sep 28, 2005 #5
    It finally dawned on me too.
     
  7. Sep 28, 2005 #6

    pervect

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    LOL, I see you read my .sig
     
  8. Sep 29, 2005 #7

    George Jones

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    Yesterday, I looked at the library's copy of Shutz, and I checked your calculation. Like pervect, I ended up with your result. Today, I looked at the library's copy of d'Inverno, which, on page 280, also confirms your result. d'Inverno also disagrees with Schutz's last connection coefficient, and with the the curvature components given by Schutz.

    Shutz and d'Inverno do agee on the vacuum field equation, though.

    My copies of Shutz and d'Inverno (and my other books) come out of storage in 2 or 3 weeks. :biggrin:

    Regards,
    George
     
  9. Sep 29, 2005 #8
    Thank both of you for spending so much time and effort on this. Emboldened by the knowledge that Schutz has the wrong value for the Riemann tensor components, here is my calculation using eqn. 6.67 on page 169.

    [tex]R^{\alpha}_{\beta \mu \nu} = \frac{1}{2}g^{\alpha \sigma}(g_{\sigma \nu,\beta \mu} - g_{\sigma \mu,\beta \nu} + g_{\beta \mu,\sigma \nu} - g_{\beta \nu,\sigma \mu})[/tex]

    so

    [tex]R^{x}_{uxu} = \frac{1}{2}g^{x \sigma}(g_{\sigma u,ux} - g_{\sigma x,uu} + g_{ux,\sigma u} - g_{uu,\sigma x})[/tex]

    Again, since the only value of [itex]\sigma[/itex] for which [itex]g^{x \sigma}[/itex] is not zero is x, so

    [tex]R^{x}_{uxu} = \frac{1}{2}g^{xx}(g_{xu,ux} - g_{xx,uu} + g_{ux,xu} - g_{uu,xx}) = \frac{1}{2f^2}(-2(\dot{f}^2 + f\ddot{f})) = -\frac{1}{f^2}(\dot{f}^2 + f\ddot{f}))[/tex]
    [tex] = -(\dot{f}/f)^2 - \ddot{f}/f[/tex]

    whereas the book has:

    [tex]R^{x}_{uxu} = -\ddot{f}/f[/tex]
     
    Last edited: Sep 30, 2005
  10. Sep 30, 2005 #9

    pervect

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    Sorry, but the computer says your textbook is right on this one

    [tex]
    R^x{}_{uxu}{} = -{\frac {{\frac {d^{2}}{d{u}^{2}}}f \left( u \right) }{f \left( u
    \right) }}[/tex]

    I'm not familiar with the particular formula you quoted, the formula I've seen (for a coordinate basis) is considerably more complicated:

    [tex]
    R^u{}_{vab} = \partial_a \Gamma^u{}_{vb} - \partial_b\Gamma^u{}_{va}+ \Gamma^u{}_{pa}\Gamma^p{}_{vb} - \Gamma^u{}_{pb}\Gamma^p{}_{va}
    [/tex]
     
  11. Sep 30, 2005 #10

    George Jones

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    Well, it's interesting stuff. Also, I have to be honest - I already had Schutz, d'Inverno, and a number of other relativity books out from the library for my own personal use. I hope I'll be able to take a closer look tomorrow.

    Very interesting! d'Inverno, using a different sign convention and interchanging x and z, and gets

    [tex]
    R_{0303} = ff''.
    [/tex]

    Schutz say that it is for a local inertial coodinate system about a point p, i.e., at p, g takes its special relativistic form, and the first derivatives of g are all zero. Unfortunately, it doesn't apply here, and (Schutz's version of) the formula Pervect gives must be used.

    BTW, I was able to use GRTensorII to do some calculations in 2+1 gravity. Thanks for the help Pervect.

    Regards,
    George
     
  12. Sep 30, 2005 #11
    This is a fine kettle of fish pervect. I used the eqn you provided (with different sub/superscripts and slightly different notation)

    [tex]
    R^{\alpha}_{\beta \gamma \delta} = \Gamma^{\alpha}_{\beta \delta,\gamma} - \Gamma^{\alpha}_{\beta \gamma,\delta} + \Gamma^{\alpha}_{\sigma \gamma}\Gamma^{\sigma}_{\beta \delta} - \Gamma^{\alpha}_{\sigma \delta}\Gamma^{\sigma}_{\beta \gamma}
    [/tex]

    and got:

    [tex]
    R^{x}_{uxu} = \Gamma^{x}_{uu,x} - \Gamma^{x}_{ux,u} + \Gamma^{x}_{\sigma x}\Gamma^{\sigma}_{uu} - \Gamma^{x}_{\sigma u}\Gamma^{\sigma}_{ux}
    [/tex]

    The first term on the right is zero, as is the second factor of the third term for all [itex]\sigma[/itex]. As for the fourth term, the only value of [itex]\sigma[/itex] for which this is not zero is x. So we have:

    [tex]
    R^{x}_{uxu} = - \Gamma^{x}_{ux,u} - \Gamma^{x}_{xu}\Gamma^{x}_{ux} = (\dot{f}/f),u - (\dot{f}/f)^2 = -\ddot{f}/f + (\dot{f}/f)^2 - (\dot{f}/f)^2 = -\ddot{f}/f
    [/tex]

    This is the same as Schutz, and the same as your computer (I assume you mean you ran GRTensorII again). But according to George, D'Inverno has something else. Can't anyone in this gang shoot straight?
     
  13. Sep 30, 2005 #12

    pervect

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    I'll just include GRTensorII's output for all the non-zero Christoffel symbols while I'm at it:

    [tex]\Gamma^x{}_{xu}=\Gamma^x{}_{ux}=
    {\frac {{\frac {d}{du}}f \left( u \right) }{f \left( u \right) }}\hspace{.25 in}
    \Gamma^y{}_{yu}=\Gamma^y{}_{uy}=
    {\frac {{\frac {d}{du}}h \left( u \right) }{h \left( u \right) }}\hspace{.25 in}
    \Gamma^v{}_{xx}=
    2\,f \left( u \right) {\frac {d}{du}}f \left( u \right) \hspace{.25 in}
    \Gamma^v{}_{yy}=
    2\,h \left( u \right) {\frac {d}{du}}h \left( u \right) \hspace{.25 in}
    [/tex]
     
  14. Oct 3, 2005 #13
    In a previous post, I used eqn. 6.67 from page 169 of Schutz. As George indicates, if I had looked closer at the text on that and the following page I would have realized that this is not coordinate invariant. Actually, the commas are a dead giveaway. The formulae that I have derived and now believe to be correct are the following:

    [tex]\Gamma^{x}_{xu} = \dot{f}/f[/tex]
    [tex]\Gamma^{y}_{yu} = \dot{h}/h[/tex]
    [tex]\Gamma^{v}_{xx} = 2f\dot{f}[/tex]
    [tex]\Gamma^{v}_{yy} = 2h\dot{h}[/tex]
    [tex]R^{x}_{uxu} = -\ddot{f}/f[/tex]
    [tex]R^{y}_{uyu} = -\ddot{h}/h[/tex]

    I went to the library to look at a copy of D'Inverno. Like Schutz, he does not have these equations exactly. In fact, as George points out, it looks like he has an equation for

    [tex]R_{uxux}[/tex]

    which may not be of interest here. I have not looked at his notational conventions yet.

    It seems that both Schutz and D'Inverno have errors (typos?), different ones, in their calculations. I have found typos elsewhere in both books.

    I am ready to forge on to calculate the vacuum field equation. I get the same answer as Schutz and D'Inverno as follows:

    [tex]R_{uu} = R^{\mu}_{u \mu u} = -(\ddot{f}/f + \ddot{h}/h)[/tex]
    [tex]R_{\alpha \beta} = 0[/tex] if [tex]\alpha \ne u[/tex] or [tex]\beta \ne u[/tex]

    [tex]R^{vv} = g^{\mu v}g^{\nu v}R_{\mu \nu} = g^{uv}g^{uv}R_{uu} = -4(\ddot{f}/f + \ddot{h}/h)[/tex]
    [tex]R^{\alpha \beta} = 0[/tex] if [tex]\alpha \ne v[/tex] or [tex]\beta \ne v[/tex]

    [tex]R = g^{\mu \nu}R_{\mu \nu} = g^{uu}R_{uu} = 0[/tex]

    [tex]G^{\alpha \beta} = R^{\alpha \beta} - \frac{1}{2}g^{\alpha \beta}R = 0[/tex]

    or

    [tex]\ddot{f}/f + \ddot{h}/h = 0[/tex]
     
  15. Oct 3, 2005 #14

    pervect

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    The symmetries of the Riemann make
    [tex]R_{abcd} = - R_{abdc}[/tex] always, and
    [tex]R_{abcd} = -R_{bacd}[/tex] whenever you have a metric (which is always in GR, perhaps not in some other applications).

    Thus [tex]R_{uxux} = R_{xuxu}[/tex]. Now [tex]R^x{}_{uxu} = g^{xo} R_{ouxu}[/tex], where you have to sum over o, but only [itex]g^{xx}[/itex] is nonzero.

    Thus [tex]R^x{}_{uxu} = g^{xx} R_{uxux}[/tex].

    I get the same results from GrtensorII.
     
  16. Oct 3, 2005 #15
    Thanks for your generous help in this matter. I forgot to apply the symmetries. Thanks also for confirming my calculations. I'm new at this and don't feel comfortable with it yet, but with your help I am gaining strength.
     
  17. Oct 4, 2005 #16

    George Jones

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    Which means that I was too hasty when I said "d'Inverno also disagrees with ..., and with the the curvature components given by Schutz.", i.e., Schutz and d'Inverno agree (up to a sign convention) about curvature since

    [tex]
    R^x{}_{uxu} = g^{xx} R_{uxux} = \frac{1}{-f^2} \left( ff'' \right).
    [/tex]

    There is a minus sign in the denominator since Schutz and dInverno use metrics with different signatures.

    It looks like Jimmy and d'Inverno have things right. Schutz has a typo/transcription error. Easy to make and easy to miss in the proofreadinding process.

    Regards,
    George
     
  18. Oct 4, 2005 #17
    Indeed!

    Extra text added to satisfy an unnecessary criterion.
     
  19. Oct 4, 2005 #18

    George Jones

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    LOL!

    I, too, was puzzled by this criterion when I first ran afoul of it.

    Regards,
    George
     
  20. Oct 4, 2005 #19
    jimmysnyder - Have you ever considered simply e-mailing Schutz? His e-mail address is - schutz@aei-potsdam.mpg.de

    pervect - I can scan that part of Schutz that jimmy is talking about and e-mail them to you if you'd like?

    Pete
     
  21. Oct 5, 2005 #20

    pervect

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    I suppose it would be good if someone told Schutz about his apparent typo, but since I don't own the book it's not going to be me.

    Did I just see a volunteer, Pete? :-).
     
    Last edited: Oct 5, 2005
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