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I Riemann Tensor

  1. Jul 14, 2014 #1
    Dear All,

    I am trying to calculate the Riemann tensor for a 4D sphere. In D'inverno's book, I have this equation [itex]R^{a}_{bcd}=\partial_{c}\Gamma^{a}_{bd}-\partial_{d}\Gamma^{a}_{bc}+\Gamma^{e}_{bd}\Gamma^{a}_{ec}-\Gamma^{e}_{bc}\Gamma^{a}_{ed}[/itex]

    But the exercise asks me to calculate [itex]R_{abcd}[/itex]. Do I lower it with a metric? if so, a metric with what indices?

    An lastly, in the above equation, there is an extra index e. Is it another dummy variable that I have to sum over with?

    :confused::cry: too much work
     
    Last edited: Jul 14, 2014
  2. jcsd
  3. Jul 14, 2014 #2

    haushofer

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    Yes, e is dummy. You should check your concention, but your metric has lower indices,

    R_{abcd}= g_{af}R^{f}_{bcd}

    The index f is again dummy.
     
  4. Jul 14, 2014 #3
    Thank you for your reply haushofer. Man this is going to take me some time. How many Riemann tensors should I get from this calculation? Or do I choose any order of indices for the tensor and work it out?

    Sorry If I am mixing up stuff.
     
  5. Jul 14, 2014 #4

    haushofer

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    Just one, with the indices shown. Work it out explicitly taking your time, and things hopefully get more clear :)
     
  6. Jul 14, 2014 #5

    haushofer

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    If you like long calculations, try supergravity next. It makes you longing for ordinary GR really soon :P
     
  7. Jul 14, 2014 #6
    :eek: maybe when I grow up :biggrin:

    Thanks for your help
     
  8. Jul 14, 2014 #7

    haushofer

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    Good luck, and take your time!
     
  9. Jul 14, 2014 #8
    So I am back and my Riemann tensor is zero for some reason :(

    Basically I have my 4D sphere (t,r,Θ,ø). And all of the Christoffle symbols I have calculate are zero (the ones that have the index t). Therefore, the Riemann tensor is zero and it should not be zero:frown:. Problem is that in the above formula for the Riemann tensor, the index a=t. And all of the Christoffle symbols contain a=t. My page contains a forest of zeros. What Am I doing wrong?

    help
     
  10. Jul 14, 2014 #9

    pervect

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    I highly recommend a symbolic algebra package of any sort. Maxima is one of the free ones.
     
  11. Jul 14, 2014 #10
    Thanks pervect. I will look for it.
     
  12. Jul 15, 2014 #11
    As an addition to post #8, I think I know where my mistake is, hopefully.

    [itex]ds^{2}=e^{\nu}dt^{2}-e^{\lambda}dr^{2}-r^{2}d\theta^{2}-r^{2}sin^{2}(\theta)d\varphi^{2}[/itex].

    I have been putting the [itex]dt[/itex] and [itex]dr[/itex] terms as 1 and -1 respectively in the metric tensor:blushing:. So when I took the derivatives when calculating the Christoffle symbols I got mostly zeros. Can this be the mistake?

    [itex] \nu[/itex] and [itex]\lambda[/itex] are functions of [itex] (t,r)[/itex]

    Should not the dt terms have c (m/s) multiplied with it? and what am I supposed to expect to get from the calculation? I have got no sources or refrences to check from. Just me and D'Inverno.

    Thanks all :redface:
     
  13. Jul 15, 2014 #12

    ChrisVer

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    As your metric is, you are supposed to get some nice algebra, but I don't think you should look for a reasonable answer. The reasonable answer appears by the time you can identify those [itex]\nu,\lambda[/itex] in some way. Why don't you try to warm yourself up by doing a calculation in smaller dimensions? Like fore example the 2 sphere....
    As for your questions:
    Obviously what you said you did, can be a mistake- partially. However you wouldn't have found the C.Symbols 0 even in this case.
    For example:
    [itex]\Gamma^{1}_{33} [/itex] I think shouldn't be zero (by setting what is in front of dr equal to -1)...

    [itex]\Gamma^{1}_{33} = \frac{1}{2} g^{11} (g_{31,3} + g_{31,3} - g_{33,1}) = \frac{1}{2} g_{33,1} = -\frac{1}{2}2r \sin^{2}(\theta)=-r \sin^{2}(\theta) [/itex]

    You don't really need to care about [itex]c[/itex] since you can always choose your units such that [itex]c=1[/itex]. In those units, length and time are equivalent.
     
    Last edited: Jul 15, 2014
  14. Jul 15, 2014 #13
    Thanks ChrisVer for the reply :smile:.

    By doing what I did (putting the dt term as 1 and dr term as -1 in the metric), I got the Christoffle symbols for a 3D sphere (6 unique symbols). But I have a 4D sphere line element in the problem. I will try again by putting the "e"'s in the metric instead of 1 and -1. This should give me another function when derived w.r.t r or t instead of zeros.

    This is an excercise in D'inverno's book page 90.

    (just saw you latex sorry, it wont affect this Christoffle symbol. It will effect the Christoffle symbols that have the index 0 or t)
     
    Last edited: Jul 15, 2014
  15. Jul 15, 2014 #14

    ChrisVer

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    I fixed what I wrote above... It's not exactly a 3D sphere, not until you reparametrize your dr.
    Nevertheless, make sure you get the Christoffel Symbols right, since I showed above that even in your case, there existed a non-zero christoffel symbol...
    The formula is:

    [itex] \Gamma^{\alpha}_{\mu \nu} = \frac{1}{2} g^{\alpha \rho} (\partial_{\nu} g_{\mu \rho}+\partial_{\mu} g_{\nu \rho}- \partial_{\rho} g_{\mu \nu})[/itex]

    [itex]\rho[/itex] is summed over... and be careful that [itex]g^{\mu \nu}[/itex] are the elements the inverse of the matrix with elements [itex]g_{\mu \nu}[/itex]... they are not the same matrices.
    in the case with the exponentials for example, [itex]g^{11} = -e^{-\lambda} [/itex]
     
    Last edited: Jul 15, 2014
  16. Jul 15, 2014 #15
    Could you please see the attached picture Chris?

    As you can see I got these symbols. Is it okay to derive the exponential to the power of lambda like in the picture?
    I am okay with calculating the symbols and such. My problem was getting zeros for the symbols that contain the index 0 or t. Edit: sorry this is how I derived e^lambda: [itex]y=e^{λ},
    y^{'}=λ^{'}e^{λ}[/itex]

    Again, thank you for taking your time to help me. I will try again as soon as crohns stops bothering me.
     

    Attached Files:

    Last edited: Jul 15, 2014
  17. Jul 15, 2014 #16

    ChrisVer

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    To avoid as many calculations as possible you just can use the symmetry of the Christoffel symbols. [itex]\Gamma^{\rho}_{\mu \nu}= \Gamma^{\rho}_{\nu \mu} [/itex].
    For someone one has to find the inverse of the metric, and start looking at:
    [itex]\Gamma^{0}_{\mu \nu} [/itex]
    [itex]\Gamma^{1}_{\mu \nu} [/itex]
    [itex]\Gamma^{2}_{\mu \nu} [/itex]
    [itex]\Gamma^{3}_{\mu \nu} [/itex]
    Keeping in mind that the metric is diagonal (so maybe you can find terms that are zero without needed to calculate them). Eg derivatives of [itex]g_{11}, g_{00}[/itex] with respect to 2,3 don't contribute anything. In fact the diagonal metric is what makes the above handy to minimize as possible the work needed.

    So for example:
    [itex]\Gamma^{3}_{\mu \nu}= \frac{1}{2} g^{33} (g_{\mu 3, \nu} + g_{\nu 3, \mu} - g_{\mu \nu,3} )[/itex]
    you don't have to care about [itex]\mu, \nu = 0,1,2 [/itex] they are identically zero... the first 2 terms because the metric is diagonal, and the last term because the elements [itex]g_{00}, g_{11},g_{22}[/itex] which could exist don't depend on [itex]\phi[/itex]. So you only need to take in account the:
    [itex]\Gamma^{3}_{3 \nu} [/itex] case ([itex]\mu=3[/itex])... and once you find it, you also have the [itex]\Gamma^{3}_{\mu 3} [/itex] because of symmetry- which would be the case to take [itex]\nu=3[/itex]. So all [itex]\Gamma^{3}_{...}[/itex]'s found.
     
    Last edited: Jul 15, 2014
  18. Jul 15, 2014 #17

    ChrisVer

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    In the same way as above, I can see that there exist a [itex]\Gamma^{0}[/itex] which is not zero.

    [itex]\Gamma^{0}_{11} \ne 0[/itex]

    [itex]\Gamma^{0}_{11} = -\frac{1}{2} g^{00} g_{11,0} [/itex] and [itex]g_{11}[/itex] depends on t.
     
  19. Jul 15, 2014 #18

    ChrisVer

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    yes that's the derivative of exp[lambda]
     
  20. Jul 15, 2014 #19
    That is if you substituted the e's in the metric which I foolishly, did not do:blushing:.

    Thanks for your help and time. Hopefully I will be able to do it now.
     
  21. Jul 15, 2014 #20

    ChrisVer

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    Your
    [itex]\Gamma^{r}_{\theta \theta} = \frac{1}{2} g^{rr} ( 2 g_{\theta r, \theta} - g_{\theta \theta,r})= -\frac{1}{2} g^{rr} g_{\theta \theta,r} = - e^{-\lambda} r [/itex]

    Well if you didn't have the exponentials, to which metric are you asking me to check the Christoffel Symbols you calculated? :P plus to this in general one with a program which can evaluate christoffel symbols can be more helpful - I can only check some of the C.S. by hand, not all that can arise (thanks the GR exam is passed for me).
    But if you are a fish in GR, you must do those calculations at least once by hand....
     
    Last edited: Jul 15, 2014
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