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

A Stress tensor in 3D Anti-De Sitter Space

  1. May 7, 2017 #1
    I am doing some mathematical exercises with 3D anti-de sitter face using the metric


    I found the three geodesics from the Christoffel symbols, and they seem to look correct to me.




    When I started calculating the Riemann and Ricci Tensor however things started to look hairy

    Rφrφr = -(1+r2)-1

    Rtφtφ = -(r+1/r)(r+r3)

    Rtrtr = -2+1/r2-(r+1/r)2

    I found the other components of the Riemann tensor to be 0 which may have been where I went wrong.

    This led me to a messy Ricci Tensor and Ricci Scalar

    Rrr= -2+1/r2-(r+1/r)2-(1+r2)-1

    Rφφ = -(r+1/r)(r+r3)

    R=guvRuv = Rrr(1+r2)+1/r2Rφφ


    This for some reason doesn't look right to me. It leads to a super complicated stress tensor as well.

    What did I do wrong here?
  2. jcsd
  3. May 7, 2017 #2

    Paul Colby

    User Avatar
    Gold Member

    I'm not sure one considers this "sporting" but the ctensor package of wxmaxima makes short work of this type of problem. I've run your case assuming that the intended metric is

    ##ds^2 = -(1+r^2)dt^2+\frac{1}{1+r^2}dr^2+r^2d\phi^2##​

    Looks based on this that your Christoffel symbols seem to agree but there is a typo in the time component geodesic equation

    ##\Gamma^t_{t r} = r+\frac{1}{r^2}##​

    yours seems to be ##r+\frac{1}{r}## which is likely just in typing your post.

    The Riemann components I get are,

    ##R^r_{t r t} = -(1+r^2)##
    ##R^\phi_{t \phi t} = -(1+r^2)##
    ##R^t_{r r t} = -\frac{1}{1+r^2}##
    ##R^\phi_{r \phi r} = \frac{1}{1+r^2}##
    ##R^t_{\phi \phi t} = -r^2##
    ##R^r_{\phi \phi r} = -r^2##​
  4. May 7, 2017 #3


    User Avatar
    Staff Emeritus
    Science Advisor

    I recommend using some sort of auotmated package at well. Maxima works and is free, though the ordering conventions are rather strange and it's a bit clunky to use. If there's interest, I could dig up Chris Hillman's file on how to use Maxima.

    Anyway, I get results similar to Pauls, though there seem to be some sign discrepancies, which I haven't tried to track down.

    $$g_{ab} = \left[ \begin {array}
    {ccc} -1-{r}^{2}&0&0\\0& \left( 1+{r}^{2} \right) ^{
    -1}&0\\0&0&{r}^{2}\end {array} \right]$$

    Geodesic equations:

    $$\frac{d^2 t}{d\tau^2} + 2 \frac{r}{1+r^2} \left( \frac{dt}{d\tau} \right) \left( \frac{dr}{d\tau} \right) = 0$$
    $$\frac{d^2 r}{d\tau^2} + r\left(1+r^2\right) \left( \frac{dt}{d\tau} \right)^2 - \frac{r}{1+r^2} \left( \frac{dr}{d\tau} \right)^2 - r\left(1+r^2\right) \left( \frac{d\phi}{d\tau} \right)^2 = 0$$
    $$\frac{d^2 \phi}{d\tau^2} + \frac{2}{r} \left( \frac{dr}{d\tau} \right) \left( \frac{d\phi}{d\tau} \right) = 0$$

    For the Riemann

    $$R^{\phi}{}_{r r \phi} = -R^t{}_{rtr} = \frac{1}{1+r^2} \quad R^t{}_{\phi t \phi} = R^r{}_{\phi r \phi} = -r^2 \quad R^r{}_{ttr} = R^\phi{}_{t t \phi} = -1-r^2 \quad $$

    Note that one needs to use the Bianci identies to get the complete set of nonzero Riemann components, for instance interchanging the last two symbols changes the sign, so that ##R^t{}_{rtr} = -R^t{}_{rrt}## - just one example of many omissions of nonzero components.
  5. May 7, 2017 #4

    Paul Colby

    User Avatar
    Gold Member

    Agreed, the index order is straight from mars but it has it's charms. wxmaxima is a workbook front end for maxima which I highly recommend. Maxima is a code written in the 60's in lisp (second in age to only fortran).

    I would be very interested in any notes you could dig up.

    I also have Mathematica which for reasons I can't quite fathom I find more obscure to use.
  6. May 7, 2017 #5


    User Avatar
    Science Advisor

    I'd also be interested in notes, @pervect. I tried to get into ctensor in maxima off the back of code in Ben Crowell's GR book, but didn't get very far.
  7. May 7, 2017 #6
    This helps immensely thanks! I must have made a typo in Riemann tensor calculation, but going back through it I got the same answers as Pervect.

    Using these tensors, how could I calculate particle path? If I had photon shot from r = 0 at t = 0, moving along a geodesic in the outward radial direction how do I calculate when would it reach r = ∞?
  8. May 8, 2017 #7


    User Avatar
    Staff Emeritus
    Science Advisor

  9. May 8, 2017 #8
    I was attempting to do this without using Maxima or the like, not to mention my coding is sub-par at best.

    I ended up finding using the metric and null geodesics to find that t(r)=tan-1(r), meaning for a photon leaving r=0 going to r=∞ it will reach ∞ at t=π/2, which is consistent with Anti-De Sitter space.

    Thanks for the help!
  10. May 8, 2017 #9


    User Avatar
    Staff Emeritus
    Science Advisor

    The geodesic equation will tell you that. In that case ##\tau## is not proper time, but an "affine parameter". Usually people use "s", but it doesn't make any difference. If you need more detail, ask.
  11. May 9, 2017 #10


    User Avatar
    Science Advisor

    Wow, thanks for that link! I just modified his first program to derive the Schwarzschild metric in Schwarzschild coordinates by messing around with the frame field, which was really neat! Something of a twofer - I learned a bit about Maxima and a bit about working in GR.

    If anyone is still in touch with Chris Hillman (I gather he took himself offline), tell him thanks from me.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted