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Ricci components in Schwarzschild solution

  1. Jun 22, 2005 #1
    Attached is a word document where I am setting out the Ricci tensor components to solve for the terms in the line element equation and get to the Schwarzschild solution. I am overlooking something as I have a sign off in the Ricci components in comparison to the many texts and literature I have referenced on this. Hopefully someone, if they have Word, can point out my error. If you do not have Word but are interested, please post a response and format you like, and I will change it. Thanks in advance.
     

    Attached Files:

  2. jcsd
  3. Jun 23, 2005 #2

    pervect

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    I tossed your metric into GRTensorII, and it spits out

    [tex]
    R_{tt} =
    -1/4\,{\frac { \left( {\frac {d}{dr}}A \left( r \right) \right)
    \left( {\frac {d}{dr}}B \left( r \right) \right) A \left( r \right)
    r-2\, \left( {\frac {d^{2}}{d{r}^{2}}}A \left( r \right) \right) B
    \left( r \right) A \left( r \right) r+ \left( {\frac {d}{dr}}A
    \left( r \right) \right) ^{2}B \left( r \right) r-4\, \left( {\frac
    {d}{dr}}A \left( r \right) \right) B \left( r \right) A \left( r
    \right) }{ \left( B \left( r \right) \right) ^{2}A \left( r \right)
    r}}
    [/tex]

    for whatever it's worth (it doesn't seem to match either one of your expressions).

    The Christoffel symbols all looked OK
     
  4. Jun 23, 2005 #3

    dextercioby

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    Why didn't you choose the more famous Ansatz

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

    It leads to nicer ODE-s.You can follow and check your calculations with Dirac's book [1].

    Daniel.

    [1]P.A.M.Dirac,"General Relativity",1975.
     
  5. Jun 23, 2005 #4
    Schwarzschild Ricci components

    Thanks for the responses. I have gone through every kind of starting point
    Ansatz, et al, but come to the same point of one term in the end being off by a sign. I attached another Word document with an abbreviated version to better identify my problem. Unfortuneately I am not around academia and am just reading GR on my own and do not have anyone to help with what seems to be a simple misunderstanding of the Ricci tensor signage. Once again thanks in advance.
     

    Attached Files:

  6. Jun 23, 2005 #5

    robphy

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    Just quickly skimming over your red equations... is there a [relative] sign-error in distributing the minus signs? In the brackets, for example, shouldn't the u^2 and uv terms have different signs (since u and v do in the preceding equation)?

    What are the definitions/sign-conventions/signature-conventions of the text you are comparing with?
     
  7. Jun 23, 2005 #6

    pervect

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    I think your expression for the Ricci is wrong.

    MTW "Gravitation" pg 340 gives the Riemann

    [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]

    where [tex]\partial_a = \frac{\partial}{\partial{x^a}} [/tex] in the original, and I've also replaced greek with latin.

    The Ricci is just

    [tex]
    R_{uv} = R^x{}_{uxv}
    [/tex]

    I don't think this set of expressions is equivalent because from the first term (the only one with a single positive Christoffel symbol), I derive the mapping from my letters to yours

    a->v
    u->sigma
    v->u
    b->sigma

    and substituting this into the left hand side, you are calculating [tex]R^\sigma{}_{uv\sigma}[/tex], which is not the Ricci, the Ricci should be [tex]R^\sigma{}_{u{\sigma}v}[/tex]
     
  8. Jun 23, 2005 #7

    George Jones

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    Either just inside the front cover, or just inside the the back cover, of MTW, there is a table of sign conventions. Unfortunately, I don't have my copy of MTW handy :frown: - does MTW list different sign conventions for Ricci?

    Regards,
    George
     
  9. Jun 23, 2005 #8

    pervect

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    There are large arrows pointing to the plus signs on the lefthand sides of the following two expressions

    [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]
    [tex]
    +R_{uv} = R^x{}_{uxv}
    [/tex]

    I interpret this as suggesting that some authors might put a minus sign where the plus sign is.
     
  10. Jun 23, 2005 #9
    Thanks,

    pervect is correct here. I lost the proper notation somewhere going back and forth between all the literature. Thanks for the MTW reference George. I bought mine when I was 13 years old as I thought it was neat and I might one day understand it. 31 years later that is starting to happen.

    Rick
     
  11. Jun 28, 2005 #10
    I'm working this out for myself (https://www.physicsforums.com/showthread.php?t=80565) and I ended up with all the same Christoffel symbols except for one:

    [tex]\Gamma^1{}_{4 4} = \frac{-A'}{2B}[/tex] (Note: If you're looking at my thread, I defined my A and B in reverse. So your A is my B and vice versa)

    Our answers for this symbol differ by a negative sign. I've looked back on it again and I think it should have a negative sign. Is this the "proper notation" you may have lost?


    (By the way... I'm an undergrad in physics, and haven't heard of this MTW. What is it exactly? It seems to be an equations manual of some sort. It sounds really useful, so would you recommend I obtain one?)
     
    Last edited: Jun 28, 2005
  12. Jun 28, 2005 #11

    dextercioby

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    MTW is also called "the telephone book" and is

    Ch.Misner,K.Thorne,J.A.Wheeler,"Gravitation",1972.

    Daniel.
     
  13. Jun 30, 2005 #12
    I had a chance to recalulate my Ricci components in light of pervect's input, which I saw would change my sign, but ultimately I only got the negative of everything I originally had.

    I am going to need a little more help. I will reference

    "An exact solution" from Math Pages" at
    http://www.mathpages.com/rr/s6-01/6-01.htm

    of which I have attached an excert ( Word attachment) with my comments in red.

    In here as everywhere else I am still not coming up with the correct sign convention on one of my expressions. If you are inclined, please look at the attached, where I have carefully substituted as shown by the author, but do not get the author's expression. It seems to be simple math, and this is not unique to this author, it is the same error I am making in comparing to all others, including Dirac, which dexercioby graciously pointed out.

    Once again thanks for input.
     

    Attached Files:

  14. Jul 1, 2005 #13

    pervect

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    I think your problem is that you incorrectly differentiated [itex]\Gamma^r{}_{tt}[/tex]

    Representing derivatives as subscripts, you want

    d/dr (e^(2u-2v) u_r), where u and v are functions of r.

    The chain rule gives us.

    e^(2u-2v) u_rr + d/dr(e^(2u-2v)) u_r

    Let x = 2u-2v, then d/dr (e^x) = d/dx (e^x) * dx/dr = e^x dx/dr

    Thus we have for the final result

    e^(2u-2v) * (u_rr + (2u_r - 2v_r)*u_r)

    Or, using Maxima
    Code (Text):

    (%i1) depends([u,v],r);
    (%o1)                [u(r), v(r)]
    (%i2) e1 :: exp(2*u-2*v)*diff(u,r);
                    du   2 u - 2 v
    (%o2)                   -- %E
                    dr
    (%i3) diff(e1,r);
                            2
            du   2 u - 2 v    du     dv    d u   2 u - 2 v
    (%o3)           -- %E          (2 -- - 2 --) + --- %E
            dr        dr     dr  2
                               dr
    (%i4)
     
     
  15. Jul 1, 2005 #14
    pervect is correct again. I failed to use the chain rule and only differentiated the first derivative again, whilst the other part of the expression is also a function of r. This gives me he correct answer for all of my various starting cases. I knew it was an obvious error which lied somewhere between arithimetic and tensor calculus.

    Thanks again.
     
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