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Ricci Tensor of FRW

  1. Mar 30, 2012 #1
    I am trying to understand FRW universe. To do so I am following the link below:

    http://www.phys.washington.edu/users/dbkaplan/555/lecture_04.pdf


    I am confused at equation 74. I got R00 but for Rij part I am always getting a[itex]\ddot{a}[/itex]. I am trying to solve it for k =0.

    Can some please expand the Rij calculation from basics?
     
  2. jcsd
  3. Mar 30, 2012 #2

    Mentz114

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    Did you do the calculation by hand ? It's hard to point out an error without seeing the working. You could list the Christoffel symbols you got.
     
  4. Mar 30, 2012 #3
  5. Mar 30, 2012 #4
    I did by hand and the significant Christoffel symbols here are:

    [itex]\Gamma^{t}_{xx}[/itex] = a[itex]\ddot{a}[/itex]

    [itex]\Gamma^{x}_{tx}[/itex] = [itex]\frac{\dot{a}}{a}[/itex]

    I am following Sean's note too. I don't know when I try to calculate R[itex]_{xx}[/itex] i.e. R[itex]^{t}_{xtx}[/itex]. I am not getting the correct answer
     
  6. Mar 30, 2012 #5

    Mentz114

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    Are you using this ?

    [tex]
    R^r_{mqs}=\Gamma ^{r}_{mq,s}-\Gamma ^{r}_{ms,q}+\Gamma ^{r}_{ns}\Gamma ^{n}_{mq}-\Gamma ^{r}_{nq}\Gamma ^{n}_{ms}
    [/tex]
     
  7. Mar 30, 2012 #6
    Yes, I used the above mentioned formula.
    where,

    r = q = t and m=s= x
     
  8. Mar 30, 2012 #7

    Mentz114

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    OK, so you've got
    [tex]
    R^t_{xtx}=\Gamma ^{t}_{xt,x}-\Gamma ^{t}_{xx,t}+\Gamma ^{t}_{nx}\Gamma ^{n}_{xt}-\Gamma ^{t}_{nt}\Gamma ^{n}_{xx}
    [/tex]

    Are you doing the summation over n ?
     
  9. Mar 30, 2012 #8
    Hang on, Christoffel symbols only contain first derivatives, what is the double dot doing there?
     
  10. Mar 30, 2012 #9

    Mentz114

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    Good point.
     
  11. Mar 30, 2012 #10
    Sorry that's a typo, it's "a[itex]\dot{a}[/itex]" only
     
    Last edited: Mar 30, 2012
  12. Mar 30, 2012 #11
    My formula is actually:
    [tex]
    R^t_{xtx}=\Gamma ^{t}_{xx,t}-\Gamma ^{t}_{xt,x}+\Gamma ^{n}_{xx}\Gamma ^{t}_{nt}-\Gamma ^{n}_{xt}\Gamma ^{t}_{nx}
    [/tex]
     
  13. Mar 31, 2012 #12

    pervect

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    Are you calculating the Ricci in a coordinate basis, or in an orthonormal frame? And it'd be helpful to get the line element (for the former) or the set of basis vectors (for the later) that you're using - IIRC there are a couple of (equivalent) ways of writing the metric for k=0.
     
  14. Mar 31, 2012 #13
    Here's my line element:

    ds2 = -dt2 + a2(t) (dx2 + dy2 + dz2)

    Can someone please show couple of steps here?
     
    Last edited: Mar 31, 2012
  15. Mar 31, 2012 #14
    [itex] R^t_{rtr} = \partial_t \Gamma^t_{rr} - \partial_r \Gamma^t_{rt} + \Gamma^t_{t \lambda} \Gamma^{\lambda}_{rr} - \Gamma^t_{r \lambda} \Gamma^{\lambda}_{tr} = \frac{\dot{a}^2+ a \ddot{a}}{1-kr^2} - 0 + 0 - \Gamma^t_{r r} \Gamma^{r}_{tr} = \frac{\dot{a}^2+ a \ddot{a}}{1-kr^2} - \frac{\dot{a}^2}{1-kr^2} = \frac{a \ddot{a}}{1-kr^2} [/itex]

    ... hopefully that's right, it's hard to get all the terms when doing calculations in latex... atleast a quick check gave the correct value for [itex] R_{rr}[/itex] so maybe it's right.
     
    Last edited: Mar 31, 2012
  16. Mar 31, 2012 #15
    if k = 0 then you get only a[itex]\ddot{a}[/itex]

    But according to the notes, we should get
    a[itex]\ddot{a}[/itex] + 2[itex]\dot{a}[/itex]2
     
  17. Mar 31, 2012 #16
    I think you're talking about the Ricci tensor: [itex] R_{rr} = R^{\mu}_{r \mu r} = \frac{a \ddot{a} + 2 \dot{a}^2 + 2k}{1-kr^2} [/itex]
     
  18. Mar 31, 2012 #17
    isn't it same as as [itex] R_{rr} = R^{t}_{r t r}[/itex]

    I am confused here. I am talking about (74) i.e Rij from:
    http://www.phys.washington.edu/users/dbkaplan/555/lecture_04.pdf
     
    Last edited: Mar 31, 2012
  19. Mar 31, 2012 #18
    No, you sum over ALL the indices!
     
  20. Mar 31, 2012 #19
    i.e t= μ and t = [itex]\nu[/itex]

    How the does Ricci tensor equation looks like then?

    [itex] R_{rr} = R^{\mu}_{r\mu r} + R^{\nu}_{r\nu r} [/itex]

    Since [itex] R^{\mu}_{r\mu r} = a\ddot{a}[/itex]

    and [itex] R^{\nu}_{r\nu r} = a\ddot{a}[/itex]

    [itex] R_{rr} = 2 a\ddot{a}[/itex]

    that's not correct. I don't know I am getting confused. I am not seeing how we get [itex]\dot{a}[/itex]2
     
    Last edited: Mar 31, 2012
  21. Mar 31, 2012 #20
    no no no, Ricci tensor is the trace of Riemann tensor, so [itex] R^{\mu}_{r\mu r} = R^{t}_{rtr} +R^{r}_{rrr} + R^{\theta}_{r \theta r} + R^{\phi}_{r \phi r} [/itex]
     
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