Ricci Tensor of FRW

  1. I am trying to understand FRW universe. To do so I am following the link below:


    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. Mentz114

    Mentz114 4,525
    Gold Member

    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. 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
  5. Mentz114

    Mentz114 4,525
    Gold Member

    Are you using this ?

    R^r_{mqs}=\Gamma ^{r}_{mq,s}-\Gamma ^{r}_{ms,q}+\Gamma ^{r}_{ns}\Gamma ^{n}_{mq}-\Gamma ^{r}_{nq}\Gamma ^{n}_{ms}
  6. Yes, I used the above mentioned formula.

    r = q = t and m=s= x
  7. Mentz114

    Mentz114 4,525
    Gold Member

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

    Are you doing the summation over n ?
  8. Hang on, Christoffel symbols only contain first derivatives, what is the double dot doing there?
  9. Mentz114

    Mentz114 4,525
    Gold Member

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

    pervect 8,157
    Staff Emeritus
    Science Advisor

    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.
  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
  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
  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
  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]
  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:
    Last edited: Mar 31, 2012
  18. No, you sum over ALL the indices!
  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
  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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