Understanding Ricci Tensor of FRW Universe: Equation 74 Explained

psimeson
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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\ddot{a}. I am trying to solve it for k =0.

Can some please expand the Rij calculation from basics?
 
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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.
 
I did by hand and the significant Christoffel symbols here are:

\Gamma^{t}_{xx} = a\ddot{a}

\Gamma^{x}_{tx} = \frac{\dot{a}}{a}

I am following Sean's note too. I don't know when I try to calculate R_{xx} i.e. R^{t}_{xtx}. I am not getting the correct answer
 
Are you using this ?

<br /> R^r_{mqs}=\Gamma ^{r}_{mq,s}-\Gamma ^{r}_{ms,q}+\Gamma ^{r}_{ns}\Gamma ^{n}_{mq}-\Gamma ^{r}_{nq}\Gamma ^{n}_{ms}<br />
 
Mentz114 said:
Are you using this ?

<br /> R^r_{mqs}=\Gamma ^{r}_{mq,s}-\Gamma ^{r}_{ms,q}+\Gamma ^{r}_{ns}\Gamma ^{n}_{mq}-\Gamma ^{r}_{nq}\Gamma ^{n}_{ms}<br />

Yes, I used the above mentioned formula.
where,

r = q = t and m=s= x
 
OK, so you've got
<br /> R^t_{xtx}=\Gamma ^{t}_{xt,x}-\Gamma ^{t}_{xx,t}+\Gamma ^{t}_{nx}\Gamma ^{n}_{xt}-\Gamma ^{t}_{nt}\Gamma ^{n}_{xx}<br />

Are you doing the summation over n ?
 
psimeson said:
I did by hand and the significant Christoffel symbols here are:

\Gamma^{t}_{xx} = a\ddot{a}

\Gamma^{x}_{tx} = \frac{\dot{a}}{a}

I am following Sean's note too. I don't know when I try to calculate R_{xx} i.e. R^{t}_{xtx}. I am not getting the correct answer

Hang on, Christoffel symbols only contain first derivatives, what is the double dot doing there?
 
clamtrox said:
Hang on, Christoffel symbols only contain first derivatives, what is the double dot doing there?
Good point.
 
  • #10
clamtrox said:
Hang on, Christoffel symbols only contain first derivatives, what is the double dot doing there?

Sorry that's a typo, it's "a\dot{a}" only
 
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  • #11
Mentz114 said:
OK, so you've got
<br /> R^t_{xtx}=\Gamma ^{t}_{xt,x}-\Gamma ^{t}_{xx,t}+\Gamma ^{t}_{nx}\Gamma ^{n}_{xt}-\Gamma ^{t}_{nt}\Gamma ^{n}_{xx}<br />

Are you doing the summation over n ?

My formula is actually:
<br /> R^t_{xtx}=\Gamma ^{t}_{xx,t}-\Gamma ^{t}_{xt,x}+\Gamma ^{n}_{xx}\Gamma ^{t}_{nt}-\Gamma ^{n}_{xt}\Gamma ^{t}_{nx}<br />
 
  • #12
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?
 
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  • #14
psimeson said:
Here's my line element:

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

Can someone please show couple of steps here?

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}

... 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 R_{rr} so maybe it's right.
 
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  • #15
clamtrox said:
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}

... hopefully that's right, it's hard to get all the terms when doing calculations in latex.

if k = 0 then you get only a\ddot{a}

But according to the notes, we should get
a\ddot{a} + 2\dot{a}2
 
  • #16
psimeson said:
if k = 0 then you get only a\ddot{a}

But according to the notes, we should get
a\ddot{a} + 2\dot{a}2

I think you're talking about the Ricci tensor: R_{rr} = R^{\mu}_{r \mu r} = \frac{a \ddot{a} + 2 \dot{a}^2 + 2k}{1-kr^2}
 
  • #17
clamtrox said:
I think you're talking about the Ricci tensor: R_{rr} = R^{\mu}_{r \mu r} = \frac{a \ddot{a} + 2 \dot{a}^2 + 2k}{1-kr^2}

isn't it same as as R_{rr} = R^{t}_{r t r}

I am confused here. I am talking about (74) i.e Rij from:
http://www.phys.washington.edu/users/dbkaplan/555/lecture_04.pdf
 
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  • #18
No, you sum over ALL the indices!
 
  • #19
clamtrox said:
No, you sum over ALL the indices!

i.e t= μ and t = \nu

How the does Ricci tensor equation looks like then?

R_{rr} = R^{\mu}_{r\mu r} + R^{\nu}_{r\nu r}

Since R^{\mu}_{r\mu r} = a\ddot{a}

and R^{\nu}_{r\nu r} = a\ddot{a}

R_{rr} = 2 a\ddot{a}

that's not correct. I don't know I am getting confused. I am not seeing how we get \dot{a}2
 
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  • #20
psimeson said:
i.e t= μ and t = \nu

How the does Ricci tensor equation looks like then?

R_{rr} = R^{\mu}_{r\mu r} + R^{\nu}_{r\nu r}

Since R^{\mu}_{r\mu r} = a\ddot{a}

and R^{\nu}_{r\nu r} = a\ddot{a}

R_{rr} = 2 a\ddot{a}

that's not correct. I don't know I am getting confused. I am not seeing how we get \dot{a}2

no no no, Ricci tensor is the trace of Riemann tensor, so R^{\mu}_{r\mu r} = R^{t}_{rtr} +R^{r}_{rrr} + R^{\theta}_{r \theta r} + R^{\phi}_{r \phi r}
 
  • #21
clamtrox said:
no no no, Ricci tensor is the trace of Riemann tensor, so R^{\mu}_{r\mu r} = R^{t}_{rtr} +R^{r}_{rrr} + R^{\theta}_{r \theta r} + R^{\phi}_{r \phi r}

So that means, for x, y and z, I have:

R^{\mu}_{x\mu x} = R^{t}_{xtx} +R^{x}_{xxx} + R^{y}_{x y x} + R^{z}_{xzx} right?

But the second term is zero and 3rd and 4th term does not have time in it so I will not "a" contribution from them
 
  • #22
I think I got it.. Thanks a lot
 
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