Understanding Ricci Tensor of FRW Universe: Equation 74 Explained

[psimeson]

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?

 

[Mentz114]

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.

 

[clamtrox]

or compare them to ones calculated here: http://arxiv.org/abs/gr-qc/9712019

 

[psimeson]

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

 

[Mentz114]

Are you using this ?

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

 

[psimeson]

Are you using this ?

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

Yes, I used the above mentioned formula.
where,

r = q = t and m=s= x

 

[Mentz114]

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 ?

 

[clamtrox]

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?

 

[Mentz114]

Hang on, Christoffel symbols only contain first derivatives, what is the double dot doing there?

Good point.

 

[psimeson]

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

 

[psimeson]

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 ?

My formula is actually:
$$R^t_{xtx}=\Gamma ^{t}_{xx,t}-\Gamma ^{t}_{xt,x}+\Gamma ^{n}_{xx}\Gamma ^{t}_{nt}-\Gamma ^{n}_{xt}\Gamma ^{t}_{nx}$$

 

[pervect]

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.

 

[psimeson]

Here's my line element:

ds² = -dt² + a²(t) (dx² + dy² + dz²)

Can someone please show couple of steps here?

 

[clamtrox]

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.

 

[psimeson]

$R^t_{rtr} = \partial_t \Gamma^t_{rr} - \partial_r \Gamma^t_{rt} + \Gamma^t_{t \lambda} \Gamma^{\lambda}_{rr} - \Gamma^t_{r \lambda} \Gamma^{\lambda}_{tr} = \frac{\dot{a}^2+ a \ddot{a}}{1-kr^2} - 0 + 0 - \Gamma^t_{r r} \Gamma^{r}_{tr} = \frac{\dot{a}^2+ a \ddot{a}}{1-kr^2} - \frac{\dot{a}^2}{1-kr^2} = \frac{a \ddot{a}}{1-kr^2}$

... 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

 

[clamtrox]

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: $R_{rr} = R^{\mu}_{r \mu r} = \frac{a \ddot{a} + 2 \dot{a}^2 + 2k}{1-kr^2}$

 

[psimeson]

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

 

[clamtrox]

No, you sum over ALL the indices!

 

[psimeson]

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

 

[clamtrox]

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 $R^{\mu}_{r\mu r} = R^{t}_{rtr} +R^{r}_{rrr} + R^{\theta}_{r \theta r} + R^{\phi}_{r \phi r}$

 

[psimeson]

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

 

[psimeson]

I think I got it.. Thanks a lot