# Understanding Ricci Tensor of FRW Universe: Equation 74 Explained

• psimeson
In summary: R_{rr} = R^{t}_{r t r}I am confused here. I am talking about (74) i.e Rij... isn't it same as as R_{rr} = R^{t}_{r t r}In summary, Sean's notes and equation 74 are confusing. He is trying to solve for Rij but keeps getting a\ddot{a}. He needs help solving for Rij.
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?

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 ?

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

Mentz114 said:
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

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 ?

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.

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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Mentz114 said:
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:
$$R^t_{xtx}=\Gamma ^{t}_{xx,t}-\Gamma ^{t}_{xt,x}+\Gamma ^{n}_{xx}\Gamma ^{t}_{nt}-\Gamma ^{n}_{xt}\Gamma ^{t}_{nx}$$

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.

Here's my line element:

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

Can someone please show couple of steps here?

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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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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

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

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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No, you sum over ALL the indices!

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

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

I think I got it.. Thanks a lot

## 1. What is the Ricci tensor of the FRW universe?

The Ricci tensor is a mathematical object that describes the curvature of spacetime in the Friedmann-Robertson-Walker (FRW) universe. It is used in Einstein's field equations to calculate the effects of matter and energy on the geometry of the universe.

## 2. How is Equation 74 related to the Ricci tensor of the FRW universe?

Equation 74 is a specific form of the Ricci tensor in the FRW universe. It takes into account the density and pressure of matter and radiation, as well as the cosmological constant, to describe the curvature of spacetime in this particular model of the universe.

## 3. Why is understanding the Ricci tensor important for studying the FRW universe?

The Ricci tensor is a fundamental tool in understanding the dynamics of the FRW universe. It allows us to calculate the expansion rate of the universe, the evolution of its density and curvature, and the effects of matter and energy on its geometry. Without a thorough understanding of the Ricci tensor, we cannot accurately model the behavior of the FRW universe.

## 4. How is the Ricci tensor related to general relativity?

The Ricci tensor is a key component of Einstein's field equations, which are the basis of general relativity. These equations describe how matter and energy interact with the curvature of spacetime, and the Ricci tensor is a crucial part of this relationship. In the FRW universe, the Ricci tensor helps us determine the overall geometry of the universe and how it changes over time.

## 5. Are there any other important equations or concepts related to the Ricci tensor in the FRW universe?

Yes, there are several other equations and concepts that are closely related to the Ricci tensor in the FRW universe. These include the Einstein tensor, which is a combination of the Ricci tensor and the metric tensor, and the Friedmann equations, which describe the evolution of the universe's density and curvature. It is important to have a solid understanding of these concepts and their relationships in order to fully comprehend the behavior of the FRW universe.

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