What are the non-zero elements of the Riemann Tensor using the FLRW metric?

[TerryW]

Homework Statement

I've been working on Exercise 14.3 in MTW. This starts with the FLRW metric (see attachment) and asks that you find the connection coefficients and then produce the non-zero elements of the Riemann Tensor.

The answer given is that there are only 2 non-zero elements vis R^t_χtχ and R^χ_θχΘ.

My problem is that I have ended up with four additional non-zero components - R^t_ΘtΘ, R^t_φtφ, R^χ_φχφ andR^Θ_φΘφ

The attachment lists the connection coefficients I've produced, a) by using the suggested methodology in the exercise (derive them from the geodesic equations) and to check my result by b) using the standard process of

Γ^μ_αβ= ½{g_μα,β + g_μβ,α - g_αβ,μ}

I found an alternative version of the FLRW metric at the url given at the bottom of the attachment and used the connection coefficients derived in that example to produce the Riemann Tensor components and find that I get the same 6 non-zero components.

Two of the four unwanted components contain $\ddot a$ which come from the derivative wrt t of one of the connection coefficients. Clearly this element will not be eliminated by subtracting the product of two connection coefficients, neither of which contain $\ddot a$ .Can anyone suggest where I might be going wrong?
TerryW

Homework Equations

See attachment

The Attempt at a Solution

See attachment




[/B]

 

[TerryW]

Don't know what happened to the attachment then.

 

[TSny]

I haven't checked all of your results for the curvature components. But I did check $R^t_{\;\theta \theta t}$ and got the same nonzero result as you.

However, I don't think this contradicts what MTW are saying in the problem. Part (b) states that if you consider components of the specific form $R^t_{\;\chi \mu \nu}$ then only $R^t_{\;\chi t \chi}$ will be nonzero. (Of course, $R^t_{\;\chi \chi t}$ would also be nonzero by antisymmetry of the last two indices.)

Likewise, out of all of the components of the specific form $R^\chi_{\;\theta \mu \nu}$, only $R^\chi_{\;\theta \chi \theta}$ (and $R^\chi_{\;\theta \theta \chi}$) will be nonzero.

They are not claiming anything about $R^t_{\;\theta \theta t}$ since this is not of the form $R^t_{\;\chi \mu \nu}$ or $R^\chi_{\;\theta \mu \nu}$ . So, your results could be OK.

Apparently, you don't need to worry about any other components. When you go to part (c), you will switch to an orthonormal frame in which you will be able to get all nonzero curvature components $R^{\hat{\alpha}}_{\;\hat{\beta} \hat{\mu} \hat{\nu}}$ from $R^\hat{\chi}_{\;\hat{\theta} \hat{\chi} \hat{\theta}}$ and $R^\hat{t}_{\;\hat{\chi} \hat{t} \hat{\chi}}$ by symmetry considerations. I haven't done this, but I think that's what's going on.

 

[TerryW]

I haven't checked all of your results for the curvature components. But I did check $R^t_{\;\theta \theta t}$ and got the same nonzero result as you.

However, I don't think this contradicts what MTW are saying in the problem. Part (b) states that if you consider components of the specific form $R^t_{\;\chi \mu \nu}$ then only $R^t_{\;\chi t \chi}$ will be nonzero. (Of course, $R^t_{\;\chi \chi t}$ would also be nonzero by antisymmetry of the last two indices.)

Likewise, out of all of the components of the specific form $R^\chi_{\;\theta \mu \nu}$, only $R^\chi_{\;\theta \chi \theta}$ (and $R^\chi_{\;\theta \theta \chi}$) will be nonzero.

They are not claiming anything about $R^t_{\;\theta \theta t}$ since this is not of the form $R^t_{\;\chi \mu \nu}$ or $R^\chi_{\;\theta \mu \nu}$ . So, your results could be OK.

Apparently, you don't need to worry about any other components. When you go to part (c), you will switch to an orthonormal frame in which you will be able to get all nonzero curvature components $R^{\hat{\alpha}}_{\;\hat{\beta} \hat{\mu} \hat{\nu}}$ from $R^\hat{\chi}_{\;\hat{\theta} \hat{\chi} \hat{\theta}}$ and $R^\hat{t}_{\;\hat{\chi} \hat{t} \hat{\chi}}$ by symmetry considerations. I haven't done this, but I think that's what's going on.

Thanks for pointing that out TSny,

I just didn't notice the detail of what was being asked and ploughed ahead working out all the components!

I don't think I'll have a problem with the rest of the exercise.One good outcome is that there is now one post on the Forum where someone can find the connection coefficients for the FLRW metric! I've searched quite a lot for a table of metrics and their connection coefficients, plus Riemann Tensor components etc without success.
Regards
TerryW

 

[TSny]

One good outcome is that there is now one post on the Forum where someone can find the connection coefficients for the FLRW metric!

Yes, that's nice.

 

[TerryW]

As I worked on the last bit of the exercise, I found a couple of errors in my workings for two the Riemann Tensor components so I've updated my original attachment. With these amended components, everything works out nicely and symmetrically.

TerryW