Undergrad Calculating the Ricci tensor on the surface of a 3D sphere

[Z3kr0m]

Hello, I'm trying to calculate Christoffel symbols on 2D surface of 3D sphere, the metric tensor is gij = diag {1/(1 − k*r2), r2}, where k is the curvature. I derived it using the formula for symbols of second kind, but I think I've made mistake somewhere. Then I would like to know which of the symbol use to calculate Ricci tensor. Hope somebody will help, sorry I'm just a beginner in relativity.

 

[pervect]

I'm not familiar with the coordinates you're using. The coordinates I usually use have a metric as in http://www.physics.usu.edu/Wheeler/GenRel/Lectures/2Sphere.pdf

I'm not sure what symbols to use for your coordinates, I chose x an y. Then your line element is:

$$g_{xx} = \frac{dx^2}{1-k\,r^2} \quad g_{yy} = r^2\,dy^2 \quad ds^2 = g_{xx}\,dx^2 + g_{yy}dy^2$$

I'm not getting good results with this, the Riemann comes out zero. I was thinking that maybe r was a function of x an y, but I don't think that makes sense.

It would also help greatly if you could use Latex to format your post. There are tutorials around on PF,
https://www.physicsforums.com/help/latexhelp/ for one.

 

[Z3kr0m]

It's metric in polar coordinates where $ds^2= \frac {dr^2 } {(1-kr^2) } + r^2d \phi^2$ , and the $k= \frac {1} {R^2 }$, where R is the radius of the sphere.

Edit: the bottom of the friction

 

[pervect]

It's metric in polar coordinates where $ds^2= \frac {dr^2 } {(1-kr^2) } + r^2d \phi^2$ , and the $k= \frac {1} {R^2 }$, where R is the radius of the sphere.

Edit: the bottom of the friction

With GRTensor, I'm getting
$$\Gamma^r{}{}_{rr} = \frac{-kr}{kr^2-1} \quad \Gamma^r{}_{\phi\phi} = (kr^2-1)r \quad \Gamma^\phi{}_{r\phi} = \frac{1}{r}$$

For the Rici

$$R_{ab} = \begin{bmatrix} \frac{-k}{kr^2-1} & 0 \\ 0 & kr^2 \\ \end{bmatrix}
$$

the sole non-zero component of the Riemann is

$$\frac{-kr^2}{kr^2-1}$$

 

[Z3kr0m]

Thank you very much! :) And can you please show how you use the Christoffels in Ricci formula? $R_{ij} = \frac {\partial \Gamma^{l}{}_{ik}} {\partial x^l} + \frac {\partial \Gamma^{l}_{il} } {\partial x^j}$ ... So which coordinate should I use as $l$?

[Mentors' note: edited to fix a small Latex formatting problem]

 

[Nugatory]

Thank you very much! :) And can you please show how you use the Christoffels in Ricci formula? $R_{ij} = \frac {\partial \Gamma^{l}{}_{ik}} {\partial x^l} + \frac {\partial \Gamma^{l}_{il} } {\partial x^j}$ ... So which coordinate should I use as $l$?

None. When an index is repeated like that, you're summing over it - google for "Einstein summation convention" for details. https://preposterousuniverse.com/wp-content/uploads/2015/08/grtinypdf.pdf includes a decent introduction.

 

[Z3kr0m]

I know the summation convention, I'm just not familiar with it in this formula. So I don't know how to use the symbols in it.

 

[pervect]

I use an automated program that shields me from the details of the claculations, which are long and error-prone.

But one can write

$$R^\rho{}_{\sigma\mu\nu} = \partial_\mu\Gamma^\rho{}_{\nu\sigma}
- \partial_\nu\Gamma^\rho{}_{\mu\sigma}
+ \Gamma^\rho{}_{\mu\lambda}\Gamma^\lambda{}_{\nu\sigma}
- \Gamma^\rho{}_{\nu\lambda}\Gamma^\lambda{}_{\mu\sigma}$$

to find the Riemann, and the Ricci is then the contraction of the first and third slots of the Riemann, i.e.

$$R_{\mu\nu} = R^{\rho}{}_{\mu \rho \nu}$$

See https://en.wikipedia.org/wiki/Riemann_curvature_tensor

You'll need to be familiar with the Einstein summation convention https://en.wikipedia.org/wiki/Einstein_notation and the concept of tensor contraction, https://en.wikipedia.org/wiki/Tensor_contraction

If you're not already familiar with these concepts, I'm not sure Wiki will explain it in enough detail for you to figure it out without a textbook (and a lot of study), but at least you'll have some idea of what to look for.

Also, I've not written out all the permutations of terms that are given by symmetries in the above. Using my automated program to calculate exactly what I wrote above, I get:

$$R^r{}_{\phi r \phi} = -R^r{}_{\phi\phi r} = kr^2$$
$$R^\phi{}_{r r \phi} = -R^\phi{}_{r \phi r} = \frac{k}{kr^2-1}$$

I tried not to make any typos, but I wouldn't guarantee it absolutely.

 

[Z3kr0m]

Okay, thank you very much! :)

 

[Nugatory]

I know the summation convention, I'm just not familiar with it in this formula. So I don't know how to use the symbols in it.

On looking again, something seems wrong in

$R_{ij} = \frac {\partial \Gamma^{l}{}_{ik}} {\partial x^l} + \frac {\partial \Gamma^{l}_{il} } {\partial x^j}$

The indices don't balance, because you have a stray $k$ on the right-hand side. What is the formula supposed to be and where did it come from?

 

[Z3kr0m]

It's a mistake, there should be $j$ instead of $k$. It's from wiki: https://en.wikipedia.org/wiki/List_of_formulas_in_Riemannian_geometry

 

[Nugatory]

It's a mistake, there should be $j$ instead of $k$. It's from wiki: https://en.wikipedia.org/wiki/List_of_formulas_in_Riemannian_geometry

OK, then you interpret a term like $\frac {\partial \Gamma^{l}{}_{ik}} {\partial x^l}$ as $$\frac {\partial \Gamma^{l}{}_{ik}} {\partial x^l}\equiv\sum_{l=0}^3\frac{\partial\Gamma^l_{ik}}{\partial{x}^l}\equiv\frac {\partial \Gamma^{0}{}_{ik}} {\partial x^0}+\frac {\partial \Gamma^{1}{}_{ik}} {\partial x^1}+\frac {\partial \Gamma^{2}{}_{ik}} {\partial x^2}+\frac {\partial \Gamma^{3}{}_{ik}} {\partial x^3}$$
Here the superscript on the $x$ is not exponentiation; it designates one of your four coordinates in whatever coordinate system you are using. For example, a common convention with Minkowski coordinates in flat space is $x^0\equiv{t}$, $x^1\equiv{x}$, $x^2\equiv{y}$, $x^3\equiv{z}$.

 

[Z3kr0m]

OK, then you interpret a term like $\frac {\partial \Gamma^{l}{}_{ik}} {\partial x^l}$ as $$\frac {\partial \Gamma^{l}{}_{ik}} {\partial x^l}\equiv\sum_{l=0}^3\frac{\partial\Gamma^l_{ik}}{\partial{x}^l}\equiv\frac {\partial \Gamma^{0}{}_{ik}} {\partial x^0}+\frac {\partial \Gamma^{1}{}_{ik}} {\partial x^1}+\frac {\partial \Gamma^{2}{}_{ik}} {\partial x^2}+\frac {\partial \Gamma^{3}{}_{ik}} {\partial x^3}$$
Here the superscript on the $x$ is not exponentiation; it designates one of your four coordinates in whatever coordinate system you are using. For example, a common convention with Minkowski coordinates in flat space is $x^0\equiv{t}$, $x^1\equiv{x}$, $x^2\equiv{y}$, $x^3\equiv{z}$.

Ok, thank you very much!