A The Metric Matrix: How Can I Invert a Non-Diagonal Matrix?

[MathematicalPhysicist]

I have the metric $ds^2=dudv+F(y,z)du^2+dy^2+dz^2$.

I have shown that for $F(y,z)=0$ it's Minkowski metric, but for $F(y,z)\ne 0$ I want to calculate Christoffel Symbols, Ricci tensor and scalar, the problem is that the metric $g_{\mu\nu}$ is not diagonal, and I need to find $g^{\mu\nu}$ in order to calculate christoffel and the rest, but how can I solve this without inverting the metric?

Thanks!

 

[Orodruin]

Why do you need to do it without inverting the metric? The metric has a straight-forward inverse.

 

[MathematicalPhysicist]

Well we have a matrix: {0,1,0,0},{1,F(y,z),0,0},{0,0,1,0},{0,0,0,1}

So it's not diagonal, so inverting it you need to use the algorithm from Linear Algebra.

 

[MathematicalPhysicist]

Ah, yes, the inverse is easy, didn't notice it.
:-D

 

[weirdoguy]

So? Is that a problem?

EDIT: Still, if it wasn't easy I see no problem in using linear algebra. That's why you learned it - to use it.

 

[Orodruin]

Also, I strongly suggest against using the typical expression in terms of the metric to compute the connection coefficients of the Levi-Civita connection. Instead, use the geodesic equations obtained from finding the optima of
$$
\int g_{\mu\nu}\dot x^\mu \dot x^\nu ds.
$$
It is the same maths, but the bookkeeping is way easier.

 

[MathematicalPhysicist]

So? Is that a problem?

EDIT: Still, if it wasn't easy I see no problem in using linear algebra. That's why you learned it - to use it.

I know, but if there are less tedious methods just in case...

 

[MathematicalPhysicist]

BTW, does someone know how to implement the calculation of Christoffel symbol, Ricci tensor and Ricci scalar in maple?

I just want to check if my calculations are right, I got:

The only Christoffel symbols that don't vanish are: $\Gamma_{uu}^y=-F'_y(y,z)/2$ and $\Gamma_{uu}^z=-F'_z(y,z)/2$; and the Ricci tensor the components that don't vanish are:
$$R^y_{uyu}=-F''_{yy}/2$$
$$R^z_{uzu}=-F''_{zz}/2$$
$$R^y_{uzu}=R^z_{uyu}=-F''_{yz}/2$$

I get that Ricci scalar vanishes since $R=R^{\alpha}_{\beta\mu\alpha}g^{\beta\mu}=0$, am I right?

I don't understand how to use this package in maple 2017, I asked for help in a suitable forum for maple, but didn't get yet an answer.

 

[Orodruin]

Did you get the point of #6? It really is the most straight-forward way of computing the Levi-Civita connection coefficients.

The only Christoffel symbols that don't vanish are: $\Gamma_{uu}^y=-F'_y(y,z)/2$ and $\Gamma_{uu}^z=-F'_z(y,z)/2$;

I get more non-zero Christoffel symbols. In particular, look for $\Gamma^v_{uy} = \Gamma^v_{yu}$ and $\Gamma^v_{uz} = \Gamma^v_{zu}$. Can you show your work?

and the Ricci tensor the components that don't vanish are:
$$R^y_{uyu}=-F''_{yy}/2$$
$$R^z_{uzu}=-F''_{zz}/2$$
$$R^y_{uzu}=R^z_{uyu}=-F''_{yz}/2$$

That's the Riemann tensor. The Ricci tensor is $R_{\mu\nu} = R^{\gamma}_{\mu\gamma\nu}$. There are also some non-zero elements of the form $R^v_{abc}$, where $abc$ are all different from $v$, but they do not affect the computation of the Ricci scalar or the Ricci tensor since they never correspond to one of the lower indices and therefore disappear in the trace. Therefore:

I get that Ricci scalar vanishes since $R=R^{\alpha}_{\beta\mu\alpha}g^{\beta\mu}=0$, am I right?

Yes. I get a vanishing Ricci scalar as well.

 

[MathematicalPhysicist]

Yes, I get that $\Gamma_{uy}^v=1/4F_y'$ and $\Gamma^v_{uz}=1/4F_z'$, I think this about sums it for Christoffel symbol.

I can scan my work, I just inverted the metric with an LA algorithm (where you augment $g_{\mu\nu}$ with the identity matrix and then get on the left the identity matrix and on the right the inverse matrix).

 

[cosmik debris]

BTW, does someone know how to implement the calculation of Christoffel symbol, Ricci tensor and Ricci scalar in maple?

Some years ago I used a Maple package called GRTensor II which I ran under the symbolic toolbox in Matlab.

Cheers

 

[Orodruin]

Yes, I get that $\Gamma_{uy}^v=1/4F_y'$ and $\Gamma^v_{uz}=1/4F_z'$, I think this about sums it for Christoffel symbol.

I got a different constant. Using the method described in #6 your work for the Christoffel symbols should not be longer than a few lines (one for the variation with respect to each coordinate). The relevant variations are the variations with respect to $u$ and $v$ of
$$
\int \underbrace{[\dot u \dot v + F \dot u^2 + \dot y^2 + \dot z^2]}_{\equiv \mathscr L} ds.
$$

 

[DrGreg]

Well we have a matrix: {0,1,0,0},{1,F(y,z),0,0},{0,0,1,0},{0,0,0,1}

So it's not diagonal, so inverting it you need to use the algorithm from Linear Algebra.

Its easy to invert$$\left(
\begin{array}{cc|c|c}
0 & \tfrac12 & 0 & 0 \\
\tfrac12 & F & 0 & 0 \\
\hline 0 & 0 & 1 & 0 \\
\hline 0 & 0 & 0 & 1
\end{array}\right)
$$provided you remember how to invert$$
\begin{pmatrix}
0 & \tfrac12 \\
\tfrac12 & F
\end{pmatrix}
$$