K.J.Healey
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Lets see. The SC metric is correct, as it is the presumptive metric for a spherically symmetric body about a mass "M", PRIOR to solution. The actual solution for the metric would be to replace the exponents with (1-2M/r) and (1-2M/r)^-1 respectively. (but that's in the limit of a point M)
Continuing, as you can see from the line element in the Kerr metric, there exists an off-diagonal component that mixes time and Phi.
Don't mix up the Einstein tensor and the metric. I assume you just made a typo. But from the metric you can get to the Einstein tensor by using that code I showed you.
The .m file is just a package you run (basically just a function), then look at the demo. It defines a metric "metric" (aka "g") and from it and the normal coordinate vector calculates the Ricci tensor & scalar (thru christoffel) automatically and spits out the Einstein tensor ("G").As for whether or not the code can handle the Kerr metric, I don't see why not. But it may not be entirely useful yet. Once you have the "G" you set up all of your equations using the GR field equation, equating it to T. You should then have a set of equations that are solvable, with the connection you listed (P vs density).
Actually, now that I'm looking at your line element. Wouldn't it make sense that your matrix form be:
<br /> g_{\mu\nu} = \left(<br /> \begin{array}{llll}<br /> e^{\nu (r)} & 0 & 0 & \frac{r_s r \alpha sin^2 \theta}{\rho^2} \\<br /> 0 & -e^{\lambda (r)} & 0 & 0 \\<br /> 0 & 0 & - \rho^2 & 0 \\<br /> \frac{r_s r \alpha sin^2 \theta}{\rho^2} & 0 & 0 & - \left(r^2 + \alpha^2 + \frac{r_s r \alpha^2}{\rho^2} \sin ^2 \theta \right) <br /> \end{array}<br /> \right)<br />
The of diagonals give that extra term. THIS ABOVE IS NOT CORRECT AT ALL. I am just showing you where to start, there needs to be off diagonals.
http://www.astro.ku.dk/~milvang/RelViz/000_node12.html
has a way. Take a look.
Continuing, as you can see from the line element in the Kerr metric, there exists an off-diagonal component that mixes time and Phi.
Don't mix up the Einstein tensor and the metric. I assume you just made a typo. But from the metric you can get to the Einstein tensor by using that code I showed you.
The .m file is just a package you run (basically just a function), then look at the demo. It defines a metric "metric" (aka "g") and from it and the normal coordinate vector calculates the Ricci tensor & scalar (thru christoffel) automatically and spits out the Einstein tensor ("G").As for whether or not the code can handle the Kerr metric, I don't see why not. But it may not be entirely useful yet. Once you have the "G" you set up all of your equations using the GR field equation, equating it to T. You should then have a set of equations that are solvable, with the connection you listed (P vs density).
Actually, now that I'm looking at your line element. Wouldn't it make sense that your matrix form be:
<br /> g_{\mu\nu} = \left(<br /> \begin{array}{llll}<br /> e^{\nu (r)} & 0 & 0 & \frac{r_s r \alpha sin^2 \theta}{\rho^2} \\<br /> 0 & -e^{\lambda (r)} & 0 & 0 \\<br /> 0 & 0 & - \rho^2 & 0 \\<br /> \frac{r_s r \alpha sin^2 \theta}{\rho^2} & 0 & 0 & - \left(r^2 + \alpha^2 + \frac{r_s r \alpha^2}{\rho^2} \sin ^2 \theta \right) <br /> \end{array}<br /> \right)<br />
The of diagonals give that extra term. THIS ABOVE IS NOT CORRECT AT ALL. I am just showing you where to start, there needs to be off diagonals.
http://www.astro.ku.dk/~milvang/RelViz/000_node12.html
has a way. Take a look.