Tolman-Oppenheimer-Volkoff equation

[K.J.Healey]

Ok, after looking at some numbers I stand corrected.

I'm currently working on something of the same process, but only in reverse. I'm trying to fit the observed data to a prediction on the rotation degradation. Unfortunately I've only just finished my first year or grad, and am getting stuck trying to come up with a metric for a rotating, compressible fluid. (Like Kerr + Tolman IV).
But then again, I just started.

 

[Orion1]

Schwarzschild metric line element:
ds^2 = e^{\nu (r)} c^2 dt^2 - e^{\lambda (r)} dr^2 - r^2 ( \sin^2 \theta d \phi^2 + d \theta^2)

In General Relativity, if the matter supports no transverse stresses and has no mass motion, then its energy momentum tensor is:
T_1^1 = T_2^2 = T_3^3 = - P(r) \; \; \; T_4^4 = \rho(r) c^2

G_{\mu\nu} = R_{\mu\nu} - {1\over2} g_{\mu\nu}R

G_{\mu \nu} + \Lambda g_{\mu \nu}= \frac{8 \pi G}{c^4} T_{\mu \nu}

R_{\mu\nu} - {1\over2} g_{\mu\nu}R = \frac{8 \pi G}{c^4} T_{\mu \nu} \; \; \; \Lambda = 0

With these expressions for the Schwarzschild metric line element and the energy momentum tensor, and with zero cosmological constant, Einstein's field equations reduce to:

\tag{3} \frac{8 \pi G P(r)}{c^4} = e^{- \lambda} \left( \frac{1}{r} \frac{d \nu}{dr} + \frac{1}{r^2} \right) - \frac{1}{r^2}

\tag{5} \frac{d \nu}{dr} = \frac{2}{P(r) + \rho(r) c^2} \left( \frac{dP(r)}{dr} \right)

How are equation solutions \tag{3},\tag{5} derived from the Einstein's field equations?
[/Color]

 

[K.J.Healey]

You have your metric, you can now calculate the rici tensor and scalar (from the metric). Then you can just write out your 4 equations with the SE-tensor.

Theres a mathematica package that will actually do this fairly easily, though its a little outdated.

http://library.wolfram.com/infocenter/MathSource/162/

It takes a metric in (in matrix form) and a corresponding 4 vector and calculates the einstein tensor (Rmn - (1/2) Rg)

You may have to modify a little for cosmo const.

 

[Orion1]

Schwarzschild metric line element:
ds^2 = e^{\nu (r)} c^2 dt^2 - e^{\lambda (r)} dr^2 - r^2 ( \sin^2 \theta d \phi^2 + d \theta^2)

Einsteintensor source code:
In[n]=

ToFileName[{$TopDirectory, "AddOns", "Applications"}]
<< einsteintensor.m
x = {t, r, theta, phi}
(metric = DiagonalMatrix[{Exp[\[Nu][r]], -Exp[\[Lambda][r]], -r^2, -r^2 (Sin[theta]^2)}]) // MatrixForm
(Einstein = Inverse[metric].Simplify[EinsteinTensor[metric, x]])
EinsteinTensor[DiagonalMatrix[{-(1 - 2 M/r), 1/(1 - 2 M/r), r^2, (r Sin[theta])^2}], {t, r, theta, phi}] // Simplify

out[n]/MatrixForm=
<br /> \left(<br /> \begin{array}{llll}<br /> e^{\nu (r)} &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; -e^{\lambda (r)} &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; -r^2 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; -r^2 \sin ^2(\theta )<br /> \end{array}<br /> \right)<br />

Einstein tensor matrix element:
G_{1 2} = \frac{e^{-\lambda (r)}}{r^2} \left(r \frac{d \nu}{dr} - e^{\lambda (r)} + 1 \right) = e^{- \lambda (r)} \left( \frac{1}{r} \frac{d \nu}{dr} + \frac{1}{r^2} \right) - \frac{1}{r^2}

\boxed{G_{1 2} = e^{- \lambda (r)} \left( \frac{1}{r} \frac{d \nu}{dr} + \frac{1}{r^2} \right) - \frac{1}{r^2}}

TOV tensor matrix element:
G_{\mu\nu} = e^{- \lambda (r)} \left( \frac{1}{r} \frac{d \nu}{dr} + \frac{1}{r^2} \right) - \frac{1}{r^2}

How is equation solution \tag{5} derived from Einstein's field equations?
\tag{5} \frac{d \nu}{dr} = \frac{2}{P(r) + \rho(r) c^2} \left( \frac{dP(r)}{dr} \right)
[/Color]
Reference:
http://library.wolfram.com/infocenter/MathSource/162/"

 

[K.J.Healey]

Intuition? Its sort of a pain. Take a loot at Tolman's book "Relativity, Thermodynamics and Cosmology"
Page 243ish. He has the three equations relating density the solutions that you have.

He says and I quote:
" furthermore, in the case of a perfect fluid, the equality between the radial stress T11 and the transverse stresses T22=T33 makes if possible to derive a very simple expression for pressure gradient. thus equating the two expressions for T11 and T22."

He then puts it as T11-T22=0
Then multiplies through by 2/r
then rewrites it in a form that he says " is seen to be equivalent to"


dP/dr + (rho_00 + P_0) v'/2 =0

where v' is dv/dt ("nu")

converts to your eq.

 

[Orion1]

...equating the two expressions for T11 and T22.

There is a partial viewing of this book online listed in reference 1, however the reference does not display p. 243 (> p. 49) and the chapter page is unavailable and radial and transverse stresses are not listed in the index.

There are tensors for a perfect fluid on p. 216.

What are Tolman's two expressions for radial stress T11 and transverse stress T22?
[/Color]
Correction, the Einstein tensor in post #54 should read:
G_{1 1} = e^{- \lambda (r)} \left( \frac{1}{r} \frac{d \nu}{dr} + \frac{1}{r^2} \right) - \frac{1}{r^2}
[/Color]
Reference:
http://books.google.com/books?hl=en...6KRE&sig=k8NQ6YmncFm4_VxCof8cm5tS-bI#PPP1,M1"

 

[Orion1]

G_{\mu \nu} = \frac{8 \pi G}{c^4} T_{\mu \nu}

Energy density:
\boxed{G_{00} = \frac{e^{-\lambda (r)} \left(r \lambda &#039;(r) + e^{\lambda (r)} - 1 \right)}{r^2}}

\frac{8 \pi G}{c^4} P(r) = \frac{e^{-\lambda (r)} \left(r \nu &#039;(r)-e^{\lambda (r)}+1\right)}{r^2}

\frac{8 \pi G}{c^4} \rho (r) c^2 = \frac{e^{-\lambda (r)} \left(r \lambda &#039;(r)+e^{\lambda (r)}-1\right)}{r^2}

Pressure radius equivalent to density radius:
\boxed{r_p = r_{\rho}}

r^2 = \left( \frac{c^4}{8 \pi G} \right) \frac{e^{-\lambda (r)} \left(r \nu &#039;(r) - e^{\lambda (r)} + 1 \right)}{P(r)} = \left( \frac{c^4}{8 \pi G} \right) \frac{e^{-\lambda (r)} \left(r \lambda &#039;(r)+e^{\lambda (r)} - 1 \right)}{\rho (r) c^2}

\frac{\left(r \nu &#039;(r) - e^{\lambda (r)} + 1 \right)}{P(r)} = \frac{\left(r \lambda &#039;(r)+e^{\lambda (r)} - 1 \right)}{\rho (r) c^2}

\boxed{\frac{P(r)}{\rho (r) c^2} = \frac{\left(r \nu &#039;(r) - e^{\lambda (r)} + 1 \right)}{\left(r \lambda &#039;(r) + e^{\lambda (r)} - 1 \right)}}
[/Color]

 

[Orion1]

\boxed{\frac{P(r)}{\rho (r) c^2} = \frac{e^{- \lambda (r)} \left( r \nu &#039; (r) + 1 \right) - 1}{e^{- \lambda (r)} \left( r \lambda &#039;(r) - 1 \right) + 1}}

 

[K.J.Healey]

First he has your two equations but I think a different sign (for 8 pi G P and rho)

You seem to be ignoring T22, which he has as:

8 \pi T_2^2 = -e^{-\lambda} \left ( \frac{\nu&#039;&#039;}{2} - \frac{\lambda&#039; \nu&#039;}{4}+\frac{\nu&#039;^2}{4}+\frac{\nu&#039;-\lambda&#039;}{2 r} \right )

That with his
8 \pi T_1^1 = -e^{- \lambda} \left ( \frac{\nu&#039;}{r} + \frac{1}{r^2} \right ) + \frac{1}{r^2}

He says for a stationary perfect fluid:

T_1^1 = T_2^2 = T_3^3 = -\rho_0

So he takes
T_1^1 - T_2^2 = 0
-e^{- \lambda} \left ( \frac{\nu&#039;}{r} + \frac{1}{r^2} \right ) + \frac{1}{r^2} +e^{-\lambda} \left ( \frac{\nu&#039;}{2} - \frac{\lambda&#039; \nu&#039;}{4}+\frac{\nu&#039;^2}{4}+\frac{\nu&#039;-\lambda&#039;}{2 r} \right ) =0
which is also
e^{- \lambda} \left ( -\frac{\nu&#039;}{r} - \frac{1}{r^2} + \frac{\nu&#039;&#039;}{2} - \frac{\lambda&#039; \nu&#039;}{4}+\frac{\nu&#039;^2}{4}+\frac{\nu&#039;-\lambda&#039;}{2 r} \right ) + \frac{1}{r^2}=0

He regroups them and multiplies thru by 2/r

<br /> e^{- \lambda} \left ( \frac{\nu&#039;&#039;}{r} - \frac{\nu&#039;}{r^2}-\frac{2}{r^3} \right ) - e^{-\lambda} \lambda&#039; \left(\frac{\nu&#039;}{r}+\frac{1}{r^2} \right) + \frac{2}{r^3} + e^{-\lambda} \left ( \frac{\lambda&#039;}{r} + \frac{\nu&#039;}{r} \right ) \frac{\nu&#039;}{2} =0

and I quote "which on comparison with 95.3 and 95.10 is seen to be equivalent to"

and 95.3 is just the set of T11 T22 and T44
and 95.10 is the t11=t22=t33=-rho and t44 = P
Where
<br /> 8 \pi T_4^4 = e^{-\lambda} \left( \frac{\lambda&#039;}{r} - \frac{1}{r^2} \right ) + \frac{1}{r^2}<br />

He takes the long equation above and says its equivalent to:
\frac{dP_0}{dr} + (\rho_{00} + P_0) \frac{\nu&#039;}{2} =0

which he says is the relativistic analouge to the Newtonian expression.

I guess it just takes some intuition to recoginse that if you differentiate one of the T's and multiply it by this and that its the same equation as what you want.
I don't see it, but I am sure he's right :)

 

[Orion1]

This is the equation for hydrostatic equilibrium:
\frac{dP}{dr} = \left( \rho + P \right) \frac{d \phi}{dr}

G_{\mu \nu} = \frac{8 \pi G}{c^4} T_{\mu \nu}

My equation for the Schwarzschild metric line element:
ds^2 = e^{\nu (r)} c^2 dt^2 - e^{\lambda (r)} dr^2 - r^2 d \theta^2 - r^2 \sin^2 \theta d \phi^2

The matrix forms of the Schwarzschild metric tensors:
G_{\mu\nu} = \left(<br /> \begin{array}{llll}<br /> e^{\nu (r)} &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; -e^{\lambda (r)} &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; -r^2 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; -r^2 \sin ^2(\theta )<br /> \end{array}<br /> \right)

T_{\mu\nu} = \left(<br /> \begin{array}{llll}<br /> \rho c^2 &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; -P &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; -P &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; -P<br /> \end{array}<br /> \right)

My equation for the Kerr metric line element:
c^{2} d\tau^{2} = e^{\nu (r)} c^{2} dt^{2} - e^{\lambda (r)} dr^{2} - \rho^{2} d\theta^{2} - \left( r^{2} + \alpha^{2} + \frac{r_{s} r \alpha^{2}}{\rho^{2}} \sin^{2} \theta \right) \sin^{2} \theta \ d\phi^{2} + \frac{2r_{s} r\alpha \sin^{2} \theta }{\rho^{2}} \, c \, dt \, d\phi

\boxed{e^{\nu (r)} = \left( 1 - \frac{r_{s} r}{\rho^{2}} \right)} \; \; \; \; \; \; \boxed{e^{\lambda (r)} = \frac{\rho^{2}}{\Lambda^{2}}}

The matrix forms of the Kerr metric tensors:
G_{\mu\nu} = \left(<br /> \begin{array}{llll}<br /> e^{\nu (r)} &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; -e^{\lambda (r)} &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; - \rho^2 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; - \left(r^2 + \alpha^2 + \frac{r_s r \alpha^2}{\rho^2} \sin ^2 \theta \right) \sin ^2 \theta<br /> \end{array}<br /> \right)

T_{\mu\nu} = \left(<br /> \begin{array}{llll}<br /> \rho c^2 &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; -P &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; -P &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; -P<br /> \end{array}<br /> \right)

Would extracting the Kerr metric matrix element G_{11} for pressure from the conceptually equivalent Mathematica source code formulas in post #54, produce an equation solution conceptually equivalent to the Schwarzschild metric matrix element G_{11} for pressure?
[/Color]
Reference:
http://en.wikipedia.org/wiki/Kerr_metric"
Schwarzschild_metric - Wikipedia
http://library.wolfram.com/infocenter/MathSource/162/"

 

[K.J.Healey]

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)} &amp; 0 &amp; 0 &amp; \frac{r_s r \alpha sin^2 \theta}{\rho^2} \\<br /> 0 &amp; -e^{\lambda (r)} &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; - \rho^2 &amp; 0 \\<br /> \frac{r_s r \alpha sin^2 \theta}{\rho^2} &amp; 0 &amp; 0 &amp; - \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.

 

[K.J.Healey]

remember your line element must be made from

<br /> ds^2=g_{\mu\nu} dx^\mu dx^\nu<br />

where dx^\mu is like (dt,dr,r d\theta, r sin\theta d\phi) I believe.

 

[Orion1]

Kerr metric...


The Kerr metric:
c^{2} d\tau^{2} = \left( 1 - \frac{r_{s} r}{\rho^{2}} \right) c^{2} dt^{2} <br /> - \frac{\rho^{2}}{\Lambda^{2}} dr^{2} - \rho^{2} d\theta^{2} - \left( r^{2} + \alpha^{2} + \frac{r_{s} r \alpha^{2}}{\rho^{2}} \sin^{2} \theta \right) \sin^{2} \theta \ d\phi^{2} + \frac{2r_{s} r\alpha \sin^{2} \theta }{\rho^{2}} \, c \, dt \, d\phi
\alpha = \frac{J}{Mc} \; \; \; \; \; \; \rho^{2} = r^{2} + \alpha^{2} \cos^{2} \theta \; \; \; \; \; \; \Lambda^{2} = r^{2} - r_{s} r + \alpha^{2}


I reduced each matrix element into the most trigonometric form.

g_{00} = 1 - \frac{r_{s} r}{\rho^{2}} = 1 - \frac{r_{s} r}{r^{2} + \alpha^{2} \cos^{2} \theta}
\boxed{g_{00} = 1 - \frac{r_{s} r}{r^{2} + \alpha^{2} \cos^{2} \theta}}
g_{03} = g_{30} = \frac{r_s r \alpha sin^2 \theta}{\rho^2} = \frac{r_s r \alpha sin^2 \theta}{r^{2} + \alpha^{2} \cos^{2} \theta}
\boxed{g_{03} = g_{30} = \frac{r_s r \alpha sin^2 \theta}{r^{2} + \alpha^{2} \cos^{2} \theta}}
g_{11} = - \frac{\rho^{2}}{\Lambda^{2}} = - \frac{r^{2} + \alpha^{2} \cos^{2} \theta}{r^{2} - r_{s} r + \alpha^{2}}
\boxed{g_{11} = - \frac{r^{2} + \alpha^{2} \cos^{2} \theta}{r^{2} - r_{s} r + \alpha^{2}}}
g_{22} = - \rho^2 = - (r^{2} + \alpha^{2} \cos^{2} \theta)
\boxed{g_{22} = - (r^{2} + \alpha^{2} \cos^{2} \theta)}

g_{\mu \nu} = \left( \begin{array}{llll} 1-\frac{r r_s}{r^2+\alpha ^2 \cos ^2(\theta )} &amp; 0 &amp; 0 &amp; \frac{r \alpha \sin ^2(\theta ) r_s}{r^2+\alpha ^2 \cos ^2(\theta )} \\ 0 &amp; - \frac{r^2+\alpha ^2 \cos ^2(\theta )}{r^2-r_s r+\alpha ^2} &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; -r^2-\alpha ^2 \cos ^2(\theta ) &amp; 0 \\ \frac{r \alpha \sin ^2(\theta ) r_s}{r^2+\alpha ^2 \cos ^2(\theta )} &amp; 0 &amp; 0 &amp; \sin ^2(\theta ) \left(-r^2-\alpha ^2 \sin ^2(\theta ) r_s r-\alpha ^2\right) \end{array} \right)

Einsteintensor source code:
In[n]=

ToFileName[{$TopDirectory, "AddOns", "Applications"}]
<< einsteintensor.m
x = {t, r, \[Theta], \[Phi]}
(metric = {{1 - (Subscript[r, s]*r)/(r^2 + \[Alpha]^2*Cos[\[Theta]]^2), 0, 0, (Subscript[r, s]*r*\[Alpha]*Sin[\[Theta]]^2)/(r^2 + \[Alpha]^2*Cos[\[Theta]]^2)}, {0, -((r^2 + \[Alpha]^2*Cos[\[Theta]]^2)/(r^2 - Subscript[r, s]*r + \[Alpha]^2)), 0, 0}, {0, 0, -(r^2 + \[Alpha]^2*Cos[\[Theta]]^2), 0}, {(Subscript[r, s]*r*\[Alpha]*Sin[\[Theta]]^2)/(r^2 + \[Alpha]^2*Cos[\[Theta]]^2), 0, 0, -(r^2 + \[Alpha]^2 + Subscript[r, s]*r*\[Alpha]^2*Sin[\[Theta]]^2)*Sin[\[Theta]]^2}}) // MatrixForm
(tensor = {{\[Rho][r]*c^2, 0, 0, Subscript[\[CapitalPhi], \[Epsilon]][r]}, {0, -P[r], 0, 0},{0, 0, -P[r], 0}, {Subscript[\[Rho], p][r], 0, 0, -P[r]}}) // MatrixForm
(Einstein = Inverse[metric].Simplify[EinsteinTensor[metric, x], TimeConstraint -> 3600]) // MatrixForm
MaxMemoryUsed[]

The resulting evaluation function required a lot of memory:
MaxMemoryUsed[] = 1745544760 bytes

Unfortunately, my computer does have enough memory to complete all the transformations and the result:[/Color]
Simplify::time: Time spent on a transformation exceeded 300 seconds, and the transformation was aborted.
General::stop: Further output of Simplify::time will be suppressed during this calculation.
[/Color]
Reference:
http://en.wikipedia.org/wiki/Kerr_metric"
http://library.wolfram.com/infocenter/MathSource/162/"

 

[Orion1]

Correction, the matrix and source code from previous post should be:
\boxed{g_{33} = \sin ^2(\theta ) \left(-r^2-\frac{\alpha ^2 \sin ^2(\theta ) r_s r}{r^2+\alpha ^2 \cos ^2(\theta )}-\alpha ^2\right)}

[/Color]
g_{\mu \nu} = \left( \begin{array}{llll} 1 - \frac{r r_s}{r^2+\alpha ^2 \cos ^2(\theta )} &amp; 0 &amp; 0 &amp; \frac{r \alpha \sin ^2(\theta ) r_s}{r^2+\alpha ^2 \cos ^2(\theta )} \\ 0 &amp; -\frac{r^2+\alpha ^2 \cos ^2(\theta )}{r^2-r_s r+\alpha ^2} &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; -r^2-\alpha ^2 \cos ^2(\theta ) &amp; 0 \\ \frac{r \alpha \sin ^2(\theta ) r_s}{r^2+\alpha ^2 \cos ^2(\theta )} &amp; 0 &amp; 0 &amp; \sin ^2(\theta ) \left(-r^2-\frac{\alpha ^2 \sin ^2(\theta ) r_s r}{r^2+\alpha ^2 \cos ^2(\theta )}-\alpha ^2\right) \end{array} \right)

Einsteintensor source code:
In[n]=

ToFileName[{$TopDirectory, "AddOns", "Applications"}]
<< einsteintensor.m
x = {t, r, \[Theta], \[Phi]}
(metric = {{1 - (Subscript[r, s]*r)/(r^2 + \[Alpha]^2*Cos[\[Theta]]^2), 0, 0, (Subscript[r, s]*r*\[Alpha]*Sin[\[Theta]]^2)/(r^2 + \[Alpha]^2*Cos[\[Theta]]^2)}, {0, -((r^2 + \[Alpha]^2*Cos[\[Theta]]^2)/(r^2 - Subscript[r, s]*r + \[Alpha]^2)), 0, 0}, {0, 0, -(r^2 + \[Alpha]^2*Cos[\[Theta]]^2), 0}, {(Subscript[r, s]*r*\[Alpha]*Sin[\[Theta]]^2)/(r^2 + \[Alpha]^2*Cos[\[Theta]]^2), 0, 0, -(r^2 + \[Alpha]^2 + (Subscript[r, s]*r*\[Alpha]^2*Sin[\[Theta]]^2)/(r^2 + \[Alpha]^2*Cos[\[Theta]]^2))*Sin[\[Theta]]^2}}) // MatrixForm
(tensor = {{\[Rho][r]*c^2, 0, 0, Subscript[\[CapitalPhi], \[Epsilon]][r]}, {0, -P[r], 0, 0},{0, 0, -P[r], 0}, {Subscript[\[Rho], p][r], 0, 0, -P[r]}}) // MatrixForm
(Einstein = Inverse[metric].Simplify[EinsteinTensor[metric, x], TimeConstraint -> 3600]) // MatrixForm
MaxMemoryUsed[]

[/Color]

 

[Orion1]

neutron star mass-radius relation...


The neutron star mass-radius relation is dependent on a particular neutron star model, however the mass-radius relation for my model based upon the Proton charge radius and Tolman mass equation solution VII:

m_n = 1.6749272928 \cdot 10^{-27} \; \text{kg} - Neutron mass
r_p = 0.8757 \cdot 10^{-15} \; \text{m} - Proton charge radius

Proton charge radius neutron density:
\rho_n = \frac{3 m_n}{4 \pi r_p^3}

Neutron star core density equivalent to Proton charge radius neutron density:
\rho_c = \rho_n

Total Tolman mass equation solution VII:
M_0(R) = \frac{8 \pi \rho_c R^3}{15} = \frac{8 \pi R^3}{15} \left( \frac{3 m_n}{4 \pi r_p^3} \right) = \frac{2 m_n R^3}{5 r_p^3}

Total mass-radius relation equation for Tolman solution VII:
\boxed{M_0(R) = \frac{2 m_n R^3}{5 r_p^3}}

Mass of a 10 km radius Tolman VII neutron star:
\boxed{M_0(10 \; \text{km}) = 9.976 \cdot 10^{29} \; \text{kg}}

\boxed{M_0(10 \; \text{km}) = 0.501 \cdot M_{\odot}}

Note that the lower limit for total radius R, is equivalent to the Schwarzschild radius and the upper limit for total mass M(R), is equivalent to the Tolman-Oppenheimer-Volkov mass limit.
[/Color]
Reference:
Neutron - Wikipedia
TOV #39 - Orion1
TOV #47 - Orion1
Schwarzschild radius - Wikipedia
Tolman-Oppenheimer-Volkoff mass limit - Wikipedia