Tolman-Oppenheimer-Volkoff equation

[Orion1]

Neutron average density:
$$\boxed{\rho_n = \frac{3}{4 \cdot 10^3 \pi N_A r_0^3}}$$

Neutron star core density:
$$\boxed{\rho_c = \rho_n}$$

Integration by substitution:
$$\frac{dP}{dr} = - \left[ \frac{3^{2/3} \pi^{4/3} \hbar^2}{5 m_n^{8/3}} \left( \left( \frac{3}{4 \cdot 10^3 \pi N_A r_0^3} \right) \left[1 - \left( \frac{r}{R} \right)^2 \right] \right)^{5/3} + \left( \frac{3}{4 \cdot 10^3 \pi N_A r_0^3} \right) c^2 \left[1 - \left( \frac{r}{R} \right)^2 \right] \right]...$$

$$...\left[ \left( \frac{4 \cdot 3^{2/3} \pi^{7/3} \hbar^2 G r^3}{5 c^4 m_n^{8/3}} \left( \left( \frac{3}{4 \cdot 10^3 \pi N_A r_0^3} \right) \left[1 - \left( \frac{r}{R} \right)^2 \right] \right)^{5/3} \right) + \frac{4 \pi G }{15 c^2} \left( \frac{3}{4 \cdot 10^3 \pi N_A r_0^3} \right) \left( \frac{3r^5}{R^2} + 2R^3 - 5r^3 \right) \right]...$$

$$... \left[ r \left( r - \frac{4 \pi G}{15 c^2} \left( \frac{3}{4 \cdot 10^3 \pi N_A r_0^3} \right)} \left( \frac{3r^5}{R^2} + 2R^3 - 5r^3 \right) \right) \right]^{-1}$$

Degenerate Fermi-TOV equation solution VII:
$$\frac{dP}{dr} = - \left[ \frac{9 \hbar^2 }{4 \cdot 10^6 m_n^{\frac{8}{3}}} \left( \frac{3}{2 \pi} \right)^{\frac{1}{3}} \left( \frac{1}{N_A r_0^3} \right)^{\frac{5}{3}} \left[1 - \left( \frac{r}{R} \right)^2 \right]^{5/3} + \frac{3 c^2}{4 \cdot 10^3 \pi N_A r_0^3} \left[1 - \left( \frac{r}{R} \right)^2 \right] \right]...$$

$$...\left[ \frac{9 \pi^{\frac{2}{3}} \hbar^2 G r^3}{10^6 c^4 m_n^{\frac{8}{3}}} \left( \frac{3}{2} \right)^{\frac{1}{3}} \left( \frac{1}{N_A r_0^3} \right)^{\frac{5}{3}} \left[1 - \left( \frac{r}{R} \right)^2 \right]^{5/3} + \frac{G}{5 \cdot 10^3 c^2 N_A r_0^3} \left( \frac{3r^5}{R^2} + 2R^3 - 5r^3 \right) \right]...$$

$$... \left[ r \left( r - \frac{G}{5 \cdot 10^3 c^2 N_A r_0^3} \left( \frac{3r^5}{R^2} + 2R^3 - 5r^3 \right) \right) \right]^{-1}$$

 

[Orion1]

Note: physicsforums only allows for 30 minutes for editing posts.

Corrections:
The third factor on post #34 should read:
$$... \left[ r \left( r - \frac{8 \pi G \rho_c}{15 c^2} \left( \frac{3r^5}{R^2} + 2R^3 - 5r^3 \right) \right) \right]^{-1}$$

The boxed solution for Neutron density on post #35 should read:
$$\boxed{\rho_n = \frac{3}{4 \cdot 10^3 \pi N_A r_0^3}}$$

The first third factor on post #36 should read:
$$... \left[ r \left( r - \frac{8 \pi G}{15 c^2} \left( \frac{3}{4 \cdot 10^3 \pi N_A r_0^3} \right)} \left( \frac{3r^5}{R^2} + 2R^3 - 5r^3 \right) \right) \right]^{-1}$$

The second third factor on post #36 should read:
$$... \left[ r \left( r - \frac{2 G}{5 \cdot 10^3 c^2 N_A r_0^3} \left( \frac{3r^5}{R^2} + 2R^3 - 5r^3 \right) \right) \right]^{-1}$$

 

[Orion1]

The attached image is my first attempt to plot the degenerate Fermi-TOV equation solution VII function based upon this model.

These are the results of a 10 km radius static model test.

The function projects semi-asymptoticly until the core radius is reached and then projects a semi-parabolic curve.

Degenerate Fermi-TOV neutron star properties:
Core Radius:
$$m = 1$$ - slope
$$sgn \left( \frac{dP}{dr} \right) = -1 \; \; \; r < r_c$$ - sign
$$\frac{dP}{dr} = 0 \; \; \; r = r_c$$
$$r_c = .3421 \; \text{km}$$

Shell 1:
$$m = 1$$ - slope
$$sgn \left( \frac{dP}{dr} \right) = 1 \; \; \; r_c < r < r_1$$ - sign
$$m = 0 \; \; \; r = r_1$$ - peak resonance
$$\frac{dP}{dr} = \text{3.638} \cdot \text{10}^{28} \; \frac{\text{N}}{\text{m}^3} \; \; \; r = r_1$$
$$r_1 = 4.8673 \; \text{km}$$

Shell 2:
$$m = -1$$ - slope
$$sgn \left( \frac{dP}{dr} \right) = 1 \; \; \; r_1 < r < R$$ - sign
$$r_2 = 10 \; \text{km}$$

Shell 3: Degenerate Iron crust

Total core pressure:
$$P_c = \int_0^R \left( \frac{dP}{dr} \right) dr$$

Reference:
http://en.wikipedia.org/wiki/Slope"
http://en.wikipedia.org/wiki/Sign_function"

 

[Orion1]

Mass integration function equation:
$$m(r) = 4 \pi \int_r^R r^2 \rho(r) dr$$

Tolman density equation solution VII:
$$\rho(r) = \rho_c \left[ 1 - \left( \frac{r}{R} \right)^2 \right] \; \; \; \rho(R) = 0$$

Integration by substitution:
$$m(r) = 4 \pi \rho_c \int_r^R r^2 \left[ 1 - \left( \frac{r}{R} \right)^2 \right] dr = 4 \pi \rho_c \left( \frac{r^5}{5R^2} + \frac{2R^3}{15} - \frac{r^3}{3} \right) = \frac{4 \pi \rho_c}{15} \left( \frac{3r^5}{R^2} + 2R^3 - 5r^3 \right)$$

Tolman mass function equation solution VII:
$$\boxed{m(r) = \frac{4 \pi \rho_c}{15} \left( \frac{3r^5}{R^2} + 2R^3 - 5r^3 \right) \; \; \; m(R) = 0}$$

Total mass function integration:
$$M_0 = 4 \pi \int_0^R r^2 \rho(r) dr$$

Tolman density equation solution VII:
$$\rho(r) = \rho_c \left[ 1 - \left( \frac{r}{R} \right)^2 \right] \; \; \; \rho(R) = 0$$

Integration by substitution:
$$M_0 = 4 \pi \rho_c \int_0^R r^2 \left[ 1 - \left( \frac{r}{R} \right)^2 \right] = \frac{8 \pi \rho_c R^3}{15}$$

Total Tolman mass equation solution VII:
$$\boxed{M_0 = \frac{8 \pi \rho_c R^3}{15}}$$


Correction:
The declaration equations for slope on post# 38 should be:
$$sgn ( m ) = -1,0,1$$ instead of $$m = -1,0,1$$

 

[Orion1]

Total Tolman mass equation solution VII:
$$M_0 = \frac{8 \pi \rho_c R^3}{15}$$

Neutron average density:
$$\rho_n = \frac{3}{4 \cdot 10^3 \pi N_A r_0^3}}$$

Core density:
$$\rho_c = \rho_n$$

Integration by substitution:
$$M_0 = \frac{8 \pi}{15} \left( \frac{3}{4 \cdot 10^3 \pi N_A r_0^3}} \right) R^3 = \frac{1}{2.5 \cdot 10^3 N_A} \left( \frac{R}{r_0} \right)^3$$

$$\boxed{M_0 = \frac{1}{2.5 \cdot 10^3 N_A} \left( \frac{R}{r_0} \right)^3}$$

Static model radius:
$$R = 10 \; \text{km}$$

Static model mass:
$$\boxed{M_0 = 3.4007 \cdot 10^{29} \; \text{kg}}$$

$$M_0 = 0.1709 \cdot M_{\odot}$$

Reference:
http://en.wikipedia.org/wiki/Sun"

 

[Orion1]

The TOV equation can only be numerically integrated using Riemann sum rules.

Total core pressure:
$$P_c = \int_0^R \left( \frac{dP}{dr} \right) dr$$

Midpoint Rule numerical integration:
$$P_c = \int_0^R \left( \frac{dP}{dr} \right) dr = \int_0^R f(r) dr = \sum_{i = 1}^n f(\overline{r_i}) \Delta r \; \; \; \Delta r = \frac{R - 0}{n} \; \; \; \overline{r_i} = \frac{1}{2} (r_{i - 1} + r_i)$$

$$\boxed{P_c = \int_0^R f(r) dr = \sum_{i = 1}^n f(\overline{r_i}) \Delta r \; \; \; \Delta r = \frac{R}{n} \; \; \; \overline{r_i} = \frac{1}{2} (r_{i - 1} + r_i)}$$

Degenerate Fermi-TOV equation solution VII
Static model test results for R = 10 km.
Total core pressure:
$$\boxed{P_c = -1.8635 \cdot 10^{15} \; \frac{\text{N}}{\tex{m}^2}}$$

Reference:
http://en.wikipedia.org/wiki/Riemann_sum"

 

[Orion1]

Correction, the numerical solution for core pressure on post # 41 is incorrect as a result of incomplete integration.
The correct solution is:

Degenerate Fermi-TOV equation solution VII
Static model test results for R = 10 km.
Total core pressure:
$$\boxed{P_c = 1.8407 \cdot 10^{36} \; \frac{\text{N}}{\tex{m}^2}}$$

 

[Orion1]

Calculating the maximum mass of a neutron star.

Neutron star total gravitational mass:
$$M_G = m(R) = \int_0^R 4 \pi r^2 \rho(r) dr$$

Schwarzschild metric spherical layer differential volume:
$$dV = \frac{4 \pi r^2 dr}{ \sqrt{\left[ 1 - \frac{r_s}{r} \right]}}$$
$$r_s$$ - Schwarzschild radius

Baryon number density:
$$n(r) = \frac{N(r)}{V(r)} = \frac{\rho(r)}{m_n}$$

Neutron star total baryon number:
$$N_B = \int_0^R n(r) dV = \int_0^R \frac{4 \pi r^2 \rho(r) dr}{ m_n \sqrt{\left[ 1 - \frac{r_s}{r} \right]}}$$

$$\boxed{N_B = \int_0^R \frac{4 \pi r^2 \rho(r) dr}{ m_n \sqrt{\left[ 1 - \frac{r_s}{r} \right]}}}$$

Neutron star total baryonic mass:
$$M_B = m_n N_B = \int_0^R \frac{4 \pi r^2 \rho(r) dr}{ \sqrt{\left[ 1 - \frac{r_s}{r} \right]}}}$$

$$\boxed{M_B = \int_0^R \frac{4 \pi r^2 \rho(r) dr}{ \sqrt{\left[ 1 - \frac{r_s}{r} \right]}}}}$$

Neutron star total binding energy:
$$B = (M_B - M_G)c^2$$

Reference:
http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?1996A%26A...305..871B&amp;data_type=PDF_HIGH&amp;whole_paper=YES&amp;type=PRINTER&amp;filetype=.pdf"
http://en.wikipedia.org/wiki/Schwarzschild_radius"

 

[malawi_glenn]

Be careful on what radius you use for the neutron.

r = r_0*A^(1/3) is an emperical formula for middle mass nuclei, not a 'law'.

It should be better to use the same radius as for the proton, i.e 0.87fm (charge radius of proton, r.m.s)

See also this thread: https://www.physicsforums.com/showthread.php?t=241465

Also, it is quite naive that a neutron star is 100% neutrons.

 

[Orion1]

Proton charge radius...


Proton charge radius:
$$r_p = \sqrt{\left< r^2 \right>} = \sqrt{-6 \lim_{Q \rightarrow 0} \frac{d}{dQ^2} F(Q)}$$

ref. 1 Proton charge radius:
r_p = 0.805 ± 0.011 and 0.862 ± 0.012 femtometer (Stein 1995).

Wikipedia Proton charge radius:
r_p = 0.875(7) fm

I found reference for the Proton charge radius, unfortunately Wikipedia does not cite a reference, which radius is more accurate?

Reference:
Stein, B. P. "Physics Update." Physics Today 48, 9, Oct. 1995.
http://scienceworld.wolfram.com/physics/Proton.html"
http://en.wikipedia.org/wiki/Proton"
http://cnr2.kent.edu/~pichowsk/IntroNuc/hw3.pdf"

 

[malawi_glenn]

Particle data group: http://pdg.lbl.gov/2007/tables/bxxx.pdf

But what you really want is the nuclear matter distribution for nucleons, i.e you want to know the 'size' of the 'quark cloud'. In the first approximation, this should be the same as the charge radius distrubution. Since the neutron is overall neutral, the charge radius is not a good measurement (it becomes negative hehe..), so the first approximation of neutron radius is the same the proton radius.

 

[Orion1]

$$r_0 = 1.25 \cdot 10^{-15} \; \text{m}$$ - empirical nuclear radius

Empirical neutron density:
$$\rho_n = \frac{3}{4 \cdot 10^3 \pi N_A r_0^3}} = 2.0297 \cdot 10^{17} \; \frac{\text{kg}}{\text{m}^3}$$

$$r_p = 0.8757 \cdot 10^{-15} \; \text{m}$$ - Proton charge radius

Proton charge radius neutron density:
$$\rho_n = \frac{3 m_n}{4 \pi r_p^3} = 5.954 \cdot 10^{17} \; \frac{\text{kg}}{\text{m}^3}$$

$$\boxed{\rho_n = 5.954 \cdot 10^{17} \; \frac{\text{kg}}{\text{m}^3}}$$

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:
$$\rho_c = \rho_n$$

Degenerate Fermi-TOV equation:
$$\frac{dP}{dr} = - \left[ \frac{3^{2/3} \pi^{4/3} \hbar^2}{5 m_n^{8/3}} \left( \frac{3 m_n}{4 \pi r_p^3} \left[1 - \left( \frac{r}{R} \right)^2 \right] \right)^{5/3} + \left( \frac{3 m_n}{4 \pi r_p^3} \right) c^2 \left[1 - \left( \frac{r}{R} \right)^2 \right] \right]...$$

$$...\left[ \left( \frac{4 \cdot 3^{2/3} \pi^{7/3} \hbar^2 G r^3}{5 c^4 m_n^{8/3}} \left( \frac{3 m_n}{4 \pi r_p^3} \left[1 - \left( \frac{r}{R} \right)^2 \right] \right)^{5/3} \right) + \frac{4 \pi G}{15 c^2} \left( \frac{3 m_n}{4 \pi r_p^3} \right) \left( \frac{3r^5}{R^2} + 2R^3 - 5r^3 \right) \right]...$$

$$... \left[ r \left( r - \frac{8 \pi G}{15 c^2} \left( \frac{3 m_n}{4 \pi r_p^3} \right) \left( \frac{3r^5}{R^2} + 2R^3 - 5r^3 \right) \right) \right]^{-1}$$

Degenerate Fermi-TOV equation charge radius equation:
$$\frac{dP}{dr} = - \left[ \frac{3^{\frac{7}{3}} \hbar^2}{4^{\frac{5}{3}} \pi^{\frac{1}{3}} m_n r_p^5} \left[ 1 - \left( \frac{r}{R} \right)^2 \right]^{5/3} + \frac{3 c^2 m_n}{4 \pi r_p^3} \left[1 - \left( \frac{r}{R} \right)^2 \right] \right]...$$

$$...\left[ \left( \frac{3^{7/3} \pi^{4/3} \hbar^2 G r^3}{5 \cdot 4^{\frac{2}{3}} c^4 m_n r_p^5} \left[1 - \left( \frac{r}{R} \right)^2 \right]^{5/3} \right) + \frac{G m_n}{5 c^2 r_p^3} \left( \frac{3r^5}{R^2} + 2R^3 - 5r^3 \right) \right]...$$

$$... \left[ r \left( r - \frac{2 G m_n}{5 c^2 r_p^3} \left( \frac{3r^5}{R^2} + 2R^3 - 5r^3 \right) \right) \right]^{-1}$$

The first attached photo is the static model based upon the empirical neutron radius, the second attachment is the static model based upon the Proton charge radius.

Replacing the empirical neutron density with proton charge radius neutron density in a R = 10 km static model resulted in a peak change in pressure increase by a factor of 8.6. The core radius increased by 15 meters and the first shell radius decreased by 112 meters.

The total core pressure became negative, which I believe is required for a neutron star with a stable core.

Total core pressure:
$$\boxed{P_c = -1.713 \cdot 10^{36} \; \frac{\text{N}}{\tex{m}^2}}$$

key:
$$\rho_c$$ - neutron star core density
$$R$$ - neutron star radius

Reference:
http://en.wikipedia.org/wiki/Neutron_star"

 

[malawi_glenn]

Orion1: the 'emperical nuclear radius' is only applicable to medium size nuclei... how many times do I have to tell you?

 

[K.J.Healey]

Maybe I'm misunderstanding this but why would the intra-nucleon distance be related to the "charge radius" of a proton? I guess I could see them being both on the order of ~1fm. But shouldn't it be a force balance between the neutrons with the strong interaction and the gravitic pressure that determines the separation distance?

Seems wrong to use anything from a single atom to describe the core of a neutron star.

 

[Orion1]

The nuclear density of nuclei is described by the empirical nuclear radius, which describes the radii of large nuclei, however it does not accurately describe the radii of light nuclei or nuclear particles, because it is not a natural law. The 'charge radius' of a proton is a natural law and used to describe the radius of a neutron in this model. (ref. 3)

Using the nuclear density properties of a single neutron to describe a neutron star nuclear core density is correct.

The neutron star itself is a balance between positive neutron degeneracy pressures generated from the Pauli exclusion principle, which states that no two identical fermions may occupy the same quantum state simultaneously and indicating repulsion between neutrons, and the negative internal and surface gravitational pressures, calculated from the TOV equation of state generated from Relativity.

Neutrons are the most 'rigid' objects known - their Young modulus (or more accurately, bulk modulus) is 20 orders of magnitude larger than that of diamond.

Reference:
http://en.wikipedia.org/wiki/Pauli_exclusion_principle"
http://en.wikipedia.org/wiki/Neutron_star"
https://www.physicsforums.com/showpost.php?p=1789174&postcount=45"

 

[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?

 

[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=
$$\left( \begin{array}{llll} e^{\nu (r)} & 0 & 0 & 0 \\ 0 & -e^{\lambda (r)} & 0 & 0 \\ 0 & 0 & -r^2 & 0 \\ 0 & 0 & 0 & -r^2 \sin ^2(\theta ) \end{array} \right)$$

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

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?

Correction, the Einstein tensor in post #54 should read:
$$G_{1 1} = e^{- \lambda (r)} \left( \frac{1}{r} \frac{d \nu}{dr} + \frac{1}{r^2} \right) - \frac{1}{r^2}$$

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:
$$\boxed{G_{00} = \frac{e^{-\lambda (r)} \left(r \lambda '(r) + e^{\lambda (r)} - 1 \right)}{r^2}}$$

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

$$\frac{8 \pi G}{c^4} \rho (r) c^2 = \frac{e^{-\lambda (r)} \left(r \lambda '(r)+e^{\lambda (r)}-1\right)}{r^2}$$

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

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

$$\frac{\left(r \nu '(r) - e^{\lambda (r)} + 1 \right)}{P(r)} = \frac{\left(r \lambda '(r)+e^{\lambda (r)} - 1 \right)}{\rho (r) c^2}$$

$$\boxed{\frac{P(r)}{\rho (r) c^2} = \frac{\left(r \nu '(r) - e^{\lambda (r)} + 1 \right)}{\left(r \lambda '(r) + e^{\lambda (r)} - 1 \right)}}$$

 

[Orion1]

$$\boxed{\frac{P(r)}{\rho (r) c^2} = \frac{e^{- \lambda (r)} \left( r \nu ' (r) + 1 \right) - 1}{e^{- \lambda (r)} \left( r \lambda '(r) - 1 \right) + 1}}$$

 

[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:

$$8 \pi T_2^2 = -e^{-\lambda} \left ( \frac{\nu''}{2} - \frac{\lambda' \nu'}{4}+\frac{\nu'^2}{4}+\frac{\nu'-\lambda'}{2 r} \right )$$

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

He says for a stationary perfect fluid:

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

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

He regroups them and multiplies thru by 2/r

$$e^{- \lambda} \left ( \frac{\nu''}{r} - \frac{\nu'}{r^2}-\frac{2}{r^3} \right ) - e^{-\lambda} \lambda' \left(\frac{\nu'}{r}+\frac{1}{r^2} \right) + \frac{2}{r^3} + e^{-\lambda} \left ( \frac{\lambda'}{r} + \frac{\nu'}{r} \right ) \frac{\nu'}{2} =0$$

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
$$8 \pi T_4^4 = e^{-\lambda} \left( \frac{\lambda'}{r} - \frac{1}{r^2} \right ) + \frac{1}{r^2}$$

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

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:
$$G_{\mu\nu} = \left( \begin{array}{llll} e^{\nu (r)} & 0 & 0 & 0 \\ 0 & -e^{\lambda (r)} & 0 & 0 \\ 0 & 0 & -r^2 & 0 \\ 0 & 0 & 0 & -r^2 \sin ^2(\theta ) \end{array} \right)$$

$$T_{\mu\nu} = \left( \begin{array}{llll} \rho c^2 & 0 & 0 & 0 \\ 0 & -P & 0 & 0 \\ 0 & 0 & -P & 0 \\ 0 & 0 & 0 & -P \end{array} \right)$$

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:
$$G_{\mu\nu} = \left( \begin{array}{llll} e^{\nu (r)} & 0 & 0 & 0 \\ 0 & -e^{\lambda (r)} & 0 & 0 \\ 0 & 0 & - \rho^2 & 0 \\ 0 & 0 & 0 & - \left(r^2 + \alpha^2 + \frac{r_s r \alpha^2}{\rho^2} \sin ^2 \theta \right) \sin ^2 \theta \end{array} \right)$$

$$T_{\mu\nu} = \left( \begin{array}{llll} \rho c^2 & 0 & 0 & 0 \\ 0 & -P & 0 & 0 \\ 0 & 0 & -P & 0 \\ 0 & 0 & 0 & -P \end{array} \right)$$

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?

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:
$$g_{\mu\nu} = \left( \begin{array}{llll} e^{\nu (r)} & 0 & 0 & \frac{r_s r \alpha sin^2 \theta}{\rho^2} \\ 0 & -e^{\lambda (r)} & 0 & 0 \\ 0 & 0 & - \rho^2 & 0 \\ \frac{r_s r \alpha sin^2 \theta}{\rho^2} & 0 & 0 & - \left(r^2 + \alpha^2 + \frac{r_s r \alpha^2}{\rho^2} \sin ^2 \theta \right) \end{array} \right)$$

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

$$ds^2=g_{\mu\nu} dx^\mu dx^\nu$$

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

 

[Orion1]

Kerr metric...


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


I reduced each matrix element into the most trigonometric form.

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

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

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:
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.

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:
$$\boxed{g_{33} = \sin ^2(\theta ) \left(-r^2-\frac{\alpha ^2 \sin ^2(\theta ) r_s r}{r^2+\alpha ^2 \cos ^2(\theta )}-\alpha ^2\right)}$$


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

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[]

 

[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.

Reference:
Neutron - Wikipedia
TOV #39 - Orion1
TOV #47 - Orion1
Schwarzschild radius - Wikipedia
Tolman-Oppenheimer-Volkoff mass limit - Wikipedia