# Tolman-Oppenheimer-Volkoff equation

## Main Question or Discussion Point

I attempted to derive the TOV equation in modern physics notation, however my equation solution does not seem to match the equation solution derived by Tolman, Oppenheimer and Volkoff. (ref.1)

Also, the equation solution listed in (ref. 1) does not match the equation solution listed on Wikipedia, which listed (ref. 1) as the source of the equation. The TOV equation listed in (ref. 1) does not contain the 'mass function' listed on the Wikipedia page (ref. 2).

The (ref. 1) paper describes how the TOV equation was derived:
$\tag{3} 8 \pi P(r) = e^{- \lambda} \left( \frac{1}{r} \frac{d \nu}{dr} + \frac{1}{r^2} \right) - \frac{1}{r^2}$
$\tag{8} e^{-\lambda} = r(r - 2u)$
$\tag{5} \frac{d \nu}{dr} = \frac{2}{P(r) + \rho(r) c^2} \left( \frac{dP}{dr} \right)$

In Eq. (3) replace $e^{- \lambda}$ by its value from (8) and $\nu '$ by its value from (5). It becomes:

Solve for: $\frac{dP}{dr}$
$\tag{10} \frac{dP}{dr} = - (P(r) + \rho(r) c^2) [4 \pi r^3 P(r) + u] [r(r - 2u)]^{-1}$
And here is my first attempt to derive the TOV equation:
$8 \pi P(r) = e^{- \lambda} \left(\frac{1}{r} \frac{d \nu}{dr} + \frac{1}{r^2} \right) - \frac{1}{r^2}$

Identity:
$e^{-\lambda} = r(r - 2u) = 1 - \frac{2u}{r} = r(r - r_s) = 1 - \frac{r_s}{r}$
$\boxed{u = \frac{r_s}{2}}$
$\boxed{e^{-\lambda} = r(r - r_s)}$
$u(r) = \frac{1}{2} r(1 - e^{-\lambda})$
$e^{-\lambda} = r(r - 2u)$
$\frac{d \nu}{dr} = \frac{2}{P(r) + \rho(r) c^2} \left( \frac{dP}{dr} \right)$
---
This is my first attempt to derive this equation.
Integration by substitution:
$8 \pi P(r) = - ( r (r - 2u) ) ( ( \frac{2}{P(r) + \rho (r) c^2} ) ( \frac{dP}{dr} ) \frac{1}{r} + \frac{1}{r^2} ) - \frac{1}{r^2}$

My equation solution:
$\boxed{\frac{dP}{dr} = -(P(r) + \rho(r) c^2) \frac{r}{2} [(8 \pi P(r) + \frac{1}{r^2})[r(r - 2u)]^{-1} - \frac{1}{r^2}]}$

TOV equation solution: ref. 1
$\frac{dP}{dr} = - (P(r) + \rho(r) c^2) [4 \pi r^3 P(r) + u] [r(r - 2u)]^{-1}$

TOV equation solution: ref. 2
$\frac{dP}{dr} = - \frac{G}{r^2} [\rho(r) + \frac{P(r)}{c^2}][m(r) + 4 \pi r^3 \frac{P(r)}{c^2}][r(r - r_s)]^{-1}$

Reference:
http://en.wikipedia.org/wiki/Tolman-Oppenheimer-Volkoff_equation" [Broken]

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Related Astronomy and Astrophysics News on Phys.org

If the the stated equation solutions are equivalent, then the mass term in ref. 2 must have originated form the first 'u' term:

$(P(r) + \rho(r) c^2) [4 \pi r^3 P(r) + u] = \frac{G}{r^2} [\rho(r) + \frac{P(r)}{c^2}][m(r) + 4 \pi r^3 \frac{P(r)}{c^2}]$

Factoring out $c^2$ from the LHS results in:
$c^2 (\frac{P(r)}{c^2} + \rho(r)) [4 \pi r^3 \frac{P(r)}{c^2} + \frac{u}{c^2}] = \frac{G}{r^2} [\rho(r) + \frac{P(r)}{c^2}][m(r) + 4 \pi r^3 \frac{P(r)}{c^2}]$

Eliminate terms:
$c^2 [4 \pi r^3 \frac{P(r)}{c^2} + \frac{u}{c^2}] = \frac{G}{r^2} [m(r) + 4 \pi r^3 \frac{P(r)}{c^2}]$

Solve for u:
$4 \pi r^3 \frac{P(r)}{c^2} + \frac{u}{c^2} = \frac{G}{c^2 r^2} [m(r) + 4 \pi r^3 \frac{P(r)}{c^2}]$

$\frac{u}{c^2} = \frac{G}{c^2 r^2} [m(r) + 4 \pi r^3 \frac{P(r)}{c^2}] - 4 \pi r^3 \frac{P(r)}{c^2}$

$u = c^2 [(\frac{G m(r)}{c^2 r^2} + 4 \pi G r^3 \frac{P(r)}{r^2 c^4}) - 4 \pi r^3 \frac{P(r)}{c^2}]$

$u = \frac{G m(r)}{r^2} + 4 \pi G r^3 \frac{P(r)}{r^2 c^2} - 4 \pi r^3 P(r)$

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$\boxed{u = \frac{G m(r)}{r^2} + 4 \pi r^3 P(r) \left(1 - \frac{G}{r^2 c^2} \right)}$
Is this equation solution correct?

Astronuc
Staff Emeritus
Still working through it, but equation 8 should be

$$\tag{8} e^{-\lambda} = r^{-1}(r - 2u)$$

so check the equations after that.

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I have not located a Mathematica command that can factor identities:
$1 - \frac{2u}{r} = r^{-1} (r - 2u)$

$\tag{8} \boxed{e^{-\lambda} = r^{-1}(r - 2u)}$
Affirmative that is correct.

$e^{-\lambda}$ is a variable in the Schwarzschild metric and I calculate it has dimensionless SI units.

Identity:
$\boxed{e^{-\lambda} = r^{-1} (r - 2u) = 1 - \frac{2u}{r} = r^{-1}(r - r_s) = 1 - \frac{r_s}{r}}$

Integration by substitution:
$8 \pi P(r) = - [ r^{-1} (r - 2u) ] \left[ \left( \frac{2}{P(r) + \rho (r) c^2} \right) \left( \frac{dP}{dr} \right) \frac{1}{r} + \frac{1}{r^2} \right] - \frac{1}{r^2}$

My equation solution:
$\boxed{\frac{dP}{dr} = -(P(r) + \rho(r) c^2) \frac{r}{2} \left[ \left(8 \pi P(r) + \frac{1}{r^2} \right) r [r - 2u]^{-1} - \frac{1}{r^2} \right]}$

Mathematic 6 solution: (at this point)
$\boxed{\frac{dP}{dr} = -(P(r) + \rho(r) c^2)(r + 4 \pi r^3 P(r) - u)[r(r - 2u)]^{-1}}$

TOV equation solution: ref. 1
$\frac{dP}{dr} = - (P(r) + \rho(r) c^2) (4 \pi r^3 P(r) + u) [r(r - 2u)]^{-1}$

TOV equation solution: ref. 2
$\frac{dP}{dr} = - \frac{G}{r^2} \left(\rho(r) + \frac{P(r)}{c^2} \right) \left(m(r) + 4 \pi r^3 \frac{P(r)}{c^2} \right) [r(r - r_s)]^{-1}$

Reference:
http://en.wikipedia.org/wiki/Tolman-Oppenheimer-Volkoff_equation" [Broken]
http://en.wikipedia.org/wiki/SI" [Broken]
http://en.wikipedia.org/wiki/Schwarzschild_metric" [Broken]

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Identity???:
$\frac{r}{2} \left[ \left(8 \pi P(r) + \frac{1}{r^2} \right) r [r - 2u]^{-1} - \frac{1}{r^2} \right]} = (r + 4 \pi r^3 P(r) - u)[r(r - 2u)]^{-1}}$

Mathematica 6 confirms identity as True.

My solution for u:
$\boxed{u = \frac{G m(r)}{r^2} + 4 \pi r^3 P(r) \left(1 - \frac{G}{r^2 c^2} \right)}$

Mathematica 6 solution for u: (ref. 1 = ref. 2)
$u = \frac{4 c^4 \pi P(r) r^5+c^4 r^3-4 G \pi P(r) r^3-c^2 G m(r)}{c^4 r^2}$

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My solution for u:
$u = \frac{G m(r)}{c^2 r^2} + \frac{4 \pi G r P(r)}{c^4} - 4 \pi r^3 P(r)$
$\boxed{u = \frac{G m(r)}{c^2 r^2} + 4 \pi r^3 P(r) \left( 1 - \frac{G}{r^2 c^4} \right)}$

Mathematica 6 solution for u: (ref. 1 = ref. 2)
$\boxed{u = \frac{4 c^4 \pi P(r) r^5+c^4 r^3-4 G \pi P(r) r^3-c^2 G m(r)}{c^4 r^2}}$

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Mathematic 6 solution: (at this point)
$\boxed{\frac{dP}{dr} = -(P(r) + \rho(r) c^2)(r + 4 \pi r^3 P(r) - u)[r(r - 2u)]^{-1}}$

$u = \frac{r_s}{2} = \frac{G m(r)}{c^2}$
$\boxed{u = \frac{G m(r)}{c^2}}$

Mathematic 6 solution: (at this point)
$\boxed{\frac{dP}{dr} = -(P(r) + \rho(r) c^2) \left(r + 4 \pi r^3 P(r) - \frac{G m(r)}{c^2} \right) \left[ r \left( r - \frac{2G m(r)}{c^2} \right) \right]^{-1}}$

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Integration by substitution:
$8 \pi P(r) = [ r^{-1} (r - 2u) ] \left[ \left( \frac{2}{P(r) + \rho (r) c^2} \right) \left( \frac{dP}{dr} \right) \frac{1}{r} + \frac{1}{r^2} \right] - \frac{1}{r^2}$

My equation solution:
$\boxed{\frac{dP}{dr} = -(P(r) + \rho(r) c^2) \frac{r}{2} \left[ \left(8 \pi P(r) + \frac{1}{r^2} \right) r [r - 2u]^{-1} - \frac{1}{r^2} \right]}$

My identity:
$\boxed{\frac{r}{2} \left[ \left(8 \pi P(r) + \frac{1}{r^2} \right) r [r - 2u]^{-1} - \frac{1}{r^2} \right] = (4 \pi r^3 P(r) + u)[r(r - 2u)]^{-1}}}$

Mathematic 6 solution:
$\boxed{\frac{dP}{dr} = -(P(r) + \rho(r) c^2)(4 \pi r^3 P(r) + u)[r(r - 2u)]^{-1}}$

$u = \frac{r_s}{2} = \frac{G m(r)}{c^2}$
$\boxed{u = \frac{G m(r)}{c^2}}$

$\boxed{\frac{dP}{dr} = -(P(r) + \rho(r) c^2) \left(4 \pi r^3 P(r) + \frac{G m(r)}{c^2} \right) \left[ r \left( r - \frac{2G m(r)}{c^2} \right) \right]^{-1}}$
TOV equation solution: ref. 1
$\frac{dP}{dr} = - (P(r) + \rho(r) c^2) (4 \pi r^3 P(r) + u) [r(r - 2u)]^{-1}$

Unresolved issues at this point:
Mathematica 6 functionally confirms 'my identity' as True, however symbolic proof unresolved.
TOV equation solution: ref. 2 listed in Wikipedia is incorrect.

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TOV equation solution: ref. 2 listed in Wikipedia is incorrect.

$\frac{dP}{dr} = - \frac{G}{r^2} \left( \rho(r) + \frac{P(r)}{c^2} \right) \left(m(r) + 4 \pi r^3 \frac{P(r)}{c^2} \right) \left( \frac{1}{r(r - r_s)} \right)$

$\frac{dF}{dL^2 \cdot dL} = \frac{dF \cdot dL^2}{dm^2 \cdot dL^2} \left( \frac{dm}{dL^3} + \frac{dF \cdot dt^2}{dL^2 \cdot dL^2} \right) \left(dm + \frac{dL^3 \cdot dF \cdot dt^2}{dL^2 \cdot dL^2} \right) \frac{1}{dL^2}$

$\frac{dF}{dL^3} = \frac{dF}{dm^2} \left( \frac{dm}{dL^3} + \frac{dF \cdot dt^2}{dL^4} \right) \left(dm + \frac{dF \cdot dt^2}{dL} \right) \frac{1}{dL^2}$

$\frac{dF}{dL^3} = \frac{dF \cdot dm}{dm^2 \cdot dL^2} \left( \frac{dm}{dL^3} \right)$

$\boxed{\frac{dF}{dL^3} \neq \frac{dF}{dL^5}}$

Reference:
http://en.wikipedia.org/wiki/Tolman-Oppenheimer-Volkoff_equation" [Broken]

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interesting...

Mathematic 6 solution:
$\frac{dP}{dr} = -(P(r) + \rho(r) c^2) \left(4 \pi r^3 P(r) + \frac{G m(r)}{c^2} \right) \left[ r \left( r - \frac{2G m(r)}{c^2} \right) \right]^{-1}$

$\frac{dF}{dL^2 \cdot dL} = \left( \frac{dF}{dL^2} + \frac{dm \cdot dL^2}{dL^3 \cdot dt^2} \right) \left( dL^3 \cdot \frac{dF}{dL^2} + \frac{dF \cdot dL^2 \cdot dm \cdot dt^2}{dm^2 \cdot dL^2} \right) \frac{1}{dL^2}$

$\frac{dF}{dL^3} = \left( \frac{dF}{dL^2} \right) \left(dF \cdot dL + \frac{dF \cdot dt^2}{dm} \right) \frac{1}{dL^2}$

$\frac{dF}{dL^3} = \frac{dF}{dL^4} \left( dF \cdot dL \right)$

Unknown extra force derivative:
$\boxed{\frac{dF}{dL^3} \neq \frac{dF^2}{dL^3}}$

Astronuc
Staff Emeritus
Well there is a problem right here

$\frac{dP}{dr} = - (P(r) + \rho(r) c^2) (4 \pi r^3 P(r) + u) [r(r - 2u)]^{-1}$

specifically

$(4 \pi r^3 P(r) + u)$

because u is dimensionless per equation (8) of Oppenheimer-Volkoff paper.

and r3P(r) has units of energy! Pressure = F/L2 = FL/L3 = energy density, where F = force = ML/T2, and Energy = Force*L.

In the Oppenheimer-Volkoff paper, I believe there is an error in equation (1), actually an omission.

ds2 = . . . . + $e^{\nu}dt^2$.

I believe it should be

ds2 = . . . . + $e^{\nu}\,c^2\,dt^2$ so that it is dimensionally correct. That is more or less the form in the Wikipedia article on TOV.

Also, in the Oppenheimer-Volkoff paper.

Also, one must be careful between the Wikipedia article and the original OV paper.

In the OV paper, in the text following OV equation (2), it states that $\rho(r)$ is the macroscopic energy density, and not the mass density, although energy density is related to mass density * c2. So this [(P(r) + \rho(r) c^2)] could be problematic.

Also is [r(r - 2u)]^{-1} correct? Remember $$\tag{8} e^{-\lambda} = r^{-1}(r - 2u)$$

One has to be careful of units, consistency of terms and errors or omissions in the literature!

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The $[r(r - 2u)]^{-1}$ term originates algebraically from 'my identity' equation which is solved symbolically in reference. 1 link below and by Mathematica 6:

$\boxed{\frac{r}{2} \left[ \left(8 \pi P(r) + \frac{1}{r^2} \right) r [r - 2u]^{-1} - \frac{1}{r^2} \right] = (4 \pi r^3 P(r) + u)[r(r - 2u)]^{-1}}}$

$$\tag{8} e^{-\lambda} = r^{-1}(r - 2u)$$

'My identity':
$\frac{r}{2} \left[ \left(8 \pi P(r) + \frac{1}{r^2} \right) e^{\lambda} - \frac{1}{r^2} \right] = \frac{4 \pi r^3 P(r) + u}{r(r - 2u)}$

Mathematica 6 solution for $e^{-\lambda}$ based upon 'my identity':
$e^{-\lambda} = \frac{r - 2u}{r}$

Reference:

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The overall SI terminology of of the TOV equation should be:
$\frac{dF}{dL^3} = \left( \frac{dF}{dL^2} \right)(dL) \left( \frac{1}{dL^2} \right)$

$\boxed{u = dL}$

Making the SI correction, the equation solution becomes:
$\frac{dP}{dr} = -(P(r) + \rho(r) c^2) \left( \frac{4 \pi r^3 P(r)}{dF} + \frac{G m(r)}{c^2} \right) \left[ r \left( r - \frac{2G m(r)}{c^2} \right) \right]^{-1}$

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The overall SI terminology of of the Wikipedia TOV equation should be:
$\frac{dF}{dL^3} = \left( \frac{dF \cdot dL^2}{dm^2} \right) \left( \frac{dm}{dL^3} \right)(dm) \left( \frac{1}{dL^2} \right)$

The only known dimensionally functional solution for the Wikipedia TOV equation:
$\boxed{\frac{dP}{dr} = - G \left( \rho(r) + \frac{P(r)}{c^2} \right) \left(m(r) + 4 \pi r^3 \frac{P(r)}{c^2} \right) \left( \frac{1}{r(r - r_s)} \right)}$

TOV = TOV Wikipedia dimensional identity:
$(P(r) + \rho(r) c^2) \left( \frac{4 \pi r^3 P(r)}{dF} + \frac{G m(r)}{c^2} \right) = G \left( \rho(r) + \frac{P(r)}{c^2} \right) \left(m(r) + 4 \pi r^3 \frac{P(r)}{c^2} \right)$

Mathematica 6 solution for $dF$:
$\boxed{dF = \frac{c^4}{G}}$

The TOV equation should be:
$\boxed{\frac{dP}{dr} = -(P(r) + \rho(r) c^2) \left( \frac{4 \pi G r^3 P(r)}{c^4} + \frac{G m(r)}{c^2} \right) \left[ r \left( r - \frac{2G m(r)}{c^2} \right) \right]^{-1}}$

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The TOV equation solution for a Neutron Star:
$\boxed{u = \frac{r_s}{2}}$

$\boxed{\frac{dP}{dr} = -(P(r) + \rho(r) c^2) \left( \frac{4 \pi G r^3 P(r)}{c^4} + \frac{r_s}{2} \right) \left[ r \left( r - r_s \right) \right]^{-1}} \; \; \; (r > r_p) \; \; \; r \neq r_p$

Integration by substitution:
$\boxed{\frac{dP}{dr} = -(P(r) + \rho(r) c^2) \left( \frac{4 \pi G r^3 P(r)}{c^4} + \frac{G m(r)}{c^2} \right) \left[ r \left( r - \frac{2G m(r)}{c^2} \right) \right]^{-1}} \; \; \; (r > r_s) \; \; \; r \neq r_s$

The TOV equation solution for a Black Hole:
$\boxed{r_s = r_p}$
$\boxed{u = \frac{r_p}{2}}$

$\boxed{\frac{dP}{dr} = -(P(r) + \rho(r) c^2) \left( \frac{4 \pi G r^3 P(r)}{c^4} + \frac{r_p}{2} \right) \left[ r \left( r - r_p \right) \right]^{-1}} \; \; \; (r > r_p) \; \; \; r \neq r_p$

Integration by substitution:
$\boxed{\frac{dP}{dr} = -(P(r) + \rho(r) c^2) \left( \frac{4 \pi G r^3 P(r)}{c^4} + \frac{1}{2} \sqrt{\frac{\hbar G}{c^3}} \right) \left[ r \left( r - \sqrt{\frac{\hbar G}{c^3}} \right) \right]^{-1}} \; \; \; (r > r_p) \; \; \; r \neq r_p$

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Astronuc
Staff Emeritus
I haven't read through the details, but it looks good.

Reflecting on my previous post, the OV paper mentions just before equation (18) that Eqs. (3), (4) and (5) from which (16) and (17) are derived are stated in "relativistic units" which apparently Tolman used. In relativistic units, c = 1, so obviously c2 = 1, and G = 1. So those factors do not show in the equations in the OV paper.

I now suspect that the c2 is in equation 1, but has value 1, so it's not explicitly written. I never like systems that use c = 1, because while they might look nicer, it's easy to make a mistake in derivations.

The Wikipedia article apparently uses SI, so c2 is explicitly used with the mass density, but then $\rho$ is mass density, not energy density.

$\boxed{\frac{dP}{dr} = -(P(r) + \rho(r) c^2) \left( \frac{4 \pi G r^3 P(r)}{c^4} + \frac{1}{2} \sqrt{\frac{\hbar G}{c^3}} \right) \left[ r \left( r - \sqrt{\frac{\hbar G}{c^3}} \right) \right]^{-1}} \; \; \; (r > r_p) \; \; \; r \neq r_p$

Planck Sphere surface pressure:
$P_p = \frac{c^7}{4 \pi \hbar G^2}$

Planck Sphere density:
$\rho_p = \frac{3c^5}{4 \pi \hbar G^2}$

Integration by substitution:
$\frac{dP}{dr} = - \left[ \frac{c^7}{4 \pi \hbar G^2} + \left( \frac{3c^5}{4 \pi \hbar G^2} \right) c^2 \right] \left( \frac{4 \pi G r^3}{c^4} \left( \frac{c^7}{4 \pi \hbar G^2} \right) + \frac{1}{2} \sqrt{\frac{\hbar G}{c^3}} \right) \left[ r \left( r - \sqrt{\frac{\hbar G}{c^3}} \right) \right]^{-1} \; \; \; (r > r_p) \; \; \; r \neq r_p$

The TOV equation solution for a Planck singularity:
$\frac{dP}{dr} = - \left[ \frac{c^7}{4 \pi \hbar G^2} + \frac{3c^7}{4 \pi \hbar G^2} \right] \left( \frac{c^3 r^3}{\hbar G} + \frac{1}{2} \sqrt{\frac{\hbar G}{c^3}} \right) \left[ r \left( r - \sqrt{\frac{\hbar G}{c^3}} \right) \right]^{-1} \; \; \; (r > r_p) \; \; \; r \neq r_p$

$\boxed{\frac{dP}{dr} = - \frac{c^7}{\pi \hbar G^2} \left( \frac{c^3 r^3}{\hbar G} + \frac{1}{2} \sqrt{\frac{\hbar G}{c^3}} \right) \left[ r \left( r - \sqrt{\frac{\hbar G}{c^3}} \right) \right]^{-1}} \; \; \; (r > r_p) \; \; \; r \neq r_p$

These equations predict 2 explosion types:
When a Neutron Star collapses into a Black Hole.
When a Black Hole collapses into a Planck singularity.

In post #18 eq. 2, the limiting values should be: (30 min. PF edit limit)
$r > r_s \; \; \; r \neq r_s$

Reference:
http://en.wikipedia.org/wiki/Planck_pressure" [Broken]
http://en.wikipedia.org/wiki/Planck_density" [Broken]

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According to Wikipedia (ref. 1, para. 2 listed below),
Wikipedia said:
the pressure exerted by degenerate matter depends only weakly on its temperature. In particular, the pressure remains nonzero even at absolute zero temperature.
Low pressure fully degenerate Fermi gas:
$$\gamma = \frac{5}{3}$$

High density quantum state relativistic degenerate Fermi gas:
$$\gamma = \frac{4}{3}$$

Polytropic degenerate Fermi gas pressure equation:
$$P(r) = K \rho(r)^{\gamma}$$

$$K$$ - particle gas properties constant

$$K = \frac{P(r)}{\rho(r)^{\gamma}} = \left( \frac{dF}{dL^2} \right) \cdot \left( \frac{dL^3}{dm} \right) = \frac{dF \cdot dL}{dm}$$
$$\boxed{K = \frac{dF \cdot dL}{dm}}$$

TOV equation solution:
$$\boxed{\frac{dP}{dr} = -(P(r) + \rho(r) c^2) \left( \frac{4 \pi G r^3 P(r)}{c^4} + \frac{G m(r)}{c^2} \right) \left[ r \left( r - \frac{2G m(r)}{c^2} \right) \right]^{-1}}$$

Polytropic degenerate Fermi gas pressure equation:
$$P(r) = K \rho(r)^{\gamma}$$

Integration by substitution:
$$\frac{dP}{dr} = -(K \rho(r)^{\gamma} + \rho(r) c^2) \left( \frac{4 \pi G r^3 (K \rho(r)^{\gamma})}{c^4} + \frac{G m(r)}{c^2} \right) \left[ r \left( r - \frac{2G m(r)}{c^2} \right) \right]^{-1}}$$

Polytropic degenerate Fermi gas pressure TOV equation:
$$\boxed{\frac{dP}{dr} = -(K \rho(r)^{\gamma} + \rho(r) c^2) \left( \frac{4 \pi G K r^3 \rho(r)^{\gamma}}{c^4} + \frac{G m(r)}{c^2} \right) \left[ r \left( r - \frac{2G m(r)}{c^2} \right) \right]^{-1}}}$$

Is this equation solution correct?

The Adiabatic indexes for a degenerate Fermi gas are listed in Wikipedia (ref. 1), however not yet listed on Wikipedia (ref. 2 ) below.

Reference:
http://en.wikipedia.org/wiki/Degenerate_matter" [Broken]
http://en.wikipedia.org/wiki/Heat_capacity_ratio" [Broken]

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According to PHYS 390 Lecture 19, the degenerate Fermi gas pressure of spin 1/2 particles is:
$$P(r) = \frac{\pi^3 \hbar^2}{15 m_n} \left( \frac{3N(r)}{\pi V(r)} \right)^{\frac{5}{3}}$$

Number density:
$$n(r) = \frac{N(r)}{V(r)} = \frac{\rho(r)}{m_n}$$
$$m_n$$ - neutron mass

Integration by substitution:
$$P(r) = \frac{\pi^3 \hbar^2}{15 m_n} \left[ \frac{3}{\pi} \left( \frac{\rho(r)}{m_n} \right)\right]^{\gamma} = \frac{\pi^3 \hbar^2}{15 m_n} \left( \frac{3}{\pi m_n} \right)^{\gamma} \rho(r)^{\gamma} = K \rho(r)^{\gamma}$$

$$\boxed{P(r) = \frac{\pi^3 \hbar^2}{15 m_n} \left( \frac{3}{\pi m_n} \right)^{\gamma} \rho(r)^{\gamma}}$$

$$\boxed{K = \frac{\pi^3 \hbar^2}{15 m_n} \left( \frac{3}{\pi m_n} \right)^{\gamma}}$$

However, I failed to resolve the SI derivation for the pressure equation:
$$\frac{dF}{dL^2} = \left( \frac{dE^2 dt^2}{dm} \right) \left( \frac{1}{dm} \right) \left( \frac{dm}{dL^3} \right) = \left( \frac{dE^2 dt^2}{dm \cdot dL^3} \right)$$

Reference:
http://www.sfu.ca/~boal/385lecs/385lec19.pdf" [Broken]

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SI derivation for the PHYS 390 Lecture 19 pressure equation:
$$\frac{dF}{dL^2} = \left( \frac{dE^2 dt^2}{dm \cdot dL^5} \right) = \left( \frac{dF^2 dL^2 dt^2}{dm \cdot dL^5} \right)$$

$$\frac{dF}{dL^2} = \frac{dF^2 dt^2}{dm \cdot dL^3}$$

Newton's second law:
$$\boxed{dF = \frac{dm \cdot dL}{dt^2}}$$

According to this solution, the PHYS 390 Lecture 19 pressure equation is correct:

$$P(r) = \frac{\pi^3 \hbar^2}{15 m_n} \left( \frac{3N(r)}{\pi V(r)} \right)^{\frac{5}{3}} \; \; \; \gamma = \frac{5}{3}$$

Reference:
http://en.wikipedia.org/wiki/Force" [Broken]
http://www.sfu.ca/~boal/385lecs/385lec18.pdf" [Broken]
http://www.sfu.ca/~boal/385lecs/385lec19.pdf" [Broken]

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According to my SI derivations, the PHYS 390 Lecture 19 pressure equation is only valid for an Adiabatic index of $$\gamma = \frac{5}{3}$$
$$\frac{dF}{dL^2} \neq \left( \frac{dE^2 dt^2}{dm} \right) \left( \frac{1}{dm} \right) \left( \frac{dm}{dL^3} \right) = \left( \frac{dE^2 dt^2}{dm \cdot dL^3} \right) \; \; \; \gamma = 1$$

$$\frac{dF}{dL^2} = \left( \frac{dE^2 dt^2}{dm \cdot dL^{3 \gamma}} \right) = \left( \frac{dF^2 dL^2 dt^2}{dm \cdot dL^5} \right)\; \; \; \gamma = \frac{5}{3}$$

$$\frac{dF}{dL^2} \neq \left( \frac{dE^2 dt^2}{dm \cdot dL^{3 \gamma}} \right) = \left( \frac{dF^2 dL^2 dt^2}{dm \cdot dL^4} \right)\; \; \; \gamma = \frac{4}{3}$$

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According to Wikipedia ref. 1 listed below, the equation for Fermi energy is:

Fermi energy equation:
$$E_f = \frac{\hbar^2}{2m_n} \left( \frac{3 \pi^2 N}{V} \right)^{2/3}$$

Fermi pressure:
$$P_f = - \frac{dE_f}{dV_f}$$

Integration by substitution:
$$P_f = - \frac{\hbar^2}{2m_n} \left( \frac{3 \pi^2 N_f}{V_f} \right)^{2/3} \frac{1}{V_f} = - \frac{\hbar^2}{2m_n} \left( 3 \pi^2 N \right)^{2/3} \frac{1}{V_f^{\frac{5}{3}}} = K \rho_f^{\frac{2}{3}}$$

$$\boxed{P_f = - \frac{\hbar^2}{2m_n} \left( 3 \pi^2 N_f \right)^{2/3} \frac{1}{V_f^{\frac{5}{3}}}}$$

Reference:
http://en.wikipedia.org/wiki/Fermi_energy" [Broken]

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