Tolman-Oppenheimer-Volkoff equation

1. Apr 7, 2008

Orion1

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:
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:
On Massive Neutron Cores, J. R. Oppenheimer and G. M. Volkoff, Physical Review 55, #374 (February 15, 1939), pp. 374–381.
Tolman-Oppenheimer-Volkoff equation - Wikipedia

Last edited: Apr 7, 2008
2. Apr 10, 2008

Orion1

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

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

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

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

Last edited: Apr 10, 2008
3. Apr 10, 2008

Orion1

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

4. Apr 12, 2008

Staff: Mentor

Still working through it, but equation 8 should be

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

so check the equations after that.

Last edited: Apr 12, 2008
5. Apr 13, 2008

Orion1

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:
On Massive Neutron Cores, J. R. Oppenheimer and G. M. Volkoff, Physical Review 55, #374 (February 15, 1939), pp. 374–381.
Tolman-Oppenheimer-Volkoff equation - Wikipedia
International System of Units
Schwarzschild metric

Last edited: Apr 13, 2008
6. Apr 13, 2008

Orion1

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)
$u = \frac{4 c^4 \pi P(r) r^5+c^4 r^3-4 G \pi P(r) r^3-c^2 G m(r)}{c^4 r^2}$

Last edited: Apr 13, 2008
7. Apr 13, 2008

Orion1

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)
$\boxed{u = \frac{4 c^4 \pi P(r) r^5+c^4 r^3-4 G \pi P(r) r^3-c^2 G m(r)}{c^4 r^2}}$

Last edited: Apr 13, 2008
8. Apr 13, 2008

Orion1

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

Last edited: Apr 13, 2008
9. Apr 16, 2008

Orion1

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.

Last edited: Apr 16, 2008
10. Apr 16, 2008

Orion1

11. Apr 19, 2008

Orion1

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:
Tolman-Oppenheimer-Volkoff equation - Wikipedia

Last edited: Apr 19, 2008
12. Apr 19, 2008

Anony-mouse

interesting...

13. Apr 19, 2008

Orion1

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:
$\boxed{\frac{dF}{dL^3} \neq \frac{dF^2}{dL^3}}$

14. Apr 19, 2008

Staff: Mentor

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!

Last edited: Apr 19, 2008
15. Apr 19, 2008

Orion1

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:
'my identity' - symbolic proof

Last edited: Apr 19, 2008
16. Apr 19, 2008

Orion1

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

Last edited: Apr 19, 2008
17. Apr 19, 2008

Orion1

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

Last edited: Apr 19, 2008
18. Apr 19, 2008

Orion1

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

Last edited: Apr 19, 2008
19. Apr 19, 2008

Staff: Mentor

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.

20. Apr 19, 2008

Orion1

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

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:
Planck pressure - Wikipedia
Planck density - Wikipedia

Last edited: Apr 19, 2008