Tolman-Oppenheimer-Volkoff equation

  • Thread starter Orion1
  • Start date
In summary, the conversation discussed the importance of setting boundaries in relationships and being assertive in communicating them. It also touched on the idea of self-worth and how it can influence one's ability to set boundaries. The participants shared personal experiences and offered advice on how to effectively communicate boundaries in different situations.
  • #1
Orion1
973
3

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:
[itex]\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}[/itex]
[itex]\tag{8} e^{-\lambda} = r(r - 2u)[/itex]
[itex]\tag{5} \frac{d \nu}{dr} = \frac{2}{P(r) + \rho(r) c^2} \left( \frac{dP}{dr} \right)[/itex]

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

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

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

My equation solution:
[itex]\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}]}[/itex]

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

TOV equation solution: ref. 2
[itex]\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}[/itex]


Reference:
http://home.comcast.net/~lambo1826/download/PHRVAO_55_4_374_1.pdf" [Broken]
http://en.wikipedia.org/wiki/Tolman-Oppenheimer-Volkoff_equation" [Broken]
 
Last edited by a moderator:
Astronomy news on Phys.org
  • #2

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

[itex](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}][/itex]

Factoring out [itex]c^2[/itex] from the LHS results in:
[itex] 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}][/itex]

Eliminate terms:
[itex] 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}][/itex]

Solve for u:
[itex] 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}][/itex]

[itex]\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}[/itex]

[itex]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}] [/itex]

[itex]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) [/itex]

 
Last edited:
  • #3

[itex]\boxed{u = \frac{G m(r)}{r^2} + 4 \pi r^3 P(r) \left(1 - \frac{G}{r^2 c^2} \right)}[/itex]
Is this equation solution correct?
 
  • #4
Still working through it, but equation 8 should be

[tex]\tag{8} e^{-\lambda} = r^{-1}(r - 2u)[/tex]

so check the equations after that.
 
Last edited:
  • #5

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

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

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

Identity:
[itex]\boxed{e^{-\lambda} = r^{-1} (r - 2u) = 1 - \frac{2u}{r} = r^{-1}(r - r_s) = 1 - \frac{r_s}{r}}[/itex]

Integration by substitution:
[itex]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} [/itex]

My equation solution:
[itex]\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]}[/itex]

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

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

TOV equation solution: ref. 2
[itex]\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}[/itex]


Reference:
http://home.comcast.net/~lambo1826/download/PHRVAO_55_4_374_1.pdf" [Broken]
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]
 
Last edited by a moderator:
  • #6

Identity?:
[itex]\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}}[/itex]

Mathematica 6 confirms identity as True.

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

Mathematica 6 solution for u: (ref. 1 = ref. 2)
[itex]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}[/itex]

 
Last edited:
  • #7

My solution for u:
[itex]u = \frac{G m(r)}{c^2 r^2} + \frac{4 \pi G r P(r)}{c^4} - 4 \pi r^3 P(r)[/itex]
[itex]\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)}[/itex]

Mathematica 6 solution for u: (ref. 1 = ref. 2)
[itex]\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}}[/itex]
 
Last edited:
  • #8

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

[itex]u = \frac{r_s}{2} = \frac{G m(r)}{c^2}[/itex]
[itex]\boxed{u = \frac{G m(r)}{c^2}}[/itex]

Mathematic 6 solution: (at this point)
[itex]\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}}[/itex]
 
Last edited:
  • #9

Integration by substitution:
[itex]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}[/itex]

My equation solution:
[itex]\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]}[/itex]

My identity:
[itex]\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}}}[/itex]

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

[itex]u = \frac{r_s}{2} = \frac{G m(r)}{c^2}[/itex]
[itex]\boxed{u = \frac{G m(r)}{c^2}}[/itex]

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

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:
  • #11

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

[itex]\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)[/itex]

[itex]\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}[/itex]

[itex]\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}[/itex]

[itex]\frac{dF}{dL^3} = \frac{dF \cdot dm}{dm^2 \cdot dL^2} \left( \frac{dm}{dL^3} \right)[/itex]

[itex]\boxed{\frac{dF}{dL^3} \neq \frac{dF}{dL^5}}[/itex]

Reference:
http://en.wikipedia.org/wiki/Tolman-Oppenheimer-Volkoff_equation" [Broken]
 
Last edited by a moderator:
  • #12
interesting...
 
  • #13

Mathematic 6 solution:
[itex]\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}[/itex]

[itex]\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} [/itex]

[itex]\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}[/itex]


[itex]\frac{dF}{dL^3} = \frac{dF}{dL^4} \left( dF \cdot dL \right)[/itex]

Unknown extra force derivative:
[itex]\boxed{\frac{dF}{dL^3} \neq \frac{dF^2}{dL^3}}[/itex]

 
  • #14
Well there is a problem right here

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


specifically

[itex] (4 \pi r^3 P(r) + u) [/itex]

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 = . . . . + [itex]e^{\nu}dt^2[/itex].

I believe it should be

ds2 = . . . . + [itex]e^{\nu}\,c^2\,dt^2[/itex] 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 [itex]\rho(r)[/itex] 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 [tex]\tag{8} e^{-\lambda} = r^{-1}(r - 2u)[/tex]


One has to be careful of units, consistency of terms and errors or omissions in the literature!
 
Last edited:
  • #15

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

[itex]\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}}}[/itex]

[tex]\tag{8} e^{-\lambda} = r^{-1}(r - 2u)[/tex]

'My identity':
[itex]\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)}[/itex]

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

Reference:
https://www.physicsforums.com/showthread.php?p=1692563#post1692563"
 
Last edited by a moderator:
  • #16

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

[itex]\boxed{u = dL}[/itex]

Making the SI correction, the equation solution becomes:
[itex]\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}[/itex]

 
Last edited:
  • #17


The overall SI terminology of of the Wikipedia TOV equation should be:
[itex]\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)[/itex]

The only known dimensionally functional solution for the Wikipedia TOV equation:
[itex]\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)}[/itex]

TOV = TOV Wikipedia dimensional identity:
[itex](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)[/itex]

Mathematica 6 solution for [itex]dF[/itex]:
[itex]\boxed{dF = \frac{c^4}{G}}[/itex]

The TOV equation should be:
[itex]\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}}[/itex]

 
Last edited:
  • #18

The TOV equation solution for a Neutron Star:
[itex]\boxed{u = \frac{r_s}{2}}[/itex]

[itex]\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[/itex]

Integration by substitution:
[itex]\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[/itex]

The TOV equation solution for a Black Hole:
[itex]\boxed{r_s = r_p}[/itex]
[itex]\boxed{u = \frac{r_p}{2}}[/itex]

[itex]\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[/itex]

Integration by substitution:
[itex]\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[/itex]
 
Last edited:
  • #19
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 [itex]\rho[/itex] is mass density, not energy density.
 
  • #20

[itex]\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[/itex]

Planck Sphere surface pressure:
[itex]P_p = \frac{c^7}{4 \pi \hbar G^2}[/itex]

Planck Sphere density:
[itex]\rho_p = \frac{3c^5}{4 \pi \hbar G^2}[/itex]

Integration by substitution:
[itex]\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[/itex]

The TOV equation solution for a Planck singularity:
[itex]\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[/itex]

[itex]\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[/itex]

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)
[itex]r > r_s \; \; \; r \neq r_s[/itex]

Reference:
http://en.wikipedia.org/wiki/Planck_pressure" [Broken]
http://en.wikipedia.org/wiki/Planck_density" [Broken]
 
Last edited by a moderator:
  • #21

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.

Adiabatic indexes:
Low pressure fully degenerate Fermi gas:
[tex]\gamma = \frac{5}{3}[/tex]

High density quantum state relativistic degenerate Fermi gas:
[tex]\gamma = \frac{4}{3}[/tex]

Polytropic degenerate Fermi gas pressure equation:
[tex]P(r) = K \rho(r)^{\gamma}[/tex]

[tex]K[/tex] - particle gas properties constant

[tex]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}[/tex]
[tex]\boxed{K = \frac{dF \cdot dL}{dm}}[/tex]

TOV equation solution:
[tex]\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}}[/tex]

Polytropic degenerate Fermi gas pressure equation:
[tex]P(r) = K \rho(r)^{\gamma}[/tex]

Integration by substitution:
[tex]\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}}[/tex]

Polytropic degenerate Fermi gas pressure TOV equation:
[tex]\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}}}[/tex]

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]
 
Last edited by a moderator:
  • #22

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

Number density:
[tex]n(r) = \frac{N(r)}{V(r)} = \frac{\rho(r)}{m_n}[/tex]
[tex]m_n[/tex] - neutron mass

Integration by substitution:
[tex]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}[/tex]

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

[tex]\boxed{K = \frac{\pi^3 \hbar^2}{15 m_n} \left( \frac{3}{\pi m_n} \right)^{\gamma}}[/tex]

However, I failed to resolve the SI derivation for the pressure equation:
[tex]\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)[/tex]

Reference:
http://www.sfu.ca/~boal/385lecs/385lec19.pdf" [Broken]
 
Last edited by a moderator:
  • #23

SI derivation for the PHYS 390 Lecture 19 pressure equation:
[tex]\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)[/tex]

[tex] \frac{dF}{dL^2} = \frac{dF^2 dt^2}{dm \cdot dL^3}[/tex]

Newton's second law:
[tex]\boxed{dF = \frac{dm \cdot dL}{dt^2}}[/tex]

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

[tex]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}[/tex]

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]
 
Last edited by a moderator:
  • #24

According to my SI derivations, the PHYS 390 Lecture 19 pressure equation is only valid for an Adiabatic index of [tex]\gamma = \frac{5}{3}[/tex]
[tex]\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[/tex]

[tex]\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}[/tex]

[tex]\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}[/tex]
 
Last edited:
  • #25

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

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

Fermi pressure:
[tex]P_f = - \frac{dE_f}{dV_f}[/tex]

Integration by substitution:
[tex]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}}[/tex]

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

Reference:
http://en.wikipedia.org/wiki/Fermi_energy" [Broken]
 
Last edited by a moderator:
  • #26
In the polytropic equation of state. the constant K is called " the entropy constant " in some literature. You have called it " the particle gas properties constant".
 
  • #27

According to Wikipedia ref. 1, para. 3 listed below:
Wikipedia said:
...where K depends on the properties of the particles making up the gas.

Wikipedia said:
where K' again depends on the properties of the particles making up the gas.

Greetings, Helios
Thanks for your important collaboration and correction.

That constant was copied from the Wikipedia ref.1, para. 3 paragraph which does not implicitly declare a key definition which states the SI name for that constant, however it does describe it twice in the same paragraph!

The Wikipedia ref. 1, still requires a lot of work with better, more implicitly written definition keys and definitional equations with more extensive referencing.

According to Wikipedia ref. 2, section 'Entropy in Astrophysics':
Wikipedia said:
In astrophysics, what is referred to as "entropy" is actually the adiabatic constant derived as follows.

[tex]PV^{\gamma} = \text{constant} = K[/tex]

Wikipedia said:
This equation is known as an expression for the adiabatic constant, K, also called the adiabat.

[tex]K[/tex] - adiabatic constant, K, also called the adiabat.

[tex]P = \frac{\rho k_{B}T}{\mu m_{H}}[/tex]

[tex]\boxed{K = \frac{k_{B}T}{\mu m_{H} \rho^{2/3}}}[/tex]
Is the adiabatic constant also called the 'entropy constant' in other literature?

Wikipedia said:
Substituting this into the above equation along with V = [grams] / ρ and γ = 5 / 3 for an ideal monoatomic gas
[tex]PV^{\gamma} = K[/tex]

[tex]\gamma = \frac{5}{3}[/tex] - ideal monoatomic gas

[tex]P = \frac{\rho k_{B}T}{\mu m_{H}}[/tex]

[tex]V = \frac{dm}{\rho}[/tex]

Integration by substitution:
[tex]PV^{\gamma} = K[/tex]

[tex]K = \left( \frac{\rho k_{B}T}{\mu m_{H}} \right) \left( \frac{dm}{\rho} \right)^{\gamma} = \left( \frac{k_{B} dm^{\gamma}}{\mu m_{H}} \right) T \rho^{1 - \gamma}[/tex]

[tex]dm = m_H[/tex]

Identities:
[tex]\frac{\rho}{\rho^{\gamma}} = \rho^{1 - \gamma}[/tex]
[tex]\frac{m_H^{\gamma}}{m_H} = m_H^{\gamma - 1}[/tex]

The adiabatic constant:
[tex]\boxed{K = \left( \frac{k_{B} dm_H^{\gamma - 1} }{\mu} \right) T \rho^{1 - \gamma}}[/tex]

Reference:
http://en.wikipedia.org/wiki/Degenerate_matter" [Broken]
http://en.wikipedia.org/wiki/Entropy" [Broken]
 
Last edited by a moderator:
  • #28

The adiabatic constant:
[tex]\boxed{K = \left( \frac{k_{B} dm_H^{\gamma - 1} }{\mu} \right) T \rho^{1 - \gamma}}[/tex]

Standard SI derivation:
[tex]K = \frac{dE \cdot dm^{\gamma - 1}}{dT} \cdot \frac{amn}{amu} \cdot dT \left( \frac{dm}{dL^3} \right)^{1 - \gamma} = \frac{dE}{dL^{3(1 - \gamma)}} \cdot \frac{amn}{amu}[/tex]

[tex]\boxed{K = \frac{dE}{dL^{3(1 - \gamma)}} \cdot \frac{amn}{amu}}[/tex]
 
  • #29
Yes, an adiabat is a curve of constant entropy on the P-V diagram. The adiabats are isentropic and the " entropy constant " is the value for one adiabat.
 
  • #30

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

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

Fermi pressure:
[tex]P_f = - \frac{dE_f}{dV_f}[/tex]

Integration by differentiation substitution:
[tex]P_f = - \frac{\hbar^2}{2m_n} \left( 3 \pi^2 N_f \right)^{2/3} \left( \frac{V^{- \frac{2}{3}}}{dV} \right)[/tex]

Differentiation identity:
[tex]\frac{V^{-n}}{dV} = -n V^{-n - 1} = \left(-\frac{2}{3} \right) V^{-\frac{2}{3} - \frac{3}{3}} = \left(- \frac{2}{3} \right) V^{-\frac{5}{3}}[/tex]

[tex]P_f = - \left(- \frac{2}{3} \right) \frac{\hbar^2}{2m_n} \left( 3 \pi^2 N \right)^{2/3} \frac{1}{V_f^{\frac{5}{3}}} = \frac{2 \hbar^2}{6 m_n} \left( 3 \pi^2 N \right)^{2/3} \frac{1}{V_f^{\frac{5}{3}}}[/tex]

[tex]\boxed{P_f = \frac{\hbar^2}{3 m_n} \left( 3 \pi^2 N \right)^{2/3} \frac{1}{V_f^{\frac{5}{3}}}}[/tex]


Post #25 requires integration by differentiation substitution.

Reference:
http://en.wikipedia.org/wiki/Fermi_energy" [Broken]
http://www.sfu.ca/~boal/385lecs/385lec19.pdf" [Broken]
 
Last edited by a moderator:
  • #31

My first attempt to integrate the Adiabatic constant and Adiabatic index with degenerate Fermi pressure.

Adiabatic constant:
[tex]K = P V^{\gamma}[/tex]

Degenerate Fermi pressure:
[tex]P = - \frac{dE}{dV}[/tex]

Degenerate Fermi energy:
[tex]E_f = \frac{\pi^3 \hbar^2}{10 m_n} \left( \frac{3 N}{\pi} \right)^{\frac{5}{3}} \frac{1}{V^{\frac{2}{3}}}[/tex]

Adiabatic constant:
[tex]K = \left( - \frac{dE}{dV} \right) V^{\gamma} = - \frac{\pi^3 \hbar^2}{10 m_n} \left( \frac{3 N}{\pi} \right)^{\frac{5}{3}} \left( \frac{V^{-\frac{2}{3}}}{dV} \right) V^{\gamma}[/tex]

Differentiation identity:
[tex]\frac{V^{-n}}{dV} = -n V^{-n - 1} = \left(-\frac{2}{3} \right) V^{-\frac{2}{3} - \frac{3}{3}} = \left(- \frac{2}{3} \right) V^{-\frac{5}{3}}[/tex]

Integration by differentiation substitution:
[tex]K = - \left( - \frac{2}{3} \right) \frac{\pi^3 \hbar^2}{10 m_n} \left( \frac{3 N}{\pi} \right)^{\frac{5}{3}} \frac{V^{\gamma}}{V^{\frac{5}{3}}} = \left( \frac{2 \cdot 3 \cdot 3^{\frac{2}{3}}}{30} \right) \frac{\pi^{ \left( \frac{9}{3} - \frac{5}{3} \right)} \hbar^2}{m_n} \left( \frac{N}{V} \right)^{\frac{5}{3}} V^{\gamma} = \frac{3^{\frac{2}{3}} \pi^{\frac{4}{3}} \hbar^2}{5 m_n} \left( \frac{N}{V} \right)^{\frac{5}{3}} V^{\gamma}[/tex]

Number density:
[tex]n(r) = \frac{N(r)}{V(r)} = \frac{\rho(r)}{m_n}[/tex]

Adiabatic volume:
[tex]V^{\gamma} = \left( \frac{m_n}{\rho(r)} \right)^{\gamma}[/tex]

[tex]K = \frac{3^{\frac{2}{3}} \pi^{\frac{4}{3}} \hbar^2}{5 m_n} \left( \frac{\rho(r)}{m_n} \right)^{\frac{5}{3}} \left( \frac{m_n}{\rho(r)} \right)^{\gamma} = \left( \frac{3^{\frac{2}{3}} \pi^{\frac{4}{3}} \hbar^2}{5} \right) m_n^{\gamma - \frac{5}{3} + \frac{3}{3}} \rho(r)^{\frac{5}{3} - \gamma} = \left( \frac{3^{\frac{2}{3}} \pi^{\frac{4}{3}} \hbar^2}{5} \right) m_n^{\gamma - \frac{8}{3}} \rho(r)^{\frac{5}{3} - \gamma}[/tex]

Degenerate adiabatic constant with adiabatic index:
[tex]\boxed{K = \left( \frac{3^{\frac{2}{3}} \pi^{\frac{4}{3}} \hbar^2}{5} \right) m_n^{\gamma - \frac{8}{3}} \rho(r)^{\frac{5}{3} - \gamma}}[/tex]

Is this equation solution correct?

Standard SI derivation:
[tex]K = \frac{dE^2 \cdot dt^2}{dm \cdot dL^5} = \frac{dF^2 \cdot dL^2 \cdot dt^2}{dm \cdot dL^5} = \frac{dF^2 \cdot dt^2}{dm \cdot dL^3} = \frac{dF}{dL^2}[/tex]

[tex]\frac{dF}{dL^2} = \frac{dF^2 \cdot dt^2}{dm \cdot dL^3}[/tex]

Newtons second law:
[tex]\boxed{dF = \frac{dm \cdot dL}{dt^2}}[/tex]

Adiabatic constant SI units:
[tex]\boxed{K = \frac{dF}{dL^2}}[/tex]

The SI adiabat is a unit of pressure:
[tex]1 \; \text{adiabat} = \frac{1 \; \text{Newton}}{1 \; \text{meter}^2}[/tex]

Reference:
http://en.wikipedia.org/wiki/Force" [Broken]
http://en.wikipedia.org/wiki/Entropy" [Broken]
http://en.wikipedia.org/wiki/Fermi_energy" [Broken]
http://en.wikipedia.org/wiki/Heat_capacity_ratio" [Broken]
http://en.wikipedia.org/wiki/Degenerate_matter" [Broken]
http://www.sfu.ca/~boal/385lecs/385lec18.pdf" [Broken]
http://www.sfu.ca/~boal/385lecs/385lec19.pdf" [Broken]
 
Last edited by a moderator:
  • #32


TOV equation solution:
[tex]\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}[/tex]

Degenerate Fermi pressure:
[tex]P(r) = \frac{3^{2/3} \pi^{4} \hbar^2 \rho (r)^{5/3}}{5 m_n^{8/3}}[/tex]

Integration by substitution:
[tex]\frac{dP}{dr} = - \left( \frac{3^{2/3} \pi^{4/3} \hbar^2 \rho (r)^{5/3}}{5 m_n^{8/3}} + \rho(r) c^2 \right) \left[ \frac{4 \pi G r^3}{c^4} \left( \frac{3^{2/3} \pi^{4/3} \hbar^2 \rho (r)^{5/3}}{5 m_n^{8/3}} \right) + \frac{G m(r)}{c^2} \right] \left[ r \left(r - \frac{2G m(r)}{c^2} \right) \right]^{-1}[/tex]

Degenerate Fermi-TOV equation solution:
[tex]\boxed{ \frac{dP}{dr} = - \left( \frac{3^{2/3} \pi^{4/3} \hbar^2 \rho (r)^{5/3}}{5 m_n^{8/3}} + \rho(r) c^2 \right) \left[ \left( \frac{4 \cdot 3^{2/3} \pi^{7/3} \hbar^2 G r^3 \rho (r)^{5/3}}{5 c^4 m_n^{8/3}} \right) + \frac{G m(r)}{c^2} \right] \left[ r \left( r - \frac{2G m(r)}{c^2} \right) \right]^{-1} }[/tex]

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

Integration by substitution:
[tex]\frac{dP}{dr} = - \left[ \frac{3^{2/3} \pi^{4/3} \hbar^2}{5 m_n^{8/3}} \left( \rho_c \left[1 - \left( \frac{r}{R} \right)^2 \right] \right)^{5/3} + \left( \rho_c \left[1 - \left( \frac{r}{R} \right)^2 \right] \right) c^2 \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( \rho_c \left[1 - \left( \frac{r}{R} \right)^2 \right] \right)^{5/3} \right) + \frac{G m(r)}{c^2} \right] \left[ r \left( r - \frac{2G m(r)}{c^2} \right) \right]^{-1}[/tex]

Degenerate Fermi-TOV equation solution VII:
[tex]\boxed{\frac{dP}{dr} = - \left[ \frac{3^{2/3} \pi^{4/3} \hbar^2}{5 m_n^{8/3}} \left( \rho_c \left[1 - \left( \frac{r}{R} \right)^2 \right] \right)^{5/3} + \rho_c 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( \rho_c \left[1 - \left( \frac{r}{R} \right)^2 \right] \right)^{5/3} \right) + \frac{G m(r)}{c^2} \right] \left[ r \left( r - \frac{2G m(r)}{c^2} \right) \right]^{-1}}[/tex]

 
Last edited:
  • #33

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

Mass integration:
[tex]m(r) = 4 \pi \int_r^R r^2 \rho(r) dr[/tex]

[tex]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)[/tex]

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

Integration by substitution:
Degenerate Fermi-TOV equation solution VII:
[tex]\frac{dP}{dr} = - \left[ \frac{3^{2/3} \pi^{4/3} \hbar^2}{5 m_n^{8/3}} \left( \rho_c \left[1 - \left( \frac{r}{R} \right)^2 \right] \right)^{5/3} + \rho_c c^2 \left[1 - \left( \frac{r}{R} \right)^2 \right] \right]...[/tex]

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

[tex]... \left[ r \left( r - \frac{2G}{c^2} \left( \frac{4 \pi \rho_c}{15} \left( \frac{3r^5}{R^2} + 2R^3 - 5r^3 \right) \right) \right) \right]^{-1}[/tex]


 
Last edited:
  • #34

Degenerate Fermi-TOV equation solution VII:

[tex]\frac{dP}{dr} = - \left[ \frac{3^{2/3} \pi^{4/3} \hbar^2}{5 m_n^{8/3}} \left( \rho_c \left[1 - \left( \frac{r}{R} \right)^2 \right] \right)^{5/3} + \rho_c c^2 \left[1 - \left( \frac{r}{R} \right)^2 \right] \right]...[/tex]

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

[tex]... \left[ r \left( r - \frac{4 \pi G \rho_c}{15 c^2} \left( \frac{3r^5}{R^2} + 2R^3 - 5r^3 \right) \right) \right]^{-1}[/tex]

key:
[tex]\rho_c[/tex] - neutron star core density
[tex]R[/tex] - neutron star radius

Reference:
http://en.wikipedia.org/wiki/Neutron_star" [Broken]
 
Last edited by a moderator:
  • #35

Neutron average density:
[tex]\rho_n = \frac{m_n}{V_n} = \frac{3 m_n}{4 \pi r_n^3}[/tex]

Neutron radius:
[tex]r_n = r_0 A_n^{\frac{1}{3}}[/tex]
[tex]r_0 = 1.25 \cdot 10^{-15} \; \text{m}[/tex]

Neutron atomic mass unit mass:
[tex]A_n = 10^3 N_A m_n[/tex]
[tex]N_A[/tex] - Avogadro's number

Integration by substitution:
[tex]\rho_n = \frac{3 m_n}{4 \pi ( r_0 A_n^{\frac{1}{3}} )^3} = \frac{3 m_n}{4 \pi A_n r_0^3} = \frac{3 m_n}{4 \cdot 10^3 \pi N_A m_n r_0^3}[/tex]

Neutron density:
[tex]\boxed{\rho_n = \frac{3 m_n}{4 \cdot 10^3 \pi N_A m_n r_0^3}}[/tex]

[tex]\boxed{\rho_n = 2.0297 \cdot 10^{17} \; \frac{\text{kg}}{\text{m}^3}}[/tex]

Neutron star density function parameters:
[tex]\rho(r) \; \propto \; \int_{\text{1E9}}^{\text{8E17}} \; \frac{\text{kg}}{\text{m}^3}[/tex]

Neutron star average density parameters:
[tex]\rho_t \; \propto \; \int_{\text{8.4E16}}^{\text{1E18}} \; \frac{\text{kg}}{\text{m}^3}[/tex]

Neutron average density:
[tex]\boxed{\rho_n = 2.0297 \cdot 10^{17} \; \frac{\text{kg}}{\text{m}^3}}[/tex]

Reference:
http://en.wikipedia.org/wiki/Neutron" [Broken]
http://en.wikipedia.org/wiki/Nuclear_size" [Broken]
"[URL [Broken] number - Wikipedia[/URL]
http://en.wikipedia.org/wiki/Neutron_star" [Broken]
 
Last edited by a moderator:
<h2>1. What is the Tolman-Oppenheimer-Volkoff equation?</h2><p>The Tolman-Oppenheimer-Volkoff equation is a mathematical equation used in astrophysics to describe the structure and properties of a static, spherically symmetric object, such as a star or planet. It takes into account the effects of gravity, pressure, and energy on the object's structure.</p><h2>2. Who developed the Tolman-Oppenheimer-Volkoff equation?</h2><p>The Tolman-Oppenheimer-Volkoff equation was developed by American physicists Richard C. Tolman, J. Robert Oppenheimer, and George M. Volkoff in the 1930s. They were studying the properties of neutron stars, which are extremely dense objects with strong gravitational fields.</p><h2>3. What is the significance of the Tolman-Oppenheimer-Volkoff equation?</h2><p>The Tolman-Oppenheimer-Volkoff equation is significant because it provides a theoretical framework for understanding the structure and properties of highly dense objects, such as neutron stars. It also helps to explain the observed properties of these objects, such as their mass and radius.</p><h2>4. How is the Tolman-Oppenheimer-Volkoff equation used in astrophysics?</h2><p>The Tolman-Oppenheimer-Volkoff equation is used in astrophysics to study the structure and properties of objects in the universe, such as stars and planets. It is also used to make predictions about the behavior of these objects under extreme conditions, such as in the presence of strong gravitational fields.</p><h2>5. Are there any limitations to the Tolman-Oppenheimer-Volkoff equation?</h2><p>While the Tolman-Oppenheimer-Volkoff equation is a useful tool in astrophysics, it does have some limitations. It assumes that the object being studied is static and spherically symmetric, which may not always be the case. It also does not take into account the effects of rotation or magnetic fields, which can have a significant impact on the structure of an object.</p>

1. What is the Tolman-Oppenheimer-Volkoff equation?

The Tolman-Oppenheimer-Volkoff equation is a mathematical equation used in astrophysics to describe the structure and properties of a static, spherically symmetric object, such as a star or planet. It takes into account the effects of gravity, pressure, and energy on the object's structure.

2. Who developed the Tolman-Oppenheimer-Volkoff equation?

The Tolman-Oppenheimer-Volkoff equation was developed by American physicists Richard C. Tolman, J. Robert Oppenheimer, and George M. Volkoff in the 1930s. They were studying the properties of neutron stars, which are extremely dense objects with strong gravitational fields.

3. What is the significance of the Tolman-Oppenheimer-Volkoff equation?

The Tolman-Oppenheimer-Volkoff equation is significant because it provides a theoretical framework for understanding the structure and properties of highly dense objects, such as neutron stars. It also helps to explain the observed properties of these objects, such as their mass and radius.

4. How is the Tolman-Oppenheimer-Volkoff equation used in astrophysics?

The Tolman-Oppenheimer-Volkoff equation is used in astrophysics to study the structure and properties of objects in the universe, such as stars and planets. It is also used to make predictions about the behavior of these objects under extreme conditions, such as in the presence of strong gravitational fields.

5. Are there any limitations to the Tolman-Oppenheimer-Volkoff equation?

While the Tolman-Oppenheimer-Volkoff equation is a useful tool in astrophysics, it does have some limitations. It assumes that the object being studied is static and spherically symmetric, which may not always be the case. It also does not take into account the effects of rotation or magnetic fields, which can have a significant impact on the structure of an object.

Similar threads

  • Special and General Relativity
Replies
2
Views
1K
Replies
1
Views
526
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
1K
Replies
4
Views
331
Replies
1
Views
1K
  • Differential Equations
Replies
3
Views
2K
  • Advanced Physics Homework Help
Replies
30
Views
1K
  • Differential Geometry
Replies
2
Views
841
Back
Top