Deriving the Chandrasekhar Limit on a Napkin

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  • #1
Hey guys,

i was wondering if someone could point me to or write out a basic mathematical derivation of this limit. All i can find are qualtative accounts or really long papers like the original published. I understand there's a lot that goes into the nitty gritty of it but i want a reasonably tight 'proof' or derivation of this limit that holds without all the specifics put in. The kind you could write on a napkin in a pub to impress people :P (and no that's not the intention )

  • #2
what if the derivation CAN'T be done with "basic mathematics"?

Also, basic mathematics is a subjective notation, it depends on who you ask..
  • #3
Ok sorry let me rephrase, i don't mean simple mathematics i mean taking the underlying principles in a qualtative argument and replacing them with their mathematical counterparts, however complex. Ie i want something other than a 40 page derviation and i know they exist because there's one in my lecture notes but i really don't like it and the lecturer is away so i can't ask him about it =(
  • #4
I think the simplest way to put it is that electron degeneracy pressure can only be sustained up to some P_max. As material is added to the white dwarf the effective gravitational pressure pushing in increases until P_grav = P_max at which point the electron degeneracy pressure cannot keep the star stable and it collapses, causing nuclei to fuse, releasing enormous amounts of energy and causing the star to explode.

Then you would start expanding the P_grav and P_max terms. Gravity is easy, P_max is less so and is probably the one causing the long derivations. I think you will need to understand the full derivation in order to write down a simple form for it. I certainly don't have this knowledge at my finger tips either.
  • #5
Chandrasekhar limit...

According to Chandrasekhar's original paper (ref. 1 pg. 116 eq. 5), the maximum mass of a neutron star:
[tex]\left( \frac{G M_c}{M'} \right)^2 = \frac{(4 K)^3}{4 \pi G} \; \; \; \; \; K = 3.619 \cdot 10^{14} \; \frac{\text{N}}{\text{m}^2}[/tex] - Adiabatic constant
[tex]M_c = 1.882 \cdot 10^{33} \; \text{g}[/tex]
[tex]M_c = .91 M_{\odot}[/tex]

However, Chandrasekhar's original paper is missing a key for [tex]M'[/tex], which is located in (1 A. S. Eddington, Internal Constitution of Stars, p. 83, eq. (57.3.)), is identified by myself as Planck's mass [tex]M' = m_p[/tex], also note that this paper does not state all the SI units!

Therefore the modern equation translation becomes:
[tex]\boxed{M_c = \frac{4}{G^2} \sqrt{ \frac{ \hbar c K^3 }{\pi} } = .311 M_{\odot}}[/tex]

According to Chandrasekhar's paper (ref. 2, pg. 462 eq. 36):
[tex]M_c = \frac{2.015 \cdot 4}{\pi^{\frac{1}{2}}} \left( \frac{K_2}{G} \right) = .92 M_{\odot} \; \; \; \; \; K_2 = \frac{h c}{8 (\mu_e m_H)^{\frac{4}{3}}} \left( \frac{3}{\pi} \right)^{\frac{1}{3}} = 3.619 \cdot 10^{14} \; \frac{\text{N}}{\text{m}^2}[/tex]

According to Chandrasekhar's next paper (ref. 3, pg. 213 eq. 58):
[tex]M_c = 4 \pi \left( \frac{2 A_2}{\pi G} \right)^{\frac{3}{2}} \frac{1}{B^2} \omega_3^0 = \frac{5.728}{\mu} M_{\odot}[/tex] - (ref. 3, pg. 214 eq. 63)

[tex]A_2 = \frac{\pi m_e^4 c^5}{3 h^3} \; \; \; \; \; B = \frac{8 \pi m_e^3 c^3 \mu H}{3 h^3}[/tex] - (ref 3 pg. 209 eq. 5)
[tex]H \; \; \; \mu[/tex] - Hydrogen mass and molecular weight

According to Chandrasekhar's next paper (ref. 4, pg. 150 eq. 43):
[tex]M_c = 4 \pi \left( \frac{K_2}{\pi G} \right) (2.018) = 0.197 \left( \frac{h c}{G} \right)^{\frac{3}{2}} \frac{1}{\mu_e H}^2 = \frac{5.76}{\mu_e^2} M_{\odot}[/tex]

According to Wikipedia, the maximum mass of a neutron star:
[tex]M_c = \frac{\omega_3^0 \sqrt{3\pi}}{2}\left ( \frac{\hbar c}{G}\right )^{3/2}\frac{1}{(\mu_e m_H)^2} \; \; \; \; \; \omega_3^0 = 2.018236[/tex] - Lane-Emden constant

[tex]M_c = 1.43 \left( \frac{2}{\mu_e} \right)^2 M_{\odot}[/tex]
However, Wikipedia does not provide any proof for the constant [tex]\omega_3^0[/tex] derived from the solution to the Lane-Emden equation.

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  • #6

My solution for the ref. 3 equation:
[tex]\boxed{M_c = \sqrt{ \frac{3}{2} } \left( \frac{ \hbar c}{G} \right)^{\frac{3}{2}} \frac{\omega_3^0}{4 \pi (\mu_e m_H)^2} = \frac{0.3639}{\mu_e^2} M_{\odot}}[/tex]

[tex]\boxed{M_c = \frac{0.3639}{\mu_e^2} M_{\odot}}}[/tex]

Lane-Emden constant:
[tex]\omega_3^0 = - \left( \varepsilon^2 \frac{d \theta_3 }{d \varepsilon} \right) = 2.018236[/tex] - (ref. 3, pg. 214 eq. 59)

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  • #7
you can derive the chandrasekhar mass limit with a back-of-the-envelope argument that i think was first presented by landau. basically, you note that the total energy in the star is roughly the sum of the fermi energies of the baryons and the gravitational binding energy; the star is unstable when the energy is positive:

[tex]E \approx E_F + E_G > 0[/tex]

here, [tex]E_G \approx -\frac{G M^2}{R}[/tex] is the usual binding energy. the fermi energy can be approximated by invoking the uncertainty principle:

[tex]\Delta x \approx n^{-1/3} \approx \left(\frac{N}{R^3}\right)^{-1/3} \approx \frac{R}{N^{1/3}}[/tex] (here [itex]n[/itex] is the number density and [itex]N[/itex] is the total number of either the degenerate particles or the particles supported by them)

[tex]\Delta p \approx \frac{\hbar}{\Delta x} \approx \frac{\hbar N^{1/3}}{R}[/tex]

[tex]E_F \approx N c \Delta p[/tex] (assuming relativistic degeneracy, which turns out to be correct)

putting all this together, noting that [itex]M = N m_1[/itex], where [itex]m_1[/itex] is the mass of the nondegenerate particle species (e.g. protons), we find that the condition for stability is

[tex]N < \left(\frac{\hbar c}{G m_1^2}\right)^{3/2} \approx \left(\frac{m_{pl}}{m_1}\right)^3[/tex]

for protons, this gives [itex]M < 1.8 M_{\odot}[/itex], which is close to the canonical value of [itex]1.4 M_{\odot}[/itex] (as given by Orion1's last two references).

however, to find this more precise value requires doing some rather nasty integrals to find the degeneracy pressure. if you are interested, the details are most likely in any graduate stellar structure book, e.g hansen, kawaler and trimble.
  • #8

Relativistic Fermi energy:
[tex]E_f = \frac{\hbar c N^{\frac{4}{3}}}{R}[/tex]

Quantum mass: (pure fermions)
[tex]M = n m_p[/tex]

Gravitational energy:
[tex]E_g = - \frac{G (N m_p)^2}{R}[/tex]

Total stellar energy condition:
[tex]E_t = E_f + E_g \geq 0[/tex]

Total stellar energy:
[tex]E_t = \frac{\hbar c N^{\frac{4}{3}}}{R} - \frac{G (N m_1)^2}{R} \geq 0[/tex]

[tex]\frac{\hbar c N^{\frac{4}{3}}}{R} \geq \frac{G (N m_p)^2}{R}[/tex]

[tex]\boxed{N \leq \left(\frac{\hbar c}{G m_p^2}\right)^{3/2}}[/tex]

Condition for stability:
[tex]N \leq \left(\frac{\hbar c}{G m_p^2}\right)^{3/2}[/tex]

[tex]N = \frac{M_c}{m_p}[/tex]

[tex]\frac{M_c}{m_p} \leq \left(\frac{\hbar c}{G m_p^2}\right)^{3/2}[/tex]

Chandrasekhar mass:
[tex]\boxed{M_c \leq \frac{1}{m_p^2} \left(\frac{\hbar c}{G}\right)^{3/2}}[/tex]

[tex]\boxed{M_c \leq 1.852 \cdot M_{\odot}}[/tex]
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  • #9

Quantum Planck mass fermion number:
[tex]N \leq \left(\frac{m_{pl}}{m_p}\right)^3[/tex]

[tex]N = \frac{M_c}{m_p}[/tex]

[tex]\frac{M_c}{m_p} \leq \left(\frac{m_{pl}}{m_p}\right)^3[/tex]

Quantum Chandrasekhar mass:
[tex]\boxed{M_c \leq \frac{m_{pl}^3}{m_p^2}}[/tex]

[tex]\boxed{M_c \leq 1.852 \cdot M_{\odot}}[/tex]

[tex]\boxed{M_c \leq \frac{m_{pl}^3}{m_p^2} = \frac{1}{m_p^2} \left(\frac{\hbar c}{G}\right)^{3/2}}}[/tex]
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  • #10
DarkEternal said:
which is close to the canonical value of [itex]1.4 M_{\odot}[/itex] (as given by Orion1's last two references).

According to my equation solution for the ref. 3 equation on post #6, the solution equation listed in Wikipedia is incorrect, which was symbolically derived incorrectly from the ref. 3 equation.

The Chandrasekhar mass for ref. 3 is:
[tex]\boxed{M_c = \frac{0.3639}{\mu_e^2} M_{\odot}}}[/tex]

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