By considering the potential energy of a degenerate star of mass $M$ and radius $R$, I can use dimensional arguments to show that the radius of the star depends on its mass as:

$$ R ∝ M^\frac{2-n}{4-3n}$$

Where in the non-relativistic regime $n = 5/3$ whilst in ultra-relativitic $n = 4/3$

By taking $n = 5/3$ I can see that non-relativistically

$$ R ∝ M^\frac{-1}{3} $$

Could someone explain to me the physical nature of what happens when $n = 4/3$ and it becomes relativistic as I am unsure what

$$ R ∝ M^{\infty}$$ means physically.

  • 1
    $\begingroup$ Have you seen en.wikipedia.org/wiki/… ? In particular, see the section titled Mass–radius relationship and mass limit. $\endgroup$ – PM 2Ring Mar 27 at 13:16

It means, that $$ M \propto R^0$$ so that if you did have a star governed by a relativistic polytrope, there is only one mass the star can have. This is the "classic" Chandrasekhar mass, at which the radius shrinks to nothing and the density increases to infinity.

In practice, the Chandrasekhar mass is itself inconsistent with a true relativistic treatment since it ignores the kinetic energy contribution to the gravitating mass. Inclusion of this in a full GR treatment leads to instability and collapse at a finite density and at a slightly lower mass.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.