Stargazing Understanding the Chandrasekhar Mass Units for Relativistic Lane Emden Equations

[MarkL]

TL;DR Summary

relativistic equation for mass give incorrect units using Lame Emden

Using Lane Emden and n = 3 (relativistic), I calculate the correct mass -- the Chandrasekhar mass (about 1.4 M_sun)
The equation goes M_total ∝ a³, because at n=3, the density, ρ, cancels out.
a² ∝ K/G = Kg² → a3 ∝ Kg³. Here K ∝ h c or Kg m³/sec² and G ∝ m³/Kg/sec²
This implies the mass, M_total units is Kg³

What is wrong, please...

 

[Ibix]

A reference would be helpful. Looking at the Wikipedia page, $K=P/\rho^{1+1/n}$, which is $P\rho^{-4/3}$ in your case. That does not seem to me to have dimensions of $\mathrm{[M]}^2$, but rather $\mathrm{[M]}^{2/3}$ if my late night algebra is correct. But that page doesn't mention a symbol $a$ anywhere, so it may be a symbol clash - which is why a reference would be helpful.

 

[MarkL]

A reference would be helpful. Looking at the Wikipedia page, $K=P/\rho^{1+1/n}$, which is $P\rho^{-4/3}$ in your case. That does not seem to me to have dimensions of $\mathrm{[M]}^2$, but rather $\mathrm{[M]}^{2/3}$ if my late night algebra is correct. But that page doesn't mention a symbol $a$ anywhere, so it may be a symbol clash - which is why a reference would be helpful.

Thank you - I discovered my mistake. I will delete this post shortly.

 

[MarkL]

a is in Lame Emden equations --- as in, r = a ξ

 

[MarkL]

I guess I can't delete...

 

[berkeman]

I guess I can't delete...

No reason to delete the thread -- it has value to others. I'll go ahead and tie it off for you.