1. Dec 5, 2006

### quasar987

In my thermo text, we arrived at an estimate of the chandrasekhar limit using the assumptions T=0, uniform mass density and.... p>>mc.

I can maaaaaybe accept that in the context of a very rough approximation, but then the text says, "A more realistic calculation, which does not suppose a uniform density, gives $M_C=1.4 M_{\bigodot}$". Thats means that in their so-called more realistic calculation, they still assume the unphysical p>>mc.

That someone explain that?

Last edited: Dec 5, 2006
2. Dec 6, 2006

### SpaceTiger

Staff Emeritus
If the momenta were not much greater than mc throughout much of the star, then the star would be sub-Chandrasekhar, practically by definition. Another way of thinking of the Chandrasekhar mass is the limit of relativistic degeneracy for a self-gravitating, electron-degenerate object.

3. Dec 6, 2006

### quasar987

Oh wait a sec. My problem is that I was thinking classically about momentum. p>>mc does not imply v>>c. So there is nothing unphysical about p>>mc.

4. Dec 6, 2006

### George Jones

Staff Emeritus
Remember, in relativity, the spatial part of momentum is given by

$$p = \frac{mv}{\sqrt{1 - \frac{v^2}{c^2}}},$$

not by the Newtonian expression $p = mv$.

Thu,s $p >> mc$ when $v >> c/\sqrt{2}$, so $v > c$ is not needed.

Edit: While I was typing, you saw the light.

Last edited: Dec 6, 2006