Timeline for Why does a star collapse under its own gravity when the gravity at its centre is zero?
Current License: CC BY-SA 4.0
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| Dec 5, 2023 at 22:59 | history | edited | joshphysics | CC BY-SA 4.0 |
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| Feb 11, 2014 at 17:06 | comment | added | ThePhysicist | @joshphysics That's about right. The exact value or expression will definitely take into account the variable density and think multivariable calculus has to be used as the density must be allowed to be a function of time along with the usual space coordinates. But I'm sure you have used Newtonian Mechanics to derive the above expression. Josh, if you don't mind could I take a look at your derivation? | |
| Feb 11, 2014 at 9:04 | comment | added | joshphysics | @Bibhu I just computed the pressure at the center of a uniformly dense sphere of mass $M$ and radius $R$ due to the gravitational interaction, and I found $\frac{3}{8\pi}\frac{G M^2}{R^4}$. You can basically derive this by considering the pressure exerted on a given mass element by all of the mass above it. I don't actually know what the density profile of a real star is, however, and with a non-constant density, the number $\frac{3}{8\pi}$ out front would generally change. | |
| Feb 10, 2014 at 13:40 | comment | added | ThePhysicist | @joshphysics but I have a book that says the gravitational pressure at the centre of a star is approx GM^2/R^4. I have been looking for the exact derivation. I know how to derive the field equation of gravity inside a star. I don't how to derive the expression for pressure. If I simply divide the force you have expressed above by the total volume of the star, I still get it in terms of R^2. | |
| Feb 10, 2014 at 13:25 | vote | accept | ThePhysicist | ||
| Feb 5, 2014 at 21:00 | vote | accept | ThePhysicist | ||
| Feb 5, 2014 at 21:09 | |||||
| Feb 4, 2014 at 3:34 | comment | added | joshphysics | @cesaruliana Cool interesting! Had never learned about the Jeans instability. Thanks for the link. | |
| Feb 4, 2014 at 3:05 | comment | added | cesaruliana | @joshphysics: I think even the newtonian argument is already good enough to get the feeling behind the colapse, although GR does change the picture a lot. For completness one could add the notion of Jeans instability en.wikipedia.org/wiki/Jeans_instability which I think shows that collapse goes on inevitably in certain conditions. For spherically symmetric cases one can map the GR argument in the newtonian one fairly easily, and thus should suffice for the Schwarzschild Black Hole | |
| Feb 3, 2014 at 23:55 | history | edited | joshphysics | CC BY-SA 3.0 |
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| Feb 3, 2014 at 20:51 | history | edited | joshphysics | CC BY-SA 3.0 |
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| Feb 3, 2014 at 19:42 | comment | added | joshphysics | @AlanSE I felt that the OP was just confused by the general idea that collapse can happen to gravitational bodies given that there is a vanishing field at the center, so I decided to give the simplistic Newtonian argument above. I certainly have not addressed details of black hole formation. If that's what the OP is actually looking for, then I certainly haven't answered the question. | |
| Feb 3, 2014 at 19:40 | comment | added | Zo the Relativist | @AlanSE: but that depends entriely on the radial dependence of the density. Assume that it falls off like $\frac{1}{r^{n}}$ for some $ n > 0$. Then, the force diverges. And the argument for why real stars collapse is based in general relativity, and the stability of stars under perturbations in full general relativity. newtonian mechanics can't help you there. | |
| Feb 3, 2014 at 19:35 | comment | added | Alan Rominger | As we're considering the limit of r to zero, I find the equation (which is a common model) to be unconvincing. As we get to the center, the force you mention tapers off to zero. However, the pressure continues to rise, and this is what stars can't maintain. As the fuel is depleted, the pressure will have no choice but to fall. I believe there is a pressure term in the momentum "currents" of the general relativity field equation. So that's certainly relevant, but still much more difficult than looking at it from the perspective of a far-away observer. | |
| Feb 3, 2014 at 19:29 | history | answered | joshphysics | CC BY-SA 3.0 |