Relativistic Roche limit? How about Newtonian Roche limit?
Hi, hasman,
Labguy said:
You can get a decent explanation at http://en.wikipedia.org/wiki/Roche_limit" .
DWR (Danger Will Robinson)! Wikipedia is unstable and lacks any effective quality control. My impression from hanging around Physics Forums for a few weeks, compared with my experience last year at Wikipedia (see
http://en.wikipedia.org/wiki/User:Hillman/Archive) is that the general relativity board at PF tends to have better quality control, at least right now.
This generic warning aside, the version of this Wikipedia article which I am looking at,
http://en.wikipedia.org/w/index.php?title=Roche_limit&oldid=91178742, does seem to give the right idea: set the magnitude of the tidal acceleration pulling apart antipodal bits of star, at a given distance from the massive object (call it a black hole if you like), equal to the magnitude of the acceleration on a bit of star at the surface of the star, due to the self-gravitation at the surface of the star. It seems appropriate to me to use Newtonian approximations for the tidal force and acceleration due to self-gravitation of the star, for reasons I'll mention in a moment.
Then we obtain m/r^2 = 4 r \, M/R^3 whence
R = \left( 4M/m \right)^{1/3} r
Here, on the left, I used the magnitude of the radial outward acceleration of a static fluid particle at radius r in a mass m Schwarzschild solution
\frac{m/r^2}{\sqrt{1-2m/r}} \approx m/r^2
and on the right I used the magnitude of the radially pulling tidal tensor component as measured by a static observer at Scharzschild radius R for a mass M Schwarzschild solution, namely 2M/R^3, times the diameter of the star.
Since gtr is a nonlinear theory, this is clearly only a rough approximation! If you wanted a truly relativistic computation, you'd need to solve the vacuum EFE for two approaching massive bodies. (There is a solution, the so-called double Kerr solution, which claims to answer to this description, but there are good reasons to be cautious about accepting this interpretation!)
Since in ignoring nonlinearities and "v^2 effects", I secretly made pretty much the same assumptions as those used to obtain the Newtonian limit of gtr, it is not surprising that I obtained the same value for the critical distance R as given in the article on the basis of Newtonian theory (allowing for the fact that even in Newtonian theory, it seems to me, one should really use the radially pulling Coulomb component of the tidal tensor, not the orthogonally compressing ones).
Plugging in values for M = 100 solar masses, m=1 solar mass, r= very roughly 1 solar diameter, or M = 150 \, \rm{km}, \; m = 1.5 \, \rm{km}, \; r = 10^6 \, \rm{km}, I find R \approx 7.4 r, or about 2200 km, which essentially agrees with your computation.
hasman said:
Because I tried it with the samples I found on internet ... but when I calculated from the formulas it does not match
Found WHERE precisely? If you found a figure at Kip Thorne's website (say) which differs drastically from our computation here, you should worry (although you might simply be misinterpreting something if so, so give a link, please). If you found a figure at some website put up my some anonymous nonprofessional physicist, you need not worry.
BTW, you can find a truly relativistic computation in this paper: http://www.arxiv.org/abs/gr-qc/0501084; a quick glance suggests that their computation agrees approximately with the back-of-the-envelope estimate above.
Zeit said:
Hello,
So, the density of the black hole is
Danger! Danger! DavidBektas and Zeit, hasman was right to be suspicious of this: one should really should avoid speaking of the "density" of a black hole. In this case, I guess this is a self-cancelling error, since we wound up using a Newtonian analysis to get a ball park figure.
Chris Hillman
