Interstellar

Originally posted by Suzianne
I always thought with a sufficiently large black hole that one may not realize when one has passed the event horizon. Not immediately, anyway.

The black hole would have to be huge. This is a hack's argument - this isn't rigorous as I'm liberally mixing results from Newtonian gravity with Einstein's theory. The formula for gravitational tidal acceleration is:

a = 2GML/R^3

This is the difference in acceleration between the top or bottom of the body and the centre (L is the half length).

M is the mass of the gravitating body - black hole in this case.
G is Newton's constant.
R is the radial distance outward.
L is the length of the tidally stretched body in the direction radially outward from the hole.

this is from Newton's theory. The Schartzschild radius is r_s = 2*G*M/c² from Einstein's theory.

We can replace 2GM with c²r_s to give:

a = c²*L*r_s/R^3

We want to know this at the event horizon so set R = r_s

a = c²*L/r_s²

r_s = c*sqrt(L/a)

Suppose L = 1 metre (about right for a human), and a = 1 Newton/kg = 1 metre per second squared ('cos it's convenient). (1/10th the gravitational acceleration at the Earth's surface)

then r_s = 3*10^8 metres or 1 light second.

This corresponds to a black hole mass of 100,000 times the mass of the sun. Saggitarius A*, the radio source at the centre of the galaxy is believe to have a mass of 4.31 million solar masses. This corresponds to a Schwartzschild radius of 40 light seconds, and the tidal acceleration would be 0.024 N/kg, so yes barely noticeable.

A stellar mass black hole on the other hand would produce a tidal acceleration of 100,000 metres per seconds squared.

Originally posted by DeepThought
The black hole would have to be huge. This is a hack's argument - this isn't rigorous as I'm liberally mixing results from Newtonian gravity with Einstein's theory. The formula for gravitational tidal acceleration is:

a = 2GML/R^3

This is the difference in acceleration between the top or bottom of the body and the centre (L is the half length).
...[text shortened]... hole on the other hand would produce a tidal acceleration of 100,000 metres per seconds squared.

10,000 g's stretching your head from your toes. That would be worse than the 14th century rack🙂 So if you instead were horizontal instead of vertical, it may only be 1000 g's separating your butt from your tummy, no big deal🙂

Isn't it a matter of choosing the right path into a black hole to give you time shift effects and so forth?

Originally posted by sonhouse
10,000 g's stretching your head from your toes. That would be worse than the 14th century rack🙂 So if you instead were horizontal instead of vertical, it may only be 1000 g's separating your butt from your tummy, no big deal🙂

Isn't it a matter of choosing the right path into a black hole to give you time shift effects and so forth?

If the black hole is rotating things are different. There is an ergosphere where space is dragged around faster than light - so to an asymptotic observer a particle on a tangential trajectory (so it can get out again), would appear to have moved faster than light.

Originally posted by DeepThought
[b]If the black hole is rotating things are different.

All black holes are rotating at enormous speeds.

Originally posted by woadman
All black holes are rotating at enormous speeds.

How do you show that?

Originally posted by sonhouse
How do you show that?

It's plausible as one would expect the progenitor stars to be rotating. A merger between two black holes where the angular momentum exactly cancels seems unlikely. The Schwartzschild metric is an idealisation, it's not realistic to expect complete spherical symmetry. There's a black hole believed to be spinning at 1,150 revolutions per second in a binary system.

http://en.wikipedia.org/wiki/GRS_1915%2B105

Shyte, I love this forum. I do. I have very little to contribute, but I love reading it. 🙂