Ray Eston Smith Jr said:
Many websites claim that someone falling into a black hole would be ripped apart by tidal forces as he crosses the event horizon. Others say that the falling observer feels nothing special as he crosses the event horizon - he doesn't get torn apart by tidal forces until he gets close to the singularity (assuming the black hole is big enough so that the tidal forces don't tear him apart way before he reaches the event horizon).
Which viewpoint is correct? Or does it depend on the frame of reference of the observer?
It depends mainly on the size of the black hole. I'd recommend Kip Thorne's book "Black Holes & Time Warps" for a good popular treatment of the topic.
For the detailed calculation
Warning: geometric units
The tidal force in the Schwarzschild basis metric works out to be 2m/r^3, and the event horizon is at r=2m, so the tidal force is 1/(4m^2).
Converting this to standard units, the tidal gradient in (meters/sec^2)/meter = 1/sec^2 will be
<br />
\frac {c^6} {(2 G m)^2}<br />
where c is the speed of light, and G is the Gravitatioanl constant
if m = 1 solar mass, I get 1e10 sec^-2, which is a billion gravities per meter.
If m = 1,000,000 solar mass, one has a comfortable .001 gravities/meter
where 1 gravity = 9.8 m/s^2 (approx 10 m/s^2).
This is a static calculation, but velocity towards the bh won't affect the tidal force.
I could use a double-check of the figures, but I think the answer is right...
