Infinitie tidal forces at event horizon?

[Ray Eston Smith Jr]

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 blackhole 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?

[jcsd]

It dpeends on the size of the black hole, for a large black hole the tidal forces will be unoticeable, for a very small black hole they beocme much more important. It alos dpeneds on the size of the frame of refernce, as obviously the larger the object, being drawn in the larger the tidal forces.

The tidal forces would not be infinite, except at the singularity I imagine.

[pervect]

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 blackhole 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


\frac {c^6} {(2 G m)^2}


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...

[Chronos]

Pervect is correct. I spent all night on that and you were really starting to make me mad.