Falling into a super massive blackhold

[PatPwnt]

Say I stick half of my body through the event horizon of a super massive black hole. I say a super massive black hole because in this case the event horizon radius will be very large and the tidal forces will not be as strong. Why can't I pull myself out if the force on the part of my body past the event horizon is not much greater than the force pulling on the part of my body that is outside? Will my body still be ripped apart?

 

[zhermes]

The 'force' IS very different across the event horizon. But its hard to even think about gravity in terms of a 'force' once you are within the event horizon (EH)---the concept of 'force' would be infinite. In General Relativity (which predicts and describes Black-Holes), gravity is not a 'force' per se but a geometry of space-time. Within the EH, space-time is so twisted and distorted that its just impossible for things not to fall inward. There is no amount of force which could stop it.

 

[bcrowell]

When people say that tidal forces at the event horizon can be made as small as you like by making a black hole big enough, the measure of tidal forces they have in mind is called the Weyl curvature tensor. John Baez has written up a nice explanation of the physical meaning of the Weyl tensor, in terms of an imaginary ball of free-falling coffee grounds that act as test particles: http://math.ucr.edu/home/baez/gr/ricci.weyl.html

The Weyl tensor is not, however, the relevant notion in your situation, because you're not free-falling. The relevant notion here would be the proper acceleration required in order to hover at a certain location. In other words, you dangle a standard mass m from a spring scale, hold the spring scale stationary at a certain radial coordinate r, and measure the force F on the spring scale. The proper acceleration needed in order to hover at r is defined as a=F/m. This quantity, unlike the Weyl tensor, does blow up at the event horizon. See http://www.mathpages.com/rr/s7-03/7-03.htm , at "the proper local acceleration of a stationary observer" for an equation giving a in terms of r.

As zhermes has pointed out, it's not just a coincidence that it works out this way. GR is basically a description of causal relationships. An event at or below the event horizon has no events in its causal future that lie farther away from the black hole.

 

[pervect]

Basically, you can't stick your hand through the event horizon of a black hole No force can hold you stationary at the surface of a black hole. There are a couple of ways of looking at this in increasing sophistication.

The least sophisticated argument, though the easiest to understand, is that you'd feel infinite gravity at the surface of a black hole,and therefore you can't stay stationary there.

A harder-to-understand approach, which I think avoids a few subtle problems with the above, is to consider the nature of the event horizon. It's not really a "place", but rather it's the worldline of a photon, or light beam. So, in any physical frame, the frame of an observer (to use the jargon,the frame of a timelike observer), the event horizon will have a relative speed of 'c',the speed of light.

So the reason you can't put your arm through the event horizon and pull it back is the same reasont hat you can't put your hand in a light beam and then pull it back. The light will always move faster than your arm.

One thing the more sophisticated approach does better than the unsophisticated one is explain how the tidal forces remain finite for an observer falling through the even horizon.

 

[Kholdstare]

Basically, you can't stick your hand through the event horizon of a black hole No force can hold you stationary at the surface of a black hole. There are a couple of ways of looking at this in increasing sophistication.

The least sophisticated argument, though the easiest to understand, is that you'd feel infinite gravity at the surface of a black hole,and therefore you can't stay stationary there.

A harder-to-understand approach, which I think avoids a few subtle problems with the above, is to consider the nature of the event horizon. It's not really a "place", but rather it's the worldline of a photon, or light beam. So, in any physical frame, the frame of an observer (to use the jargon,the frame of a timelike observer), the event horizon will have a relative speed of 'c',the speed of light.

So the reason you can't put your arm through the event horizon and pull it back is the same reasont hat you can't put your hand in a light beam and then pull it back. The light will always move faster than your arm.

One thing the more sophisticated approach does better than the unsophisticated one is explain how the tidal forces remain finite for an observer falling through the even horizon.

Actually Stephen Hawking describes that the gravitational field around a black hole is not only intense but varying with distance. So the outer part of your body (outside event horizon) will be torn apart due to the difference in the attractive force felt by several region of your body. The length might be so small but the difference might be very large. Thus it will be torn and spread apart over the space cleared by the receding event horizon of the black hole.

For the lower part of your body i can't be sure but i guess it will be attracted with infinite gravity and will meet the singularity with zero radius. However in that case you might be turned into a soup of elementary particles haha...

Also the event horizon is difficult to reach as it seems to be continually accelerating from an observer. A pulse train sent from event horizon to a stationary observer would reach there with more and more delay in time.

 

[pervect]

I think you've misread something there. At the central singularity, one will be torn apart by tidal forces. However, one will NOT be torn apart by tidal forces at the event horizon.

For a popularization of this, see for instance Kip Thorne's excellent "Black Holes and Time-Warps." The prolouge, in particular, describes the experiences of various observers in and around black holes of various sizes. The ones falling into small enough black holes get ripped apart before they reach the event horizon, but the observers falling into larger black holes can easily withstand the tidal forces.

For a more technical reference, MTW's big black book, "Gravitation", computes the tidal forces at the event horizon, and shows them to be finite and equal to 2GM/r^3 (for the radial component of the force, this calculation is done on on pg 860, I've added back the coefficient of G , which has a value of 1 in geometric units but is needed to use standard units). Since the event horizon is located at r= 2GM/c^2, you get the tidal force at the event horizon equal to 2GM / (2GM/c^2)^3 =

$$\frac{c^6}{\left(2\,G\,M\right)^2}$$

where the units are meters per second^2 / meter. So you can see the tidal force at the event horizon is finite,and varies inversely with the square of the mass of the black hole you're falling into, thus the larger the better.

The T in MTW is Thorne, by the way, the author of the popularization I mentioned previously.