# Shouldn't objects entering black holes be instantly shredded?

1. May 21, 2013

### qwertyflatty

Nothing can escape a black hole, so as an object enters the event horizon, the atoms currently inside the black hole wouldn't be able to bond with the atoms currently outside of the black hole, causing it to sever. But I've never heard of any physicist saying that this would happen, so I'm probably wrong.

Could someone explain why an object wouldn't get shredded immediately from entering a black hole's event horizon?

2. May 21, 2013

### ModusPwnd

3. May 21, 2013

### qwertyflatty

I mean instantly, as it's entering the event horizon. Spaghettification may happen sometime after passing through the event horizon.

4. May 21, 2013

### Integral

Staff Emeritus
Actually Spaghettification happens as you approach the event horizon due to the tremendous gravitational gradient present.

5. May 21, 2013

### Chronos

For a supermassive black hole, you would not even notice you had entered its event horizon [unless you tried to escape]. Tidal forces are virtually insignificant. So, nothing would appear out of sorts for an object in free fall entering the event horizon of a supermassive black hole - until later.

6. May 22, 2013

### qwertyflatty

I'm not talking about tidal forces.

Anything that crosses the event horizon can't get out. When an object crosses the event horizon, it doesn't all cross at the exact same time. Only parts of it cross at any instant. If nothing inside the event horizon can get out, then the atoms of the object currently inside the event horizon shouldn't be able to bond to the atoms of the objects outside the event horizon, causing it to be severed.

7. May 22, 2013

### phinds

No, as chronos pointed out, large black holes have such a small gravity gradient near the event horizon that the EH is a "non-event" horizon.

The EH has NOTHING to do with spahettification. It can happen before or after the EH depending on the size of the BH and for it to happen AT the EH is possible but extremely unlikely as it would require a precise confluence of BH size and particular strength of material entering.

8. May 22, 2013

### Chronos

Actually, this idea is very reminiscent of that proposed by Polchinski last July: Black Holes: Complementarity or Firewalls?, http://arxiv.org/abs/1207.3123. Obviously, there is no way to test such an idea.

Last edited: May 22, 2013
9. May 22, 2013

### willem2

Objects just outside of the event horizon, will need a very large acceleration to get out. It is the force necessary for the acceleration that will tear the object apart, but the same force would also tear the object apart without a nearby black hole.

10. May 22, 2013

### phinds

Huh? I don't get what you're saying

11. May 22, 2013

### Spourk

Speaking of tidal forces, has anyone read the shortstory, "Neutron Star" by Larry Niven? It describes what it might be like if a pilot of an indestructable ship gets close enough to a Neutron star to experience tidal effects.

Poor Larry Niven though, he has been chastized into the floor by all kinds of students and physicists alike, especially for Ringworld.

Niven writes: "I keep meeting people who have done mathematical treatments of the problem raised in the short story 'Neutron Star', .... Alas and dammit, Shaeffer can't survive. It turns out that his ship leaves the star spinning, and keeps the spin."

12. May 22, 2013

### willem2

Any object very close to the event horizon of a black hole needs to accelerate very quickly to escape from the black hole. It is the force needed for this acceleration that will tear an object apart that has gone halfway past the event horizon. If you don't try to accelerate the object it will fall in entirely and won't get torn apart. (as long as tidal forces aren't too big)

13. May 22, 2013

### WannabeNewton

The force that must be exerted at infinity to not even accelerate backwards but keep stationary a unit test mass in the limit as one approaches the event horizon is given by the surface gravity $\kappa$.

In particular, it can be shown that if $\xi^{a}$ is the killing vector field normal to the (null) killing horizon that represents the event horizon of a stationary black hole, then $\kappa^{2} = \lim \{\frac{-(\xi^{b}\nabla_{b}\xi^{c})(\xi^{a}\nabla_{a}\xi_{c})}{\xi^{d}\xi_{d}}\}$ where the limit is as one approaches the event horizon. It is easy to show that the local proper acceleration of an object stationary in this limit is just $a^{c} = \frac{(\xi^{b}\nabla_{b}\xi^{c})}{(-\xi^{d}\xi_{d})}$ so, letting $V = (-\xi^{a}\xi_{a})^{1/2}$, the above turns into $\kappa^{2} = \lim(Va)$.

For a static black hole, $V$ is nothing more than the redshift factor so the above limit just gives the force that must be exerted at infinity in order to keep a unit test particle in place as one approaches the event horizon. The reason why what you're saying, regarding the object being halfway into the black hole, makes no sense is because the local acceleration $a$ diverges to infinity at the event horizon whereas the redshift factor vanishes at the event horizon (the horizon is a null surface and the killing vector field is normal to it hence it is null on the horizon so it has zero length) meaning the force that must be exerted at infinity to keep the object in place once parts of it have gone through the event horizon isn't even well-defined, let alone a force that could accelerate it backwards.

14. May 22, 2013

### qwertyflatty

I'm not talking about tidal forces, and I'm not talking about hovering near an event horizon.

When an object enters the event horizon, the entire object can't enter at the same instant. The same is true for a person passing through any point. Ex. A person diving into a swimming pool would enter hands first, then head, then the rest of his body.

Assume the black hole is massive enough that it's still intact as it reaches the event horizon. Any object that passes through the event horizon can't transmit information outside of the event horizon. Ergo, the atoms currently inside the event horizon can't bond with the atoms outside during the instances where some parts of object are inside. Ergo, the atoms inside should become severed from the outside.

Do you understand my question now?

15. May 23, 2013

### Staff: Mentor

To put it another way (using your pool analogy) - if you are half submerged in the black hole, you can't feel your legs, as the signal from the legs is not able to get to the brain.

It makes me think about different, related scenario. Imagine two oppositely charged point like objects, spinning around their center of mass (let's assume objects are large enough to be described by the classic mechanics). They approach the event horizon and one of them falls below before the other. What happens to the trajectory of the object that is not yet below the event horizon?

16. May 23, 2013

### Fredrik

Staff Emeritus
I think you would be going much faster than the nerve signal, so the rest of your body will be inside before the signal reaches your brain.

Now you're probably thinking about ways to go slowly through the event horizon. I wouldn't recommend it. Keep in mind that the force required to keep you at a fixed coordinate "altitude" goes to infinity as you approach the horizon. I haven't done the math on this, but I think it's safe to say that slowing you down to the speed of a nerve signal will require forces that are many orders of magnitude greater than what it would take to crush you.

17. May 23, 2013

### Staff: Mentor

I feel like there is some kind of contradiction here.

18. May 23, 2013

### VantagePoint72

No, they are agreeing with each other. The surface gravity at an event horizon is huge, but you don't feel surface gravity when you're in free fall. You only feel tidal forces when in free fall, which are not large at an event horizon. To feel to the absolute value of the gravitational acceleration, you need to resist it and slow your descent—Frederik's point. I agree with him that this force would almost certainly be enough to crush you. On the other hand, if you start free falling towards the event horizon from a far enough distance that you could have maintained a fixed height without being crushed, then by the time you reach the horizon you will be going extremely fast. Much faster than the propagation of nerve signals in the pool analogy.

Last edited: May 23, 2013
19. May 23, 2013

### Staff: Mentor

OK, so the analogy was poor. Still, the analogy was there only to explain what the original question was about.

What about trajectory of the charged object?

20. May 23, 2013

### phinds

Absolutely not true as a generality. Tidal forces are so huge at the EH of a small BH that you would be ripped apart WAY before you even get to the EH.