# Can we hit black hole's surface without dying?

1. Dec 3, 2006

### chandubaba

me and my cousin rikkhon speculate that if we enter a black hole in an unimaginable velocity(synchronising with the acceleration due to gravity of the black hole )then isn't it possible that we may make a successful hit at the surface of the black hole without dying?

2. Dec 3, 2006

### DaveC426913

1] When you say "unimaginable velocity", you mean less than c, right?
2] You do not have to have a high velocity to synchronize with acceleration due to gravity. Gravity will ensure you are at the correct velocity to be in freefall. But it's not the gravity you have to worry about, it's the tides (i.e. the differences in gravity between your head and your feet).
3] Black holes do not have surfaces.

I'm not sure what your setup is meant to accomplish.

3. Dec 3, 2006

### AlphaNumeric

You can pass through the event horizon of a sufficently massive black hole without noticing it. The tidal forces descrease at the event horizon as the mass of the black hole increases. So for one which is billions of times the mass of the Sun, you can pass the event horizon without problem. For a 'light' black hole, which some theories hypothesise exist from the big bang, the tidal forces would rip you apart well before you got to the event horizon.

In either case, you'd be dead well before you got to the singularity because the tidal forces always increase to the point where they'll rip you apart before you get to the singularity.

4. Dec 3, 2006

### Chris Hillman

I'd add to what AlphaNumeric wrote the comment that you wouldn't notice anything in particular happening to your spaceship as you passed through the horizon.

Chris Hillman

5. Dec 3, 2006

### AlphaNumeric

^ Yep, that too. You'd notice the view out your porthole/window looking pretty weird as the light bending effects become so powerful, but if you were falling in a windowless lift you'd not know when you passed the event horizon.

Of course this is all from a test particle's point of view. The human body is a bit different. If you were falling slow enough, you'd reach a point where your feet were inside the EH and your head not. Then you wouldn't be able to see your feet. From then on, your feet would be invisible, as would anything below your eyes. Infact, all bodily functions would stop since blood wouldn't be able to go up from your heart to your head or nerve impulses. You'd die at the event horizon, but your body's general shape would not be turn apart till you hit strong enough tidal forces. Of course, even if you did survive past the event horizon, you'd be doomed to death anyway.

6. Dec 3, 2006

### Chris Hillman

Hi, AlphaNumeric,

Agreed: observers falling freely into a black hole, or hovering outside a black hole, or otherwise using a rocket engine to manuever in a strong gravitational field, would experience various interesting optical effects. The simulations at http://casa.colorado.edu/~ajsh/schw.shtml are potentially a bit misleading in the present context, but serve to give some indication.

Consider the Schwarzschild vacuum solution. Speaking somewhat loosely, the "slowfall frame field" models the physical experience of infalling observers who fire their rocket engine radially inwards with just the right thrust to oppose the attraction of the hole, according to Newtonian theory, namely $$m/r^2$$, which of course is less than the corresponding quantity in gtr, namely $$m/r^2/\sqrt{1-2m/r}$$, with the result that these observers slowly fall radially inwards.

More precisely, in the ingoing Eddington chart
$$ds^2 = -(1-2m/r) \, du^2 + 2 \, du \, dr + r^2 \, \left( d\theta^2 + \sin(\theta)^2 d\phi^2,$$
$$-\infty < u < \infty, \; 0 < r < \infty, \; 0 < \theta < \pi, \; -\pi < \phi < \pi$$
consider the frame field
$$\vec{e}_0 = \partial_u - m/r \, \partial_r$$
$$\vec{e}_1 = \partial_u + \left( 1 - m/r \right) \, \partial_r$$
$$\vec{e}_2 = 1/r \, \partial_\theta, \; \vec{e}_3 = 1/r/\sin(\theta) \, \partial_\phi$$
Here the acceleration vector of the world lines (integral curves of the timelike unit vector $$\vec{X} = \vec{e}_0$$) is $$\nabla_{\vec{X}} \vec{X} = \frac{m}{r^2} \, \vec{e}_1$$. These world lines are given by $$u-u_0 = -r^2/2/m$$.

Careful, that's not true as stated. I think you meant to say: "your view of your feet would be somewhat out of date". Likewise for electrical signals from that region of your body. Of course--- and this is rather the point--- this would be true in any region of any spacetime.

You already know that nothing special happens locally at the event horizon, so you already know that this can't be quite right. But it is true that anyone falling sufficiently "slowly" in the sense of slowly decreasing Schwarzschild radial coordinate would have to be accelerating outward, and accelerating sufficiently hard can certainly break a human body, as in the example of a skydiver whose parachute fails catastrophically.

I am not sure what you are thinking of here, but as MTW remark, the Coulomb form of the tidal tensor appears not only for static observers in the exterior, but also for freely and radially infalling observers--- and for the slowfall observers defined above. Specifically, wrt the above frame field, the expansion tensor (of our timelike congruence) is
$$\theta[\vec{X}]_{\hat{a} \hat{b}} = m/r^2 \, \rm{diag} \left( 1, \, -1, \, -1 \right)$$
and the tidal tensor is
$$E[\vec{X}]_{\hat{a} \hat{b}} = m/r^3 \, \rm{diag} \left( -2, \, 1, \, 1 \right)$$
(Hatted indices are often used to stress that components refer to a specific frame field, not the coordinate basis; see MTW.)

Chris Hillman

Last edited: Dec 3, 2006
7. Dec 4, 2006

### chandubaba

All I could understand is that it is impossible to enter a black hole singularity without damaging ourselves.But maybe we were not articulate in post #1.What my little brother had asked me was that if we give an EXTRA force towards the black hole's force,nullifying the black hole's gravity effect by inertia(I suppose)then may be we survive.....I am all the more confused ,please clarify.

8. Dec 4, 2006

### DaveC426913

Here's some quite sophisticated animations of what you'd see as you fell into a black hole - it shows what the companion star, the surface of the event horizon and your probes look like. It is rather bizarre.

Last edited: Dec 4, 2006
9. Dec 4, 2006

### Staff: Mentor

That sounds like regular freefall to me...

10. Dec 4, 2006

### Chris Hillman

Singularity =/= horizon

Hi, chandubaba,

Don't confuse the curvature singularity deep inside the event horizon with the event horizon itself. These are completely different. A human could in principle survive falling past the event horizon of a supermassive black hole, because the tidal forces are much weaker there than they would be at the horizon of a stellar mass black hole. But anything which falls past the event horizon cannot avoid continuing to fall until it encounters (after a finite lapse of time as measured by an ideal clock) the curvature singularity, and nothing can survive that encounter!

(Weaker tidal forces at the horizon for more massive holes? That probably sounds weird, but it wasn't a typo and it's not hard to understand: the tidal forces scale like $$m/r^3$$ where m is the mass of the hole and r is the Schwarzschild radial coordinate, which at least outside the hole is a bit like the radial coordinate in a polar spherical chart, which you are probably accustomed to in flat euclidean space. The event horizon is located at $$r=2m$$, so the tidal forces at the horizon scale like $$1/m^2$$, i.e. much weaker for a million solar masses than for one solar mass.)

I am not sure I understand, but it sounds like he recognizes that to avoid falling in you need to fire your rocket toward the hole, i.e. to accelerate away from the world line you would otherwise follow, a timelike geodesic which would fall into the hole, which is what I guess he wants to avoid.

Again, I am not sure I understand the question, but if he was asking whether an observer outside the horizon can avoid falling in by firing his rocket engine to move radially away, the answer is yes. Same principle as the Apollo moon rocket climbing away from the Earth.

Chris Hillman

Last edited: Dec 4, 2006
11. Dec 5, 2006

### Jheriko

I think that what he wants to know is equivalent to: Is it possible to land on the surface of an object whose physical radius is equal to its Schwarzchild radius by using a rocket (or rocket like engine) to slow our approach so that we don't die on impact?

Although it also sounds like he might want to enter then exit the event horizon of a black hole by accelerating through it somehow...

It would be nice if the question came with a diagram of some sort. Maybe you could open mspaint and scribble one out and stick it on imgshack?

12. Dec 5, 2006

### chandubaba

inertia nullifying the black hole's gravity

I wonder how much "EXTRA FORCE" will be required.May be I am wrong.

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13. Dec 5, 2006

### DaveC426913

I think what Chandu is assuming is that gravity is "pulling them down", as if they were standing on its surface. i.e. they'd get flattened.

And Chandu's thinking that the rocket thrust would cancel this. Correct?

Chandu, what you and your brother are missing (I suspect) is the concept of freefall.

Think about this: astronauts in the space station do not feel Earth's pull on them. They float. Even if the space station were falling directly toward Earth (in the same manner as your black hole adventurers in their rocket) they would NOT feel the pull of gravity. They are still in free fall.

The only time you would feel the pull from with Earth or from the BH, would be if you STOPPED allowing yourself to fall freely.

14. Dec 5, 2006

### chandubaba

......but sir why dont we feel the pull from the earth or black hole in free fall.Ulimately there is some force towards the centre of the BH.Is it related to relativity(a ball you throw in a moving train would come back to your hand,even though the train is moving).Please explain in a way I can comprehend both the concept of the train and free fall ,why do we observe such wiered phenomenon!

15. Dec 5, 2006

### AlphaNumeric

Thansk for the explainaton Chris. I sort of went from what I knew to be right into a touch on the side of wild speculation. I've done enough GR that I should have realised I was doing a bit off course with my second paragraph :)

16. Dec 6, 2006

### Jheriko

I would guess that you mean bouncing a ball of the wall. The only time this works is if the train is moving at a constant velocity, if the train is accelerating we get the same affect as we get with gravity (think about the jolt you recieve when you are sitting in the train and it first starts to move).

If the train was in space and we accelerated forwards at 9.8m/s we would feel a ficticious force "pulling" us towards the back of the train with the same approximate strength as gravity on earth.

If we imagine an observer in a train and in free fall (imagine he is unlucky enough to be way up in space trapped in a train), then he is accelerating towards the source of gravity with acceleration equal to the acceleration due to gravity. The observer does experience the force due to gravity, however because he is in a train accelerating in the same direction as the gravitational force is acting, which creates the appearance of a force in the opposite direction (remember accelerating forwards pushes you into the back of your seat), he feels nothing, since the two observed forces cancel out each other's effects exactly.

Hope this helps make it a bit clearer why free fall works how it does.

So... in answer to your original question (judging from the diagram) the rocket requires no thrust to nullify the effects of the black hole's gravity.

Of course since the gravitational field is dependent on distance we would still have tidal effects where there are small differences in the gravity acting across the rocket, and these would infact get stronger as you approached the black hole, eventually tearing the rocket apart.

17. Dec 6, 2006

### DaveC426913

If you are in a high-speed elevator that is descending at 10m/s^2, your feet will come off the floor. You will be in free fall. Though there is still a force pulling you towards the centre of the Earth, you do not feel it.

Can you elaborate on the above? I'm confused. Are you throwing the ball straight up, or are you throwing it forward?

See elevator experience, above.

18. Dec 6, 2006

### Chris Hillman

Is it possible to enter a singularity?

Hi, chandubaba,

Is it possible that when you say "singularity" you mean "event horizon"? (The Schwarzschild coordinate chart which is most often used in simple discussions of black holes is not valid on or inside the event horizon, so this is often called a "coordinate singularity", but this is not to be confused with "curvature singularity", which is a completely different concept.)

Chris Hillman

19. Dec 7, 2006

### chandubaba

1 As sir jheriko points out "If the train was in space and we accelerated forwards at 9.8m/s we would feel a ficticious force "pulling" us towards the back of the train with the same approximate strength as gravity on earth." What my point is why not add to this acceleration i.e(9.8m/s + rocket engine's acceleration)nullifying the black holes gravity completely.
2 Dav sir ,I refer to ball thrown up STRAIGHT.but after much thought I got the answer.It is because of the pseudo force due to CONSTANT velocity of the train that the ball comes back to your hand.
3 Chris sir I dont know what is schwarzchild coordinate.All I know is singularity is the mid point of the black hole and event horizon is the volume from where u cant escape the black hole's gravity.

20. Dec 7, 2006

### DaveC426913

I think this statement is the crux of what I believe is your misunderstanding.

Let's turn off the engine as we near the black hole. We are now falling freely towards the black hole, correct? Say we're still a few hundred thousand miles from the event horizon. What kinds of forces do you think we would feel while falling freely towards the black hole? What kinds of forces do you think we'd feel as we got within a few thousand miles?