# Spaghettification near event horizon

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1. Aug 16, 2015

### Stephanus

Dear PF Forum,
Can we avoid spaghettification for some times, once we're inside event horizon?
I choose a rather massive black hole, so the tidal force won't be so big at EH.
Mass: 1 trillion solar mass
What is the gravitational force at EH?
I calculate: $g = \frac{GM}{r^2} = 3 m/s^2$
But
So, once we're inside the event horizon minus 1 cm, acceleration increases without bound? But if acceleration is infinity then once we're inside EH, no matter how big massive the black hole is we'll arrive at the singularity at once (in the observer inside EH frame)?

2. Aug 16, 2015

### phinds

That doesn't seem to make sense. There is, as far as I am aware, nothing in particular that happens as you cross the event horizon of a supermassive black hole. You don't even notice it.

3. Aug 16, 2015

### Staff: Mentor

Yes. Spaghettification only has to happen as you approach the singularity, not the horizon.

4. Aug 17, 2015

### Stephanus

Don't even notice it?
But...
I'm sorry I can't imagine it.
Is the acceleration force approaching EH very high? It's not the speed that I ask. The acceleration.
Thanks for explain it to me.

5. Aug 17, 2015

### Stephanus

But you said before, near EH the acceleration is very, very high. I cannot picture that. And object can withstand high velocity (even at 99% speed of light, right?) But how can object withstand high acceleration?

6. Aug 17, 2015

### Jorrie

A person free-falling through an EH does not feel the acceleration, just tidal acceleration (difference in acceleration between body parts). This is small over human sizes when the BH is large.

7. Aug 17, 2015

### Stephanus

https://www.physicsforums.com/threa...on-of-a-large-black-hole.827183/#post-5200300
But there things that I have to learn before I can understand that thread. And rather than interfere with someone's thread, I think it's better that I create my own
So I think I study acceleration inside EH. About the spaghettification, I agree with you.
And I remember something a mentor posted months ago in my own thread.
And this is the FIRST time I argue about a mentor post. Is the acceleration inside EH is extremely big?

8. Aug 17, 2015

### Jorrie

It is problematic talking about "acceleration inside the EH", because relative to what shall one specify it?
The equation that you have quoted is acceleration relative to an observer momentarily at constant radial parameter R, but there cannot be such observers at or inside the EH - everything is traveling towards the central singularity (whatever that may be).

Do you already know the Schwarzschild geometry? If not, you will have to master that before thinking about the 'inside'. A good introduction with the various embedding diagrams is available here: http://casa.colorado.edu/~ajsh/schwp.html

9. Aug 17, 2015

### Staff: Mentor

If you're hovering at a constant altitude above the horizon, yes. But that doesn't cause spaghettification. It just causes you to have to withstand a higher and higher g force, until finally your body is crushed by it.

If you're free-falling into the hole, you feel no acceleration at all.

10. Aug 17, 2015

### A.T.

Depending on how the force is applied to you during hover, you might experience spaghettification or pizzafication.

11. Aug 17, 2015

### PAllen

Ah, yes, if you want to be pizza, don't sit on a grate.

12. Aug 17, 2015

### Stephanus

Yes, we won't feel acceleration at all. The spaghettification will be felt near singularity. How near? Should have calculated it.
But before I read about spaghettification, I would to know about the speed between EH and the singularity.
For example there's a massive black hole, out side earth.
An object fall just inside EH. The acceleration is very high. So the instant (more likely before entering EH) is the speed of the object will be very high if it free falling?
For example: If an object is as far as the moon from earth, it will be attracted to earth at 2GM/r^2 = 0.00276m/s2. Rather slow, but the velocity acceleration is accelerating toward the earth, right.?
Okay, anyway according to your equation. If something is hovering at Event Horizon then suddenly we release it. It doesn't crawl at a couple of meter per seconds.
Not at $a = \frac{GM}{R^2}$ but at $\frac{a}{\sqrt{1-\frac{2GM}{R^2}}}$
No, wait. It still has some value in it. It's not undefined.
$R_s = \frac{2GM}{c^2}$,
so
$a = \frac{GM}{{R_s}^2\sqrt{1-\frac{2GM}{{R_s}^2}}}$
Let me simplify this.
$a = \frac{GM}{{{\frac{2GM}{c^2}}}^2\sqrt{1-\frac{2GM}{{\frac{2GM}{c^2}}^2}}}$
$a = \frac{c^2}{2GM\sqrt{1-\frac{c^4}{2GM}}}$
I tought the speed is undefined. But is this the simplified formula for acceleration in black hole (or any object after adjusting with relativity)?

13. Aug 17, 2015

### Staff: Mentor

You don't have a choice about that; if you're hovering, the force is being applied radially outward.

14. Aug 17, 2015

### stevebd1

The following is taken from this thread-
You might also find these two link of interest-

Spacetime Geometry Inside a Black Hole
http://www.jimhaldenwang.com/black_hole.htm

Falling Into and Hovering Near A Black Hole
http://mathpages.com/rr/s7-03/7-03.htm

15. Aug 17, 2015

### A.T.

16. Aug 17, 2015

### Staff: Mentor

There is no invariant way to assign a "speed" to an object inside the horizon. Outside the horizon, you can use observers "hovering" at constant altitude above the horizon to define an invariant notion of "being at rest", and define "speed" as speed relative to those observers. By this definition, the speed of an object free-falling into the hole approaches the speed of light as the object approaches the horizon (but never quite reaches $c$ as long as the object is above the horizon).

At and inside the horizon, however, there are no "hovering" observers, and there is no other family of observers that can be viewed as "being at rest", so there's no fixed standard from which to define "speed".

Nothing can hover at the event horizon; it would have to be moving outward at the speed of light to do so.

You've done your algebra wrong. It's a lot easier to adopt "geometric" units, in which $G = c = 1$; then your equations don't get cluttered with a lot of extra coefficients.

$$a = \frac{M}{r^2 \sqrt{1 - 2M / r}}$$

Now plug in $r = 2M$. It should be obvious that the factor under the square root in the denominator goes to zero, making $a$ undefined, but let's do the algebra anyway:

$$a = \frac{M}{r \sqrt{r^2 - 2M r}}$$

So at the horizon, we have

$$a_{EH} = \frac{M}{2M \sqrt{4M^2 - 4M^2}} = \frac{1}{2 \sqrt{4M^2 - 4M^2}} = \frac{1}{4M \sqrt{1 - 1}}$$

This is obviously undefined.

17. Aug 17, 2015

### Stephanus

Thanks for any responses and answer.
I think I'm ready to my original question.
Okay, now after I gather (or rather read) the facts, there's an interesting problem at the other thread.
https://www.physicsforums.com/threa...ast-the-horizon-of-a-large-black-hole.827183/
Here let me quote some post at those thread that indirectly address my issue here.. I create this "spaghetti" thread, because I focus on the acceleration. Not trying to derail that thread.
If this violates PF rule, please let me now.
So if the black hole is big enough, say 1 trillion solar mass. (10 times the mass of milky way?) Is it possible for a star in iron burning process inside the event horizon to become a black hole before it gets "spaghettificated"?
No wait, once it become a black hole itself, it can't get streamed, right?
But is it possible for an object to become a black hole inside the event horizon of a black hole?

18. Aug 17, 2015

### Stephanus

@stevebd1
He did call that the "ouch" radius for 1 g different from about 1.7 m? Talk about some fancy names like Chandrasekhar limit, Planck length, Hubble volume, this one just "ouch"?

19. Aug 17, 2015

### Stephanus

I'm sorry what is G here? $G = c = 1$
Is it Newton G: $6.674×10_{−11} N⋅\frac{m^2}{kg^2}$ or
g as acceleration 9.m/s2
But looking at your equation G is speed of light.
Schwarzchild radius is $R_s = \frac{2GM}{c^2}$
Is G in $G = c = 1$ the same as
G in $R_s = \frac{2GM}{c^2}$ ?

20. Aug 17, 2015