Entering supermassive black holes

[Terrr]

I often see, when people talk about black holes, someone claim that since there are no huge tidal effects at the supermassive black hole's event horizon, that someone on a spaceship falling into such a black hole would not be destroyed and would survive, continuing inwards, until, after some time (some say minutes, some say days) they would get spaghettified by getting closer to the singularity. After all, the biggest supermassive black hole out there is about 10 light days across.

Yet, AFAIU, the general relativity equations show that the gravitational acceleration, as you approach the Schwarzschild radius, approaches infinity, and as you cross the event horizon, your speed should be c.

If that's true, then due to time dilation, the subjective time on the spaceship in question should stop completely, and the destruction, from the subjective view point, should be instantaneous. Am I right?

 

[Drakkith]

Your speed as you approach the event horizon will increase towards c, but will never actually be c. The time on the spaceship, as viewed from an external frame away from the black hole would slow and come to a near stop with the spaceship frozen in space very near the event horizon. On board the ship you would see the universe slow down and come to a near stop as your velocity becomes close to c and in-falling light has a hard time catching up to you. On board the ship the process of falling in and the resulting destruction would happen in real time.

 

[DaveC426913]

On board the ship the process of falling in and the resulting destruction would happen in real time.

Real time would be that 10 light-day fall from EH to singularity - modified by length contraction. i.e. if you were doing .99c when you crossed the EH, the distance to the singularity would be length-compressed by a factor of 7 - so only 1.5 days to fall.

Now, even that's not right, because you're still accelerating, so as you fell, the length contraction to the singularity would shrink even more.

 

[Drakkith]

Thanks Dave! Always backing up my posts with numbers! Knew I could count on you! Now, back to my Physics Workbook for Dummies...(yes, I am doing that right now actually)

 

[Terrr]

Real time would be that 10 light-day fall from EH to singularity - modified by length contraction. i.e. if you were doing .99c when you crossed the EH, the distance to the singularity would be length-compressed by a factor of 7 - so only 1.5 days to fall.

Well, the radius would be 5 days. So - if you were doing 0.999999995c, it would be 43 seconds (not taking into account further acceleration). And if you were doing 0.9999999999995 etc...

But the whole point of "approaching c as you approach the event horizon" is that when you do REACH it, your speed is, actually, c. So you cannot be "doing .99c when you crossed the EH".

 

[DaveC426913]

Well, the radius would be 5 days. So - if you were doing 0.999999995c, it would be 43 seconds (not taking into account further acceleration). And if you were doing 0.9999999999995 etc...

Right.

But the whole point of "approaching c as you approach the event horizon" is that when you do REACH it, your speed is, actually, c.

No. You do not reach c.

See this recent thread, where the same question was asked:
https://www.physicsforums.com/showpost.php?p=3782823&postcount=17

 

[Jolb]

The Schwartzchild radius is the distance from the center of the black hole to the point at which the escape velocity is c. (Ironically you get the same Schwartzchild radius if you compute it using Newtonian gravity.) So Dave is right, your speed at the event horizon is less than c.

The one issue I have with what was said is that it's not known for sure whether you and your ship are destroyed after falling into a black hole, granted that there are negligible tidal forces around a supermassive black hole. You feel no forces if you are in free fall, whether or not you're beyond the event horizon. The "spaghetti" effect is due to tidal forces, so it wouldn't happen. Are you guys implying that once you reach the singularity, you're destroyed? What would the mechanism be?

I'm no expert on GR, but I've heard that there are a bunch of nonintuitive results related to falling into a black hole. For example, if you spiral into a black hole (you have some angular velocity about the center), in your reference frame you're completely fine, but in the reference frame of an inertial observer at infinity, you'd appear to be stretched out into an accretion disk around the black hole. Is this true?

 

[Drakkith]

The one issue I have with what was said is that it's not known for sure whether you and your ship are destroyed after falling into a black hole, granted that there are negligible tidal forces around a supermassive black hole. You feel no forces if you are in free fall, whether or not you're beyond the event horizon.

I thought it was simply that the effects were negligible until your gradiant is extremely large like it is near a small black hole.

 

[DaveC426913]

The one issue I have with what was said is that it's not known for sure whether you and your ship are destroyed after falling into a black hole, granted that there are negligible tidal forces around a supermassive black hole. You feel no forces if you are in free fall, whether or not you're beyond the event horizon. The "spaghetti" effect is due to tidal forces, so it wouldn't happen. Are you guys implying that once you reach the singularity, you're destroyed? What would the mechanism be?

You will be destroyed by tidal forces, it's just in a supermassive BH, you'll last longer before the tidal forces become appreciable. But ultimately, they will still exceed the structural strength of any known material as it nears the singularity.

Adn yes the singularity will destroy what's left. It's strong enough to collapse atoms - it's certainly strong enough to destroy a ship.

 

[chill_factor]

I am completely ignorant in general relativity so let me ask a question:

If you were facing backwards, towards the outside, as you fell into a supermassive black hole, what would you see as you were falling in, crossing the event horizon, and then inside? What would the universe outside look like?

 

[DaveC426913]

I am completely ignorant in general relativity so let me ask a question:

If you were facing backwards, towards the outside, as you fell into a supermassive black hole, what would you see as you were falling in, crossing the event horizon, and then inside? What would the universe outside look like?

There's actually quite a variety of animations online. Google to your heart's content.

 

[Jolb]

You will be destroyed by tidal forces, it's just in a supermassive BH, you'll last longer before the tidal forces become appreciable. But ultimately, they will still exceed the structural strength of any known material as it nears the singularity.

Adn yes the singularity will destroy what's left. It's strong enough to collapse atoms - it's certainly strong enough to destroy a ship.

I get the idea that any real black hole is not infinitely massive and thus has nonzero tidal forces. But I'm thinking of a limiting case: basically its mass and EH radius go to infinity. (I guess the EH would be an infinite plane in space). Clearly this isn't a physically realistic situation but please don't blame me for trying gedanken experiments in the realm of relativity. Here's my line of thought, let me know if it seems to make sense.

In that case (infinite black hole) there would be NO tidal forces. So you wouldn't be destroyed. I guess in this case, the singularity would also be infinitely far from the event horizon, so that wouldn't destroy you.

But I guess there's a contradiction now: the singularity couldn't be any unique point in space (by symmetry). And singularities are by definition a point in space, right? If so, this gedanken experiment is a contradiction. So (theorem): there can't be any infinitely massive black holes, and every black hole must have tidal forces, so you die in any black hole.

Does this make sense?

 

[chill_factor]

There's actually quite a variety of animations online. Google to your heart's content.

i can only find ones facing inwards.

 

[Terrr]

Right.


No. You do not reach c.

See this recent thread, where the same question was asked:
https://www.physicsforums.com/showpost.php?p=3782823&postcount=17

I read it. I still don't understand it.

"As you approach the EH, your speed approaches c" - is that a correct statement or not?

If it is correct, then you either never reach EH or your speed actually gets to c. If neither of those is correct, what is the answer to: "At what speed do you cross the EH"?

 

[Jolb]

I read it. I still don't understand it.

"As you approach the EH, your speed approaches c" - is that a correct statement or not?

[STRIKE]Not correct.[/STRIKE]

I read it. I still don't understand it.

"At what speed do you cross the EH"?

[STRIKE]Any allowed (<c) speed.[/STRIKE]

Edit: Dave's next post is correct. My bad.

 

[DaveC426913]

I read it. I still don't understand it.

"As you approach the EH, your speed approaches c" - is that a correct statement or not?

Correct.

If it is correct, then you either never reach EH or your speed actually gets to c.

Neither.

If neither of those is correct, what is the answer to: "At what speed do you cross the EH"?

At a speed arbitrarily close to, yet still less than c.

What is the sum of the sequence 1/2 + 1/4 + 1/8 + 1/16 ... up to 1/n, where n is a finite number? The sum is arbitrarily close to, but still less than, 1.

If you get in a spaceship and accelerate at 1g for 1 million years, your speed will have reached very close to, but not actually, c. If you continue to accelerate for another billion years, your speed will have reached EXTREMELY close to, but not actually, c.

 

[Jolb]

Why does your speed necessarily approach c?

Suppose you were on a satellite in a circular orbit around a black hole just beyond the schwartzchild radius. Nonrotating black holes have an event horizon radius equal to their schwartzchild radius, right? You jump off the satellite in the direction opposite to its orbital velocity such that you now have zero angular velocity and zero radial velocity and fall straight into the black hole. Since your distance to the event horizon is nearly zero, you accelerate negligibly before crossing the EH. Right?

Edit:
Wrong. Your acceleration is infinite in the region close to the event horizon. So you do have a nonzero change in velocity.

 

[Terrr]

At a speed arbitrarily close to, yet still less than c.

What is the sum of the sequence 1/2 + 1/4 + 1/8 + 1/16 ... up to 1/n, where n is a finite number? The sum is arbitrarily close to, but still less than, 1.

If you get in a spaceship and accelerate at 1g for 1 million years, your speed will have reached very close to, but not actually, c. If you continue to accelerate for another billion years, your speed will have reached EXTREMELY close to, but not actually, c.

For the purposes of figuring out what the subjective time is on that ship that it takes to get to the singularity, "speed arbitrarily close to, yet still less than c" means you get wiped out instantly. Or "in time arbitrarily close to, yet still more than 0" if you prefer. Right?

 

[Terrr]

Why does your speed necessarily approach c?

Suppose you were on a satellite in a circular orbit around a black hole just beyond the schwartzchild radius. Nonrotating black holes have an event horizon radius equal to their schwartzchild radius, right? You jump off the satellite in the direction opposite to its orbital velocity such that you now have zero angular velocity and zero radial velocity and fall straight into the black hole. Since your distance to the event horizon is nearly zero, you accelerate negligibly before crossing the EH. Right?

That would be correct in a Newtonian gravity system. The event horizon and anywhere close to it isn't.

Let me put it this way - for the biggest black hole found so far (18B solar masses) according to Newton's equations the gravitational acceleration at the event horizon is only around 100g. But since this is the black hole, obviously 100g is not enough to escape - no acceleration is enough. The gravitational acceleration at the event horizon is infinite.

Here is someone's post that contains the formula for the gravitational acceleration of a black hole: https://www.physicsforums.com/showthread.php?t=174994

 

[Terrr]

If you get in a spaceship and accelerate at 1g for 1 million years, your speed will have reached very close to, but not actually, c. If you continue to accelerate for another billion years, your speed will have reached EXTREMELY close to, but not actually, c.

You need infinite acceleration in order for a mass-possessing object to reach c. How fortunate that the gravitational acceleration at the event horizon is infinite.

 

[Jolb]

I stand corrected! You do go arbitrarily close to c.

The derivation is somewhat involved. You can find it here.
http://mathpages.com/rr/s6-07/6-07.htm
and more background is here
http://mathpages.com/rr/s6-04/6-04.htm

 

[Terrr]

What is the sum of the sequence 1/2 + 1/4 + 1/8 + 1/16 ... up to 1/n, where n is a finite number? The sum is arbitrarily close to, but still less than, 1.

1. "As you approach infinite number of members of the sequence, the sum approaches 1."

2. "As you approach the event horizon, the speed approaches c."

Since you cannot ever REACH infinite number of members in the sequence, the sum never reaches 1.

If you posit that you cannot ever REACH the event horizon, then the speed can never reach c. But you can reach it. So...

 

[DaveC426913]

For the purposes of figuring out what the subjective time is on that ship that it takes to get to the singularity, "speed arbitrarily close to, yet still less than c" means you get wiped out instantly. Or "in time arbitrarily close to, yet still more than 0" if you prefer. Right?

The subjective time aboard ship is a matter of the radius of the BH from EH to singularity. That could be light minutes or light hours. This will be modified relativistically, since distances along your path will be contracted. But still, it's a finite time. you could fall for many minutes.

You need infinite acceleration in order for a mass-possessing object to reach c. How fortunate that the gravitational acceleration at the event horizon is infinite.

It isn't.

1. "As you approach infinite number of members of the sequence, the sum approaches 1."

2. "As you approach the event horizon, the speed approaches c."

Since you cannot ever REACH infinite number of members in the sequence, the sum never reaches 1.

If you posit that you cannot ever REACH the event horizon, then the speed can never reach c. But you can reach it. So...

That does not mean the the EH is analogous to infinity. You're taking the analogy too literally.

 

[Terrr]

The subjective time aboard ship is a matter of the radius of the BH from EH to singularity. That could be light minutes or light hours. This will be modified relativistically, since distances along your path will be contracted. But still, it's a finite time. you could fall for many minutes.

Not if your speed is infinitesimally close to c.

It isn't.

According to the formulas, it is.

 

[DaveC426913]

Not if your speed is infinitesimally close to c.
According to the formulas, it is.

You are quoting a thread that is full of errors and then locked. Get a better source.

 

[Terrr]

You are quoting a thread that is full of errors and then locked. Get a better source.

In your opinion, at what speed do you cross EH and what is the gravitational acceleration at EH?

 

[skeptic2]

Now, even that's not right, because you're still accelerating, so as you fell, the length contraction to the singularity would shrink even more.

It is my understanding that in a gravitational field, the amount of length contraction is determined by √ 1 - (v^2/c^2) where v is the escape velocity at the point in question. This means that length contracts to zero at the EH, not at the singularity.

 

[Terrr]

It is my understanding that in a gravitational field, the amount of length contraction is determined by √ 1 - (v^2/c^2) where v is the escape velocity at the point in question. This means that length contracts to zero at the EH, not at the singularity.

Such a simple concept that some are having really hard time understanding, as per OP.

 

[skeptic2]

One wonders why an object, free falling from infinity, shouldn't have an infalling velocity the same as the escape velocity (but different in sign) at every point along its trajectory.

 

[skeptic2]

"As you approach the EH, your speed approaches c" - is that a correct statement or not?

Correct.

If it is correct, then you either never reach EH or your speed actually gets to c. If neither of those is correct, what is the answer to: "At what speed do you cross the EH"?

Neither.
At a speed arbitrarily close to, yet still less than c.

From the book, Exploring Black Holes, Introduction to General Relativity by Edwin F. Taylor and John Archibald Wheeler, page 2-22:
"A stone far from a black hole and initially at rest with respect to it begins to move toward the black hole. Gradually the stone picks up speed, finally plunging to the center. With what speed v does this stone pass a spherical shell at radius r?
[snip]
The resulting speed is v = (2M/r)^(1/2). [17]

Surprisingly equation 17 is correct in general relativity too, but only when the speed is interpreted as the speed of the in-falling object as measured by the shell observer."Could we be confusing reference frames?