Conditions necessary for a human to survive event horizon

[nomadreid]

In various explanations of the event horizon which do not invoke the existence of a firewall (thereby upholding the dictum that an observer would not notice any difference upon passing the event horizon until she looked out the window), one uses the concept of a theoretical observer passing the event horizon. The more careful explanations (sorry, I know I'm not giving references, but I am sure the readers will be familiar with them) mention that the observer is some sort of specially constructed robot who would survive the tidal forces of the black hole. However, what would be the conditions necessary (minimum size of black hole, speed of spaceship, etc) , if any exist, for a human being to survive the passage into the inner part of the space defined by the event horizon (at least for a short time -- before he gets too close to the singularity)? I am referring to his own perception of his survival (I know that an outside observer would never see her enter) even though the event horizon may be defined by an outside observer. Thanks.

 

[Nugatory]

Try this thread, and especially post #5 on the "ouch radius": https://www.physicsforums.com/threads/how-far-from-event-horizon-are-you-safe.771596/#post-4857627

 

[nomadreid]

Thanks, Nugatory. I am a little puzzled: the "ouch radius" given in that post is given by the cube root of 4GM/g, g ≈9.8 m/s². If I leave aside the issue of the rotation of the black hole, and put in the value of 10 million solar masses given for the same "comfort" in
https://en.wikipedia.org/wiki/Supermassive_black_hole, I get indeed a radius that is well within the Schwarzschild radius given by 2GM/c². Fine. Nonetheless, trying to get the very rough minimum mass by using the ouch radius = the Schwarzschild radius, that is solving (4GM/g) ^1/3, ≈2GM/c² for M, I get about 9 x 10³⁴, or under a hundred thousand solar masses, two orders of magnitude smaller than the value given in the Wiki article. Even given that this is an approximation, this sounds like a big difference. Is the lower density enough to account for such a big difference?

[mentor's note - edited to clean up the formatting, as some browsers do funny things with the rich-text editor and URLs]

 

[newjerseyrunner]

Here's the thing: the event horizon is nothing special, it's just the point of no return. If you could get the Vostok 1 to the supermassive black hole at the center of the galaxy, a cosmonaut would survive just fine, they wouldn't even notice that they went through the event horizon. It's just the point at which the escape velocity of the black hole > c. It's an insane amount of gravity, but gravity doesn't kill you, it doesn't even hurt you, its tidal gravity that'll rip you to pieces.

Remember that at that point though, survival is limited to the time between when you enter and when you get to close, it's not possible to ever increase your distance from the black hole. There are no conditions (as we currently understand the universe) where you could enter an event horizon and come back out as anything other than slow Hawking radiation.

 

[nomadreid]

thanks for the input, newjerseyrunner. I understand that survival depends on the tidal forces, and that once one has passed the event horizon, one will inescapably be toast. I am interested in the minimum mass of a black hole where a human could survive the event horizon (or apparent horizon, as Hawking would prefer, and assuming no firewall), even if that survival is of a small duration.

 

[newjerseyrunner]

You want the size of a black hole where the event horizon is equal to the distance as which the tidal forces would rip you apart?
Well, first determine how much force it would take, you'd have to do some research on that, I have no idea. Then use that to as the Tidal force of the Roche limit for humans.

F_Tidal = 2GM_{Black hole}M_HumanHeight_Human / d³

The only variables in here are now M_{Black hole} and d. Since you want to know where the Roche limit equals the event horizon, substitute the Schwarzschild radius of M_{Black hole} for d.

r_{Schwarzschild} = 2M_{Black hole}G/c^2

F_Tidal = 2GM_{Black hole}M_HumanHeight_Human / (4M_{Black hole}G/c^2)³

Solve for M_{Black Hole}

 

[nomadreid]

Thanks, again, newjerseyrunner. Using your equation, with mass*height of human as 100 kg-m, and the F_tidal = 10 m/s² (The Roche limit is rather extreme, since a human would die long before his body is actually torn apart... so I take the same "comfort zone" as the Wikipedia article cited, even though one can take more before dying, but it is about the same order of magnitude, and anyway my interest is comparing my calculations to the conclusion of the Wiki article), I get very roughly the same result as I got with the other equation: about 10⁵ solar masses, not the 10¹⁰ solar masses that the Wiki article claimed. I wouldn't expect my calculation to be exactly the same as the Wiki article, since the lower density comes into play, but five orders of magnitude is a bit much, no?

 

[newjerseyrunner]

If you got an answer of F_tidal = 10m/s^2 you must have done the math wrong; your units should be in kg*m/s^2

 

[jartsa]

Wikipedia:
"a person on the surface of the Earth and one at the event horizon of a 10 million M☉ black hole experience about the same tidal force between their head and feet"

Is that the part where the Wikipedia is supposedly talking about the smallest black hole whose event horizon a human can cross without too much pain?

 

[newjerseyrunner]

Can someone help me? I tried to do the math myself, but ended up with a mismatch in units.


Get an equation for tidal forces due to gravity: https://en.wikipedia.org/wiki/Roche_limit#Derivation_of_the_formula
F_Tidal = 2GM_{Black hole}M_HumanHeight_Human/d³

Rearrange it do get d by itself, leave the cube for now
d³ = 2GM_{Black hole}M_HumanHeight_Human/F_Tidal

Since we want the object to come apart at the event horizon, we need the formula for that (diameter, not radius)
d_{Schwarzschild} = 4M_{Black hole}G/c²

Since we want the roche limit and event horizon to be the same, place formula above in for d
(4M_{Black hole}G/c²)³ = 2GM_{Black hole}M_HumanHeight_Human/F_Tidal

Expand the left side
64M_{Black hole}³G³/c⁶ = 2GM_{Black hole}M_HumanHeight_Human/F_Tidal

Move eveything except mass of the black hole to the right
64M_{Black hole}³ = 2M_{Black hole}M_HumanHeight_Humanc⁶/F_TidalG²

Getting confused with all the super and sub-scripts, so I'm going to replace with actual values now, assume:
Height_Human = 2m
Mass_Human = 70kg
64M_{Black hole}³ = 280_kg*mM_{Black hole}c⁶/F_TidalG²

Simplify (don't forget the units!)
M_{Black hole} = (35/2)^1/2c³ kg^1/2m^1/2 / 2G^1/2F_Tidal

Replace constants and variables (to allow units of variables to be presented)
c = 300000000 m/s
G = 6.674*10^-11m³kg^-1s^-2
M_{Black hole} = x kg
F_Tidal = y kg *m/s

x kg = (35/2)^1/2(300000000m/s)³ kg^1/2m^1/2 / 2(6.674*10^-11m³kg^-1s^-2)^1/2y(kg*m/s)

Simplify the units
x kg = (256033 * 2.7*10²⁵ / y) m^1/2/(kg s)Okay, I'm lost, my units don't match up, the order of magnitude seems low too. 10^25kg is a little bigger than Earth. The units aren't even close to right, I'm off by a factor of (kg²s)/m^1/2?

 

[pervect]

Can someone help me? I tried to do the math myself, but ended up with a mismatch in units.


Get an equation for tidal forces due to gravity: https://en.wikipedia.org/wiki/Roche_limit#Derivation_of_the_formula
F_Tidal = 2GM_{Black hole}M_HumanHeight_Human/d³

Rearrange it do get d by itself, leave the cube for now
d³ = 2GM_{Black hole}M_HumanHeight_Human/F_Tidal

Since we want the object to come apart at the event horizon, we need the formula for that (diameter, not radius)
d_{Schwarzschild} = 4M_{Black hole}G/c²

If, for the sake of my sanity, we let M be the mass of the black hole, m be the mass of the human, H be the height of the human, and F be the tidal force, we should get

$$F = \frac{2G\,M\,m\,H}{r^3} \quad r^3 = \frac{2 G\,M\,m\,H}{ F}$$

Note that r is correct, d (diameter) is not correct in this formula. Furthermore, I don't see the relevance of Roche's limit, which is defined as:

The Roche limit, sometimes referred to as the Roche radius, is the distance within which a celestial body, held together only by its own gravity, will disintegrate due to a second celestial body's tidal forces.

But a human body is held together by means other than gravity, so a celestial body will disintegrate at a different point than the human body. You need to decide which case you're analyzing, the title of your thread implies to me that you're interested in the human body case, in which case the Roche limit should be irreleveant.

 

[nomadreid]

newjerseyrunner: first, thanks for pointing out my typo (that's all it was) in my units for the tidal force. Secondly, I am trying to find where you got your formula for the Roche limit from that page, because already in that formula the units don't seem to pan out (but correct me if I am wrong; I am in a hurry just now, and will come back to this later in the day):
(eliminating the subscripts)
F = 2GM*M*Height /d³
units:kg*m*s^-2 =?=[m³*kg^-1*s^-2]*kg*m*m^-3
kg*m*s^-2 ≠ s^-2*m

 

[nomadreid]

@ jartsa: yes, that is the passage I was referring to.

 

[Ibix]

@ jartsa: yes, that is the passage I was referring to.

In that case, I think you are mis-reading Wiki. It says that the tidal effect at the surface of the ten million solar mass black hole is comparable to the tidal effect on the surface of the Earth - i.e. strong enough to shove liquid around but totally negligible in terms of the human body. The ouch radius is the radius where the tidal effect is painfully significant - when the difference in g force between your head and toes is 1g.

I think that's consistent with what you found.

 

[nomadreid]

Thank you, Ibix. That makes sense. Therefore, if I am now reading that correctly, the answer to my original question is around 10⁵ solar masses (given the simplifying assumptions made).

 

[jartsa]

Let's believe all the physicists that tell us that a person free falling through an event horizon of a supermassive black hole does not feel anything special.

Now let's consider the distance between that falling person's head and toes. If nothing odd has happened, the distance is very short, because the person is moving very fast.

So the situation when the falling person does not feel too much tidal stretching occurs when the person is ridiculously short, I mean the person is a normal person shortened by the Lorentz contraction.

What if the distance between the head and the toes was something normal, like 2 m? Well then the tidal forces would be much larger, and the person would be torn apart.

 

[Nugatory]

So the situation when the falling person does not feel too much tidal stretching occurs when the person is ridiculously short, I mean the person is a normal person shortened by the Lorentz contraction.

What if the distance between the head and the toes was something normal, like 2 m? Well then the tidal forces would be much larger, and the person would be torn apart.

Lorentz contraction has nothing to with it. In the infaller's own frame he is not contracted, and if he's not torn apart in one frame he won't be torn apart in any.

What matters is the gravitational gradient, the difference between the gravitational acceleration of his head and his feet. That gradient will be smaller for a larger black hole, which is why everyone is calculating the necessary size above.

 

[jartsa]

Lorentz contraction has nothing to with it. In the infaller's own frame he is not contracted, and if he's not torn apart in one frame he won't be torn apart in any.

What matters is the gravitational gradient, the difference between the gravitational acceleration of his head and his feet. That gradient will be smaller for a larger black hole, which is why everyone is calculating the necessary size above.

Let's say that near the event horizon of a supermassive black hole there is some kind of small space barn hovering. A long flagpole falls very fast trough the barn, there are two open doors at both ends of the barn, the doors are closed simultaneously when the pole is inside the barn, but the barn is so short that the doors hit the two ends of the flagpole, and some door pieces get stuck on the pole. Tidal forces are pulling the pieces apart, like tidal forces were pulling the doors apart when the barn was still intact. It's a small barn, so the tidal forces are small. The flagpole is experiencing small tidal forces.And now let's say that near the event horizon of a supermassive black hole there is some kind of large space barn hovering. A long flagpole falls quite slowly trough the barn, this flagpole has the same rest length as the previous flagpole, but it falls much slower. There are two open doors at both ends of the barn, the doors are closed simultaneously when the pole is inside the barn, but the barn is not so long that the doors would not hit the two ends of the flagpole, and some door pieces get stuck on the pole. Tidal forces are pulling the pieces apart, like tidal forces were pulling the doors apart when the barn was still intact. It's a large barn, so the tidal forces are large. The flagpole is experiencing large tidal forces.

 

[Nugatory]

some kind of small space barn hovering.

there is some kind of large space barn hovering.

The frame in which an extended object like a barn is hovering in a gravitational field (black holes and event horizons are irrelevant here) is not inertial so the Lorentz transformations don't work. Even if they did, the spacetime is curved so none of the assumptions of special relativity hold - in particular length contraction is derived using a particular simultaneity convention that doesn't work here.

It is true that for any given gradient, the tidal forces are larger when the object is larger (tidal forces produced by the moon do nothing to a glass of water, but move enormous amounts of water across the 12,500 kilometer diameter of the earth) but the length that you're using in this calculation has nothing to do with the Lorentz-contracted length.

 

[newjerseyrunner]

But a human body is held together by means other than gravity, so a celestial body will disintegrate at a different point than the human body. You need to decide which case you're analyzing, the title of your thread implies to me that you're interested in the human body case, in which case the Roche limit should be irreleveant.

I agree that the Roche limit is useless. I didn't use a Roche limit formula, I only used the formula to derive the tidal force on an object. I didn't mean to say I'd use the Roche limit, and I didn't do so in my maths.

 

[sweet springs]

Another way for human to survive event horizon may be a case of micro black hole. I suppose micro black hole could pass through human body without damaging it. I am sorry that I have no calculation done for saying it.

 

[nomadreid]

sweet springs, if a micro black hole were small enough to pass through a human body, it would be emitting so much Hawking radiation that it would burn the human up, if I understand correctly (but I am happy to be corrected)

 

[jartsa]

The frame in which an extended object like a barn is hovering in a gravitational field (black holes and event horizons are irrelevant here) is not inertial so the Lorentz transformations don't work. Even if they did, the spacetime is curved so none of the assumptions of special relativity hold - in particular length contraction is derived using a particular simultaneity convention that doesn't work here.

It is true that for any given gradient, the tidal forces are larger when the object is larger (tidal forces produced by the moon do nothing to a glass of water, but move enormous amounts of water across the 12,500 kilometer diameter of the earth) but the length that you're using in this calculation has nothing to do with the Lorentz-contracted length.

I see.

This is speculation about the possibility of us laypeople misunderstanding something:

The physicists are telling us that we are not ripped apart when we fall through the event horizon of a supermassive black hole. Well, maybe that's because we would be ripped apart if we slowed down the falling speed, but it's impossible to slow down the falling speed at the event horizon? Maybe we should not conclude that we are not ripped apart above the event horizon, from the fact that we are not ripped apart at the event horizon?

 

[Ibix]

The point is that it's not the falling that rips you apart. It's that your head and toes are trying to fall at different rates and the intermolecular forces in your body must hold you together.

Look at this in a purely Newtonian way: $$F=-\frac {GMm}{r^2} $$The difference in the g force on your head and your toes is just the gradient of the force times the length of your body: $$\frac {dF}{dr}L=2\frac{GMm}{r^3}L $$ Now if you substitute in the Schwarzschild radius (completely illegitimately because Newton isn't appropriate at that radius) you find that the difference in force at the Schwarzschild radius is proportional to $M^{-2}$. So the more massive the black hole the lower the tidal force at the Schwarzschild radius. If you want to make this a legitimate Newtonian calculation, simply substitute some multiple of the Schwarzschild radius and trust Peter when he says there's a similar relation in the relativistic regime.

I must admit that I'm not sure how to carry out the equivalent calculation in GR. Is the following about right? I pick the locally flat radially infalling frame at my waist. There must be two radially infalling geodesics passing through my head and toes that are instantaneously at rest in this frame, which are the paths my head and toes will try to follow. Get the geodesic deviation between these and its magnitude is analogous to the Newtonian $(dF/dr)L/m$?

 

[PeterDonis]

I pick the locally flat radially infalling frame at my waist. There must be two radially infalling geodesics passing through my head and toes that are instantaneously at rest in this frame, which are the paths my head and toes will try to follow. Get the geodesic deviation between these and its magnitude is analogous to the Newtonian $(dF/dr)L/m(dF/dr)L/m$?

This is pretty much it; the only caveat is that, to assess geodesic deviation, you will need to go beyond the confines of a single local inertial frame, since by definition curvature (i.e., geodesic deviation) is negligible inside a single LIF.

 

[Ibix]

This is pretty much it; the only caveat is that, to assess geodesic deviation, you will need to go beyond the confines of a single local inertial frame, since by definition curvature (i.e., geodesic deviation) is negligible inside a single LIF.

Indeed. I'm only using an LIF because it let's me be talk about "falling at the same velocity on two different geodesics" without having to worry too much about what I mean by that. I shall have a crack at actually doing it when I'm a bit more awake.

 

[Ilja]

Here's the thing: the event horizon is nothing special, it's just the point of no return.

I think one has to add here an "if GR is true".

It is, I would say, quite typical for alternative theories of gravity that they may agree more or less with GR almost everywhere but start to disagree near the horizon, and then in such a strong way that a horizon will not be formed at all.

 

[Ilja]

sweet springs, if a micro black hole were small enough to pass through a human body, it would be emitting so much Hawking radiation that it would burn the human up, if I understand correctly (but I am happy to be corrected)

This would require that there is such an animal as Hawking radiation. This is far from being clear. The standard derivation makes completely unreasonable assumptions about the range of validity of existing theory.

Just to clarify: All the theory we have is not even quantum gravity, but semiclassical gravity. Quantum gravity is asssumed to be necessary for distances below Planck length. Hawking radiation lasting more than seconds after the collapse requires the validity of this theory up to distances of $e^{-10^5}$ Planck length.

 

[jartsa]

The point is that it's not the falling that rips you apart. It's that your head and toes are trying to fall at different rates and the intermolecular forces in your body must hold you together.

My point is something like this:

A long rope whose one end hangs near one planet, while the other end hangs near another planet, is under stress because of tidal forces.

In a small laboratory between the planets we manufacture a rope whose rest length equals the first rope's rest length, the only difference between the ropes is the velocity and the relativistic length. (This rope fits inside the factory because the rope moves relative to the factory)

Somehow I'm sure the second rope does not feel as much tidal forces as the first rope, although the gravity source is the same, and the ropes are identical ropes in different frames. Somehow the velocity of an object affects the tidal forces felt by the object.(let's say those planets are very far away from each other, so that we can have some very flat space between the planets, we put the laboratory in that very flat space)