# I Conditions necessary for a human to survive event horizon

Tags:
1. Apr 1, 2016

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.

2. Apr 1, 2016

### Staff: Mentor

3. Apr 1, 2016

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]

Last edited by a moderator: Apr 1, 2016
4. Apr 1, 2016

### 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.

5. Apr 1, 2016

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.

6. Apr 1, 2016

### 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.

FTidal = 2GMBlack holeMHumanHeightHuman / d3

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

rSchwarzschild = 2MBlack holeG/c^2

FTidal = 2GMBlack holeMHumanHeightHuman / (4MBlack holeG/c^2)3

Solve for MBlack Hole

7. Apr 1, 2016

Thanks, again, newjerseyrunner. Using your equation, with mass*height of human as 100 kg-m, and the Ftidal = 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?

8. Apr 1, 2016

### newjerseyrunner

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

9. Apr 1, 2016

### 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?

10. Apr 1, 2016

### 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

FTidal = 2GMBlack holeMHumanHeightHuman/d3

Rearrange it do get d by itself, leave the cube for now
d3 = 2GMBlack holeMHumanHeightHuman/FTidal

Since we want the object to come apart at the event horizon, we need the formula for that (diameter, not radius)
dSchwarzschild = 4MBlack holeG/c2

Since we want the roche limit and event horizon to be the same, place formula above in for d
(4MBlack holeG/c2)3 = 2GMBlack holeMHumanHeightHuman/FTidal

Expand the left side
64MBlack hole3G3/c6 = 2GMBlack holeMHumanHeightHuman/FTidal

Move eveything except mass of the black hole to the right
64MBlack hole3 = 2MBlack holeMHumanHeightHumanc6/FTidalG2

Getting confused with all the super and sub-scripts, so I'm going to replace with actual values now, assume:
HeightHuman = 2m
MassHuman = 70kg

64MBlack hole3 = 280kg*mMBlack holec6/FTidalG2

Simplify (don't forget the units!)
MBlack hole = (35/2)1/2c3 kg1/2m1/2 / 2G1/2FTidal

Replace constants and variables (to allow units of variables to be presented)
c = 300000000 m/s
G = 6.674*10-11m3kg-1s-2
MBlack hole = x kg
FTidal = y kg *m/s

x kg = (35/2)1/2(300000000m/s)3 kg1/2m1/2 / 2(6.674*10-11m3kg-1s-2)1/2y(kg*m/s)

Simplify the units
x kg = (256033 * 2.7*1025 / y) m1/2/(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)/m1/2?

Last edited: Apr 1, 2016
11. Apr 1, 2016

### pervect

Staff Emeritus
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:

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.

12. Apr 1, 2016

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

13. Apr 1, 2016

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

14. Apr 2, 2016

### Ibix

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.

15. Apr 2, 2016

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).

16. Apr 2, 2016

### 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.

Last edited: Apr 2, 2016
17. Apr 2, 2016

### Staff: Mentor

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.

18. Apr 2, 2016

### jartsa

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.

19. Apr 2, 2016

### Staff: Mentor

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.

20. Apr 2, 2016

### newjerseyrunner

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.