# Crossing the event horizon of a black hole

I am struggling with an understanding on what the longest proper time an observer can spend before he will be destroyed into the singularity. How should I approach this problem?

Dale
Mentor
2020 Award
Try http://arxiv.org/PS_cache/arxiv/pdf/0705/0705.1029v1.pdf" [Broken].

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Ich
Hey, I just looked for exactly this paper with keywords "proper time singularity" in the abstract. Nothing.

The word singularity means division by zero. This is not allowed. Therefore, the solution of the GR equations for the observer crossing the event horizon is not valid. Instead, we observe someone falling down the black hole, and note that it takes forever to reach the horizon. The horizon is like the end of the universe. GR tells us that mass changes geometry.

JesseM
The word singularity means division by zero. This is not allowed. Therefore, the solution of the GR equations for the observer crossing the event horizon is not valid. Instead, we observe someone falling down the black hole, and note that it takes forever to reach the horizon. The horizon is like the end of the universe. GR tells us that mass changes geometry.
There is no physical singularity at the event horizon of a black hole. Schwarzschild coordinates do go to infinity there, but you can pick different coordinate systems (like some of the ones mentioned on this page) where there is no coordinate singularity at the event horizon, and you can show it only takes the infalling observer a finite proper time to pass the event horizon. The singularity at the center of the black hole is a real physical one though, since infinities appear there no matter what coordinate system you choose.

Yes, the singularity is at the center of the black hole. However, the center does not exist. There is no such thing as the inside of the black hole, as it takes forever to reach the surface, i.e., the event horizon. When we talk about the "inside", we are referring to the solution of the General Relativity (GR) equations from the point of view of the observer falling down. However, this solution is not valid, due to the existence of the singularity. This is like boundary conditions restricting which solutions can be allowed; this is the basis of the physics of music. In other words, since it is impossible to observe an object crossing the horizon, then nothing can cross the horizon. Can you travel past the end of the universe? Remember, geometry is not Euclidean near the horizon. The horizon is an end of the universe, as it takes forever to get there.

Ah, but great physicists have discussed this singularity! So what! They are wrong! Very simple! Division by zero is not allowed!

JesseM
Yes, the singularity is at the center of the black hole. However, the center does not exist. There is no such thing as the inside of the black hole, as it takes forever to reach the surface, i.e., the event horizon.
No, not for the infalling observer it doesn't--it takes only a finite proper time (time as measured by a clock they're carrying) for them to cross the event horizon. See What happens to you if you fall into a black hole? for example.
aranoff said:
In other words, since it is impossible to observe an object crossing the horizon
It's quite possible, if you're willing to dive in after it.
aranoff said:
Can you travel past the end of the universe?
No, although it would be a bit silly to claim that nothing exists beyond the edge of the visible universe (a sphere centered on Earth with a radius of about 50 billion light years--see here) just because light from those regions wouldn't have had time to reach us since the Big Bang.
aranoff said:
Ah, but great physicists have discussed this singularity! So what! They are wrong! Very simple! Division by zero is not allowed!
If you are not familiar with this board's policy on claims which contradict mainstream physics, please read the IMPORTANT! Read before posting message which appears at the top of the board.

tan(x) has a singularity at x = 90°. The tan function is not defined here. It is not infinity, but not defined. A singularity is simply a point where the function or equation is not defined.

stevebd1
Gold Member
I am struggling with an understanding on what the longest proper time an observer can spend before he will be destroyed into the singularity. How should I approach this problem?
This is probably in the paper posted by DaleSpam but one equation that I've seen a number of times for the fall-in time for Schwarzschild black holes is-

$$\tau_{max}\text{[seconds]}=\frac{\pi M}{c}=\frac{\pi Gm}{c^3}\ \equiv\ 1.548\times10^{-5}\ \times\ \text{sol mass}$$

where τ is the wristwatch time (proper time) in seconds, M is the gravitational radius (M=Gm/c^2), G is the gravitational radius, m is mass and c is the speed of light.

For a 10 sol mass black hole, the maximum free-float horizon to crunch time is 1.548x10^-4 seconds or 0.155 milliseconds, for a 3 million sol mass black hole, the time is ~46 seconds.

The maximum free-float horizon to crunch distance is-

$$\tau_{max}\text{[metres]}=\pi M$$

JesseM
tan(x) has a singularity at x = 90°. The tan function is not defined here. It is not infinity, but not defined. A singularity is simply a point where the function or equation is not defined.
The limit of tan(x) as you approach 90 is certainly infinity, but you're right, a singularity can be any undefined point. Anyway, the fact remains that you can find perfectly good coordinate systems where all physical quantities have well-defined finite values on the event horizon, so there is no physical singularity there.

The limit of tan(x) as you approach 90 is certainly infinity, but you're right, a singularity can be any undefined point. Anyway, the fact remains that you can find perfectly good coordinate systems where all physical quantities have well-defined finite values on the event horizon, so there is no physical singularity there.
Again, I repeat, the singularity is at the center of the black hole. The equation of motion which is the solution of GR, is not valid at this point. Is it valid near the singularity? I say no. I view the singularity as a boundary condition saying that this solution is not valid.

JesseM
Again, I repeat, the singularity is at the center of the black hole. The equation of motion which is the solution of GR, is not valid at this point. Is it valid near the singularity? I say no. I view the singularity as a boundary condition saying that this solution is not valid.
I agree with this, but it's not what you seemed to be saying before. Before you seemed to be saying the event horizon was a singularity, and that an observer could never really pass it. Your words:
The word singularity means division by zero. This is not allowed. Therefore, the solution of the GR equations for the observer crossing the event horizon is not valid. Instead, we observe someone falling down the black hole, and note that it takes forever to reach the horizon. The horizon is like the end of the universe.

The singularity is at the center. This means that the equation of motion, the solution of GR, is not valid at the center. This means the equation is not valid anywhere inside the black hole. The singularity acts like a boundary condition, restricting the validity of equations. A valid solution of the wave equation must satisfy the boundary conditions.

But wait! How can a black hole, which in the simplest case, can be imagined as a sphere, not have a center? Answer: the geometry near the event horizon is not Euclidean.

JesseM
The singularity is at the center. This means that the equation of motion, the solution of GR, is not valid at the center. This means the equation is not valid anywhere inside the black hole.
Uh, how do you figure? It's not valid right at the singularity, but what about, say, halfway between the singularity and the event horizon? You have no justification for saying that the equations cease to give valid predictions at the event horizon just because GR breaks down at the singularity, that's a total non sequitur.

Well... where is located the mass of the black hole? If it is inside the event horizon, how its gravitational atraction acts upon anything outside the horizon? Wouldnt it involve travel faster than light?

stevebd1
Gold Member
There seems to be plenty of maths that supports the fact that there is an 'inside' to the event horizon, basically put $s^2<0$ exists inside the event horizon where $s^2=c^2\Delta t^2-\Delta r^2$ (or $c^2\Delta t^2<\Delta r^2$) while $s^2>0$ exists in space outside the EH (or outside the ergosphere as in the case of a rotating black hole). There seems to be plenty of metric out there that support this with SR taking care of infinities that crop up at the EH. I cannot remember who said the following but 'the event horizon is not where GR ends but where GR begins to end as it starts to unravel towards the singularity'. Of course, the idea of GR unravelling might change as a theory of quantum gravity is established and the 'singularity' is better understood.

Steve

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Uh, how do you figure? It's not valid right at the singularity, but what about, say, halfway between the singularity and the event horizon? You have no justification for saying that the equations cease to give valid predictions at the event horizon just because GR breaks down at the singularity, that's a total non sequitur.
This is a sequitur! Boundary conditions (BC) are very basic in mathematics and physics. The singularity at the center means that the solution of GR inside the hole is not valid at the center. Therefore, it is not valid period. The motion of a vibrating string is an example where possible solutions are rejected due to BC.

I suggest you do some research on BC. The concept of BC is very sophisticated in mathematics.

JesseM
This is a sequitur! Boundary conditions (BC) are very basic in mathematics and physics. The singularity at the center means that the solution of GR inside the hole is not valid at the center. Therefore, it is not valid period. The motion of a vibrating string is an example where possible solutions are rejected due to BC.

I suggest you do some research on BC. The concept of BC is very sophisticated in mathematics.
Sorry, no, you're talking nonsense here. Of course I'm familiar with the idea of boundary conditions, a basic idea in physics which is not particularly "sophisticated" at all, but it does not somehow allow you to say that nothing inside the event horizon is valid, physicists only believe that GR becomes significantly wrong in the immediate neighborhood of the singularity, not the entire region inside the event horizon. Perhaps you are confusing the physical boundary of the black hole (the event horizon) with the notion of "boundary conditions", but they are unrelated, boundary conditions are just the conditions at the boundary of whatever region of spacetime you wish to consider when you're setting up the problem, they have nothing to do with the event horizon. Nor is there any notion in physics that a singularity at one point in a solution invalidates the solution as a whole.

The solutions of GR are continuous functions. If we allow the solution of GR inside the BH, then there is a discontinuity, which invalidates this solution.

JesseM
The solutions of GR are continuous functions. If we allow the solution of GR inside the BH, then there is a discontinuity, which invalidates this solution.
Totally illogical. First of all, physicists have no such rule about throwing out solutions containing discontinuities; as I've said before, it is thought that the singularity is a sign that we need quantum gravity to get accurate predictions about the immediate neighborhood of the black hole's center, but that GR can be trusted far from the Planck scale. And if you think it "invalidates this solution", it's completely arbitrary for you to say it only invalidates the region inside the black hole's event horizon, but not outside it. What do you think is so special about the event horizon? A Schwarzschild spacetime is a solution containing a singularity, period, we don't use separate "solutions" for the region outside the event horizon and the region inside. Likewise, all the cosmological models in GR contain singularities at the Big Bang, would you say that we should therefore throw these cosmological solutions out, including their predictions about expanding space long after the Big Bang which have had quite a lot of experimental confirmation?

The nature of GR is such that discontinuous solutions are not allowed. A discontinuity is not physical.

There is another point. A physical theory is a mathematical system (a collection of arbitrary self-consistent statements) which can be verified by observations and experiments. Since it is impossible to perform observations inside the event horizon, this solution does not exist.

So that's it. No solution inside. By the definition of a physical theory. By the requirement that the solution exist everywhere inside the BH.

As we watch things falling down the BH, we note that it takes forever to get there. We also note that it takes forever to go straight in a Euclidean line. Therefore, the event horizon is geometrically the same as a straight line. Remember the equation of GR: G=T.

Speaking about the inside of the BH is just as wrong as speaking about outside the universe. Geometrically impossible.

JesseM
The nature of GR is such that discontinuous solutions are not allowed. A discontinuity is not physical.
So why don't you reject the entire solution instead of just arbitrarily rejecting the region inside the event horizon? The horizon does not mark the boundary between two different "solutions", the whole spacetime is one GR solution!!! And do you also reject the entirety of cosmological models which contain an initial Big Bang singularity?

Hurkyl
Staff Emeritus
Gold Member
The solutions of GR are continuous functions. If we allow the solution of GR inside the BH, then there is a discontinuity, which invalidates this solution.
Where is this alledged discontinuity? Is it at a point of space-time? Or is it merely at some point in a faux-coordinate chart that doesn't correspond to a point of space-time?

And no, continuity really isn't a hard requirement of the theory -- it (and sufficient differentiability) only a requirement for the use of the simplest mathematical framework.

As we watch things falling down the BH, we note that it takes forever to get there. We also note that it takes forever to go straight in a Euclidean line. Therefore, the event horizon is geometrically the same as a straight line.
This is patently absurd.

JesseM