Crossing the event horizon of a black hole

In summary: No, you cannot cross this horizon." This is like a boundary condition in physics saying... "No, you cannot go faster than light." In summary, the conversation discusses the concept of a singularity in relation to black holes and the validity of the General Relativity equations near the event horizon. It also explores the idea of boundaries and limitations in physics, such as the speed of light and the concept of infinity.
  • #1
MasterD
13
0
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?
 
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  • #2
Try http://arxiv.org/PS_cache/arxiv/pdf/0705/0705.1029v1.pdf" [Broken].
 
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  • #3
Hey, I just looked for exactly this paper with keywords "proper time singularity" in the abstract. Nothing.
 
  • #4
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.
 
  • #5
aranoff said:
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.
 
  • #6
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!
 
  • #7
aranoff said:
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.
 
  • #8
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.
 
  • #9
MasterD said:
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-

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

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-

[tex]\tau_{max}\text{[metres]}=\pi M[/tex]
 
  • #10
aranoff said:
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.
 
  • #11
JesseM said:
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.
 
  • #12
aranoff said:
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.
 
  • #13
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.
 
  • #14
aranoff said:
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.
 
  • #15
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?
 
  • #16
There seems to be plenty of maths that supports the fact that there is an 'inside' to the event horizon, basically put [itex]s^2<0[/itex] exists inside the event horizon where [itex]s^2=c^2\Delta t^2-\Delta r^2[/itex] (or [itex]c^2\Delta t^2<\Delta r^2[/itex]) while [itex]s^2>0[/itex] 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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  • #17
JesseM said:
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.
 
  • #18
aranoff said:
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.
 
  • #19
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.
 
  • #20
aranoff said:
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?
 
  • #21
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.
 
  • #22
aranoff said:
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?
 
  • #23
aranoff said:
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.
 
  • #24
I wonder if aranoff is not confusing the idea of continuity in the spacetime manifold on which the coordinate system and tensor fields are defined with the idea of continuity in the tensor fields themselves. I imagine the first probably is a requirement, but not the second. For example, take a simple example like a planet with a sharply-defined surface--doesn't the surface already mark a type of discontinuity in the matter field?
 
  • #25
JesseM said:
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?
Actually, the complete Schwarzschild solution has two event horizons. They are connected by two different universes or through a wormhole (Einstein-Rosen bridge) in spacetime.
 
  • #26
JesseM said:
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?

According to the external observer, there is no spacetime inside the BH.

We currently have no theory of cosmology, but hypotheses. A theory is a rigorous mathematical framework plus observations and experiments. Since all current cosmological models contain an initial singularity, they are not rigorous, and so do not qualify as a theory.
 
  • #27
JesseM said:
I wonder if aranoff is not confusing the idea of continuity in the spacetime manifold on which the coordinate system and tensor fields are defined with the idea of continuity in the tensor fields themselves. I imagine the first probably is a requirement, but not the second. For example, take a simple example like a planet with a sharply-defined surface--doesn't the surface already mark a type of discontinuity in the matter field?

The gravitational field is continuous as one passes through the surface. Consider a uniform mass. As the test particle enters the mass, we draw a sphere, center the planet, radius the distance to the test particle. We calculate the mass of the sphere, and use this mass to calculate the field.
 
  • #28
aranoff said:
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.

Clarification: There is no end to the universe, as one can imagine going in a straight line forever. No matter where we are, we can continue further. As we move outward, time marches on. This is what infinite means. The set of integers is infinite. Pick any number. We can find a larger number by adding one. There is no last integer. Same here. There is no last position in the universe.

The same is what happens as we approach a BH, due to the geometry. Remember that the geometry is not Euclidean. We can imagine falling forever towards the BH. No matter where we are, we can continue further. As we fall, time marches on. The distance to the BH is infinite. No matter how long we wait and how close we get to the event horizon, we can always imagine waiting longer and getting closer. There is no end point. This is what I meant by my statement "the event horizon is geometrically the same as a straight line." Sorry for the confusion. The topic is confusing enough!
 
  • #29
JesseM said:
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

And I answered you. You say we need another theory, which you call quantum gravity. This means you agree that GR alone is not valid inside the BH. We agree.

I am not interested in discussing possible future developments in physics. I am focused in understanding current theories as they are. My point is GR (today's GR, not the future GR) is not valid inside the BH.
 
  • #30
MasterD said:
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?

You are a complicated entity. You have cells and organs. There is a lot of information about you. The only information about a black hole is its mass, angular momentum, and charge. Once an object falls into the BH, all the information is lost. The only information about you is your mass.
 
  • #31
aranoff said:
The gravitational field is continuous as one passes through the surface. Consider a uniform mass. As the test particle enters the mass, we draw a sphere, center the planet, radius the distance to the test particle. We calculate the mass of the sphere, and use this mass to calculate the field.
I wasn't talking about the gravitational field (curvature tensor), I was talking about a discontinuity in the values of the stress-energy tensor which defines the distribution of matter and energy in the spacetime, and which is related to the curvature tensor by the Einstein field equations in GR. My point is that although the manifold on which the tensors of GR are defined may be required to be continuous, there is definitely no requirement that the values of the tensors themselves should vary continuously. I challenge you to find a single textbook or paper by a professional physicist which states that a spacetime with a discontinuity in the values of the tensor should be thrown out. Of course you won't, because that would imply physicists throw out the interior regions of black holes as well as cosmologies including a Big Bang singularity (which is not separated from the rest of the universe by an event horizon), while any GR textbook will contain detailed discussions of these solutions.
aranoff said:
According to the external observer, there is no spacetime inside the BH.
Irrelevant to your previous argument, you said we should throw out any "solution" which includes a singularity, and at a mathematical level the entire BH spacetime (inside and out) is a single solution to the Einstein field equations, the fact that the region inside the horizon is not observable by anyone who doesn't cross it doesn't make that region a separate "solution". You are just inventing ad hoc ways of justifying your prejudices using poorly-defined terminology, not making any rigorous argument against the validity of GR anywhere inside the horizon.
aranoff said:
The same is what happens as we approach a BH, due to the geometry. Remember that the geometry is not Euclidean. We can imagine falling forever towards the BH. No matter where we are, we can continue further. As we fall, time marches on. The distance to the BH is infinite. No matter how long we wait and how close we get to the event horizon, we can always imagine waiting longer and getting closer. There is no end point. This is what I meant by my statement "the event horizon is geometrically the same as a straight line." Sorry for the confusion. The topic is confusing enough!
Nonsense, any textbook on black holes will tell you that for an observer falling into the BH, the proper time (time as measured by a clock falling with them) to cross it is finite. The coordinate time in Schwarzschild coordinates to cross it is infinite, but there is nothing special about Schwarzschild coordinates, one can use other coordinate systems such as Eddington-Finkelstein coordinates where the time to cross it is finite, it is really only proper time that has physical significance (one could even come up with a coordinate system where it takes an infinite coordinate time for you to cross from one end of your room to the other).
aranoff said:
You say we need another theory, which you call quantum gravity. This means you agree that GR alone is not valid inside the BH. We agree.
Wrong, I simply echo physicists in saying GR is not valid at the Planck scale. The energy densities would only approach the Planck density very close to where GR predicts a singularity (probably in the neighborhood of one Planck length from it), in regions of the interior far from the singularity there's no such reason for thinking GR would significantly disagree with quantum gravity (except perhaps in the case of the inner horizon of a rotating black hole, which calculations suggest would see infinitely blueshifted light from outside, again creating energy densities greater than the Planck density).
aranoff said:
I am not interested in discussing possible future developments in physics. I am focused in understanding current theories as they are. My point is GR (today's GR, not the future GR) is not valid inside the BH.
Indeed, this forum is not a place to debate interpretations of GR which are widely accepted among physicists, as you are doing by rejecting everything inside the horizon--you will find no textbooks which agree with you, or that say that solutions with discontinuities in tensor fields should automatically be rejected, this is just stuff you're making up with no justification. Please read the IMPORTANT! Read before posting thread at the top of the forum, attempting to disprove mainstream views on GR is explicitly forbidden, if you continue to do so I'll report your posts.
 
  • #32
Okay, let me understand you. We agree that T can be discontinuous. We agree that if we reject the singularity, then G is discontinuous. You therefore say that it is acceptable in GR for G to be discontinuous? What are the implications of this? Any other examples of G being discontinuous?

" challenge you to find a single textbook or paper by a professional physicist which states that a spacetime with a discontinuity in the values of the tensor should be thrown out."
Again, you agree that the solution of GR at the center of the BH is not valid. We agree! Since it is not valid there, it is not valid anywhere. If you do not like my continuity argument, okay. I actually view it as a boundary condition saying this solution is not valid. Why do you reject my boundary condition reasoning?

"a Big Bang singularity (which is not separated from the rest of the universe by an event horizon), while any GR textbook will contain detailed discussions of these solutions."
It is fine to have discussions. It is not fine to say that cosmology including the Big Bang singularity is a valid physical theory, as it contains a mathematical inconsistency. Instead, it is a working hypothesis, which hopefully will become part of a future theory. BTW, have you heard of Rosen's cosmology, in which there is no infinite big bang?

We must focus on the basic principles of physics. We cannot say just because this is what physicists say it must be so. Physicists said von Laue current is valid. I had to fight this to get my paper published in 1972. If you were the group moderator then, you would not have permitted my paper to appear on the site. I was amazed at the wrong ideas most physicists had, 75 years after SR, which is a simple theory.

What I insist is clear statements of basic principles and the conclusions drawn from them.

Why are you talking about Planck, when we are discussing GR, not QM?

What about the principle that if something cannot be observed in principle, it does not exist? The "finite time to cross the event horizon", as predicted by GR, cannot be observed in principle!
 
  • #33
aranoff said:
According to the external observer, there is no spacetime inside the BH.
That statement is wrong as the existence of any region of spacetime is observer independent.

The event horizon simply signifies that no signals from events inside can be observed from the outside of the event horizon. The event horizon functions like a one-way membrane.
 
  • #34
When I said spacetime does not exist inside the BH, I am relying on the fundamental principle of physics which states that something that cannot be observed in principle does not exist. This principle is more basic than GR or QM or whatever.
 
  • #35
aranoff said:
the fundamental principle of physics which states that something that cannot be observed in principle does not exist.
Well... it can be observed in principle (and by a non-hypothetical observer, too), it's just that those observations cannot be communicated back.
 

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