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MasterD
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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?
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.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.
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: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.
It's quite possible, if you're willing to dive in after it.aranoff said:In other words, since it is impossible to observe an object crossing the horizon
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:Can you travel past the end of the universe?
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.aranoff said:Ah, but great physicists have discussed this singularity! So what! They are wrong! Very simple! Division by zero is not allowed!
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?
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.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.
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.
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: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.
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.
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.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.
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.
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.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.
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?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.
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?aranoff said:The nature of GR is such that discontinuous solutions are not allowed. A discontinuity is not physical.
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?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.
This is patently absurd.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.
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.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?
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?
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?
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.
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
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?
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: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.
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:According to the external observer, there is no spacetime inside the BH.
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: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!
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: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.
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.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.
That statement is wrong as the existence of any region of spacetime is observer independent.aranoff said:According to the external observer, there is no spacetime inside the BH.
Well... it can be observed in principle (and by a non-hypothetical observer, too), it's just that those observations cannot be communicated back.aranoff said:the fundamental principle of physics which states that something that cannot be observed in principle does not exist.