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A black hole thought experiment

  1. Nov 7, 2006 #1
    Suppose I dive head first into a black hole of size 100 light second from far away. When I reckon that I will cross the event horizon in 1 second, I throw a stone to my right at 1/100 the speed of light, which pushes me to my left at 1/10000 the speed of light. Here are my questions:

    a. will a distant observer perceive the stone to depart me almost tangentially or radially?
    b. using my proper time, will the stone and I still cross the event horizon in about 1 second?
    c. will the points where the stone and I cross the event horizon be about 1/100 light second and 1/10000 light second away from the original expected position, or will we be smeared along a long track so that there are no definite points of entry?
    d. will the stone fly by (or hit) me before we cross the horizon?
    e. will a distant observer see the stone revolve once, revolve indefinitely, frozen somewhere, or form a blurred ring (if he can see very redshifted light and fast moving objects)?

    I think the frozen star theory will have very different predictions to the black hole theory.

    Wai Wong
  2. jcsd
  3. Nov 8, 2006 #2
    Another black hole thought experiment:

    A scientist tries to build a black hole without singularity as follows.

    He mounts a ping-pong ball at the center of a balloon of radius 10km in space. He then pumps small neutral particles, say neutrons, into the balloon. He expects a black hole to be formed when enough mass (I reckon a bit over 3 solar masses) has been pumped in. But it turns out that he is unable to pump the last few particles inside - they do move, but just seem to take for ever to enter the balloon.

    So he dives into the balloon. He reckons that once he is inside, the total mass will exceed the critical mass for a black hole, so he must witness the formation of a black hole.

    Let us just consider gravity forces and ignore other kind of forces. Here are my questions:

    a. since the ping-pong ball is empty, there is no gravity inside so its surface will not experience much pressure all the way, is that right?
    d. to the scientist, will the ping-pong ball's size seem to increase or decrease as he nears the balloon boundary?
    c. according to his proper time, will the scientist ever cross the balloon boundary?
    d. if so, a black hole must be born some time. Will it be born with the size of the balloon or born with a smaller size and then grow rapidly to engulf the whole thing?
    e. what happens to the ping-pong ball at the instant the black hole is formed? Does it just shrink to zero size or implode in zero time?

    My intuition says he will never reach the balloon boundary. The balloon boundary seems to be at finite distance, but never gets much closer, just like the end of a rainbow. I know that is against all established deductions, but maybe, just maybe, I will prove them wrong one day.

    Wai Wong
  4. Nov 8, 2006 #3


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    The scenario is a bit involved, but I'll tell you that the equations you need (differential equations) to solve for black hole orbits can be found online at http://www.fourmilab.ch/gravitation/orbits/

    Of course you'll need some calculus to be able to do anything with them. This particular formulation is taken from Misner, Thorne, Wheeler "Gravitation", but most introductory GR books will discuss this.
  5. Nov 8, 2006 #4


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    The ping poing ball will experience severe pressures. While the gravity inside the ping pong ball would be zero, the pressure on the ping pong ball will be enormous. Think of taking a ping pong ball to the bottom of the sea, or better yet to the center of the Earth or the center of Jupiter. There will be a LOT of pressure on the ping pong ball.

    The pressure at the center of a static Schwarzschild solution becomes infinite just at the point where the solution become a black hole.

    The way I read your problem, the scientist won't be able to see the ping-pong ball until after he crosses the horizon, or rather there will be an indeterminate delay for the light signals, depending on just how close to a black hole the object becomes. So I don't think the problem is well defined.

    Yes, I think we've been over this before. Look at the collapse of a dust cloud into a black hole, for instance for a simpler example, on the websites I mentioneed.

    A black hole is a global phenomenon, not a local one. What will actually happen is that the ping pong ball will hold out as long as it can, but it will eventually be crushed. It would be best to imagine dropping very very small amounts of matter into the almost-black hole, and then allowing the system to reach equilbrium, to make the problem simple to analyze. Eventually, the pressure on the ping pong ball (possibly aided by dynamic phenomena such as shock waves from the infalling matter) will fail, and the ball will crush. This will happen before a black hole is formed. The pressure will basically increase without limit.

    The strongest possible "ping pong ball" made out of normal matter would be a ball of pure radiation. This would have a pressure of 1/3 of its energy density.

    For completeness, we should consider the possibility that the ping-pong ball made out of "exotic matter". In this case, the ping pong ball would actually have a negative mass. In this odball case the negative mass of the ping-pong ball could make the total mass of the system lower than the critical limit. But if you have a given ping pong ball of a certain configuration, even if it is made out of exotic matter, the pressure on it will incrrease without bound, and if you pile enough stuff on top of it, the pressure will increase without limit until it collapses.

    And it's still wrong :-(.
  6. Nov 9, 2006 #5


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    There's one other counter-intuitive result I ought to mention from the second problem.

    The amount of mass increase when lowering an object into a black hole is not the mass m, of the object, but the "energy at infinity" of the object.

    If one imagines lowering an object, very slowly, into an "almost black hole", via some sort of cable, the black hole's mass will increase by less than the masas of the added object. In the limit, lowering a 1kg mass into the almost-black hole will cause virtually no increase in the mass of the black hole, as all of the energy will essentially be transmitted out via the cable. Of course the cable has to be almost infinitely strong for this to work. In actuality, infinitely strong cables, like infinitely strong ping pong balls, don't exist.
  7. Nov 9, 2006 #6
    Don't both names refer to the same theory?
  8. Nov 9, 2006 #7
    I wished I had the math skills to solve the differential equations. I am trying, but I am not sure if it will take finite or infinite time. B:)

    I hope some experts out there have a solution. A very approximate one is fine.

    All references I read tell me I won't notice anything unusual when I cross the event horizon of a large black hole, but my intuition tells me otherwise: the stone and I will disintegrate before reaching the event horizon and each atom nucleus follows its own 'geodesic' to spread over the event horizon but never cross the event horizon. There may be enough time for the stone to fly by me a number of times before my head starts to disintegrate, depending on the strength of the chemical bonds that hold it. I may have disintegrated before I am 1 second from the event horizon.

    Wai Wong
  9. Nov 9, 2006 #8
    When the first few neutrons enter the balloon, they will be attracted by the minute gravity of the ping-pong ball and deposit evenly on its surface. Only when the entire surface is tightly packed with neutrons will there be a second layer if the neutrons behave like solid spheres. After a considerable number of layers are deposited, there will be significant gravity attraction between neutrons, which will try to pull each neutron towards the centre so the ping-pong ball will start to shrink. Once the ping-pong ball shrinks a little bit, it will be separated from the first layer of neutrons. Will the first layer of neutrons move inwards? They cannot. If they move inwards, the distance between neutrons will have to be reduced. Since they are tightly packed, moving inwards means the neutrons have to shrink, and internal forces like the strong force will oppose that. Thus the inward pressure is counteracted by the tangential force(s) between adjacent neutrons. In reality I think they will shrink a little, but not by much. If the neutrons shrinks by 50%, the ping-pong ball just shrinks 50%; it won't collapse. The neutron layers are somewhat like the hard egg shell which shields the embryo inside from the heavy mother sitting on top.

    The Schwarzschild solution ignores interaction between particles other than gravity, so it is just an approximation. Near the center such forces can't be ignored.

    After a long thought, I come up with the following analysis.

    Before the scientists dives in, there is no black hole yet, and thus there is no event horizon to hide the ping-pong ball. So the scientist can always see the ping-pong ball. There is no gravity at the center, so I think there is no relativistic contraction before the dive. The indeterminate delay will approach infinity, which makes the ping-pong ball appear to be receding to infinity and getting redshifted more and more. During his fall outside the balloon, I guess his acceleration is never enough to cancel the receding speed of the ping-pong ball so the latter will keep receding and getting more redshifted just the same. Anyway the ping-pong ball will be practically unobservable before he enters the balloon.

    As far as observing the ping-pong is concerned, my analysis yields a result not much different from a true black hole, so I agree it is not a good question to ask here.

    Not quite. What we have been over is that if there is an event horizon, it will grow. Once the event horizon grows, it can explain certain phenomena. To me, that is circular argument, because it assumes the existence of an event horizon in the first place. My scenario of proto black hole (see below) can also explain the collapsing dust cloud so the score is 0:0.

    This experiment tries to find out at what radius the event horizon actually starts. If my argument is valid and the ping-pong is of finite size before a black hole is formed, then regardless of where the event horizon starts, the ping-pong ball will have to experience instant space-time change if the Schwarzchild solution is exact. I think instant space-time change is not allowed, because the differential equations are continuous. If the ping-pong ball implodes in finite time, then the Schwarzchild solution can't be exact, and one has to find the exact solution in order to understand the singularity.

    The problem is the system never reaches equilibrium. So at any time, there will be 'proto event horizons', which I define as follows:

    For any particle, the proto event horizon is the event horizon of the black hole that can be formed from the particle plus all the mass equally close or closer to the center.

    Before a black hole is formed, each layer of particles will have a different proto event horizon. The proto event horizon does not grow, because new infalling matter, assumed to be evenly distributed, does not exert gravity on particles inside. So infalling particles will approach their own new and bigger proto event horizons, while the inner layers are always approaching the same old proto event horizons. There is an outermost proto event horizon, but it is never a real event horizon. For new infalling particles, they can never catch up with the previously outermost layer, so they will experience the gravity of the same mass. Whether that mass is a true black hole or a proto black hole makes no difference as fas as gravity is concerned.

    My hypothesis fits the initial condition - no black hole exists initially - while the black hole theory does not fit the initial condition and as far as I know no one has been able to describe how the transition can take place.

    As time approaches infinity, my scenario will yield a sphere with layers of particles reaching their proto event horizon. So M(r)=rc^2/2G. The density at radius r will thus be D(r)=c^2/8πGr^2. There won't be any singularity at the center though, because the density will plateau at the density of neutrons.

    Light from the ping-pong ball can always cross the proto event horizons in finite time, although it may take trillions of years. So information is never lost in such a proto black hole.

    I understand my claim is an extraordinary claim, but I don't have extraordinary proof. If any of you can prove my claim, it will be shocking news in the physics world. In that case, don't forget to mention my name. B:)

    Wai Wong
    Last edited: Nov 9, 2006
  10. Nov 9, 2006 #9
    This depends on the size of the black hole. If it is big enough, you can pass through the event horizon without suffering from the tidal effect (yet!). And why would the atom nucleus spread over the event horizon instead of crossing it? :confused:
  11. Nov 9, 2006 #10
    In the black hole theory there is a definite event horizon where matter can cross in finite proper time and there are all sorts of strange phenomena inside.

    I don't know much details of other people's frozen star theory, but according to my interpretation, the event horizon can never be reached in finite proper time and thus there is no singularity, no infinite time dilation, no worm holes etc. The inside of a frozen star is dense matter given by my density formula


    Near the center, the density is limited by the density of neutrons.

    Wai Wong
  12. Nov 9, 2006 #11
    That is what all the other people say and I suspect they are wrong. They are wrong because they forget about the space-time curvature caused by the infalling object.

    My understanding of space-time geometry is minimal, so my following account is entirely a guess. Correct me if I am wrong.

    When you combine two dents in the space-time curvature, the contours will no longer be round but sort of dumb bell shape, not unlike the accretion disk around a star and a black hole.

    For ordinary objects, the dumb bell is very lopsided, but it is not round nonetheless. Thus the Schwarzchild solution can no longer be applied. Near the supposed entry point, the space-time curvature of the original event horizon is reduced by the gravity of the infalling object, and thus it is no longer a part of the event horizon. What will become of the event horizon? It is ill-defined in that case. The space-time curvature is not symmetrical in all directions. I think that is called a saddle. There are points on the space-time surface where light can escape in certain directions but not in other directions. Thus there is no definite event horizon.

    I suspect the atoms on the surface of my torso will find it easier to move sideways then forward, just like light bending around a black hole. At speed close to c, an atom aren't much different from a photon after all.

    Wai Wong
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