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What happens when you enter a black hole

  1. Jul 31, 2009 #1
    According to some members on this site and other facts I've found, it would be possible to travel towards Earth with such a velocity that Earth would perceive the traveller as a black hole.
    Inversely at that same speed, the travelleler would see Earth as the black hole.

    What would happen when the traveller entered the black hole.

    My guess, is that space has a limit to how much mass it can contain. The size of the black hole compared to Earth, will be just large enough to contain all the relative mass of Earth without breaking that limit.
    In other words, its center of gravity will be larger than Earth itself.
    I believe what would happen is that the traveller would think hes going straight in, and would end up on the other side of the black hole. This seems to violate less things, because if he could crash into the planet, well, idunno.
  2. jcsd
  3. Jul 31, 2009 #2


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    What? Who said that.
  4. Jul 31, 2009 #3
    When you enter a black hole, you lose all hopes of exiting..
  5. Jul 31, 2009 #4
    It's either a black hole or it isn't. Either light can escape from it or it can't. If you're not a black hole in your own reference frame, no one can "perceive" you to be one in theirs.
  6. Jul 31, 2009 #5
    They were wrong. It is the rest mass of a black hole (or of anything, for that matter) that gives it its gravitational field. Note that the location of kinetic energy is relative to the reference frame of the observer. For example, would the abovementioned traveller also see the Earth as a black hole but no change to himself? Which thing is a black hole and which thing isn't, is not a relative concept. As ZikZak said, you're either a black hole or you aren't. In this case, neither the Earth nor the traveler are black holes. However, if there is kinetic energy contained in a system, then that relativistic mass becomes rest mass, and there is a gravitational field gained from it. So if that traveler, when he hit the Earth, were trapped in a box, then the Earth-traveler system would become a black hole if there had been enough energy to begin with.
  7. Jul 31, 2009 #6
    When you approach a black hole, you get burned by the Hawking's radiation.
    Whent you get inside, you are burned by the light that could not escape from it.:smile:
  8. Jul 31, 2009 #7

    George Jones

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    No. According to the book Quantum Fields in Curved Space by Birrell and Davies, pages 268-269,
    This finite amount is negligible for observers freely falling into a black hole.
    I assume that you're referring to the infinitely blue-shifted radiation at the Cauchy horizon inside a rotating black (or charged) hole, which creates a weak (not crushing) curvature singularity at the Cauchy horizon. Some researchers believe that this is a non-destructive singularity.

    Roughly, if components of g (the metric) are continuous but "pointy" (like the absolute value function), then first derivatives of g have step diiscontinuities (like the Heaviside step function), and second derivatives of g (used in the curvature tensor) are like Dirac delta functions. If a curvature singularity blows up like a Dirac delta function, then integration produces only a finite contribution to the tidal deformation of an object, which, if the object is robust enough, it can withstand. See, for example,

    http://physics.technion.ac.il/~school/Amos_Ori.pdf [Broken],

    particularly pages 15 (starting at "Consequence to the curvature singularity at the IH: (IH = Inner Horizon)), 16, 19 (for physical implications of the weak singularity), and 24.
    Last edited by a moderator: May 4, 2017
  9. Jul 31, 2009 #8
    I know this might be a rough explanation... but if your trying to get into a BH, you'll need to get pass the gravitational gaps and the huge amounts of disintegrating space junk traveling close to c first :biggrin:
    Last edited: Jul 31, 2009
  10. Aug 1, 2009 #9
    Well, thanks you guys, it seems from your counter explanations, that what I have in mind may be possible. As vin said, you cannot escape a black hole, and as Zikzak said, you would be looking at yourself from your own reference frame and therefor only have rest mass.
    So, although Earth see's the traveller as a black hole, the traveller sees the Earth as a black hole.
    Remember that this mass is only relativistic and just like the light that is blue shifted in front and red shifted behind ,the mass and therefor gravitational field would be the same.

    As the traveller began passing through the black hole, he would look behind him and not see a black hole, because relativisticly his "gravity" has redshifted behind him and would not see a black hole. In that possible example, exiting the other side of this black hole would be easy, because it doesn't exist from the other direction.

    This would be more like a wormhole than a conventional black hole I suppose, and although it may be pointless to argue at this point, I think because of the relativistic gravity, relative faster than light speed could be established in this manner.
  11. Aug 1, 2009 #10


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  12. Aug 1, 2009 #11
    That web page does not seem like it contradicts this argument.
    It uses the fact that a black hole would not form around its rest frame as evidence it would not form in any other frame
    "since its mass and volume haven't changed in its rest frame, it should not form a black hole in that frame--and therefore not in any other frame either"
    Just because it doesn't exist from its rest frame doesn't mean it doesn't exist in another frame.
    Clearly from Earths frame, light is different, blue shifted, in accordance with its velocities. So what your visually experiencing when on Earth is definitely changing.

    What types of particles give off Gamma Radiation? Do you think that if you could visually see the actual particles they would appear normal? Just a normal heated element in a flash light, giving off Gamma Radiation?, probably not. If you could actually see the element it would probably look so condensed, so dense that a gamma ray would be expected from the object.

    It would take very powerful releases of energy to create Gamma Rays, usually coming from very dense sources, our perception of something relies soley on the Energy that it emits or reflects. Since we know that our perception of this rocket will change based on its velocity, why can't it be a black hole, in realitly all that would change would be how it looks.
  13. Aug 1, 2009 #12


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    You missed the critical point:

    "this is based on a particular static solution to the Einstein field equations of general relativity, and ignores momentum and angular momentum as well as the dynamics of spacetime itself. In general relativity, gravity does not only couple to mass as it does in the newtonian theory of gravity. Gravity also couples to momentum and momentum flow; the gravitational field is even coupled to itself".

    The source of gravity in GR is not mass, but the stress-energy tensor. When you have a moving mass then the stress energy tensor not only has a relativistic mass term but also has momentum and momentum flow terms. These violate the assumptions of the Schwarzschild solution and prevent an event horizon from forming.

    Again, you do not turn into a black hole by going fast, nor does something else turn into a black hole by going fast in your frame. If there is no event horizon in the rest frame then no event horizon forms in any other frame. It is always possible to send light signals and they always have finite Doppler shift.
  14. Aug 1, 2009 #13
    Ok, So, I agreed with you, for the sake of the argument that gravity is not affected by your relative mass.

    However, that does not mean it will not necessarily form a black hole, because, thats assuming that gravity is a necessary component of a black hole. That may not be true, nothing would be violated necessarily if a visual black hole was created with the same rest mass, or gravity it always had.
    Light would still reach you as it always would, with it's finite doppler shift, it just will seem to come from around the event horizon instead of directly from the flashlight.

    This seems like it is possible without violating any laws does it not?
    The question remains however, would the "visual" black hole be represented according to relative mass? Or would the size of the black hole be the size of the rockets matter compressed to shwartzwald density? I think it would make more sense if the second were true.

    Also, I guess what I'm saying essentially, the rocket as it approaches definitely appears condensed, normally we think of this as being a 1 dimensional compression of depth, maybe its a 2 dimensional compression of height and width (if the rocket is heading directly towards you) and a 1 dimensional expansion of depth. Which, if were true would mean you could travel fast enough that you could go straight through a planet without hitting a particle (if you were going fast enough).
    Last edited: Aug 1, 2009
  15. Aug 1, 2009 #14


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    Then it is not a black hole. End of story.
  16. Aug 1, 2009 #15
    Wow, you paraphrased me so badly you changed the meaning of what I said.
    Lets see if I can paraphrase a different part of it and get a different answer

    Quote myself " Light would still reach you ....it just will .....come from around the event horizon instead of directly from the.....rocket"

    It must be a black hole, end of story.
    Hmm now it's different
  17. Aug 2, 2009 #16


    Staff: Mentor

    In your frame you can send a light signal to the earth, therefore in the earth's frame you can also. In the earth's frame the earth can send a light signal to you, therefore in your frame the earth can also. There is no event horizon for the light to go around. No event horizon in either frame so no black hole in either frame.

    Let's see how you randomly assert that your idea is still tenable :rolleyes:

    PS I avoided the "around the event horizon" part because there is no event horizon as the other part indicated. Your post was self-contradictory.
  18. Aug 2, 2009 #17
    gravity is what causes a black hole to form. Dalespam stated the situation as clearly as possible...
  19. Aug 2, 2009 #18


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    Gravitational gaps? :confused:
  20. Aug 2, 2009 #19
    As your traveling further and further into the gravitational well, there must be huge gravitational gaps among each millimeter taken, especially as your getting closer to the event horizon :smile:
  21. Aug 2, 2009 #20
    Not true, your comparing two different frames.
    From the rockets frame it can send light to Earth, and from Earths frame it can send light to the rocket. That is what you meant, that is the comparable frame. What your saying is like comparing one time frame with a distance frame from a different location.
    More of the same.
    Avoiding part of the argument may make it easier for you to feel right, that doesn't mean you are.
    Obviously experimental data shows that the event horizon is comparable to the mass of the black hole only, but how many fast moving black holes has anyone witnessed? Would be hard to take notice of, considering its mass wouldn't change, only its size.
  22. Aug 2, 2009 #21


    Staff: Mentor

    OK, here is a thought experiment for you to help you see how your idea is self-contradictory.

    Dr. Doom is inertially flying by earth at relativistic speeds (time dilation factor 1E6) in the SS Doomsday. In the ship's frame the ship is 1E3 Schwarzschild radii in size, and in the earth's frame the earth is 7E8 Schwarzschild radii in size. Thus, according to you, the ship is a black hole in the earth's frame, but the earth is not a black hole in the ship's frame. Dr. Doom sends a signal to his Doomsday Device (located on the earth and equipped with a sufficiently broadband reciever) which will destroy earth. In the ship's frame the signal reaches the earth and the earth blows up. (According to you) in the earth's frame the signal cannot cross the event horizon and so the earth does not blow up. Your idea thus leads to a paradox.

    A more succinct way to say this is that if there exists a null geodesic from A to B in one reference frame then there exists a null geodesic from A' to B' in any other reference frame. Again, the bottom line is that you do not become a black hole if you travel fast enough. Full stop. None of your justifications are sound.
    Last edited: Aug 2, 2009
  23. Aug 2, 2009 #22
    Though I appreciate your well thought out experiment, in it's beginning you claimed I said something, which I did not.

    I am not claiming the ship is a black hole in Earths frame but the Earth is not a black hole in the ships frame,
    I am claiming the Earth is a black hole from the ships frame, and the ship is a black hole from the earths frame.

    So in your example, the signal sent from the ship would make it as far as the event horizon as far as you could tell, and from Earth the signal would not escape the event horizon of the ship, atleast until the two collided.
  24. Aug 3, 2009 #23


    Staff: Mentor

    No, your claim is that relative velocity can cause an object to turn into a black hole in some frames due to the increase in relativistic mass. Since the relativistic mass increase is symmetric and since the formation of a black hole depends on density if two objects have different proper densities then it is always possible to formulate such a scenario. In any case, my more succinct phrasing (if there exists a null geodesic from A to B in one reference frame then there exists a null geodesic from A' to B' in any other reference frame) works across event horizons so it includes the case of equal proper density objects.

    You still have nothing supporting your argument. If you wish to proceed further then simply solve the EFE for a fast moving point mass and show the formation of this supposed event horizon.
  25. Aug 3, 2009 #24
    Incorrect. No one sees anyone as a black hole. Black holes are regions of spacetime from which light cannot escape to arbitrary distances. I can send a light ray out from me to arbitrary distance, and no matter what frame you are in, that photon recedes an arbitrary distance from me. Therefore, since the photon recedes an arbitrary distance from in in all frames, I am not a black hole in any of them.

    It makes no difference where the light flash "appears" to come from, and in any case, there is an event where the light flash is adjacent to my body that can be observed by anyone and everyone. Thus the light flash appears to come from me, it does in fact come from me, and does recede to arbitrary distance. I am not a black hole. Period.

    Same as what? Mass does not redshift. It does not blueshift. Even the "relativistic mass" does not do these things. You seem to be confused by the concept of Relativistic Mass, which is a very confusing and therefore deprecated concept. You CANNOT just plug in the Relativistic Mass into, say, Newton's theory of gravity and expect get any kind of physical result. Newtonian Gravity is Wrong. Relativity is not GMM/r^2 with relativistic mass. In Relativity, the source of gravity is stress-energy, a tensor, and the "relativistic mass" doesn't enter into it. What goes into it for the purposes of this problem is mass and energy flux, which gravitate in different ways. Yes, as seen in a frame in which the body moves, its gravity is different than the frame in which it is at rest, but not simply stronger. The gravity induced by the kinetic energy is not purely attractive (in fact, it is mostly skew to the line of velocity, like magnetism) and NEVER, EVER generates black holes.

    YES IT DOES. Because if a photon can escape from the body in its rest frame, then it is observed to escape in ANY frame. If a photon can escape, it is not a black hole.
  26. Aug 9, 2009 #25
    One way you might look at this is by starting with a simple form of the GR equation for gravity-


    where [itex]g[/itex] is gravity, [itex]\rho[/itex] is density and [itex]P[/itex] is pressure.

    Pressure can be described as confined kinetic energy (in fact, the units for P and KE can be described as synonymous) so P in the above equation can also loosely represent KE. If we apply the above to an object that is moving in the direction of x, the Px would be positive, Py would be negative (i.e. negative pressure behind the moving object) and Pz would be zero. This means while the object increases in relativistic mass, its gravity might remain unchanged.

    While pressure and kinetic energy can only be described as synonymous when considering the non-relativistic equation for kinetic energy [itex](E_k=\frac{1}{2} m v^2)[/itex], the same might apply in principle at relativistic speeds.
    Last edited: Aug 9, 2009
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