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Can objects 'fall faster than c' in a gravity field?

  1. Jun 10, 2008 #1
    Can objects 'fall faster than c' in a gravity field??

    Here is my question, which has been eating away at me...

    According to GR, objects in deep space can receed from one another at faster than c, only given that they are sufficiently far apart. This is because Hubble velocities faster than c are allowed because the velocities are not peculiar in their local reference frame. This seems to be the opinion of most cosmologists (Australian camp). If this is true, and if the expansion is not a true velocity, restricted by c, but instead a result of geometrical expansion, then is gravity itself the same? Do objects bound to reference frames falling into black holes approach, and in fact, exceed c?? Is gravity also just geometrical distortion, not associated with a 'peculiar' velocity, and therefore not limited by c???

    I apologize if my language is not correctly stated. I am an amateur. Any input is greatly appreciated.
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  3. Jun 11, 2008 #2


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    Only in the inertial coordinate systems of flat SR spacetime, or "locally inertial" coordinate systems in curved GR spacetime (i.e. coordinate systems in a very small region where the curvature approaches zero), is the speed of light guaranteed to be c...if you pick a non-inertial coordinate system, the speed of light can be anything. Even in flat SR spacetime with no gravity whatsoever, if you choose an accelerating coordinate system then the coordinate speed of light in this system may be something other than c, and may even vary from one location to another.
  4. Jun 11, 2008 #3
    Hi Jesse, thanks for the feedback. Just to clarify... GR curvature is not limited by c?? Meaning, if an objects falls into an sufficiently strong gravity field, according to GR, it's "falling velocity" is not necessarily limited by c?? I think that is what I'm hearing. Sorry for the laymens terms.
  5. Jun 11, 2008 #4
    You are thinking of this backwards, look at how the Shapiro Delay works.
    http://en.wikipedia.org/wiki/Shapiro_delay" [Broken]
    As observes in a frame far away from the signals passing near the massive sun we see light slowing down not speeding up or going faster.

    Remember that is only WRT our frame of reference and only while the light is within the nearby influence of the massive sun does the speed of light slow down to allow for the Shapiro Delay.
    From the frame of reference nearby the sun where the light is actually traveling; time is “slower” and distances “shorter” WRT to frames not under the same gravitation influence. But the rules of physics and the speed of light in that frame is still “c”.

    In the same manner from the perspective of that frame near the sun, distant frames like us (as observers being observed) are seen as having “faster” time and “longer” distances WRT the nearby sun reference frame – thus distant light (near us) would be observed as traveling faster than “c” from that frame of ref.
    But the actual speed in the local space where light is traveling is still of course “c”.
    Last edited by a moderator: May 3, 2017
  6. Jun 11, 2008 #5
    Ok, I think I'm getting it. This is like the astronaught with the clock falling into the blackhole. We watch his clock come ever closer to 12:00 but we never get past that last 11:59:59, right? The clock becomes a 'still' image and just fades away. Then, what you are saying is if we watch him execute a speed-of-light experiment, we see his results come in slower than if we were to commit the same experiment locally, near us. Therefore, c is slower, for US, as viewed from outside, looking into HIS frame.

    So, does that mean that his motion, quickly approaching the black hole, also seems to slow down? As if he never quite gets there? Or, does his 'still' and fading image simply accelerate away from us very quickly toward the center. Will he witness the "end of the universe"?

    Again, I realize this is all relative, so I might be missing the point. This will be my last post. Thanks all.
  7. Jun 11, 2008 #6


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    In the local inertial coordinate system of a freefalling observer right next to the object at some moment, the object's velocity at that moment will always be c or less. But outside of inertial coordinate systems, velocity can be anything you want it to be, depending on what coordinate system you choose. Again, this has nothing to do with gravity, it's true of non-inertial coordinate systems in flat SR spacetime too.
  8. Jun 11, 2008 #7


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    When we say that galaxy A and galaxy B are moving apart with speed v, it really means that "the proper distance between A and B along the shortest possible path in the hypersurface of constant time coordinate, changes by v units of length for each unit of time that we change the time coordinate". I don't know if there's any other situation in GR except for cosmological expansion where a speed defined this way can be >c.
  9. Jun 12, 2008 #8


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    If starter of the topic has abandoned it maybe I can ask his question in a different manner.

    If for example radar signal is sent to some massive object and received after reflecting from surface time measured would be more than in the same situation without gravity.
    However the time necessary for forward trip is less than for backward trip, right?
    I will assume that answer is yes and will ask further (if answer is no then of course next question is pointless).
    Is it possible that for outside observer calculated speed for radar signal at some moment of forward trip exceeds c?
  10. Jun 12, 2008 #9
    No, the 'inside' observer calculates the "exceeds c" effect WRT that inside RF.

    Also there is no forward or backward - the Shapiro Delay works by sending signal close to a mass with return passing the same way compared to the same trip without the mass. Start with the Wiki link.
  11. Jun 12, 2008 #10
    Hi Again. I know I was finished but I have a new question regarding this problem. I've heard interpretations of GR and black holes which say that if an observer falls into a black hole, the whole history of the universe will display itself to them before they hit the middle. This view seemed to be founded on the idea that time slows for a traveler who is accelerated to near light-speed. Here is my question...Even though time slows for this observer (therefore the space outside the blackhole seems to speed up to them), does their arrival at the center of the blackhole take an eternity? Or, Do they immediately hit the center? Here is a mind experiment to demonstrate...

    Imagine a black hole, with the mass of 1 billion suns, sitting far away from any other objects. Now this hypothetical blackhole is, you guessed it, made of anti-matter (bear with me). Now, by some unknown event 1 billion suns all tavel toward this black hole from all directions. They are stable stars, shinning light, and they are all made of regular matter. As observers we wish to witness all of them simultaneously colide with the black hole. There is zero spin with this entire scenario (I understand this is impossible). Here is the question...
    As we watch the closing sphere of suns fall into the gravity-well of the blackhole, do the suns experience "the entire life-time of the universe" and therefore we, as observers outside, never see them colide with the anti-matter, or, Do they all smash the center and anihilate the black hole right in front of us at that moment?
    This would, I assume, unleash enormous amounts of radiation (coliding matter with anti-matter). Would this radiation be able to escape it's own gravitational field??
  12. Jun 12, 2008 #11


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    An object falling into a black hole will hit the singularity within a finite time (measured by its own clock) after crossing the event horizon. For a black hole with a small mass (only a few solar masses), this is typically a very short time. I did the calculation as "homework" once, but I don't remember the result. I think it was something like 10-5 seconds.

    There is no difference between a black hole that formed from collapsing anti-matter and one that formed from collapsing normal matter. A black hole that isn't rotating has no other properties than its mass and its total electric charge, so they are a lot like (enormous) elementary particles.

    To an observer at a constant distance from the black hole, those falling stars would appear to never reach the event horizon (if I remember what I read correctly, I haven't calculated this myself). They will seem to slow down and stop at the event horizon. Photons from those stars reaching the external observer will be more and more redshifted (less energetic) and arrive less and less frequently.
  13. Jun 13, 2008 #12
    Thanks Fredrik. That's the best explanation Ive heard of it. Thanks for the clarity. I'm curious to hear what you would say about the black hole being made of anti matter and the resultsas witnessed by the stationary observer. Cheers.
  14. Jun 13, 2008 #13


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    Of course we can not talk about measured time for one way trip but only for round trip.
    Therefore question was about calculated speed, not about measured delay.

    As I understand calculated speeds will depend on reference frame that is chosen for calculations. I don't know if there is definition how local reference frame can be extended to include events in distant places with different gravity. It seems natural for me to take reference frames that are analogous to surface reference frame with adjusted time (matching observers rapidity of time).
  15. Jun 13, 2008 #14
    I guess that is the source of my confusion with this problem. I'm guessing you simply can't have any global understanding of velocity in the first place, therefore to ask "can an object fall faster than light" is not appropriate, hence it's all relative. I was hoping to rid myself of that by introducing anti-matter, somehow exposing the hidden result within the black hole (I believed that the anti-matter/matter mixture would instantaneously vanish all matter within the blackhole, reducing it to a measuable explosion of photon radiation which could be understood and witnessed by the outside observer, hence evaporating the blackhole at a measurable moment by any outside and distant observer. On second thought, the energy from the anti-matter/matter mixture, even if reduced to photons, would still be massive and therefore maintain a massive gravitational field, now twice what it was before. Is that correct?). Thanks all for the replys. - HJ
  16. Jun 14, 2008 #15
    IMH non expert opinion that is correct.

    Also, don't forget that while the shell of stars is collapsing towards the black hole event horizon, the event horizon is expanding, so the time to arrive at the event horizon may not be what is calculated by assuming a static target.
  17. Jun 14, 2008 #16
    It would be twice that was before due to the "matter" involved and also twice that was before due to the "anti-matter" involved, so the net after would be the same as the total before.
  18. Jun 15, 2008 #17
    The Schwarzschild radius of the Earth is about 5.9E-13 meters. That does not mean there is a small black hole at the centre of the Earth because not all the mass of the Earth is contained within that radius. Although the size of a black hole is usually defined as the Schwarzschild radius a better definition of the size of a black hole is the radius of the event horizon (where dtau/dt=0) which just happens to coincide with the Schwarzchild radius (Rs) for a static stable and non rotating black hole.

    In units where G=c=1 if we start with a black hole of radius R=1 and mass M = Rc^2/(2G) =0.5 and add a shell of stars also of mass 0.5 at say radius =10 then the new Rs is 2 but there is no longer any event horizon anywhere within the combined mass because not all the mass is contained within the Rs. At this point the central black hole is no longer a black hole because it does not have an event horizon. When the shell has collapsed to about Rs*9/8 depending on mass distribution, the event horizon reappears at the centre of all the combined mass and then rapidly moves outwards as the outer shell collapses towards the new Rs.

    The equation for the interior Schwarzchild solution is:

    [tex] dtau/dt = 3/2\sqrt{1-R_s/R_m}-1/2\sqrt{1-R_sR_o^2/R_m^3}[/tex]

    Rs is the Schwarzchild radius,
    Rm is the radius that includes all the mass,
    Ro is the radius where the gamma factor is being measured

    and it is assumed the observer is at infinity (as in the exterior Schwarzchild solution).

    The location of the event horizon (where dtau/dt=0) is given by:

    [tex] R_o = \sqrt{9R_m^2-8R_m^3/R_s}[/tex]

    and it can be seen if total mass has a surface radius Rm that is greater than Rs*9/8 that the event horizon is imaginary and so does not exist. In short the gravitational gamma factor at a given radius is not only determined by the mass below that radius but also by the mass above the radius and the presence of the spherical shell beyond the black hole effectively destroys the property of the black hole having an event horizon so technically it is no longer a black hole until the shell collapses and new black hole is formed with twice the radius of the original black hole without the mass of the additional outer shell.

    The combination of the interior and exterior Schwarzschild solutions allows the formation of a black hole to be analysed dynamically and by analysing the gravitational acceleration at various radii it seems that a black hole with a singularity of infinite density is unstable and does not in fact form. It seems that the final stable form of a black hole is a thin hollow shell just outside the Schwarzschild radius. Such a black hole would look just like a singularity black hole. Radiation from the physical surface would be highly redshifted and look just like Hawking radiation. Its entropy is contained in the physical surface area. The information paradox is not a issue because the information is never lost to another universe but is contained in the physical surface area of the physical shell black hole.
    Last edited: Jun 15, 2008
  19. Jun 16, 2008 #18
    I have no clue what your objection is.
    Herbascious J was saying the a Black Hole where significant portion of the internal mass in normal matter was to interact with an equivalent amount of mass in the form of anti-matter and claiming that the destruction of the matter through conversion into energy would give a “gravitational field, now twice what it was before” as though the mass of the black hole was twice what it was before. All I said was that the mass inside the black hole IS twice what it was before – in order to get normal matter of amount mass “1” to interact with anti-matter, a mass “1” of anti-matter must be added into the Black Hole. The matter anti-matter interaction does not increase or change the total mass of the Black Hole as observed by its gravitational field.
    By adding anti-matter stars into the black hole his problem has simply added 1 +1 =2 for total mass that has doubled. The effective mass inside the black hole is equal to all the mass added to the black hole regardless of the type of mass (matter or anti-matter).
    If hypothetically you where to added antimatter mass to a black hole the matter anti-matter reactions, unseen to anyone outside, would have no effect on the observed new total mass of the black hole.
    Are you somehow disagreeing with 1+1=2 in this example?
  20. Jun 16, 2008 #19
    I am not disagreeing with your statements in this post. I think there has been a misunderstanding about what Herbascious was saying. I believe all Herbascious meant to say is that the mass of the black hole after adding an equivalent mass antimatter would be twice what it was before the anti matter was added. I do not believe he meant to imply that when matter and antimatter are both present that the annihilation process someone doubles the mass. He was considering that maybe after the annihilation process there would be no black hole left because all the mass might have escaped in the form of radiation, but I think he quickly realised that the radiation energy would have a gravitational field equivalent to the total original mass and would not escape.

    I mistakenly assumed you were saying that when a shell of distant stars is added, with a mass equal to the mass of the black hole that the mass of the system would instantly double (true), that the Schwarzschild radius of the total system would instantly double (true) and that the radius of the event horizon of the black hole (or system) would instantly double (false). I was just pointing out that the event horizon moves dynamically, first dissapearing when the distant mass shell of stars is added to the system, then reappearing at the centre and finally settling down to double the original radius when the shell has collapsed. Sorry for any confusion.
  21. Jun 17, 2008 #20
    Hi Kev, this is correct. I literally realized this mistake mid-typing. I was attempting to construct an hypothetical event where the evaporation of the blackhole could be measured and therefore skirt-around some of the hidden aspects of the measurement of time locked within the blackhole, but I guess there is not tricking GR! Thanks for the great responses, this has been my first post at PF and it's been great reading the feedback. Cheers!
  22. Jun 17, 2008 #21


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    Doesn't the no-hair theorem imply that, at least from the point of view of observers outside the horizon, there can be absolutely no way to distinguish a black hole which formed out of matter and a black hole which formed out of a collection of antimatter with equal mass and charge? (for example, a black hole formed from a collection of protons vs. a black hole formed from a collection of positrons with the same mass) I suppose this theorem is classical so it's not certain it would still apply in quantum gravity...
  23. Jun 17, 2008 #22
    I think that is the conclusion we have all reached in this thread. That a black hole composed entirely of antimatter would be indistinguishble from one composed of normal matter.
  24. Jun 18, 2008 #23
    The above statement has intrigued me. I'm curious... Would the collapsing shell of matter (stars), falling into the blackhole, momentarily 'relax' the internal pressure of the black hole, simply due to the presence of the approaching matter outside of the event horizon. You mention that the event horizon might disapear. Does this mean that the energy trapped within the black hole might begin to escape in some way, due to the powerful opposing gravity field (stars)? If these stars have not yet collapsed, can radiation slip past them, allowing the entire system to loose energy from some kind of outward push? Considering the anti-matter scenario we have been suggesting, could there be anihilations that are now observable, even if only brief?
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