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2 Masses accelerating to eachother

  1. Dec 3, 2007 #1
    Suppose we have two equal masses, say at least the mass of our sun separated by a distance of 1 ly and devoid of all outside forces. These objects would keep accelerating towards eachother until they approach the speed of light and as their mass increases their acceleration decreases but however due to their increased mass the acceleration due to gravity increases and it keeps going faster and faster towards eachother.

    What am I missing that would counteract this.
     
    Last edited: Dec 3, 2007
  2. jcsd
  3. Dec 3, 2007 #2
    sorry, but why would they approach the speed of light? as the body's mass increases it takes more to accelerate the body, causing no real acceleration, just a constant attraction- am i right in thinking this?
     
  4. Dec 3, 2007 #3

    HallsofIvy

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    No, his point is a valid one- though mistaken. As the body's mass increases, it requires more force to maintain constant acceleration. However, his point is that since gravitation force also involves mass, the gravitational FORCE itself increases.

    Denton, there are others here who are better qualified than I am, but I think the difficulty is that you have to set up a coordinate system and "observe" from that coordinate system. If you take one of the masses as your "frame of reference", it is, by definition, motionless in that frame of reference. You apply the Lorentz transformations to one body only.
     
  5. Dec 3, 2007 #4

    Dale

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    If two bodies are initially at rest wrt each other and fall towards each other then their velocity will always be less than the escape velocity at that separation. It will not generally (except in the case of black holes) approach any terribly large velocity let alone velocities approaching c.
     
  6. Dec 3, 2007 #5

    LURCH

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    captain Obvious asks, "What about two Black Holes, then? Their escape velocity is greater than c, so what would be expected to happen as they approach one another? Aslo, their initial velocity doesn't need to be at rest relative to one another; how 'bout if they were moving rapidly toward each other to begin?"
     
  7. Dec 3, 2007 #6
    The escape velocity at the surface of a gravitational body is equal to the final velocity of a test particle dropped from infinity to the surface of that body. In principle and by definition a test particle falling from infinity to a black hole would impact the surface (event horizon) of the black hole at light speed. However all motion is relative to the local speed of light in the gravitational field. From the point of view of a distant observer time and motion slows down deep in a gravitational well. Near the event horizon the speed of light aproaches zero. A consequence of this is that a falling object moving at relativistic speeds actually decellerates as it approaches an event horizon. This is required to prevent a falling object exceeding the local speed of light. Another consequence is that a falling photon also decellerates as it approaches the event horizon (from the point of view of a distant observer).

    So, yes .. an object falling from infinity to a black hole would achieve the local speed of light but it just so happens that the local speed of light is zero (or undefined).

    An object with a high initial velocity towards the black hole would also be subject to decalleration at some point in its fall to be consistent with nothing exceeding the local speed of light.

    Warning..this is my interpretation as layperson armchair physics enthusiast ..don't quote me or use this in your homework ;)
     
    Last edited: Dec 3, 2007
  8. Dec 3, 2007 #7

    pervect

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    Basically, yes, in the sense that the velocity of the infalling particle would approach 'c' as measured by a stationary observer near the event horizon of a black hole.

    Note, however, that there really isn't any such thing as a stationary observer at the event horizon. So 'c' is a limiting velocity as one considers a sequence of stationary observers who get closer and closer to the event horizon.

    In the sense that the rate of change of the Schwarzschild 'r' coordinate with respect to the Schwarzschild 't' coordinate appraoches zero (for light and also for an infalling object), this is also correct. However, this sort of velocity has very little physical significance.

    A good observation, because someone who took some of these remarks about deacceleration too seriously could wind up seriously confused :-(. Using physically meaningful defintions of velocity, there is no "deacceleration" of objects approaching a black hole, i.e. a black hole's gravity always attracts objects, it never repels them.
     
    Last edited: Dec 3, 2007
  9. Dec 3, 2007 #8
    Good point
     
  10. Dec 3, 2007 #9

    JesseM

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    Gravitation in general relativity is not simply a function of relativistic mass--if it were, then an object would become a black hole at sufficiently large velocities in my frame. But even if we imagine a universe where the gravitational force is [tex]\frac{Gmm}{r^2 * (1 - v^2 / c^2 )}}[/tex] (i.e. plugging [tex]m(v) = m/\sqrt{1 - v^2/c^2}[/tex] into the Newtonian gravity formula for two objects of mass m), it is not at all obvious to me that the velocity would approach c in the limit as the distance between two objects approached zero...it really depends on what function you'd get when you solved the differential equation:

    [tex]a(t) = \frac{Gmc^2}{(4x(t)^2 * (c^2 - v(t)^2)}[/tex]

    ...where x(t) is the distance from one mass to the midpoint between them (so r(t) = 2x(t)), and where v(t) is the first derivative of x(t) with respect to t, and a(t) is the second derivative of x(t) with respect to t. Once you had this function, you could check if v(t) would approach c in the limit as x(t) approached 0, for any choice of the initial value of x...it might very well not do so, since there is only a finite time before they hit each other.
     
  11. Dec 3, 2007 #10
    I agree that the black hole's gravity never repels. A stationary object will always accelerate towards the black hole when released. However it would appear that an object released from a greater height would experience a lesser acceleration than a recently released object when they are at the same altitude from the point of view of a distant observer. In other words the gravitational acceleration at a given altitude appears to be velocity dependant in my (maybe confused) interpretation.


    This quote

    "As determined by the B-observer a light signal emitted from A with the initial velocity (14) will be slowing down (with − g) as it ”falls” in the Earth’s gravitational field and will arrive at B with a final velocity exactly equal to c. Therefore an observer at B will agree with an observer at A that a light signal will accelerate with an acceleration g on its way from B to A and will decelerate while ”falling” in the Earth’s gravitation"

    is from an ARXIV document http://arxiv.org/PS_cache/gr-qc/pdf/9909/9909081v7.pdf (where A is an observer above B in a falling elevator)

    It seams reasonably authoritive. Would anyone care to comment on its accuracy and is it all helpful in this discussion?
     
  12. Dec 3, 2007 #11
    In a 1948 letter to Lincoln Barnett, Einstein wrote
    "It is not good to introduce the concept of the mass M = m/(1-v2/c2)1/2 of a body for which no clear definition can be given. It is better to introduce no other mass than `the rest mass' m. Instead of introducing M, it is better to mention the expression for the momentum and energy of a body in motion." See http://math.ucr.edu/home/baez/physics/Relativity/SR/mass.html

    It seams Einsten wanted to distance himself from the concept of relativistic mass. (ie mass increasing with increasing relative motion). One reason relativistic mass became a problem is that a large body could appear to be a black hole to an observer with relative motion while it would appear to be perfectly normal to an observer at rest with the body. One way out of this "paradox" would be to conclude that only inertial mass increases with relative motion but not gravitational mass. This introduces another problem as it is implicit in the equivelance principle which is the foundation of GTR that inertial and gravitational mass are exactly equivelent. To find out more do a search for the Eötvös experiment. The solution by Einstein and various text books is banish the concept of relativistic mass. Einstein wished only to discuss the energy and momentum of a body. However, in the general theory of relativity, energy has an effect on the spacetime geometry (gravity) due the mass energy relationship (E=Mc^2) so the matter is not entirely resolved or clear IMHO.
     
  13. Dec 3, 2007 #12
    Denton I haven't calculate but I suppose that's impossible as it requires an initial infinite distance of the bodies and an infinite time as well.
     
    Last edited: Dec 3, 2007
  14. Dec 3, 2007 #13
    Right, I forgot completely that both would have relativistic effects, but only one would when observed by the other. However would it not yeild the same results?

    I have thought so too.

    So you're saying that the concept of relativistic mass increase is just an illusion or doesn't exist?
     
  15. Dec 3, 2007 #14

    pervect

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    Why? It's just an arxiv preprint, it's never been published in a peer-reviewed journal. There are lots of strange and fringe papers "published" on arxiv that never see the light of day anywhere else.

    The name Petkov is familiar - ah yes, I remember now, he's a philosopher with some rather *strange* ideas about relativity, most of which (including this paper) have not been accepted for publication.

    Definitely not "authoritative" as far as physics goes, by any reasonable definition of "authoritative".
     
  16. Dec 5, 2007 #15
    So can i get confirmation that relativistic mass increase does not affect gravitational strength?
     
  17. Dec 5, 2007 #16

    pervect

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    It is incorrect to replace the mass M in Newton's gravitational formula GM/r^2 with relativistic mass. But it is also incorrect to say that the gravitational field of a moving particle is the same as a stationary one. For instance, we know that if we have a pressure container containing an ideal gas made out of atoms*, and we heat up the gas so that the atoms move around faster, that the mass M increases due to the added heat energy. But the rest mass of the atoms doesn't change when we heat up the mass, the only difference is that the atoms in one case are moving faster, and in a different case are moving more slowly.

    There is unfortunately no easy way to set up an experiment to measure the "gravitational field" of a single moving particle as most people think of the term gravitational field. What most people would envision by the term "gravitational field" is, I think, the amount of force needed to hold a body "stationary" while another, massive body, "whizzed by" - the familiar formula from Newton.

    The problem is, how do you define "holding a body stationary"? You might think to use the fixed stars as a reference, but it's clear that the light from them will be distorted by the passing body and its changing gravity. So simply keeping the fixed stars in a fixed observed position won't work. In fact, it's not particularly clear how to solve this problem, or if it has a solution. (I'll skip over some not particularly noteworth proposals that are not peer-reviewed).

    There is a fairly simple way around this issue, which is to look at tidal gravity, the rate of change of the gravitational field. This has an exact physical and mathematical interpretation that is quite straightforwards.

    The tidal gravity around a moving mass is not, however, spherical - it's compressed so that the field is strongest transverse to the direction of motion.

    In the limit of a very rapidly moving particle, the gravitational field (defined as the tidal gravitational field) of a moving mass becomes that of a gravitational wave. This is known as the Aichelberg-Sexyl solution. It's rather similar to how the electric field of a rapidly moving charge approaches an impulsive electrical wave. (But note that to be precise we are really considering a different sort of field in the gravitational case as I mentioned earlier, a tidal gravitational field)

    For the (highly technical) details, see for instance http://arxiv.org/abs/gr-qc/0110032

    If you're not familiar with the electric field of a rapidly moving charge, you might want to look at

    http://www.phys.ufl.edu/~rfield/PHY2061/images/relativity_15.pdf

    for the detailed equations see
    http://www.phys.ufl.edu/~rfield/PHY2061/images/relativity_14.pdf

    *atoms are simpler to deal with than molecules and the associated gas is more nearly ideal.
     
    Last edited: Dec 5, 2007
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