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Addition of velocities for General Relativity

  1. Feb 3, 2009 #1
    I know about the addition of velocities for SR v = (v1+v2)/(1-v1*v2/c^2). This equations is derived using Lorentz transformations. How do I derive the addition of velocity formula for particles in a non-inertial frame? Can I just add gammas? u = (u1+u2)/(1-u1*u2/u0^2)
    u1 = gamma*v1
    u2 = gamma*v2
    u0 = gamma*c

    Or is this too naive? Thanks.
  2. jcsd
  3. Feb 3, 2009 #2

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    I'm not sure what you mean by adding velocities in a non-inertial frame, as velocities that are constant in an inertial frame are not constant in a non-inertial frame. (This is true even in Newtonian mechanics) I think you need to be more careful in describing exactly what it is you want to calculate.

    That said, your equations are not correct - they have velocities greater than light.
  4. Feb 4, 2009 #3
    A third party observer can observe an object moving to the left at .75c and an object moving to the right at .75c for a separation speed of 1.5c. The left and right object will see themselves moving apart at .96c. Is that what you are asking?
  5. Feb 4, 2009 #4
    Sorry typed the wrong sign in the denominator. But the point is how to calculate the relative velocity in a gravity well (i.e. near big astrophysical body). The equation for SR is found everywhere, but how do we derive the relation for GR? The equations for SR are in a flat spacetime. How do we understand what two particles see as they move near the horizon of a black hole? I have tried following the method of SR by just changing dt to dtau but it doesn't work algebraically. Also the Lorentz transformations also assume flat spacetime. So I am stuck mathematically and philosophically. Hope this clears up everything. Thanks.
  6. Feb 4, 2009 #5


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    What you asked in the OP is special relativity, not general.

    Substituting each velocity for "gamma times the same velocity" looks like a very strange thing to do, and it's definitely incorrect.

    Speeds can be >c if we allow accelerating (or arbitrary) frames.

    The original question isn't well-defined. The first problem is that you didn't really define what velocities you're talking about. For example, the velocity of (the origin of) frame F' in F can't possibly be a relevant quantity. We can however pick a specific point on the world line of the object F'', and define the "velocity at p" of F' in F to be 1 divided by the slope in F, at p, of a curve of constant x' that passes through p. With this definition, it might seem that there's still some hope that we'll be able find a velocity addition law of the form

    [tex]v_{F''F}(p)=v_{F''F'}(p)\oplus v_{F'F}(p)=\mbox{?}[/tex]

    where [itex]\oplus[/itex] is an operation yet to be determined. However, the velocity at p of F'' in F' isn't determined just by the shape of a curve of constant x' that passes through p. We also need to know the shape of a curve of constant t' that passes through p. The problem isn't that we don't have that information. It's that it doesn't appear in any form on the right-hand side of the equation above. So the velocity at p of F'' in F can't be a function of the other two velocities.

    You could try to remedy the situation e.g. by specifying that we're dealing with frames that are specified by a timelike curve and the radar time synchronization procedure (which gives us a way to construct the entire coordinate system just from knowledge of the time axis). Unfortunately, this doesn't work either, since we still need to specify a segment of the t' axis to fully determine the velocity at p of F'' in F'. So the velocity we seek isn't just a function of two other velocities in this (less general) case either.
  7. Feb 4, 2009 #6


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    This is obviously a much more difficult problem. It's hard enough just to figure out how one particle "sees" it's surroundings.
  8. Feb 4, 2009 #7
    That is pretty much what I am talking about, but in the vicinity of a black hole. I was looking at the explaination for the Lorentz velocity transformations and played around with trying to get an equation for the gravity field using dtau. I was hoping someone had an idea of how to start using GR, but I do appreciate the input thanks.

  9. Feb 4, 2009 #8


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    To expand on what Fredrik said, there isn't enough information to give an answer.

    With inertial frames in special relativity, it's sufficient to specify the relative velocity of the two observers and that's enough (with some assumptions about aligned axes) to specify the two frames.

    In general relativity (or even accelerating frames in special relativity), you need more information, as you are free to choose just about any coordinate system you like.

    Consider, for example, two coordinate systems [itex](t,x)[/itex] and [itex](T,X)[/itex] (in one space dimension only to keep it simple -- extend in the obvious way for more dimensions).

    Then we can write, for a particle moving along a worldline with velocity [itex] v = dx / dt [/itex] in the first frame,

    [tex]\frac{dX}{d\tau} = \frac{\partial X}{\partial x} \frac{dx}{d\tau} + \frac{\partial X}{\partial t} \frac{dt}{d\tau} = \left( v \frac{\partial X}{\partial x} + \frac{\partial X}{\partial t} \right) \frac{dt}{d\tau} [/tex] ......(1)

    [tex]\frac{dT}{d\tau} = \frac{\partial T}{\partial x} \frac{dx}{d\tau} + \frac{\partial T}{\partial t} \frac{dt}{d\tau} = \left( v \frac{\partial T}{\partial x} + \frac{\partial T}{\partial t} \right) \frac{dt}{d\tau} [/tex] ......(2)​

    You can then get the velocity in the second frame [itex] V = dX / dT [/itex] by dividing (1) by (2). The answer you get will depend on what exactly the formulas are that express T and X as functions of t and x, assuming you can find such formulas for the problem in question.

    In the case of inertial frames in special relativity, the formulas would be just the Lorentz transform. In the case of one frame accelerating relative to an inertial frame in special relativity, you would use the Rindler coordinate equations. In the the case of a hovering observer and a freefalling observer near a black-hole, you could use the Schwarzschild-to-Kruskal-Szekeres transformation (I think?).
  10. Feb 4, 2009 #9


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    If you were adding velocities at the same event then I would think you would just transform to your local Minkowski space and use the usual velocity addition formula. If you are adding velocities at different events then I would think that you couldn't do it because of parallel transport issues etc.
  11. Feb 5, 2009 #10


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    As with many things in relativity, it depends what you mean by "velocity". My analysis in post #8 applies to "coordinate velocity", a notional value [itex]dx/dt[/itex] that is highly dependent on your choice of coordinates, and which needn't bear much relationship with what some people call "physical velocity" i.e. velocity measured by a local, free-falling observer using locally-inertial coordinates. If you are talking about velocity in this latter sense, then DaleSpam is right.

    The "physical" velocity of light is always c, but the coordinate velocity need not be.

    (I'm sure DaleSpam knows all this, I'm pointing it out for the benefit of other readers.)
  12. Feb 5, 2009 #11
  13. Feb 5, 2009 #12
    I was satisfied with DaleSpam's post but now you have me thinking. :smile: Am I correct to say that as long as you are observing from the same reference frame all laws will apply and c is c? If you could observe from outside the gravitational effect of the black hole the light would appear to travel slow?
  14. Feb 5, 2009 #13


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    Only in the infinitesimally small local inertial rest frames of observers in freefall is the speed of light c in GR...in any coordinate system in a large region of spacetime where curvature is non-negligible, the laws of physics will look different from the laws seen by inertial observers in SR, and light may move at coordinate speeds other than c.
  15. Feb 5, 2009 #14
    Lets make sure I understand what you are saying. In SR the curvature of space can be consistent with speed over long distances. In GR the gravitational effects change over short distances so the reference frame is small.
  16. Feb 5, 2009 #15


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    There's no curvature in SR, so your reference frame will be valid everywhere. For freely-falling observers a SR mini-space exists around them, the range of validity depends on the curvature there.

    The tangent and cotangent vector spaces of a worldline can be used as frame. Provided it is not rotating this frame will be flat enough over a certain region to be SR like.
  17. Feb 5, 2009 #16


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    You can have a large coordinate system in GR, but it just won't be an inertial coordinate system where the usual SR laws of inertial frames apply. This is related to the equivalence principle which relates the laws seen by a freefalling observer in a small region of curved GR spacetime to the laws seen by an inertial observer in SR spacetime--see this article for a good summary (the last section of the article explains why it only works in arbitrarily small regions of GR spacetime).
  18. Feb 5, 2009 #17
    I probably should have said "In SR the the lack of curvature of space can be consistent with speed over long distances. Thanks for your reply. Things slowly come into focus.
  19. Feb 5, 2009 #18
    Last edited by a moderator: Apr 24, 2017
  20. Feb 5, 2009 #19
    In a local free falling frame a particle approaching a black hole crosses the event horizon, disappears, and is torn apart as it approaches what is believed to be a singularity of immense gravitational strength. From the perspective of an outside stationary observer at great distance ,the particle appears to never reach the event horizon.

    Locally everything seems approximately flat like special relativity; over greater distances curvature makes things appear different as time dilates and space contracts (General relativity).
    Last edited: Feb 6, 2009
  21. Feb 10, 2009 #20
    Sorry, that I took a hiatus. I was reading to get a handle on my question. Okay, the general idea is that general relativity gives us ds^2 = g_{ij} dx^1 dx^j where g is the spacetime metric. If within the metric we set m=0, or r -> infty, then we get the flat spacetime which gives the gamma from SR that we all know and love.

    So I am thinking that maybe the first thing is to see if there is a way to derive Lorentz transformation from the Minkowski metric. However, this very difficult because I need to find the Jacobian for the transformation, g' = J^T g J. My thinking is that if I use this equation I can solve for the Jacobian elements and see if it gives the Lorentz transformation equations. If it doesn't work then that is a real problem. Because the Lorentz transformation is derived using simple algebra assuming the constancy of "C", but shouldn't the Minkowski metric also contain that info?

    Hopefully, my pencil explodes before my head does. Thanks for the replies they have provided great inspiration.
  22. Feb 10, 2009 #21


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    I don't know if you can derive the Lorentz transformation from the Minkowski line element alone. If you draw a 1+1 space-time diagram, the rest frame has a vertical world-line ( I have t running vertically). A boosted world-line is inclined off the vertical. To make the boosted world-line the rest frame, one performs a hyperbolic rotation to make it vertical. This makes the original world-line inclined. A rotation in the space-time plane is a Lorentz transformation.
  23. Feb 10, 2009 #22
    I see where you are going with that. I tried something like that, but using
    [tex]ds' - cd\tau' = \lambda (ds - cd\tau)[/tex] and
    [tex]ds' + c d\tau' = \mu (ds + cd\tau)[/tex]
    I then did the usual algebra to get Lorentz type transformations. By defining a four-velocity u as \frac{ds}{d\tau} which is defined as (u_x,0,0,c). My "transformation" equation simplified to one, thus proving that the reference frames move through time at the speed of light. Duh.

    So I am hoping that the matrix approach yields better results.
  24. Feb 10, 2009 #23


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    The transformation I referred to can be written ( [itex]\beta[/itex] is the rapidity, which is an angle)

    [tex]\left[ \begin{array}{cc}
    cosh(\beta) & sinh(\beta) \\\
    sinh(\beta) & cosh(\beta) \end{array} \right][/tex]

    which is like a rotation except it's symmetric and has hyperbolic trigs.
  25. Feb 10, 2009 #24


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    If you're going to derive the Lorentz transformation from the Minkowski metric, I think you're going to have to understand the concept of an "isometry" (and also "pullback" and "push-forward"). What you have to show is that the group of isometries of the Minkowski metric is isomorphic to the group of functions of the form

    [tex]x\mapsto\Lambda x+a[/tex]​

    from [itex]\mathbb R^4[/itex] into itself, where [itex]\Lambda[/itex] is a linear operator that satisfies


    This would prove that two common definitions of the Poincaré group are equivalent. I can show you how to do that if you want to (I LaTeXed this proof a year ago, but I might have to clean it up a little), but I don't see what this has to do with what you asked earlier in this thread.
  26. Feb 11, 2009 #25


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    I should think that it would be reasonable to start by writing the geodesic equation for your two particles and then calculating the geodesic deviation. See Gravitation and Spacetime by Ohanian and Ruffini section 6.6 for the details.
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