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Gravity and time and space translation asymmetry

  1. Jul 15, 2014 #1
    Suppose we have a test mass located at some point above the earth. It is allowed to free fall down toward the earth. It passes through a closed surface. Prior to entering the surface it has an initial momentum pi and initial kinetic energy kei. After leaving the closed surface it has acquired a greater momentum pf and a greater kinetic energy kef. Clearly, the momentum and kinetic energy of the test mass has not been conserved.

    Could we choose one of two explanations for this? The first one is there was no time and space translation symmetry during the transit of the test mass through the closed surface, and therefore, it follows that momentum and energy were not conserved.

    Or during the transit within the closed surface the gravitational field imparted an impulse on the test mass and simultaneously did work on the test mass, and therefore, momentum and energy were conserved when we include the effects of the gravitational (curving of space-time) field.

    If the second explanation is the correct one, how can this be if the test mass is weightless as general relativity asserts? That is, there was no force acting on the test mass during its transit within the closed surface, yet it still acquired momentum and kinetic energy.
     
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  3. Jul 15, 2014 #2

    PeterDonis

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    What kind of closed surface will do this? I don't understand what you're describing here. Are you just trying to say that an object free-falling under gravity increases its momentum and kinetic energy as it falls?
     
  4. Jul 15, 2014 #3
    Assume the closed surface is an imaginary sphere. The problem is addressing what happens to the momentum and kinetic energy of the test mass while it passes through the interior of the sphere. I am framing this question in terms of a continuity relation or equation. The placement of this sphere is arbitrary, as long as it is above the surface of the earth and the test mass is influenced by the gravitational field of the earth within the volume space of the sphere.

    Yes, the kinetic energy and momentum of the test mass increases as it free falls to the earth. But the question deals with a small portion of space and time of the test mass's trajectory toward the earth. I am asking what is the correct way to describe the kinematics and/or dynamics of the test mass for the incremental time and space displacement within the sphere.

    The essential question can be watered down to this: Is the explanation that there was no time and space translation symmetry during the transit of the test mass within the sphere sufficient to explain why the momentum and kinetic energy of the test mass were not conserved. That is, is the explanation of the effect of the gravitational field on the test mass redundant, a post hoc explanation to force the conservation of momentum and energy.

    Remember, general relativity asserts masses in free fall are weightless, that is, there is no force on a mass in free fall. So how can the momentum and kinetic energy of the test mass increase, if there were no forces acting on the test mass to impart an impulse and to do work on the test mass? My understanding is it's because the background of space and time has ceased to have the properties of time and space translation symmetry. And this is because the space-time properties have been "curved" by the presence of the mass-energy of the earth as described by the field equation of gr.
     
    Last edited: Jul 15, 2014
  5. Jul 15, 2014 #4

    WannabeNewton

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    There is no need for all the extraneous setup and detail in your post. It is all irrelevant. As Peter noted your question is simply one of how the gravitational field increases the kinetic energy and momentum of freely falling particles in the framework of GR. First of all, kinetic energy and momentum are frame-dependent quantities. It only makes sense to talk about their increase or decrease relative to some extended reference frame in curved space-time. Then the answer is simple because the freely falling particle accelerates relative to an extended reference frame ##O## fixed with respect to spatial infinity, assuming an asymptotically flat space-time, and so its kinetic energy and momentum clearly change at each point in ##O## that the particle travels through. In the particle's own rest frame ##O'##, that is the freely falling frame centered on the particle, it has zero momentum and zero kinetic energy and its total energy is just its rest mass.

    What causes the freely falling particle to accelerate relative to ##O##? The absolute acceleration of the frame ##O## manifests itself as a fictitious "acceleration due to gravity" acting on the freely falling particle in the opposite direction as observed in ##O## but the true acceleration is that of ##O## due to it not being a freely falling frame. In space-times with time translation symmetry, the natural choice for ##O## is the one defined by the time-like Killing field.
     
    Last edited: Jul 15, 2014
  6. Jul 15, 2014 #5

    PeterDonis

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    Ok, that makes it clearer. The closed surface is irrelevant; all you really need to say is what's in the above quote.

    It depends on what you want to describe those dynamics relative to.

    If you're describing those dynamics relative to a local inertial frame, the test mass is moving in straight line at a constant speed, just as in special relativity. It feels no force and experiences no acceleration, and its momentum and energy do not change.

    If you're describing those dynamics relative to a global (non-inertial) frame in which the Earth is at rest, then the test mass increases its kinetic energy and decreases its potential energy as it falls; its total energy remains constant. This works basically the same as it does in Newtonian gravity. (Note that the concept of "potential energy" is not always well-defined in GR, but this is one of the special cases in which it is.)

    You may ask, what about the test mass's momentum? See below.

    This is partly right. In general, spacetimes in GR do not have to have any kind of symmetry at all, and in such a general, non-symmetric spacetime, there is no meaningful concept of energy or momentum conservation in a global sense. (Energy and momentum are always conserved locally in GR, because the covariant divergence of the stress-energy tensor is always zero. But that's a separate issue from the one we're discussing.)

    However, it just so happens that, if we idealize the Earth as a static, spherically symmetric mass, the spacetime around it *does* have time translation symmetry. That is why there *is* a meaningful concept of energy conservation relative to a frame in which the Earth is at rest, as I described above. (Basically, the time translation symmetry lets you define a useful notion of "potential energy", which then lets you treat kinetic + potential energy as conserved, as you can in Newtonian gravity.)

    But there is *no* space translation symmetry in the spacetime around the (idealized) Earth, so there is no meaningful concept of momentum conservation. That's why there is no way to fudge up a global concept of "momentum" that is conserved as the test mass falls. (Note that there *is* spatial rotation symmetry in the spacetime, because the idealized Earth is spherical, so there is a meaningful concept of *angular* momentum that is conserved for a freely falling test mass. In the simple case of a test mass falling radially, the angular momentum is zero.)
     
  7. Jul 15, 2014 #6

    ChrisVer

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    Another way I think, is to look at the geodesics which the particle will follow.

    [itex] \ddot{x}^{\mu} + \Gamma^{\mu}_{\nu \sigma} \dot{x}^{\nu} \dot{x}^{\sigma} =0 [/itex]

    From this you can also see that there is no mass need for a particle to feel "gravity", as long as the gravity itself is affected by the spacetime geometry (this is encoded in the Christofel Symbols).
     
  8. Jul 15, 2014 #7
    Thanks for your answers. There is much to ponder.
     
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