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Killing vectors and momentum conservation

  1. May 18, 2015 #1
    Consider the flat FRW metric in Cartesian spatial co-moving co-ordinates:

    ##ds^2=-dt^2+a(t)^2(dx^2+dy^2+dz^2)##

    As I understand it, since the metric does not depend on the the spatial co-ordinates, there exist Killing vectors in the ##x##,##y##,##z## directions.

    Does this imply that the 3-momentum of particles travelling on geodesics is conserved (when one refers to the co-moving co-ordinates ##x##,##y##,##z##) ?

    I have read elsewhere that the 3-momentum of free particles goes like ##1/a(t)## instead.

    Is the difference due to proper co-ordinates being used instead of the co-moving co-ordinates ##x##,##y##,##z##?
     
  2. jcsd
  3. May 18, 2015 #2

    bapowell

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    Yes, the proper velocity redshifts as [itex]1/a[/itex].
     
  4. May 18, 2015 #3
    If one has a gas of particles in a rigid box would that gas cool down as the Universe expands?
     
  5. May 18, 2015 #4

    bapowell

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    Yes. The energy density declines by [itex]1/a^3[/itex] for nonrelativistic particles.
     
  6. May 18, 2015 #5
    Are you assuming that the box expands with the Universe so that its volume is proportional to ##a^3##?

    I'm assuming that the box is rigid - does the gas cool in that case?
     
  7. May 18, 2015 #6

    Chalnoth

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    Yes, it cools. But to a very good approximation matter has a temperature of zero (because its temperature is usually much, much less than the rest mass of electrons, let alone protons).
    I believe the conserved quantity in this case is ##a^2 \dot{x}##.
     
  8. May 19, 2015 #7

    timmdeeg

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    Jcap is talking about a rigid box. So, the density of the gas within it should be constant over time. If correct, I don't understand, that the gas cools.

    How about an initial state, where the gas particles are in rest to each other, means T = 0 K. In order to follow the Hubble flow, the distances should increase then and by bouncing against the rigid walls (which are not co-moving), the particles should gain kinetic energy. Is this reasoning correct?
     
  9. May 19, 2015 #8

    Chalnoth

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    Yes, I was responding to the OP and didn't catch that part. Certainly the only change in temperature of a gas inside a rigid box will stem from an exchange of energy with its surroundings. So the temperature change would be related to the temperature of its surroundings and how efficiently it exchanges energy with them.
     
  10. May 20, 2015 #9

    timmdeeg

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    Thanks. Would the temperature of the gas inside an adiabatically isolated box increase (without isochoric work being done on it) in expanding spacetime?
     
  11. May 20, 2015 #10

    Chalnoth

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    If the box is isolated from exchanging heat with its surroundings, its temperature would, by definition, remain unchanged.

    Basically, it's a rigid box. The expanding universe component is irrelevant when you have a rigid box. There is no way that an expanding universe can change any conclusions at all about the behavior of matter inside such a rigid box (except in terms of how the expansion impacts the temperature of the surroundings.
     
  12. May 21, 2015 #11

    timmdeeg

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    Sorry, I still have a problem. Initially, the center of the box shall be co-moving and the matter particles inside too. So, locally close to the walls, there should be relative velocities between those and the particles, resulting in peculiar velocities of the particles after bouncing against the walls. After some time all particles except those in the center of the box should possess peculiar velocities.
    But it seems I am wrong. Could you kindly comment?
     
  13. May 21, 2015 #12

    timmdeeg

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    But the kinetic energy in the frame of the box doesn't change during bouncing. I think I had a silly misconception.
     
  14. May 21, 2015 #13

    Chalnoth

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    Regardless of the initial conditions, the gas inside the box will rapidly reach equilibrium.
     
  15. May 21, 2015 #14

    timmdeeg

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    Understand, thanks.
     
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