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I What does "expansion of the universe" mean?

  1. Jun 23, 2016 #1
    [Moderator's note: thread spun off from post in a different thread quoted below.]

    That seems to me a rather extraordinary claim.

    Could you demonstrate how for instance cosmological redshift can be explained in a coordinate independent fashion?
     
    Last edited by a moderator: Jun 27, 2016
  2. jcsd
  3. Jun 23, 2016 #2

    Dale

    Staff: Mentor

    It is pretty straightforward. The comoving congruence is coordinate independent, as is the fact that the resulting expansion tensor is expanding.

    That is just calculating a null geodesic between two lines in the comoving congruence.
     
  4. Jun 23, 2016 #3
    The expansion tensor is not a spacetime tensor but a tensor on some 3+1 foliation.

    Let's stay scientific!
     
  5. Jun 23, 2016 #4

    Dale

    Staff: Mentor

    The expansion tensor is a perfectly valid tensor, both theoretically and operationally.
     
  6. Jun 23, 2016 #5
    It is misleading to use this tensor as an argument!

    :frown:
     
  7. Jun 23, 2016 #6

    Dale

    Staff: Mentor

    Why? It is coordinate independent and it shows that the universe expanding. It therefore directly addresses what you considered to be an extraordinary claim.

    Simply calling it misleading or unscientific is hardly a well reasoned argument. Your aggressive tone to pervect is not well considered, nor is your current tone with me.
     
  8. Jun 23, 2016 #7

    PeterDonis

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    2016 Award

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    Incorrect. The definition of the expansion tensor uses the projection tensor ##h_{ab} = g_{ab} + X_a X_b## which projects out the part of an arbitrary vector or tensor field that is orthogonal to the vector field ##X##. But such projected vectors, tensors, etc. are still 4-vectors, 4-tensors, etc.

    This is incorrect, not only because of the above, but because it is not always possible to even find such a 3+1 foliation in a spacetime containing a given timelike congruence, such that every 3-surface in the foliation is orthogonal to every timelike curve in the congruence. It happens to be possible for the congruence of comoving observers in FRW spacetime, but you can't depend on it as a general property.
     
  9. Jun 23, 2016 #8
    Let me ask you this would you use the expansion tensor as an argument to prove coordinate independence?
     
  10. Jun 23, 2016 #9

    PeterDonis

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    You don't need an argument to prove coordinate independence; coordinate independence is an obvious fact about any observable quantity in GR, since any such quantity must be expressible as a scalar invariant. So I don't understand what your issue is here.

    A photon emitted by a particular comoving observer has a 4-momentum vector determined at the event of emission by the photon's frequency as measured by the emitter--this emitted frequency is just the contraction of the photon's 4-momentum with the emitter's 4-velocity at the event of emission. The photon's 4-momentum vector determines a unique null geodesic from the event of emission to the event of reception; the photon's 4-momentum is parallel transported along this null geodesic to the event of reception. At the event of reception, the comoving observer who measures the photon's frequency has a 4-velocity, which is contracted with the photon's 4-momentum to give the measured frequency (modulo a factor of Planck's constant if you are not using natural units). This entire process is manifestly coordinate independent. The fact that the process results in a redshift is easily derived from the fact that the congruence of comoving observers has a positive expansion scalar, i.e., that its expansion tensor has a positive trace. This derivation is also manifestly coordinate independent.
     
  11. Jun 23, 2016 #10
    Cosmological redshift is not a coordinate dependent quality, ok it seems I have a lot to learn!

    :rolleyes:
     
  12. Jun 24, 2016 #11

    Dale

    Staff: Mentor

    This is true, but as @PeterDonis mentions this isn't what is usually meant by the term pseudo tensor in GR.

    To your point however, if the trace of the expansion tensor is positive then the congruence is separating. If you time reverse everything (manifold and congruence) then you get a negative trace indicating that the congruence is contracting.

    So the one additional thing that we have to do to say that the universe is expanding in a frame invariant manner is to invariantly identify which direction along the timelike congruence is the positive direction. For that we can simply appeal to features of the universe like entropy or the weak force interaction.
     
  13. Jun 24, 2016 #12
    Sure, but that appeal in no way can affect how a tensor and a pseudotensor are defined mathematically, and yet you claimed it was a tensor, wich is an object defined in mathematical terms only.
     
  14. Jun 24, 2016 #13
    How about observers who are not comoving?
    Or are those perhaps discounted as non relevant to general relativity?
     
  15. Jun 24, 2016 #14

    Dale

    Staff: Mentor

    Words are defined however a group of people chooses to define them, and often different groups of people use the same word to mean different things.

    My comment that it is a valid tensor is correct, as is your comment that it is a pseudotensor. We were just using the definitions of different groups of people.
     
  16. Jun 24, 2016 #15

    Dale

    Staff: Mentor

    What about them? Why would they be discounted in any way?
     
  17. Jun 24, 2016 #16
    Well how would this tensor work for non-comoving observers?
     
  18. Jun 24, 2016 #17

    Dale

    Staff: Mentor

    The same way it works for comoving observers. There is no restriction or requirement that an observer only use congruences that include their worldline.

    For example, both a rotating and an inertial observer may use the same congruence to study a rotating disk.
     
    Last edited: Jun 24, 2016
  19. Jun 24, 2016 #18

    PeterDonis

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    Non-comoving observers would be a different congruence of worldlines, so they would have a different expansion tensor. The reason the comoving observers are used to define the expansion "of the universe" is that those observers are picked out by a particular physical property: they all see the universe as homogeneous and isotropic for all time. No other congruence of worldlines has this property.
     
  20. Jun 24, 2016 #19
    Exactly!
     
  21. Jun 24, 2016 #20

    Dale

    Staff: Mentor

    Exactly what?

    They do form different congruences, but nothing prevents a non-comoving observer from identifying the comoving congruence and calculating their expansion tensor.

    The identification of the comoving congruence does not depend on the motion of the observer doing the identification , nor does it depend on the choice of coordinates.
     
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