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Universal expansion velocity SR or GR?

  1. May 27, 2008 #1
    If I want to consider the velocity of distant receding galaxies which may be receding at velocities close to the speed of light can I use SR? I know you are not suppose to use SR with when velocities are close to the speed of light. If a velocity is due to universal expansion does that make a difference? It would seem not, but I am not sure if a velocity due to universal expansion is really a velocity, just like relitivistic mass is not really mass.
    It is a poor example, I know, but it helps with the point.
     
  2. jcsd
  3. May 27, 2008 #2

    CompuChip

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    No, I don't think you're just allowed to do that, because actually the galaxies are not (only) moving away from us, but the universe expands ("space is being created in between"). A famous analogy is that of coins on a balloon: if you inflate the balloon, the distance between the coins increases, while they are not actually moving relative to each other (if you'd draw a grid on the balloon, the coordinates of the coins would not change). But to an observer on one of those coins, it would seem that the other ones were moving away from him (in all directions, making it seem he was himself at the center of the expansion). In fact, it is possible to have them receding faster than the speed of light, I believe, because there is not actual "motion" involved in the normal sense.

    I think you said that the wrong way 'round (or I just understood it that way). SR is exactly meant for the regime where v ~ c.

    As I explained, AFAIK it's not. But not in the same what that "relativistic mass is not really mass".
     
  4. May 27, 2008 #3
    A consequence of GR is that you cannot compare velocities at different spacetime points. The term "speed of recession" is not what it sounds like.
     
  5. May 27, 2008 #4
    Well then GR is out and SR is Ok.
    Another question.

    In SR, can I calculate gravitational force from a distant mass by using its rest mass or its relativistic mass in Newton’s Gravitation equation?
    I don’t think I am suppose to do this.

    How can I calculate an acceleration or a force caused by a very distant mass?
    Is it possible?
     
  6. May 27, 2008 #5

    Dale

    Staff: Mentor

    I think you must use GR because, IIRC, distant receding galaxies can be receding at velocities much greater than the speed of light. This is possible in GR, but not in SR.
     
  7. May 27, 2008 #6

    Ken G

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    Gold Member

    I think the problem with using SR is that the spacetime has to be flat, and in the cosmology we observe it is not. In a flat spacetime, say where you only had a "Big Bang" of test particles with no appreciable gravity, you might be able to get away with SR and just note that your perceptions of distant regions are highly length contracted relative to what they perceive locally (a la the "Milne" model), so if you wanted to have a test-particle equivalent of a cosmological principle, you'd have to include that carefully. You would be using a coordinatization where the recession speeds are never superluminal. Note that whether they are faster than c in our own universe is also a coordinate dependent issue-- I believe this was genneth's point that the concept of velocity is not terribly meaningful at distantly separated points, it is just a choice of coordinates. But in our universe, if you choose comoving-frame coordinates where the local coordinate charts follow the gradual separating of the galaxy clusters, the recession rate is faster than c for the most distantly observed galaxies. Could you get that by using comoving frame coordinates in flat spacetime? Perhaps you could.
     
    Last edited: May 27, 2008
  8. May 27, 2008 #7
    Thanks for mentioning the Milne Model.
    I have read several articles about it.
    Although it may not be a serious contender for the real universe condition it made the difference between SR and GR results much easier to understand.
     
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