Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I Age of the Universe as Measured by a Non-Co-Moving Observer

  1. Jun 28, 2016 #1

    Drakkith

    User Avatar

    Staff: Mentor

    In a recent thread I incorrectly stated that an observer moving at a high velocity relative to a co-moving observer would measure the age of the universe as more than the co-moving observer would. Apparently that is incorrect, and the age of the universe measured by the first observer would be less than that measured by the co-moving observer. I'm just curious as to how this works.

    Is it possible to explain this using just SR and moving reference frames, or do we need to invoke GR?
     
  2. jcsd
  3. Jun 28, 2016 #2

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    One's initial intuitive guess would be that SR couldn't answer this question since we are dealing with a curved spacetime. However, I think there is a sense in which we can sort of use SR to see why it's true.

    First, a key fact about comoving observers (which does require GR to derive, since you need to show that the FRW metric is a solution of the EFE): all comoving observers experience the same amount of proper time between any two surfaces of constant FRW coordinate time. These surfaces are picked out by the fact that they are homogeneous and isotropic; no other family of spacelike hypersurfaces in FRW spacetime has this property.

    Now consider any non-comoving observer's worldline, and ask how much proper time will elapse along it between two surfaces of constant FRW coordinate time. In standard FRW coordinates, the worldline of this non-comoving observer will have some nonzero spatial displacement, which we can consider to be purely radial. For simplicity we consider the case of a spatially flat FRW universe. The proper time along the non-comoving worldline will be the integral of the line element, which we can write as:

    $$
    \tau = \int \sqrt{dt^2 - a^2(t) dr^2}
    $$

    Of course we can't fully evaluate this integral without knowing the specific dynamics of the scale factor ##a## and the function ##dr/dt## that gives the non-comoving observer's spatial motion. However, we don't need to do any of that to see that the above integral must give a value that is smaller than the corresponding value for a comoving observer between the same two surfaces of constant FRW time:

    $$
    \tau_o = \int dt
    $$

    Technically, we do need one other premise to complete the argument: we need to recognize that the "age of the universe" is evaluated starting at some particular surface of constant FRW coordinate time. The usual convention is to use the "Big Bang" surface, i.e., the surface of constant FRW coordinate time that marks the end of inflation and the beginning of the hot, dense, rapidly expanding state that became the universe we observe.

    So far I've phrased everything purely in GR terms. But of course the two integrals above look very much like the corresponding integrals for a stationary vs. a moving observer in a particular inertial frame in SR. So we could make a similar sort of argument, heuristically, in the local inertial frame of a comoving observer. But I think we would still need extra information from GR to really nail it down.
     
  4. Jun 29, 2016 #3

    Drakkith

    User Avatar

    Staff: Mentor

    Thanks Peter. One question right now: What's a "surface of constant FRW coordinate time"?
     
  5. Jun 29, 2016 #4

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    A spacelike hypersurface of constant coordinate time ##t## in FRW coordinates. :wink:

    Physically, it's a spacelike hypersurface that is homogeneous and isotropic. FRW spacetime can be foliated by a family of such hypersurfaces, and standard FRW coordinates are constructed such that each hypersurface in the family is labeled by a unique value of the coordinate time ##t##. This time also turns out to be the proper time of comoving observers, i.e., observers that always see the universe as homogeneous and isotropic; the worldlines of such observers are everywhere orthogonal to the family of hypersurfaces just described.
     
    Last edited: Jun 29, 2016
  6. Jun 29, 2016 #5

    Chalnoth

    User Avatar
    Science Advisor

    To be really technical, the time of the Big Bang is the singularity that exists in the Big Bang model, while not considering pre-Big Bang models such as inflation or a bounce cosmology. This is close enough to reheating that they're essentially indistinguishable, however (the difference is something of the order of [itex]10^{-30}[/itex] seconds, if I recall).

    So, basically the Big Bang time is easy an easy to calculate moment that's close enough to the events that kicked off our observable universe that we generally aren't concerned about any discrepancies.
     
  7. Jun 29, 2016 #6

    Drakkith

    User Avatar

    Staff: Mentor

    I don't think my Calculus 2 class prepared me for this... :rolleyes:
     
  8. Jun 29, 2016 #7
    That's easy: every observer that is comoving with the hubble flow (in other words at rest relative to the cmb) observes the universe to be 13.8 billion years old. The time dilation between two observers is then just calculated via the v relative to the cmb (the local proper velocity), not the relative recessional velocity between them. Fraser Flav did a short video on this which is also good for laymen: What Time Is It In The Universe?

    The term "hypersurface" sounds wild, but it is only a volume in comoving coordinates. Just imagine the famous surface of an expanding raisin bread. Every raisin is sitting on the 2D surface of the bread; now just add a 3rd dimension and watch the raisins expand. In this metaphor every raisin has the same proper time.

    I think nobody who doesn't already know the answear could understand this hardcore formulation (:
     
    Last edited: Jun 29, 2016
  9. Jun 29, 2016 #8

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    Chapter 8 of Carroll's online lecture notes gives a good presentation, explaining in more detail what the terms I used mean. He also gives a good introduction to differential geometry in chapters 2 and 3, which helps when you encounter "hardcore" formulations like the one I posted. :wink:
     
  10. Jun 29, 2016 #9

    Drakkith

    User Avatar

    Staff: Mentor

    Thanks all!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Age of the Universe as Measured by a Non-Co-Moving Observer
Loading...