There is a similar question on here, but I figured this focuses on a more specific aspect. My apologies if it is seen as a duplicate. I am not a cosmologist. I do have a background in physics, but cosmology is not my forte, or preference.

The twin paradox is well known. To summarize (my version at least) ,in mathematically minimal terms one twin goes on an expedition to outer space or some distant place and eventually returns to the same location as the other twin where eventually they correspond their experiences. The twin that remained stationary has experienced more time than the twin that has traveled. From initial event, the traveling twin has experienced y amount of time. The stationary twin has experienced x amount of time. $x>y$. If the universe, cosmologically speaking is $z$ years old at the event when the traveling twin decided to venture out, how "old" is the universe when the twins reconcile? It would seem that for one twin the universe would be $x+z$ years old, while for the other twin the universe would be $y+z$ years old. I have always loved relativity, and even QM, but cosmology and or Astronomy I reluctantly admit is baffling to me at many times. Does anyone have a fulfilling explanation? Thanks.

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    $\begingroup$ Yes, the age will depend on observer's frame. However, due to global features of the Friedman-Lemaître-Robertson-Walker solution to the Einstein equations there exists a privileged frame comoving with the expanding universe, and the standard cosmic time is defined in it. Cosmological dates are given according to that time. $\endgroup$ – Conifold May 7 '17 at 23:24
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    $\begingroup$ The first paragraph is an apology, and I have lately observed that in many other questions. Do the moderators here intimidate you? $\endgroup$ – user126422 May 8 '17 at 0:16
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    $\begingroup$ If there were clocks that existed from the origin of the universe, then yes, in most cases they will disagree with each other $\endgroup$ – user126422 May 8 '17 at 0:19
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    $\begingroup$ An even more extreme example: A photon that was present at the Big Bang will tell you, as it enters your eye, that the Universe is zero years old. Obviously you disagree. $\endgroup$ – WillO May 8 '17 at 0:36
  • $\begingroup$ And all the people that ever entered an eternal black hole, regardless of when they have entered (to you), they will agree that they entered at the same time. $\endgroup$ – user126422 May 8 '17 at 0:47

The answer is clearly yes - the age will clearly depend on the timekeeper's frame (if we make a thought experiment where a "timekeeper" can exist since the Big Bang). You clearly understand the twin paradox, so an obvious refinement of the question is can the Universe have different ages for inertial observers: still the answer is yes if you imagine extending geodesics out from your position indefinitely into the past with different tangents. All of these geodesics will end up at the Big Bang, and the proper time along them - their "length" since the Big Bang - will clearly depend on their relative velocity to you as they reach you. This notion is summarized by WillO's pithy comment (which should itself be an answer):

"An even more extreme example: A photon that was present at the Big Bang will tell you, as it enters your eye, that the Universe is zero years old. Obviously you disagree."

if you allow a bit of poetic license (for anthropomorphic photons who might disagree with you).

But, as also note in the comments, there is a notion of "universal" or "cosmic" time that we believe is well defined in our Universe: this is the parameter $t$ in the Friedman-Lemâitre-Robertson-Walker metric. It is by this parameter that cosmologists make statements such as the age of the Universe is 13.8 thousand million years old. It has the following thought-experimental meaning: suppose we meet a timekeeping observer that has been comoving with the so-called Hubble flow since the Big Bang. Then the age of the Universe as measured by $t$ is the same as the proper time that has elapsed for this observer. The Hubble flow is defined by points that have constant spatial co-ordinates as defined by the FLRW metric: relative motion between objects in the Hubble flow is owing to the change of the scale factor $a(t)$ with $t$ alone. An observer moving with the Hubble flow would see perfect isotropy in the Cosmic Microwave Background Radiation. On Earth we do not; we see a very faint dipolar anisotropy, and so Earth is believed to have a peculiar velocity of approximately $371{\rm km\,s^{-1}}$ towards the constellation Leo.

It should also be noted that this possibility of Universal time only arises in spacetimes which have timelike Killing vector fields and which can be foliated so as to be decomposed into a partition of sheets of constant $t$, just like flat Minkowski spacetime can be. It is an experimental fact that our Universe seems to be consistent with the theory that our spacetime, over large time/length scales, is homogeneous and isotropic and thus modelled by the FLRW metric. Naturally, the FLRW metric is an example of a metric that can be foliated or sliced up in this way, but this is not true of arbitrary solutions of the Einstein field equations. In general solutions, no such universal time is possible. So the fact of a universal "cosmic" time is not a given, but rather it is only believed to be true by dint of experimental confirmation of the FLRW metric postulate.

  • $\begingroup$ However, the "frame" in this case is the same for each twin. While they "had" ventured on different paths, at the "reconciling" event they are no longer in different frames, they are at the same relative speed, at the same location. One twin has accelerated away from the other and then returned to the location of the other, once again decelerating back to the same frame and location as the "stationary" counterpart. $\endgroup$ – IntuitivePhysics May 9 '17 at 16:36
  • $\begingroup$ BTW, your "refinement" as can the Universe have different ages for inertial observers is pretty awesome. $\endgroup$ – IntuitivePhysics May 9 '17 at 17:23

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