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In a previous Phys.SE question, Does a spaceship travelling at near lightspeed see the universe aging slow or fast?, the answer (which was followed by a proof involving co-moving reference frames) was given as

The short answer is that yes, an astronaut moving relative to the cosmic microwave background would measure a shorter time since the Big Bang than an observer stationary wrt to the CMB.

However, an observer in such a spaceship will consider the time of any object which is at the CMBR co-moving reference frame to be moving slower than itself. Is this not a conflicting result?

For example, let's say the spaceship and a CMBR Earth communicate as they pass by each other. Each would have an estimate of the age of the universe, and each would have an estimate of the measured age of universe that the other would give, based on their own measurement of the age and the time dilation that they assume the other would experience. Here are the results

Expected and Obtained Values

The CMBR observer is fine - both his estimate of the universe's age that the spaceship would give and the estimate actually given by the spaceship match and are less than his own estimate of the universe's age. However, the spaceship expects the CMBR observer to have a lower estimate of the age of the universe because their clock is (according to spaceship observer) ticking slower than his own. What the spaceship observer does not expect is that the CMBR observer's estimate is larger than his own estimate of the age of the universe, yet that is what happens. How is this resolved without implying a preferred reference frame?

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  • $\begingroup$ The spaceship observer knows exactly how fast he is going relative to the CMB since he sees it as a red-blue shifted dipole, just like we do. He can derive the actual age of the universe from the dipole measurement. Where is the conflict? The two observers are simply not "seeing" the universe the same. $\endgroup$
    – CuriousOne
    Mar 30, 2016 at 1:02
  • $\begingroup$ But that approach is presuming a preferred reference frame. $\endgroup$
    – matscienceman
    Mar 30, 2016 at 1:12
  • $\begingroup$ It's not presuming anything. It simply acknowledges what the observer in a rocket will see. See CMB dipole: astronomy.swin.edu.au/cosmos/C/…. $\endgroup$
    – CuriousOne
    Mar 30, 2016 at 1:20
  • $\begingroup$ On first read, it seems to me that the OP is reasoning from a global inertial reference frames perspective. $\endgroup$
    – Alfred Centauri
    Mar 30, 2016 at 1:33
  • $\begingroup$ @CuriousOne But what you said above was that the age of the universe was derived from the dipole reference frame (ie the CMBR rest frame). If that is the only frame that can correctly determine the age of the universe, how is that not a 'special reference frame'? (See comment in the original question) $\endgroup$
    – matscienceman
    Mar 30, 2016 at 1:36

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How is this resolved without implying a preferred reference frame?

That you asked this implies you are thinking that reference frames have universal extent. While that is true in special relativity, it is not the case in general relativity. Reference frames are local in general relativity.

That said, there is a frame in which cosmologists prefer to work, and that is a frame locally at rest with the cosmic microwave background. This is the frame that yields the longest proper time for the light from the surface of last scattering. That does not mean that this is the "preferred frame" (with all the baggage that goes along with that term). It is just the frame in which cosmologists prefer to work.

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  • $\begingroup$ I hadn't realised that GR used a subtly altered the meaning of Reference Frame, limiting its applicability to "local". How local is "local"? Does this mean that it is problematic to consider SR on a universal scale? Does this mean that some of the classic paradoxes (twins, Bells etc) are also outside the range of "local"? Is the use of co-moving reference frames (with an expanding universe) automatically putting the question into the realm of GR and not SR? If not, how could this be resolved using SR without reference to GR concepts? $\endgroup$
    – matscienceman
    Mar 30, 2016 at 6:51
  • $\begingroup$ @matscienceman: 'local' is 'completely local' in GR: the only reason you can treat reference frames as nonlocal at all in special relativity is because you can treat the spacetime manifold as a vector space, which GR does not. $\endgroup$
    – user107153
    Mar 30, 2016 at 7:01
  • $\begingroup$ @matscienceman - To a mathematical physicist, "local" has the mathematical meaning of an infinitesimal ball about the origin. To a more practical physicist, "local" has the slightly broader meaning of being within the limits of what sensors can detect. The GOCE satellite used to map the Earth's gravitational was not a local experiment in the context of Einstein's elevator car thought experiment. GOCE used six high-precision accelerometers arranged orthogonally in pairs separated by 50 cm, making each pair an arm of a gravity gradiometer. $\endgroup$
    – David Hammen
    Mar 30, 2016 at 11:28
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While the spaceship accelerates the time on earth is also running faster in the reference frame of the spaceship. The fact that everybody sees the other's clock tick slower is only true for uniform motions. The time dilation coming from gravity and acceleration, which are equivalent, is absolute. So if the spaceship does the calculation right it can easily transform into the system which is at rest to the CMB and find out the age of the universe in that frame of reference.

If the spaceship is created at rest (relative to the CMB) it will have to accelerate to gain velocity in the direction of the earth. Even if the spaceship is created at motion from the beginning there is the so called "relativity of simultaneity".

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  • $\begingroup$ The bits about acceleration don't apply to this question - the spaceship doesn't come from CMBR Earth. And transforming the coordinate into the CMBR rest frame is suggesting a preferred rest frame. $\endgroup$
    – matscienceman
    Mar 30, 2016 at 1:24
  • $\begingroup$ Maybe the second part I just added might answer your question $\endgroup$
    – Yukterez
    Mar 30, 2016 at 1:25
  • $\begingroup$ Could you elaborate how it might answer the question - assume the spaceship was created at motion. I understand that simultaneity is different from different reference frames, when viewed without a Lorentzian approach, but how does that resolve the conflict? Thanks. $\endgroup$
    – matscienceman
    Mar 30, 2016 at 2:10

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