I put my assistant in a spaceship and accelerate it to near the speed of light. 100 years from now (in my time), my assistant is travelling with speed $0.99c$. At that time I put up a super sophisticated WMAP-like probe and measure the age of the universe very precisely to be 13.8 billion years (to several decimal places). My assistant has a similar probe and performs the same measurement. What age will she calculate for the age of the universe?
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3 Answers
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Ignoring the initial acceleration here she would measure it to be exactly the same as from her frame of reference she wasn't moving at all, and it was in fact you that was moving (except for the acceleration at the start).
However for you she would measure the 100 years of time passed to in fact be $$\frac{100}{\sqrt{1-0.99^2}}$$ or $708.88$ years, which whilst being a very large time difference for us as humans, is nothing compared to the age of the universe, and as such it wouldn't affect the age up to a few decimal places in, and uncertainty in the measurement is much larger than this anyway.
answered Aug 7, 2017 at 15:18
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$\begingroup$ Good. That's what I thought. So why do I read that there's no simultaneous events in Special Relativity? It appears to me that no matter what reference frame you're in, no matter how fast you're going, every experiment will agree on the age of the universe based on CMB or SNE Ia data. $\endgroup$– user32023Aug 7, 2017 at 15:28
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$\begingroup$ There's no simultaneity because light has the same speed to all observers. Say you're in a car travelling at 0.5c and someone not in motion sees two lightning strikes at the same time both in front and behind you. The one behind you takes longer to reach you as you are travelling away from it, the same as the one in front takes less time. Every experiment will give the same results for the age of the universe, but only because there's not enough time dilation to allow for anything near a significant difference. $\endgroup$ Aug 7, 2017 at 15:31
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1$\begingroup$ CMB travels at the speed of light in a vacuum relative to all observers, so yes what you're saying should hold true. I suppose one small detail is that the faster you're going the more distances are contracted and hence the wavelengths of the CMB would appear slightly smaller. This would in turn make the measurement of the universes age slightly smaller as well - if our instruments had that kind of precision. Apologies before I don't think I quite understood what you were getting at. $\endgroup$ Aug 7, 2017 at 16:07
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4$\begingroup$ I had thought that redshifts were corrected to be in the comoving rest frame of the CMB. The correction is very small. The dynamics of the universe are not governed by Special Relativity. $\endgroup$– ProfRobAug 7, 2017 at 20:08
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3$\begingroup$ Important comment by @RobJeffries. The age of the Universe is defined as the time passed since Big Bang by a comoving observer, and so doesn't care about the speed of your assistant, nor about the speed of Earth, in any reference frame. $\endgroup$– pelaAug 9, 2017 at 9:40
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If shortly after the Big Bang you send your assistant on a trip around a closed universe at a constant speed, she would return $\gamma$ times younger than you. So in her frame the universe would be $\gamma$ times younger than in yours.
The WMAP measurements are made with the CMB dipole removed (speed subtracted), so they would always agree with your clock, but not with hers.
This scenario is not possible in an open universe. The assistant would never return to compare the clocks.
answered Aug 27, 2018 at 20:33
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The answer is going to be very model-dependent.
First of all, for the notion of "the age of the Universe" to make any sense, you need to assume a Universe that's foliated into three dimensional spacelike leaves, with some preferred global time coordinate.
Next, you seem to be supposing that you and your probe make your measurements "at the same time". It's not clear whether this means a) at the same time according to you, b) at the same time according to your probe, or c) at the same time according to the preferred global time coordinate.
So let's see what happens in a very simple model:
Spacetime is the region in the upper half-plane bounded by $t=-Kx$ and $t=Kx$, you travel along the worldline $x=0$, until time $T-100$, at which you launch your probe (which gets up to speed immediately). It travels along the blue worldline to $C$, where it makes its measurement. You make your own measurement at $B$, which is simulataneous with $C$ according to both you and the preferred global time coordinate.
You, therefore, declare the age of the Universe to be $T$.
Measured from the origin to $C$, we have $\Delta t=T$ and $\Delta x=99$. In your probe's coordinates, the time interval Lorentz transforms to $\Delta t'\approx 7.09(T-98)$ When $T$ is on the order of $13.8$ billion, we can ignore the $98$ and say that your probe will measure the Universe to be about 7 times as old as you do; call it about 98 billion years.
(On the other hand, if we've all agreed in advance that "the age of the Universe" is to be defined with reference to the preferred global time coordinate, then the probe will of course reject its own measurement and accept yours.)
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$\begingroup$ There is a contradiction in your scenario between having a preferred frame and the probe making a measurement in the frame of the probe. Either you have a preferred frame or you don't. If you do, then you measure the age in this frame, period. If you don't, then you can measure the age in any frame, but cannot compare the measurements between frames, because the universal simultainety cannot be established. Wat you measure in your scenario is not the age of the universe in the frame of the probe. If the probe is launched at the Big Bang, when it reaches you its clock will show less than 14 Gy. $\endgroup$ Aug 28, 2018 at 18:33