Does a spaceship travelling at near lightspeed see the universe aging slow or fast?

Question

Now I know this probably is well trodden territory, but this question has bugged me for some time, and I couldn't find a similar question in the archives (although plenty about relativistic time dilation and the twin paradox, which is closely related but not quite the same). The theory of special relativity suggests that two objects moving relative to each other (at near light speeds) will observe the other as moving at a slower time. However, both of these reference frames are local reference frames - they are not a global perspective that the universe is. So I'll give a little background before asking about the scenario, to show the difference between this and previous questions seen here (although I doubt I have been exhaustive in my search here).

At the time of Einstein, no one knew a means of measuring speed relative to the universe that was applicable everywhere in the universe (short of Mach's principle which Einstein supplanted in GR). Neither did they know the universe had a finite age, let alone a means of measuring it. But now we have both - we can measure the age of the universe by the wavelength of the MBR, and we can "in principle" measure the rate of change of the universe by the slow change in the MBR's wavelength. (If we really want to, we can also measure historic elements such as the rate and duration of supernova and the expansion of the universe.) Additionally our velocity relative to the universe can be known by measuring the variation in the MBR due to the Doppler Effect (assuming the MBR is more or less isotropic). That is, in the direction of travel, the MBR would be blue-shifted, while behind, it would be red-shifted, and the speed can be deduced by the magnitude of the wavelength shift. So we now have "in principle" methods of measuring both our velocity with respect to the universe and a form of universal time.

So if an alien is traveling at relativistic speeds to the universe, and measuring the MBR and it's wavelength change with respect to the internal clock of his ship, will it observe the universe to be aging faster or slower than we would observe?

Answer 1

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 this vague statement needs stating more carefully to make it useful.

If we ignore minor irritants such as inflation and quantum mechanics then the geometry of the expanding universe is described by the FLRW metric:

$$ ds^2 = -c^2dt^2 + a^2(t)d\Sigma^2 $$

For the purposes of this question let's assume the universe is flat, and we'll consider only radial motion. This allows us to simply the metric to get the expression for the proper time:

$$ d\tau^2 = dt^2 - \frac{a^2(t)}{c^2}dr^2 \tag{1} $$

The quantity $d\tau$ is the proper time, which is equal to the time measured by a clock carried by a freely falling observer. The radius $r$ is measured in comoving units, which aren't the same as the distance measurements we make. The distances we measure are $r$ multiplied by the scale factor $a(t)$, and it's the increase in $a(t)$ with time that we see as the expansion of the universe. In these units stationary observers, i.e. at constant $r$, are moving away from us as the universe expands.

Where the CMB comes in is that the CMB is roughly isotropic for stationary (in comovong coordinates i.e. constant $r$) observers. NB the CMB does not define the comoving frame, it approximately coincides with the comoving frame because of the way it was created. See Assuming that the Cosmological Principle is correct, does this imply that the universe possess an empirically privileged reference frame? and Is the CMB rest frame special? Where does it come from? for more on this.

Anyhow, if you're a comoving observer then your $r$ coordinate is constant and therefore $dr=0$ and the metric simplifies to:

$$ d\tau^2 = dt^2 $$

And this immediately integrates to give $\tau = t$. The age of the universe $\tau$ is just the time $t$ shown on the clock you've been carrying since the Big Bang. this applies to all comoving observers so all comoving observers agree on how long it is since the Big Bang.

But suppose you're not a stationary observer. Suppose you are moving at some comovingg velocity $v(t)$ so:

$$ \frac{dr}{dt} = v(t) $$

then:

$$ dr = v(t)dt $$

and we can substitute this into our equation (1) above to get:

$$ d\tau^2 = dt^2\left(1 - a^2(t)\frac{v^2(t)}{c^2}\right) $$

Again we get the time since the Big Bang by integrating:

$$ \tau = \int_0^t \sqrt{1 - a^2(t')\frac{v^2(t')}{c^2}} dt' $$

Actually doing the integration is hard even for constant $v$ because $a(t)$ is a complicated function of time, but we don't need to do the integration to see that $\tau < t$. The quantity $a^2(t')v^2(t')/c^2$ is positive because it's a square, and that means the function $\sqrt{1 - a^2(t')v^2(t')/c^2}$ is less than one if $v > 0$. So whatever the precise form of $v(t)$ the integral is always going to give the result:

$$ \tau < t $$

Since we comoving observers found $\tau = t$ that means an observer who is moving in comoving coordinates must measure a time since the Big Bang that is less than we measure.