# Does the accelerating expansion of spacetime mean that the pace of time is changing?

Space expands everywhere, also here. And time is inseparable from space. Does this mean that time also "expands" as in changing its pace? Is the changing rate of time also astronomically observable? How did time behave during the radical inflation shortly after Big Bang?

• Pace of time--relative to what? You can compare two different clocks to each other (e.g., as in time dilation), but it's very unclear what comparison you're trying to make here. – Stan Liou Feb 28 '16 at 6:25
• @StanLiou In a similar way that the expansion of space is compared relative to, well, to itself I suppose. – LocalFluff Feb 28 '16 at 9:13
• I've often thought about this myself. I don't really get what 10E-30 sec during the big bang really means when time is intricately tied up with what is "banging". – Jack R. Woods Feb 28 '16 at 15:37

To talk about 'the rate of time', we essentially need at least two different time coordinates. For example, this happens in special-relativistic time dilation, which is equivalent to $\mathrm{d}t'/\mathrm{d}t$ across two different inertial frames. Fortunately, we can do something similar here.

Space expands everywhere, also here. And time is inseparable from space. Does this mean that time also "expands" as in changing its pace? ... In a similar way that the expansion of space is compared relative to, well, to itself I suppose.

A spatially isotropic and homogeneous universe has the metric in the form $$\mathrm{d}s^2 = -\mathrm{d}t^2 + a^2(t)\mathrm{d}\Sigma^2\text{,}$$ where $a(t)$ is the scale factor and $\mathrm{d}\Sigma^2$ is the metric of an isotropic and homogeneous Riemannian manifold: the 'open' hyperbolic $3$-plane, the flat Euclidean $3$-space, or the 'closed' $3$-sphere (or real projective $3$-space, but that's usually not considered because it's non-orientable). If the scale factor is ever zero in the past, the cosmological time for this is conventionally chosen to be $t = 0$.

The cosmological time measures the proper time of an observer at rest relative to the bulk of the matter in the universe, so in some sense it's the most intuitive choice of a time coordinate, but like all coordinates, it's not sacred. We can, for example, define a conformal time coordinate $\eta$ such that $\mathrm{d}\eta = \mathrm{d}t/a$, in which the metric takes the form $$\mathrm{d}s^2 = a^2(\eta)\left[-\mathrm{d}\eta^2 + \mathrm{d}\Sigma^2\right]\text{,}$$ and so all of the dimensions of spacetime are affected by cosmic expansion in the same way. Therefore, I think conformal time satisfies the requirements in your question, although it is not measured by any local clock.

Is the changing rate of time also astronomically observable?

The scale factor is astronomically observable, and $\mathrm{d}\eta/\mathrm{d}t = 1/a$, so yes.

How did time behave during the radical inflation shortly after Big Bang?

The conformal time essentially uses the particle horizon as a measure of time, i.e. the furthest distance from which an ideal lightlike signal could have travelled since $t = 0$ in order to reach the observer by the present time. During inflation, the particle horizon rapidly expanded.

• Did the horizon really expand fast during the inflation? Didn't it instead move closer, with things earlier within the horizon inflating to beyond it, the visible universe shrinking. As if time went backwards. – LocalFluff Feb 28 '16 at 16:17
• @LocalFluff The particle horizon is the boundary of the region that the observer could receive signals from by some specific time, e.g. present. It's basically the observable universe in principle (the practically observable universe is smaller, of course). The event horizon is the boundary of the region that observer could send a causal signal to, even if one waits forever. In some sense they're the opposites of each other. Both of them are also different from the Hubble sphere, the place at which recession velocity is $c$ from the observer. – Stan Liou Feb 28 '16 at 22:12