A person is four miles up in the atmosphere, and freefalls towards earth. Another person is standing on earth, watching them come down. During this period of freefall, which person experiences more time dilation?
What I am wondering is - is the person freefalling through the gravitational field (effectively experiencing zero g) more 'at rest' than the person standing on earth (who is effectively accelerating upwards with a force of 1g)?
Calculations like the one you propose are somewhat tedious and not very illuminating, so let me address what I think you're really asking. If we look at an observer on the Earth's surface and one who is (momentarily) at the same point but falling at some velocity $v$ then how do their time dilations compare?
We will be calculating the time dilation from the point of view of someone floating in space far from the Earth. This is (approximately) a Schwarzschild observer, and we'll be measuring times and distances using the Schwarzschild coordinates. We do the calculation using the Schwarzschild metric:
$$ c^2d\tau^2 = \left(1-\frac{r_s}{r}\right)c^2dt^2 - \frac{dr^2}{1-\frac{r_s}{r}} - r^2d\theta^2 - r^2\sin^2\theta d\phi^2 \tag{1} $$
where $r_s$ is the Schwarzschild radius $r_s = 2GM/c^2$. The quantity $d\tau$ is the proper time, which is the time measured on a clock carried by the observer on the surface or falling. $t$ and $r$ are time and radial distance measured by the Schwarzschild observer. The time dilation is the ratio $d\tau/dt$.
The time dilation for the observer stationary at the surface of the Earth is easy because since the radial and angular coordinates aren't changing we have $dr = d\theta = d\phi = 0$ and the metric simplifies to:
$$ c^2d\tau^2 = \left(1-\frac{r_s}{r}\right)c^2dt^2 $$
and a quick rearrangement gives us:
$$ \frac{d\tau}{dt} = \sqrt{1-\frac{r_s}{r}} \tag{2} $$
The calculation for the falling observer is more complicated. We'll assume the observer is falling radially so $d\theta = d\phi = 0$. If the radial velocity is $v$ that means:
$$ \frac{dr}{dt} = v $$
or:
$$ dr = v\,dt $$
Now we substitute for $dr$ in the metric and this gives us:
$$ c^2d\tau^2 = \left(1-\frac{r_s}{r}\right)c^2dt^2 - \frac{v^2dt^2}{1-\frac{r_s}{r}} $$
and rearranging gives:
$$ \frac{d\tau}{dt} = \sqrt{1-\frac{r_s}{r}}\sqrt{1 - \frac{v^2}{c^2}\frac{1}{\left(1-\frac{r_s}{r}\right)^2}} \tag{3} $$
If we denote the time dilation of the stationary observer by $T_s$ and the time dilation of the falling observer by $T_f$ then comparing equations (2) and (3) gives us:
$$ T_f = T_s \sqrt{1 - \frac{v^2}{c^2}\frac{1}{\left(1-\frac{r_s}{r}\right)^2}} $$
so we conclude that:
$$ T_f < T_s $$
Time flows more slowly for the falling observer than for the stationary observer i.e. the time dilation is greater for the falling observer than for the stationary one.
This happens because the gravitational time dilation is determined by the spacetime curvature not by the acceleration the observer feels. It doesn't matter whether the observer is moving or not, the spacetime curvature is the same so the gravitational time dilation is the same. However the falling observer has an extra time dilation due to their velocity, so the total time dilation for the falling observer is greater.
If time is dilated then one second is longer; the opposite would be contraction and a second would be shorter.
Now consider in the classical picture an object that is dropped above the surface of the earth loses gravitational potential energy, and gains in kinetic energy; it moves faster.
Consider too, that time is an aspect of motion - we measure time by motion after all: pendulums, light-years or light-metres.
Consider also that falling is inertial motion, it is motion which experiences no force; so physically nothing changes for this object in its frame.
But the man above sees it gain in kinetic energy, gain in motion; time is an aspect of motion; so more happens in one second; which, from an alternative perspective, he could say its because there time has contracted.
