One of the most frequently misunderstood points about the expanding universe cosmology concerns the idea that space is "stretching" or that it is somehow dragging or pulling matter apart.

Nothing is further from the truth, and it is perhaps instructive to see why.

It all begins with the quantity $a(t)$ in the metric $d\tau^2=dt^2-a(t)^2dr^2$. This quantity is *a component of the metric*. That is to say, it characterizes how the gravitational field changes over time in this expanding universe.

True, it also happens to be a "scale factor" in the following sense: If this FLRW universe is a universe of dust particles (no pressure, equation of state $w=0$), then (and only then!) the average distance between particles will increase at the rate $a(t)$, and the density will be proportional to $a(t)^{-3}$.

However, no force is acting on any particle if it is at rest with respect to the isotropic reference frame. In fact, let us look at the geodesic equation:

$$\frac{d^2x^\mu}{d\tau^2}+\Gamma^{\mu}_{\alpha\beta}u^\alpha u^\beta=0.\tag{1}$$

Consider a particle at rest with respect to the isotropic reference frame: $u^t=1$, $u^r=0$. We find that all accelerations are identically zero.

The geodesic equations for a particle moving with velocity $u^r=dr/d\tau=v$, however, become

$$\begin{align}

\frac{d^2t}{d\tau^2}&=-a^2Hv^2,\tag{2}\\

\frac{d^2r}{d\tau^2}&=-2H\frac{dt}{d\tau}v,\tag{3}\end{align}$$

where $H=\dot{a}/a$ is the Hubble parameter. Given $H=2/3(1+w)t$ and $a=t^{2/3(1+w)}$ (with suitable choices for the beginning of time $t_0$ and the unit of time), we have

$$\begin{align}

\frac{d^2t}{d\tau^2}&=-\frac{2}{3(1+w)}t^{(1-3w)/(3+3w)}v^2,\tag{4}\\

\frac{d^2r}{d\tau^2}&=-\frac{4}{3(1+w)t}\frac{dt}{d\tau}v,\tag{5}\end{align}$$

or

$$\begin{align}\frac{d^2t}{d\tau^2}&=-\frac{2}{3(1+w)}t^{(1-3w)/(3+3w)}v^2,\tag{6}\\

\frac{dv}{d\tau}&=-\frac{4}{3(1+w)t}\frac{dt}{d\tau}v.\tag{7}\end{align}$$

Equation (7) is easily rewritten as

$$\begin{align}

\frac{1}{v}\frac{dv}{d\tau}&=-\frac{4}{3(1+w)t}\frac{dt}{d\tau},\tag{8}\\

\frac{d(\log v)}{d\tau}&=-\frac{4}{3(1+w)}\frac{d(\log t)}{d\tau},\tag{9}\\

\log v&=-\frac{4}{3+3w}\log t+\log C,\tag{10}\\

v&=Ct^{-4/(3+3w)},\tag{11}\end{align}$$

which leads to (6) in the form

$$\frac{d^2t}{d\tau^2}=-\frac{2}{3+3w}C^2t^{-(7+3w)/(3+3w)}.\tag{12}$$

Not very useful, mind you, but this is just a characterization of time dilation, after all.

No, the important equation is (11), which tells us how the speed of the particle decreases over time. It's not a force: It is a result of the particle moving ever closer to a location in space where its velocity matches the velocity of the comoving frame, in which the universe appears isotropic.