Lightspeed.

[Wyrm]

Er... in the comoving frame (i.e., comoving with the bulk of the matter of the universe), r' = Hr, where H is the (time-dependent) Hubble parameter, which has units of 1/time, so the separation velocity r' does have units of distance over time.

The Hubble constant the expansion rate, as I understand it. At least that's how MTW treats it. Yes, recesslional velocity does have units of distance over time, more on this in a sec.

You must have misunderstood. The Hubble constant isn't a rate of expansion, as it would have to have dimensions of [L/T] — and it doesn't: the usual units of the Hubble constant are (km/s)/Mpc, which has dimensions [(L/T)/L] = [1/T]. It's reciprocal is the Hubble time, which is to first order the age of the universe. (I find it unlikely that the Phone Book made such an elementary blunder.)

The true rate of expansion of the observable universe is the rate at which our absolute past expands.

[Paolo]

You must have misunderstood. The Hubble constant isn't a rate of expansion, as it would have to have dimensions of [L/T] — and it doesn't: the usual units of the Hubble constant are (km/s)/Mpc, which has dimensions [(L/T)/L] = [1/T]. It's reciprocal is the Hubble time, which is to first order the age of the universe. (I find it unlikely that the Phone Book made such an elementary blunder.)

Went back and checked: "One parameter close to the observations is the Hubble expansion rate today, i.e., the 'Hubble constant...' (p.772, MTW, emphasis theirs)

The true rate of expansion of the observable universe is the rate at which our absolute past expands.

Not sure what you mean by "absolute past," but if you're saying what I think you're saying then we should note that Hubble's parameter depends on comoving time and you can compute the conformal time coordinates by dividing out the time derivative of the scale factor and integrating from t = 0 to now in cosmological time. Correct?

[Kuroneko]

The Hubble constant the expansion rate, as I understand it.

The Hubble parameter is H = a'/a, where a = a(t) is the scale factor, and ' denostes differentiation with respect to cosmological time. This found in any other textbook; you almost-quote this from MTW later.

Went back and checked: "One parameter close to the observations is the Hubble expansion rate today, i.e., the 'Hubble constant...' (p.772, MTW, emphasis theirs)

That's correct. What I don't understand is why you perceive this to be a problem.

I just finished reading Davis and Lineweaver on this matter, and I guess they're talking about me and my textbooks, but my understanding is that we only arrive at superluminal velocities by dividing proper distance today by cosmological time, even though proper and comoving distance differ in the past.

Let me explain procedurally what I mean and then you can see if you mean something else. By 'at rest', I mean at rest relative to the bulk of the matter in the universe (which is simple in FRW models, as they're homogenous), and by 'cosmological time', I'm referring to time as measured by an observer at rest. Suppose we have two particles, both at rest. Then, at any instant of cosmological time t, we can calculate the length of the shortest spacelike geodesic between the two particles, and call it 'proper distance' R = R(t).
Result: R(t) = [a(t)/a(t₀)]R(t₀).
Conclusion: R'/R = a'/a = H(t), where H(t) is the Hubble parameter.
Addendum: "Separation velocity" between particles at rest would be v_r(t) = H(t)R(t). "Peculiar velocity" adds to this for particles that are not at rest with the bulk of the matter in the universe.

I'm not sure what you're doing; it sounds like you're taking R/t instead of dR/dt, which makes no sense. (Although that would explain why you thought that it had very little physical meaning.) Note that all of the above still works even in FRW models that had no Big Bang, and can even be generalized to any spacetime that can be foliated by spacelike surfaces, except that we lose the scale factor in that case.