I'm still wading through the paper, but I remain skeptical. A key section, I think, is near eq 30:
This is the key equation. The complexity of this velocity expression as computed by observers external to the distribution of matter is in very sharp contrast to the simplicity of the proper velocity form \beta = \sqrt{F/r} as witnessed by local observers. However, it is the former that's relevant for astronomical observers.
(emphasis in original).
While I can appreciate that the expression for dr/dt in the particular coordinate system that C&T chose might look more complicated, if they are correct in their approximations there is no reason it should be numerically significantly different from the simpler expression they computed for a local observer.
Furthermore, conceptually, the external observer won't actually be measuring dr/dt. He'll rather be measuring some redshift factor, z, or possibly some sort of "apparent angular width". Let's assume, however, that he measures z for simplicity. This is what people usually measure when they measure rotation curves. One doesn't actually take some radar measurement to measure distance as a function of time and, even if one did this, it still wouldn't be exactly the same thing as a measure of the r coordinate, it would only be approximately the same. So let's calculate what we actually measure, z.
It's worth noting that we're measuring radial velocities in this problem, and orbital velocities in the more usual case of rotation curves. This doesn't have a major effect, except that we have to distinguish between different portions of the cloud by some means other than angular position.
Then if the local velocity at the edge of the infalling cloud is sqrt(2m/r) (see the text near eq 8) which I'll quote in part:
The geodesic solution for dr/dt and the metric coefficeints g_00 and g_11 of (1) are used to evalutate the proper radial velocity.
This equal sqrt(2m/r) in magnitude for particles released from rest at infinity and is seen to approach 1, the speed of light, as r approaches 2m. ... However, for asymptotic observers who reckon radial distance and time increments as dr and dt, the measured velocity is
\frac{dr}{dt} = -(1-\frac{2m}{r}) \sqrt{\frac{2m}{r}} \hspace{1 in} (8)
But, as I noted, our observer at infinity will not actually measure 8)., What he'll (probably) actually measure is the redshift, z. We can compute z by finding the redshift due to the local velocity relative to a local stationary observer multiplied by the gravitational redshift from the local stationary observer to the distant stationary observer:
i.e.
1+z = \left( \frac{1}{\sqrt{g_{00}}} \right) \left( \sqrt{\frac{1+\sqrt{\frac{2m}{r}}}{1 - \sqrt{\frac{2m}{r}}}} \right) = \frac{1+\sqrt{\frac{2m}{r}}}{1-\frac{2m}{r}}<br />
where the first term is the gravitational redshift, with g_00 = 1-2m/r, and the second term is the relativistic doppler shift due to the local velocity. The algebraic simplification is done by multiplying the numerator and denominator inside the square root by (1+sqrt(2m/r)) and simplifying.
Comparing this formula for 1+z to 8), we see that as we would expect, z-> infinity in the strong field case. If we series expand our expression for 1+z, we find that
z+1 \approx 1 + \sqrt{\frac{2m}{r}} +\left(\sqrt{\frac{2m}{r}}\right)^2
and the second term can be ignored for small enough v, being quadratic in the square root.
We should also note that if we series expand the formula for relativistic doppler shift:
z + 1 = \sqrt{\frac{1-v}{1+v}} in a series for small v, we get z+1 = 1 - v + v^2/2
which explains why measuring z is equivalent to measuring v for small z (or small v).
The most important point is that the local velocity determines the local redshift (from our infalling observer to a local stationary observer), and that the redshift is multiplicative, so that the total redshift to our distant observer is the local doppler redshift multiplied by the gravitational redshift from our local observer to our distant observer.
Thus, if the gravitational redshift is negligible because we are in a weak field, all the redshift must be due to the local velocity. Thus playing games with the coordinates can't make the distant velocity different from the local velocity unless we have significant gravitational redshift - but it has been assumed by C&T that we do not have significant gravitational redshift.
Redoing the analysis in terms of 'z' instead of coordinates puts the problem in slightly closer touch to what is happening physically IMO, and to my mind it makes it very clear that the local redshift must be the same as the redshift at infinity if we ignore the gravitational component of the redshift.
