Paul Colby said:
There is no way for an isolated mass to acquire a coordinate velocity (in these coordinates).
More precisely, a test object that starts out at rest in these coordinates, and always moves on a geodesic (i.e., always feels no force), will never acquire any coordinate velocity in these coordinates; it will stay at the same spatial coordinates forever. So any change in its
proper distance from any other similar test object, also at rest in these coordinates and moving on a geodesic, must be due to a change in the metric itself, i.e., to a GW passing (assuming that's the only kind of spacetime curvature we're considering).
Paul Colby said:
What I find confusing is how this (coordinate) force could become proportional to the second derivative of the metric strain
Ah, I see; you're wondering how the second derivative of the coordinate ##x^\mu## gets "translated" into the second derivative of the metric strain. See below.
Paul Colby said:
the geodesic equation I quoted above actually appears in non-geodesic motion as,
$$
\frac{dx^{\mu}}{ds^{2}}+\Gamma{^\mu}_{\alpha\beta}\frac{dx^\alpha}{ds}\frac{dx^\beta}{ds}=F^\mu
$$
where ##F^\mu## is the applied 4-force
The first term here has a typo, it should be ##d^2 x^\mu / ds^2##. With that correction, yes, this equation is the "force equation" for non-geodesic motion, given a choice of coordinates so that the ##\Gamma##s are well-defined. (The only caveat is that the "force" here is actually force per unit mass, i.e., acceleration.) And at the start, when the GW is just starting to pass by and there is no strain, the term in ##\Gamma## is zero (because the only nonzero 4-velocity component is the ##t## component and there are no nonzero ##\Gamma##s with lower indexes ##tt##). So the "force equation" is just
$$
\frac{dx^{\mu}}{ds^{2}} = F^\mu
$$
What is ##dx^\mu / ds^2##? Let's suppose that we put one test mass at ##x^i = 0## for ##i = 1, 2, 3##, and a second test mass at ##x^1 = L##, ##x^2 = x^3 = 0##; i.e., we align the two masses and the spring along the ##x## axis. The proper distance between the two masses is then ##D = \sqrt{g_{xx}} L##, which before the GW passes is just ##L##, since ##g_{xx} = 1## if no GW is present.
Now,
if the spring were not there, so that the test masses were both moving geodesically, then the proper distance between them while the GW is passing would be ##D = \sqrt{1 + h_{xx}} L##. For small ##h## we can write this as ##D = \left( 1 + \frac{1}{2} h_{xx} \right) L##. The second derivative of this is then ##d^2 D / ds^2 = \frac{1}{2} L d^2 h_{xx} / ds^2##. If we set ##h_{xx} = h_+ e^{i \omega t}##, then we get the expression on the RHS of the equation you referred to in your OP.
Now consider what happens if we put the spring in. The expression for ##d^2 D / ds^2## above defines what geodesic motion would be; i.e., it defines the motion that the test masses would have if there were no spring. So with the spring present, that same expression defines the "driving force" that the spring is resisting. For example, if the spring had an infinite spring constant, so that it maintained the two test masses at exactly the same proper distance ##L## no matter what, then the equation above would give the coordinate acceleration of the masses, in the coordinates we're using, where geodesically moving test masses would have zero coordinate acceleration.
This is all heuristic, but to delve much deeper into it would require more math than is appropriate for a "B" level thread.
