yuiop said:
This expression is the integrated distance between the Shwarzschild radial coordinates ro and ri. <snip>
OK, thanks. I saw this huge, unwieldy expression and had no idea where it came from. It does appear to look right - Maple spits out something that appears to be equivalent, though it looks different on first glance.
Now if the rod is moving with respect to the Schwarzschild observer at infinity then the integrated length of the rod with equal Schwarzschild coordinate times, is the Schwarzschild coordinate length of the rod, but it will not agree with the proper length of the rod as could be measured by infinitesimal rulers attached to the moving rod. Now I think it is reasonable to assume that if the rod is falling and the rod is sufficiently rigid to overcome tidal forces stretching it (to an acceptable accuracy or assume a hypothetical infinitly rigid rod), then the proper length of the falling rod should not change as it falls, but its integrated coordinate length will change. Agree?
You mentioned the "barn and the pole" paradox in another post, and this as well, so I think you're seeing the point.
When you calculate the integrated distance of a moving object that has one end at Schwarzschild coordinate R1 and the other end at R2 at the same Schwarzschild cordinate time ... you are in fact calculating the proper length of a stationary rod that spans those two coordinates.
Yes, this is what you are calculating.
Both you and Passionflower seem to be very enamored of coordinates. So perhaps it's worth noting how one would create locally Lorentz coordinates at a point.
Suppose we have some metric, say the Schwarzschild metric, g_{ij}, and we introduce new coordinates (r,t) -> (r',t') Because g_ij is a covariant (i.e. subscripted) tensor , it transforms in the following way, where I've written out everything longhand for clarity rather than using the Einstein convention
<br />
g_{r'r'} = g_{rr} \frac{\partial r}{\partial r'} \, \frac{\partial r}{\partial r'} + g_{rt} \frac{\partial r}{\partial r'} \, \frac{\partial t}{\partial r'} + g_{tt} \frac{\partial t}{\partial r'} \, \frac{\partial t}{\partial r'}<br />
<br />
g_{r't'} = g_{rr} \, \frac{\partial r}{\partial r'} \, \frac{\partial r}{\partial t'} + g_{rt} \frac{\partial r}{\partial r'} \, \frac{\partial t}{\partial t'} + g_{tt} \frac{\partial t}{\partial r'} \, \frac{\partial t}{\partial t'}<br />
<br />
g_{t't'} = g_{rr} \frac{\partial r}{\partial t'} \, \frac{\partial r}{\partial t'} + g_{rt} \frac{\partial r}{\partial t'} \, \frac{\partial t}{\partial t'} + g_{tt} \frac{\partial t}{\partial t'} \, \frac{\partial t}{\partial t'} <br />
Because the Schwarzschild metric is diagonal, it's easy to find the expression r = alpha r' and t = beta t' to make it an identity matrix, creating a local Lorentz frame in terms of the new coordinate values.
i.e.
<br />
g_{r'r'} = g_{rr} \alpha^2<br />
<br />
g_{t't'} = g_{tt} \beta^2<br />
and r' = r / alpha, t' = t/ beta, we then can think of r' and t' as being the "local" coordinates.
Finding the proper linear relationship between (r,t) and (r',t') to represent the "local coordinate system of the moving particle" is also possible, but more work. The point is to diagonalize the metric tensor in the "local" coordinates.
Again, this is an approximation type thing - the transformation makes the metric Lorentzian only over a small region of space-time. Because space-time is curved, you can introduce coordinates to make it Lorentzian everywhere.