I Proper distance between surfaces of constant r

1. Oct 29, 2016

timmdeeg

How about proper length in curved spacetime?
Let's consider the radial distance between two spherical shells in Schwarzschild spacetime. The proper distance between them follows from the spacelike form of the metric with $dt=0$ for simultaneity. So I think having the r-values of the shells the proper distance calculated by this should equal the rest length of a rod between them, correct? But how would two observers, one on each shell, measure the proper distance between them? Is that possible with the knowledge of the frequency shift they see each other?

2. Oct 29, 2016

Staff: Mentor

If the rod is at rest relative to the shells, yes.

One way would be to have a whole family of observers, each at rest at some slightly different value of $r$ between the shells, and each measuring the proper length of the small piece of the rod centered on them. Then you add up the lengths of all the small pieces.

If you know the mass of the central object and the $r$ values of the two ends of the rod, that's sufficient to calculate the proper length of the rod. The frequency shift alone is not enough; that tells you the ratio of the $r$ coordinates (assuming you know the mass of the central object), but not their absolute values, which is what you need to calculate the proper length. If you know the frequency shift of both ends of the rod relative to infinity, or to some known finite $r$ value, that is enough, since from that you can calculate the $r$ values of the two ends of the rod.

3. Oct 30, 2016

timmdeeg

Understand. Another possibility to know the ratio of the $r$ coordinates independently from the frequency shift would be that the two shell observers measure the free-fall velocity $dr/dt$ (from rest at infinity) of objects, right?

4. Oct 30, 2016

Staff: Mentor

The observers can't measure $dr/dt$, at least not directly; then can only measure the velocity of a free-falling object in their local inertial frame. The ratio of those velocities, by itself, is not sufficient to even know the ratio of the $r$ coordinates (let alone their absolute values, which is what you really need).

Also, $dr/dt$ for an object falling in from rest at infinity does not actually mean what you appear to think it means; it is a coordinate speed, not an actual relative speed. It does not continue to increase as $r$ gets smaller. (I am assuming that we are using Schwarzschild coordinates.) It starts increasing as we go inward from infinity, but then it stops increasing and starts to decrease, and approaches zero as $r$ approaches $2M$, the event horizon. This is because of the behavior of Schwarzschild coordinates near the horizon.

5. Oct 30, 2016

timmdeeg

Hmm, sorry, I meant something different, not coordinate speed, but the speed of the free-fall objects relative to the respective shell as measured locally by the $s$hell observers $dr_s/dt_s=-\sqrt(2M/r)$, whereby in this case $dr_s=d_\sigma$. So, the ratio of the speeds should yield the ratio of the $r$-coordinates of the two shells, if I see it correctly.

6. Oct 30, 2016

Staff: Mentor

This is not correct as you write it; $- \sqrt{2M / r}$ is the rate of change of $r$ with respect to the proper time of the infalling object, not the shell observer. Also, you don't want the rate of change of $r$; you want the rate of change of distance in the shell observer's local inertial frame, which is not the same as $r$.

7. Oct 30, 2016

timmdeeg

I have seen this formula for radial velocity 'as measured by the shell observer' in "Exploring Black Holes" page 3-15, but obviously misinterpreted it. Thank you for your comment and help.

8. Oct 30, 2016

Staff: Mentor

Your reference to Exploring Black Holes, which is a reliable source, made me go back and check the math, and it turns out I was mistaken. Let me run briefly through the math. We start with the equation for $dr / dt$ in Schwarzschild coordinates for an object free-falling from rest at infinity (I am using units where $c = 1$):

$$\frac{dr}{dt} = - \left( 1 - \frac{2M}{r} \right) \sqrt{\frac{2M}{r}}$$

Now we rewrite the LHS as follows, using $r_s$ and $t_s$ to denote coordinates in the shell observer's local inertial frame:

$$\frac{dr_s}{dt_s} \frac{dr}{dr_s} \frac{dt_s}{dt} = - \left( 1 - \frac{2M}{r} \right) \sqrt{\frac{2M}{r}}$$

From the Schwarzschild metric, we have $dr / dr_s = dt_s / dt = \sqrt{1 - 2M / r}$. (Note carefully the position of the subscript $s$ in each of the derivatives.) Plugging this into the equation above cancels the factor of $1 - 2M / r$ and leaves us with the expression you wrote down, which is therefore correct. I apologize for the mixup on my part.

9. Oct 30, 2016

timmdeeg

Ah, I see, thanks for explaining.