The Schwarzschild Metric - A Simple Case

[starthaus]

Correct. If you have a long enough rod, rather than integrate

$$\int_{r1}^{r2} \frac{dr}{\sqrt{1-1/r}}$$

There is no justification for such an integral for the problem in the OP.

and then multiply the result by gamma,

Why would you do such a meaningless thing?

which varies, you want to integrate instead

$$\int_{r1}^{r2} \gamma\left( r \right) dr = \int_{r1}^{r2} \frac{dr}{1-1/r}$$

since in this case $\gamma$ = 1/sqrt(1-v^2) = 1/sqrt(1-1/r)

I hope you realize that the multiplication of the integral and the multiplication of the integrand followed by integration produce different results, so the two computations produce different results notwithstanding that both calculations are equally meaningless.
Is this some sort of test to see how many things we can detect in a post?

 

[yuiop]

I am running some numbers to verify I understand it.

If we take the front clock at R=40 we have 180.9850160 time units left.
Now let's take an epsilon of 13 time units, this will give us a back coordinate value of R=41.98100155. Thus the coordinate distance between the front and the back is 1.98100155 which gives a proper distance (for a static observer) of 2.005621996. The two are not identical due to the curvature of spacetime.

I calculated r for t=-60-0 for the front and t=-73-13 for the back and plotted the results. I also calculated the coordinate and proper distance (for a static observer).

I wonder if you would like to run some numbers using the same parameters I used in post 64 quoted below:

As I said before, it is probably easier to relax the fall from infinity requirement (although we might try that later) and assume a drop from initially at rest from a finite height.

After a coordinate time of 5 the leading clock will have fallen to 9.99437 and in the same coordinate time the trailing clock will have fallen to 10.953 and the coordinate gap will have increased from 1 to 1.009 if both particles were initially at rest and released at the same coordinate time from heights 10 Rs and 11 Rs.

It is a bit more interesting if you let the particles fall further.

If we start with the same release heights and let the leading particle fall to height 2 Rs then the coordinate fall time is 2.64 and trailing clock is at 4.64 Rs.

I am using the equations given in post 48 to obtain these results, but I have to numerically solve them as the mathematical software I use is unable to find a symbolic solution for r when t is know.

Here is a list of results for various initial and final heights for the trailing clock, all with the leading clock starting one coordinate unit lower and end point determined by the leading clock arriving at 2 Rs.

11 : 4.64
101 : 9.81
1001 : 21.86
10001 : 48.42
100001 : 105.987
1000001 : 230.22
10000001 : 498.02
100000001 : 1075.06
1000000001 : 2318.31

Very approximately, each tenfold increase in the initial drop height results in a two fold increase in the coordinate separation at the end point.

No, it is not that simple. All we have done is determined the coordinate separation of the unconnected free falling particles. We have still not determined the proper separation of the freely falling particles in the rest frame of the particles and I not even sure at this point if it is possible to define what that is. It is as you probably already aware, very difficult to operationally define the proper length of the falling particles for non infinitesimal separations. It does however give an indication of the tidal stretching in coordinate terms.

The attached diagram is a graph in Schwarzschild coordinates of the objects falling from 10 Rs and 11 Rs.

and see how the results compare. The method I used was essentially based on using equal coordinate falling times so the results should be similar to yours and Pervect's.

Now I do see a stretch both in coordinate and proper time. Is this the tidal stress? One problem with this approach is that we cannot run this past the event horizon, the major spaghettification happens closer to r=0, rather than changing coordinate charts is there any way we can use proper time for this?

For equal proper time you could try the method outlined by Mentz in post 147 or even use the equation you gave earlier for "time till doom"

$$tau = -(2/3)\sqrt{r^3/rs}$$

which is basically the same thing. As you know, you can use those equations from infinity all the way to through the event horizon to the singularity.

 

[pervect]

It would be interesting to compare with whatever comes out as "distance" via this method to something more standard, like Fermi or Riemann normal coordinates (MTW, pg 285, 332).

I probably haven't stressed enough the fact that there are many "distances" in GR - depending for the most part on the definition of simultaneity, which is a coordinate dependent concept.

Unfortunately, the equations (as usual) are rather intractable, but there's enough for numerical analysis.

Normal" coordinates, whether they be Fermi-normal or Riemann-normal coordinates, measure distance along a space-like geodesic passing through some specified point. Fermi normal coordinates would probably be the best choice for this situation since we have a worldline. So in the region of applicability, for Fermi normal coordinates we use the time along the worldline for the time coordinate, and the distance along the space-like geodesic as the space coordinate.

We can paramaterize a space-like geodesic as a curve t(s), r(s), where s is an affine parameter.

If we require this space-like geodesic to pass through some point (r0,t0), and to be orthogonal to the worldline of an object "falling from infinity" at that point, I get the following equations which determine the geodesic r(s), t(s).

I've assumed that r_s = 1.

$$\frac{dr}{ds} = -\sqrt{1 + \frac{1}{r0} - \frac{1}{r}}$$

$$\frac{dt}{ds} = \frac{r}{\sqrt{r0} \, \left( r -1 \right) }$$

though I don't think we need dt/ds, actually.

and of course at t=t0, r= r0 - it's simplest to take t0 = 0.

Finding the value of s at which r=1 should be the "distance to the event horizon in Fermi normal coordinates", s is an "affine paramaterization" along the geodesic which measures total length.

This is a solution for ingoing geodesics because dr/ds is negative - for outgoing geodesics, just change the sign - making s negative should also work.