The length of a falling elevator

[yuiop]

and solve for the coordinate distance (dr) between two events with the same coordinate time t so that dt = 0:

$$dr = dS \sqrt{1-r_s/r}$$

From the above we see that the coordinate length (dr) of an infinitesimal rod is smaller that the proper length (dS) of the same rod by the gravitational gamma factor. This is in effect, gravitational length contraction.

No, for a correct derivation see Rindler again , section 11.2. Rindler shows why :

$$dr = dl \sqrt{1-r_s/r}$$

where $$dl$$ is the "radial ruler distance".

Solve for dS. We get:

$$dS = dr / \sqrt{1-r_s/r}$$

Solve for dl. We get:

$$dl = dr / \sqrt{1-r_s/r}$$

Therefore we conclude dl = dS.

 

[starthaus]

Solve for dS. We get:

$$dS = dr / \sqrt{1-r_s/r}$$

Solve for dl. We get:

$$dl = dr / \sqrt{1-r_s/r}$$

Therefore we conclude dl = dS.

The question was, how did you end up with the incorrect formula for calculating $$\Delta S$$?

 

[Passionflower]

In SR if an object with proper length L=1 is moving with velocity 0.8c relative to an inertial observer, the coordinate length of the moving object is L' = 0.6

The proper length remains constant but different observers measure different coordinate lengths. This is "velocity length contraction".

In GR (Schwarzschild coordinates) if a short object has proper length L=1 and it is located at say R=2Rs then its coordinate length is L' = sqrt(1-Rs/R) = sqrt(2) = 0.7071.

The proper length remains constant but its coordinate length varies upon where it is located. This is the "gravitational length contraction".

If a local observer measures the velocity of the short falling object to be 0.8c as it passes and the local observer is located at R = 2Rs, then the coordinate length of the falling object at R=2Rs according to the Schwarzschild observer at infinity is L' = 1*0.6*0.7071 = 0.4243. i.e the object is subject to both gravitational and velocity length contraction. (The length of the falling object as measured by the local observer as it passes him, is simply 0.6 i.e exactly the same as in SR)

All the above assumes an infinitesimal falling object. For an extended non-infinitesimal object, integrated distances have to used.

Ok, I see what you are saying, however you suggest we use the radar distance for this. Now can that be right? Take for instance the Rindler case, here radar and ruler distance are not equal, are you suggesting that in the Schwarzschild case ruler and radar distance are identical? Clearly if they are not, we must use a different formula.

 

[yuiop]

$$dS^2 = -(1-r_s/r) c^2 dt^2 + (1-r_s/r)^{-1} dr^2$$

and solve for the coordinate distance (dr) between two events with the same coordinate time t so that dt = 0:

What does $$dt=0$$ mean for you?

From the above I thought it was clear that by dt=0 I meant the two coordinates had the same coordinate time as measured by an observer at infinity and the events are simultaneous in those coordinates.

 

[yuiop]

The question was, how did you end up with the incorrect formula for calculating $$\Delta S$$?

The equation you are using is for a stationary object. In this thread we are considering a falling (moving) object.

 

[starthaus]

The equation you are using is for a stationary object.

It was a simpler example meant to illustrate how to do a correct derivation.

In this thread we are considering a falling (moving) object.

Well understood. You still got it wrong. That was the point.

 

[yuiop]

Ok, I see what you are saying, however you suggest we use the radar distance for this. Now can that be right? Take for instance the Rindler case, here radar and ruler distance are not equal, are you suggesting that in the Schwarzschild case ruler and radar distance are identical? Clearly if they are not, we must use a different formula.

I was not actually using radar distances, but ruler distances. I accept that they are different over extended distances and depending on where the radar measurement is made from.

We can relate ruler measurements to extended radar measurements like this. Take a very short ruler. Measure its radar distance from both ends. If the ruler is sufficiently short, the radar length measured from either end is approximately the same. If the error is unacceptable, make the ruler shorter. Make lots of such rulers. Lay them end to end from r1 to r2. This is the ruler length from r1 to r2 and is (I think) equivalent to the proper distance from r1 to r2 and the same as what you obtain when you calculate the integrated distance from r1 to r2. This distance is not the same as the total radar measurement if you send a signal from r1 to r2 and divide by 2 and the total radar measurement from the other end will be different again.

The measurement of a short falling rod by a local observer can be done like this. the local observer holds his ruler vertically. He notes the time the leading part of the falling rod passes the top of his vertical ruler. He notes the time the leading part of the falling rod passes the lower part of his ruler. The divides the ruler length by the time take to obtain the local velocity of the ruler. He can do the same measure of velocity for the trailing edge of the falling rod. He may obtain a slightly higher reading for this, but for a sufficiently short falling rod the difference is small and he can use an average velocity. Now that he know the velocity of the falling rod, he can measure how long it takes between the front of the rod and the back of the rod to pass the same clock at one end of his ruler. From v*t he can obtain the length of the passing rod. For a sufficiently short rod and a sufficiently short measuring ruler, the difference in proper clock rates between the clock at the top of his ruler and the the bottom of the his ruler is negligible. For example consider the difference in proper clock rates between two clocks that are vertically separated by 1 millimetre in the Earth's gravitational field. This would be almost impossible to detect. If that is not good enough, make the distance shorter. This is (sort of) how calculus works.

 

[Passionflower]

I was not actually using radar distances, but ruler distances.

This is the formula you suggest we use right?

$$\Delta S = \int^{r2}_{r1} \frac{1}{(1-r_s/r)} dr \qquad = \qquad (r2-r1) + r_s \ln \left(\frac{r2-r_s}{r1-r_s}\right) \qquad = \qquad \Delta r + r_s \ln \left(\frac{r2-r_s}{r1-r_s}\right)$$

Now are you saying this is not the radar distance formula?

 

[yuiop]

This is the formula you suggest we use right?

$$\Delta S = \int^{r2}_{r1} \frac{1}{(1-r_s/r)} dr \qquad = \qquad (r2-r1) + r_s \ln \left(\frac{r2-r_s}{r1-r_s}\right) \qquad = \qquad \Delta r + r_s \ln \left(\frac{r2-r_s}{r1-r_s}\right)$$

Now are you saying this is not the radar distance formula?

That's correct. It is the ruler distance from r1 to r2 of a physical calibrated stationary ruler that extends from r1 to r2, where r1 and r2 are the simultaneous Schwarzschild radial spatial coordinates of a falling object with proper length Delta S. The proper length is not changing as it falls, just the coordinate length. The radar length measured by an observer falling with the rod is more complicated, but I would be interested in eventually finding out what that would be.

 

[Passionflower]

That's correct. It is the ruler distance from r1 to r2 of a physical calibrated stationary ruler that extends from r1 to r2, where r1 and r2 are the simultaneous Schwarzschild radial spatial coordinates of a falling object with proper length Delta S. The proper length is not changing as it falls, just the coordinate length. The radar length measured by an observer falling with the rod is more complicated, but I would be interested in eventually finding out what that would be.

I see, so then we have to conclude that the radar distance formula between two stationary objects is the same formula as the ruler distance formula for falling from infinity. As an extension we could get the complete formula that includes an initial velocity and one where the fall is from a given r coordinate value.

And indeed, a radar distance formula for with the same options as for the ruler distance.

 

[starthaus]

This is fine if we are measuring the coordinate length of a stationary object. If the object is moving we have to be careful to specify equal coordinate times for the events locating the ends of the object.

Again, I think this is fine if the measured object is not moving with respect to the Schwarzschild coordinates.

I suspect that when all the dust settles, we will find that the proper length of a falling object is the product of the gravitational and velocity related length contraction factors. The proper length $\Delta S$ of a falling object that extends from Schwarzschild coordinate (r2,t) to (r1,t) at coordinate time t, should then be:

$$\Delta S = \int^{r2}_{r1} \frac{1}{(1-r_s/r)} dr \qquad = \qquad (r2-r1) + r_s \ln \left(\frac{r2-r_s}{r1-r_s}\right) \qquad = \qquad \Delta r + r_s \ln \left(\frac{r2-r_s}{r1-r_s}\right)$$

... if I have got it right and my educated guess is correct.

This is not a derivation, this is just a wild guess. Besides, it contradicts your other "guess" from this thread where you claim (again, with no proof) that:

Again this the coordinate length of short stationary object. For a short moving object the equation is:

$$ds=\frac{dr}{\sqrt{1-\frac{2GM}{rc^2}}\sqrt{1-\frac{v^2}{c^2}}}$$

where v is the local velocity.

Physics is a precise science, not a guessing game.