# The length of a falling elevator

• yuiop
In summary: A Schwarzschild observer at infinity:The Schwarzschild observer at infinity measures the length of the falling elevator to be 0.2.2) A local observer at r=4m when the falling elevator is passing:The local observer measures the length of the falling elevator to be 0.2 as well. This is because the elevator is passing the observer at a velocity close to c (since it is falling towards the black hole), causing significant length contraction in the direction of motion.3) A free-falling observer inside the elevator using radar measurements from the top of the elevator:The free-falling observer at the top of the elevator measures the length of the elevator to be 0.2. This is because
yuiop
This relates to the thread "https://www.physicsforums.com/showthread.php?t=431712"" started by Passionflower, but is hopefully even simpler (but non trivial).

The basic question, is what is the length of a falling elevator (or ruler) that is reasonably rigid, according to:

1) A Schwarzschild observer at infinity.
2) A local observer at r=4m when the falling elevator is passing.
3) A free-falling observer inside the elevator using radar measurements from the top of the elevator.
4) A free falling observer inside the elevator using radar measurements from the bottom of the elevator.
5) How do any of the above measurements compare to two particles initially co-located with the top and bottom of the elevator and released at the same time as the elevator, but unconnected to the elevator and allowed to free-fall naturally.

By "reasonably rigid", I mean the forces that hold the elevator together are significantly stronger than than the tidal forces pulling it apart and the length of the elevator is short enough to minimise the tidal forces.

In trying to analyse this question (as a first step to answering Passion's question in the other thread) I strated by assuming these two statements are true, "Spacetime is locally Minkowskian for a local stationary (accelerating) observer in Schwarzschiuld coordinates" and "An observer in a free-falling elevator would be unable to detect if they are free falling or stationary in flat space, without reference to information external to the elevator".

My initial investigation suggests that the two statements are incompatible and one of them has to be rejected. The second statement has the condition "exluding tidal effects" and I suspect it this condition that allows us to reject the second statement.

I have already asked a lot of questions and I do not expect anyone to answer all of them, but if someone could supply a definitive answer and/or equation to anyone of the questions, that would be a significant start.

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Well, to measure the length of anything, you integrate the lorentz interval, ds, along some curve. The trick is - what curve do you use?

If you have a coordinate system in mind, the answer is easy. You just pick the position of the head of the ruler and the tail of the ruler "at the same time", and integrate along the coordinate axis of constant time joining the head and tail of the ruler, and that's your length.

The question of how to determine the "proper" length of a ruler is a bit trickier, at least for a long ruler.. The way I'd suggest is this. Take the ruler, and mark out its subdivisions. If the ruler is rigid, the subdivisions always are the "same distance" apart. If you make the subdivisions small, you can use radar, or the notion of distance in the local Lorentz frame, to determine the distance between the subdivision. I think it's already been noted in previous threads that the radar distance back-front and front-back varies for a long accelerating ruler, hence the importance of "dividing it up" in this manner. You'll know your subdivisions are small enough when your different notions of distance (front radar distance, back radar distance, distance in one of the two possible local frames) all agree with each other. If there is a significant discrepancy, you need more subdivisions.

If you are thinking of a "rigid" ruler, this should be good enough. You might also want to "sketch out" the curve that you are measuring your length along on a space-time diagram. You are ultimately measuring the interval along some curve, if you use the radar method, for instance, the curve goes from the midpoint of the sent and received signals on the emitter worldline to the time of arrival of the radar pulse on the reflecting worldline. Sketching out the curve in this manner is quite helpful in seeing what's going on.

Hi Pervect,

I realize there many "fine points" to resolving this question. However, we know that a "very" short stationary ruler with length = 1 measured locally will have length = 1/sqrt(1-2GM/(Rc^2)) as measured by a Schwarzschild observer at infinity and we are here assuming a "rigid" ruler that does not collapse under its own weight. We should in principle be able to lower a ruler that has been calibrated to have length 1 at infinity and slowly lower to the low down ruler and when the two rulers are alongside each other and at rest wrt to each other, they should both have the same length. Agree?

Now is it possible to determine (to first order and for a sufficiently short ruler) whether or not a falling ruler length contracts (as measured the Schwarzschild observer at infinity) by a factor proportional to the velocity related time dilation factor of the falling ruler or by a factor proportional to the gravitational time dilation, or by the product of both velocity and gravitational factors (as is the case for the ticking rate of a falling clock)?

yuiop said:
Hi Pervect,

I realize there many "fine points" to resolving this question. However, we know that a "very" short stationary ruler with length = 1 measured locally will have length = 1/sqrt(1-2GM/(Rc^2)) as measured by a Schwarzschild observer at infinity

I think we're already have differing notions of what "length" means, alas.
It appears to me that you are taking the Schwarzschild coordinate of one end of the ruler, subtracting the Scwharzschild coordinate of the other end, and calling the difference in coordinates the "length". Or am I mistaken? How do you justify the above statement about the length of the ruler?

yuiop said:
3) A free-falling observer inside the elevator using radar measurements from the top of the elevator.
4) A free falling observer inside the elevator using radar measurements from the bottom of the elevator.
Edited: Upon reflection I think the solution below applies to a stationary elevator not an elevator free falling.

Ok, let's assume we have the following situation:

Elevator Height: 0.2
Bottom has a lower R value than the top of the elevator.

So let's calculate for when the elevator is at RBot = 3

First we need to know where the top is

We solve:

$$\sqrt {{\it ro}\, \left( {\it ro}-2 \right) }-\sqrt {3}+2\,\ln \left( {\frac {\sqrt {{\it ro}}+\sqrt {{\it ro}-2}}{\sqrt {3}+1}} \right) = 0.2$$

Which gives us: 3.117636346 for R0

If spacetime would have been flat we would have expected a coordinate of 3.2, but it is curved.

So since now we have RTop we can calculate the radar roundtrip time:

We use:

$$2\,x-6+4\,\ln \left( x-2 \right)$$

Substituting 3.117636346 for x gives us: 0.6801368916

But this is coordinate time!

Now we need to translate that into proper time for RBot and RTop

We multiply by:

$$\sqrt{1-{2 \over r}}$$

So for RBot we get: 0.3926772175
And for RTop we get: 0.4072244662

Assuming I did not mess up, perhaps the next step should be to determine the roundtrip times for a free falling elevator with a velocity both equal and not equal to escape velocity. Who can help?

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Hi.
I restate yuiop's question below in order to know whether my understanding of the problem is correct or not.

Lc: the world line of particle representing ceiling of elevator.
Lf: the world line of particle representing floor of elevator.
To make it simple, there is no interaction between the two particles. Both particles move on geodesic line. You can choose initial condition of particles as you think it reasonable when drawing lines.

Ef: the event that particle representing floor touches the sphere r=2m
Observer in system S draw time axis from Ef and finds the cross point with Lf, say Ec(S). We need label (S) because event Ec(S) do not coincide among the systems.

Question ; Find the space-time interval between events Ef and Ec(S) for some systems.

Regards.

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Passionflower said:
The elevator travels at escape velocity (e.g. free fall from infinity).

Assuming I did not mess up, perhaps the next step should be to determine the roundtrip times for a free falling elevator with a velocity not equal to escape velocity. Who can help?

Does this imply a limitation of freefall coordinate velocity acquired due to acceleration, dependant on the large mass but independant of the distance fallen??

Austin0 said:
Does this imply a limitation of freefall coordinate velocity acquired due to acceleration, dependant on the large mass but independant of the distance fallen??
If freefalling from infinity then the answer is yes.

When the freefalling object already has an initial speed then the acceleration of the gravitational field will speed it up even more, but up to a limit, this limit is sqrt(2), above this limit the gravitational field will actually slow the object
down, just as in the case of light.

By analogy think of the gravitational field as a river. The river current (gravitational strength) gradually increases towards a point where there is a big waterfall (event horizon) and boats fall vertically (point of no return). All the way at the beginning of the river, two boats, a small distance apart, not tied to the shores start to drift slowly due to the very slow current (fall from infinity). Farther down these boats start to drift apart in relation to each other due to the fact that the front current is stronger than the back (tidal acceleration). Many boats go on the river, some boats are motorized in an attempt to go faster (speed above escape velocity), that works up to a point but if they go very fast the current will actually slow them down. And then there are the beacons, boats that go down the river and try to resist the current so that they are standing still in the river (stationary observers). Sometimes the people at the end of the river like to send out probes to discover what is far down end of the river (radar signals). But the farther they send out the probes the longer it takes for them to come back (racing against the current), and if they send them farther than one certain point they have never returned (past the event horizon). Apparently something stops them from going back, even while they have a very strong motor aboard (inertial acceleration becomes infinite). River scientists speculate that once a boat reaches the waterfall, the boat really does not flow on the river anymore (timelike becomes spacelike) and that eventually the boats will crash on the hard rocks at the bottom of the waterfall (singularity).This is obviously a rather poor analogy but it can be helpful to imagine the different cases.

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Passionflower said:
If freefalling from infinity then the answer is yes.

When the freefalling object already has an initial speed then the acceleration of the gravitational field will speed it up even more, but up to a limit, this limit is sqrt(2), above this limit the gravitational field will actually slow the object
down, just as in the case of light.
OK good analogy. To make my question more specific:
Freefalling from infinity towards a massive black hole what would the acquired speed limit be relative to light??
Would it possibly attain a speed approaching c and would then be slowed down just as in the case of light
or would there be an inherent limit much less than c?
I am unclear on the meaning of sqrt(2), in this case?
Thanks

We're getting into definitional issues again, I think. Let's look at the Earth for an example of why I think coordinates are being missused here. I was hoping to do this interactively, but, apparently that's not working out.

We define locations on the Earth by specifying a pair of coordinates, lattitude and longitude. And, we note that a nautical mile is approximately one arc minute of lattitude, and one arc minute of longitude at the equator.

However, we don't measure length as number of degrees of lattitude or longitude. And an arc-minute of longitude does not always correspond to the same distance - it's a smaller difference near the poles.

So we don't use arc-minutes of longitude to measure distance, we use nautical miles. Similarly, we don't use "arc-minutes of longitude per second" to measure velocities. If we did, we'd say that naval vesels mysteriously traveled "faster" near the north pole, and "slower" near the equator. But instead, we say that the vessels travel the same speed everywhere, because we measure speed as distance vs time.

How do we calculate distances? The metric of the Earth's surface defines it for us. Given the metric, we find the small interval of distance, ds, between two nearby points. To find the distance between more distant points, we integrate this differental distance along some curve to find the distance. Usually, we want the shortest distance between two points, so we usually integrate along a "great circle", which is a geodesic path. But in any event, we find the distance by integrating the differential along some curve.

The schwarzschild coordinates are coordinates, just like lattitude and longitude. And they aren't the only choice by any means. Isotropic coordinates are also frequently used, and there are many more choices as well (Painleve coordinates, Novikov coordinates, Kruskal-Szerkes, etc, etc, etc.).

Thus, subtracting coordinates to define distances doesn't have any physical significance, because one can, and does, change coordinates "at will. Coordinates are a human invention, like markings on a map they shouldn't affect anything physical.

Relativity does throw one "twist" into this perscription that distances should be independent of coordinate choices. We need to separate space from time to define the notion of distance at all. So to this extent, we need to make some sort of coordinate choice - the choice of how to split space and time - before we can measure it. This explains why moving rulers have a different length than stationary ones - it's because of the space-time split. Once we do the split, though, it's pretty clear (at least I think it shouldl be clear!) that we measure the distance by using the metric, and by integrating the length of some curve in which all points have the same time coordinate.

pervect said:
I think we're already have differing notions of what "length" means, alas.
It appears to me that you are taking the Schwarzschild coordinate of one end of the ruler, subtracting the Scwharzschild coordinate of the other end, and calling the difference in coordinates the "length". Or am I mistaken? How do you justify the above statement about the length of the ruler?

pervect said:
We're getting into definitional issues again, I think. Let's look at the Earth for an example of why I think coordinates are being missused here. I was hoping to do this interactively, but, apparently that's not working out.

We define locations on the Earth by specifying a pair of coordinates, lattitude and longitude. And, we note that a nautical mile is approximately one arc minute of lattitude, and one arc minute of longitude at the equator.

However, we don't measure length as number of degrees of lattitude or longitude. And an arc-minute of longitude does not always correspond to the same distance - it's a smaller difference near the poles.

So we don't use arc-minutes of longitude to measure distance, we use nautical miles. Similarly, we don't use "arc-minutes of longitude per second" to measure velocities. If we did, we'd say that naval vesels mysteriously traveled "faster" near the north pole, and "slower" near the equator. But instead, we say that the vessels travel the same speed everywhere, because we measure speed as distance vs time.

How do we calculate distances? The metric of the Earth's surface defines it for us. Given the metric, we find the small interval of distance, ds, between two nearby points. To find the distance between more distant points, we integrate this differental distance along some curve to find the distance. Usually, we want the shortest distance between two points, so we usually integrate along a "great circle", which is a geodesic path. But in any event, we find the distance by integrating the differential along some curve.

The schwarzschild coordinates are coordinates, just like lattitude and longitude. And they aren't the only choice by any means. Isotropic coordinates are also frequently used, and there are many more choices as well (Painleve coordinates, Novikov coordinates, Kruskal-Szerkes, etc, etc, etc.).

Thus, subtracting coordinates to define distances doesn't have any physical significance, because one can, and does, change coordinates "at will. Coordinates are a human invention, like markings on a map they shouldn't affect anything physical.

Relativity does throw one "twist" into this perscription that distances should be independent of coordinate choices. We need to separate space from time to define the notion of distance at all. So to this extent, we need to make some sort of coordinate choice - the choice of how to split space and time - before we can measure it. This explains why moving rulers have a different length than stationary ones - it's because of the space-time split. Once we do the split, though, it's pretty clear (at least I think it shouldl be clear!) that we measure the distance by using the metric, and by integrating the length of some curve in which all points have the same time coordinate.

O.K. I think I understand what you are getting at, but we may be differing in semantics. I have highlighted the part that you consider to be the most fundamental definition of length above.

In your example, nautical miles would presumably be a "proper length" and is what a tape measure would presumably measure. Now if for some reason reason we chose to measure the distance in arc minutes of longitude we would presumably call that a "coordinate distance" because it is dependent on the choice of coordinates.

In the interests of moving along I will accept your definition of "coordinate length" as stated in the bolded sentence at the end of your quote. Just out of curiosity, what is the name for the "distance" $\Delta r = r2-r1$ where r1 and r2 are Schwarzschild spatial radial coordinates?

pervect said:
We're getting into definitional issues again, I think. Let's look at the Earth for an example of why I think coordinates are being missused here. I was hoping to do this interactively, but, apparently that's not working out.
...
This explains why moving rulers have a different length than stationary ones - it's because of the space-time split. Once we do the split, though, it's pretty clear (at least I think it shouldl be clear!) that we measure the distance by using the metric, and by integrating the length of some curve in which all points have the same time coordinate
Two of the 5 questions concern radar distance, would you disagree that radar distance measured by a clock is a coordinate independent concept? So we can start with that right? Let's put down some formulas? Do you know how to calculate it? If so, then please take the effort in writing them down.

I at least made an attempt to help solving this, if I make mistakes, fine point them out so that I and others can learn. But point them out directly, say exactly what is wrong and then state the right form. Just saying it is wrong is not very helpful.

yuiop said:
Just out of curiosity, what is the name for the "distance" $\Delta r = r2-r1$ where r1 and r2 are Schwarzschild spatial radial coordinates?
As far as I know there is no name for this 'distance'. It really is no distance, in fact the R value relates to the curvature.

But we can get a proper distance by integrating between these r values:

$$\Delta \rho = \int_{r_1}^{r_2} g(r) dr = \int_{r_1}^{r_2} {1 \over \sqrt{1 - {r_s \over r}}} dr$$

That gives us:

$$\Delta \rho = \sqrt {{\it r_2}\, \left( {\it r_2}-{\it r_s} \right) }-\sqrt {{\it r_1}\, \left( {\it r_1}-{\it r_s} \right) }+{\it r_s}\,\ln \left( {\frac { \sqrt {{\it r_2}}+\sqrt {{\it r_2}-{\it r_s}}}{\sqrt {{\it r_1}}+\sqrt {{ \it r_1}-{\it r_s}}}} \right)$$

We can also get the area and the volume. All three are finite all the way up to the Schwarzschild radius. In principle one can also calculate the proper distance from 0<R<=Rs.

Attached is a graph showing the difference in volume between increasing R and R+1 slices in a Schwarzschild metric of Rs=1 and flat spacetime. As you can see in the graph the volume is a bit higher in the Schwazschild metric, basically one can place more objects of a unit length into a Schwarzschild space compared to a flat space. It is a bit similar to being able to place more unit rulers along a rotating disk compared to a disk that does not rotate. Also you can see at R=1 it is all over for the Schwarzschild metric because R=1 is the Schwarzschild radius or also called the sphere that represents the event horizon.

Interesting questions to ponder:
- Can we find a relation between the increase of volume compared to flat space and energy?
- Is there a simple analytic formula describing the ratio between VSchwarzschild and VFlat?

#### Attachments

• Schwarzschild Volume.jpg
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Hi.
sweet springs said:
You can choose initial condition of particles as you think it reasonable when drawing lines.

Measuring wavelength of light coming from infinity where wavelength = lambda0 is enough for our purpose, I think. The height of elevator equals to ( constant times of ) the light wavelength initially, it keeps on so in any frame of reference if we disregard tidal force at ceiling and floor of the elevator.
Regards.

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Passionflower said:
Edited: Upon reflection I think the solution below applies to a stationary elevator not an elevator free falling.
Agree.

Ok, let's assume we have the following situation:

Elevator Height: 0.2
Bottom has a lower R value than the top of the elevator.

So let's calculate for when the elevator is at RBot = 3

First we need to know where the top is

We solve:

$$\sqrt {{\it ro}\, \left( {\it ro}-2 \right) }-\sqrt {3}+2\,\ln \left( {\frac {\sqrt {{\it ro}}+\sqrt {{\it ro}-2}}{\sqrt {3}+1}} \right) = 0.2$$

Which gives us: 3.117636346 for R0

If spacetime would have been flat we would have expected a coordinate of 3.2, but it is curved.

So since now we have RTop we can calculate the radar roundtrip time:

We use:

$$2\,x-6+4\,\ln \left( x-2 \right)$$

Substituting 3.117636346 for x gives us: 0.6801368916

But this is coordinate time!

Now we need to translate that into proper time for RBot and RTop

We multiply by:

$$\sqrt{1-{2 \over r}}$$

So for RBot we get: 0.3926772175
And for RTop we get: 0.4072244662

Assuming I did not mess up, perhaps the next step should be to determine the roundtrip times for a free falling elevator with a velocity both equal and not equal to escape velocity. Who can help?
Your numerical results here seem OK ( using the symbolic version of radar distance as given here https://www.physicsforums.com/showpost.php?p=2902350&postcount=93 ).

The next step you describe is quite a big step. The next "baby step" on the way might be to determine the coordinate and proper times for a falling object. There are some equations that prove useful in post #21 of this old thread: https://www.physicsforums.com/showthread.php?t=249722&highlight=tunnel&page=2

yuiop said:
The next step you describe is quite a big step. The next "baby step" on the way might be to determine the coordinate and proper times for a falling object. There are some equations that prove useful in post #21 of this old thread: https://www.physicsforums.com/showthread.php?t=249722&highlight=tunnel&page=2
In the semi-free falling case when a light beam leaves the bottom of the elevator at RBot to reach the top, the top R coordinate will decrease in the mean time. So we will have to take both the equation of motion for the top and the light beam to see where they meet. Then we have the new R coordinate, and from there we can do a light reflection in the other direction the same way but now for the top R coordinate Then we have the total coordinate time which can be converted to proper time, although this conversion may have a caveat.

Now the difficulty is that I do not know the equation of motion when we assume the elevator stays rigid. To keep things simple perhaps we can assume that the center of the elevator is free falling and both the top and bottom are adjusting to keep it rigid. Now Pervect gave a diff equation in the other topic related to the tidal effect that we perhaps can use as a first approximation for such an adjustment? The equation of motion of light is no problem.

Are you following or does the above approach make no sense to you?

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Well, let's see if I understand the question. First let me make one urgent suggestion - use a SHORT elevator.

The basic question, is what is the length of a falling elevator (or ruler) that is reasonably rigid, according to:

1) A Schwarzschild observer at infinity.

Operationally, it's not clear what measurements the observer at infinity is making to perform the measurement. If you go to theory, though, you'd just want to integrate along the curve representing the ruler at some constant Scwarzschild time.

2) A local observer at r=4m when the falling elevator is passing.

While the space-time of the Schwarzschild metric is curved, if you have a small enough ruler, you can ignore the curvature. Then you simply have the problem of a ruler with a known proper length, falling past you at some velocity. You already know how to solve this problem from SR, the problem is not essentially different. All that remains is to compute the velocity as measured by a local observer. The good news is that I worked this out once. The bad news, is I can't find out where. I did find http://arxiv.org/PS_cache/gr-qc/pdf/0111/0111073v1.pdf though, which has a formula.

3) A free-falling observer inside the elevator using radar measurements from the top of the elevator.

4) A free falling observer inside the elevator using radar measurements from the bottom of the elevator.

If you specify a sufficiently short elevator, then the radar methods will give the proper length, which won't change. Why make life difficult for yourself? It won't illuminate anything conceptually to use a long elevator - in fact, it'll obscure some of the actual physics.

Your radar distances will be good as long as the assumption that the coordinate-speed of light is constant and equal to 'c' is satisfied. This will be true near the origin of whatever coordinate system you set up, in fact you'll get the distances by measuring a certain number of wavelengthss of a specified frequency of light if you use the SI defintion.

One final thought - the Novikov coordinate system might be a nice choice, an observer free-falling from infinity has a worldline of R* = constant, t = tau. The metric of the Novikov coordinate system really should tell you everything you'd want to know about local measurements - well, I'm not sure everyone else here shares my viewpoint that "the metric tells all", alas, but it does. really. Unfortunately, Novikov coordinates are computationally rather intractible.

In aid of finding the trajectory of something that's "following you into a black hole a constant distance behind you".

We can take advantage of the fact that we know the four-velocity of an object falling into a black hole. Let R be the maximum height, then the 4-velocity at r is

$$\frac{dr}{d\tau} = - \sqrt{\frac{2M}{R} - \frac{2M}{r}}$$

$$\frac{dt}{d\tau} = \frac{ \sqrt{ 1 - \frac{2M}{R} } } {1 - \frac{2M}{r}}$$

The above equations ASSUME that the lead particle is free-falling into the black hole and has some constant energy E at infinity, E = sqrt(1-2M/R). You can work out the problem without this assumption, but then you need to find the four-velocity on your own.

We know that the time direction of the infalling observer is just it's 4 velocity. To put it another way, if we look at some proper time $\tau$+$\epsilon$ to the first order the particle's position will be

$$r = r + \epsilon \frac{dr}{d\tau}$$

$$t = t + \epsilon \frac{dt}{d\tau}$$

We need to solve the metric equations for the vector that's orthogonal to the time vector of our infalling particles worldline to find the particle's notion of a space-like vector.

That is we want p and q to represent the components of the vector (p,q) such that the dot product of (p,q) and the four-velocity vector (dt/dtau, dr/dtau) is zero, i.e.

$$g_{tt}\, p\, \frac{dt}{d\tau} + g_{rr} \, q \, \frac{dr}{d\tau} = 0$$

We also want our vector to be of unit length (it will be space-like).

$$g_{tt}\, p^2 + g_{rr} \, q^2 = 1$$

here g_tt and g_rr are the metric coefficients, i.e.

$$g_{tt} = -1 + \frac{2M}{r}$$

$$g_{rr} = \frac{1}{1 - \frac{2M}{r}}$$

Then the coordinates of our following particle should be:

$$t = t + \epsilon*p$$
$$r = r + \epsilon*q$$

We have to adjust both t and r, because the infalling particle's notion of simultaneity is different from that of the Schwarzschild observer.

when I solve these I get
$$p = \frac{\sqrt{\frac{2M}{R} - \frac{2M}{r}}}{1 - \frac{2M}{r}}$$

$$q = \sqrt{1-\frac{2M}{R}}$$

I'm not aware of having made any mistakes, but I'd strongly advise double checking this

This should get the people who really want to crunch through everything started.

And let me stress again - epsilon MUST BE SMALL! How small? You don't want the metric coefficients to vary appreciably between the following particle and the lead particle. How much is "appreciably" depends on the accuracy you want.

If you want , for whatever masochistic reason, to find a particle following you at a larger distance, you need to break it up into a chain of following particles, and redo the analysis using the appropriate metric coefficients at each point. Note that you'll need to find the four-velocity yourself, because the third particle in the chain will be following the second, which won't be following a geodesic, so we can't use the formulae I provided above for the four-velocity.

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pervect said:
We can take advantage of the fact that we know the four-velocity of an object falling into a black hole. Let R be the maximum height, then the 4-velocity at r is

$$\frac{dr}{d\tau} = - \sqrt{\frac{2M}{R} - \frac{2M}{r}}$$

$$\frac{dt}{d\tau} = \frac{ \sqrt{ 1 - \frac{2M}{R} } } {1 - \frac{2M}{r}}$$]
Do you assume dr/dt= 0 at R?

Then isn't the correct formula:

$$\frac{dr}{d\tau} = -{\frac { \left( r-2\,M \right) \sqrt {2\,M} \sqrt {R-r}}{{r}^{3/ 2}\sqrt {R-2\,M}}}$$

And

$$\frac{dt}{d\tau} = {r\sqrt {{\frac {R-2\,M}{R}}} \over r-2\,M }$$

Added: See Reyder - Introduction to General Relativity (2009) Chapter 5.12

pervect said:
The above equations ASSUME that the lead particle is free-falling into the black hole and has some constant energy E at infinity, E = sqrt(1-2M/R). You can work out the problem without this assumption, but then you need to find the four-velocity on your own.
I missed this on initial viewing of your posting.

I am not sure what this is: constant energy at infinity?
Can you explain?

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dr/dtau should be zero at r=R, so dr/dt should also be zero at r=R

I believe my expressions are right: MTW 25.27 (pg 663) gives

$$d \tau = \frac{dr} { \sqrt{\frac{2M}{r} - \frac{2M}{R}} }$$

$$R = \frac{2M}{1-E^2}$$

And 25.18 (pg 657) gives the expression for dt/dtau
$$\frac{dt} {d\tau} =\frac{E}{ 1-\frac{2M}{r} }$$

http://www.fourmilab.ch/gravitation/orbits/

has similar equations online (they took them from MTW, I"m pretty sure). If you set L=0 (for a radial infall), and solve for E given dr/dtau = 0 at r=R from the equations at the above URL you should also get the above.

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Passionflower said:
In the semi-free falling case when a light beam leaves the bottom of the elevator at RBot to reach the top, the top R coordinate will decrease in the mean time. So we will have to take both the equation of motion for the top and the light beam to see where they meet. Then we have the new R coordinate, and from there we can do a light reflection in the other direction the same way but now for the top R coordinate Then we have the total coordinate time which can be converted to proper time, although this conversion may have a caveat.

Now the difficulty is that I do not know the equation of motion when we assume the elevator stays rigid. To keep things simple perhaps we can assume that the center of the elevator is free falling and both the top and bottom are adjusting to keep it rigid. Now Pervect gave a diff equation in the other topic related to the tidal effect that we perhaps can use as a first approximation for such an adjustment? The equation of motion of light is no problem.

Are you following or does the above approach make no sense to you?

pervect said:
While the space-time of the Schwarzschild metric is curved, if you have a small enough ruler, you can ignore the curvature. Then you simply have the problem of a ruler with a known proper length, falling past you at some velocity. You already know how to solve this problem from SR, the problem is not essentially different.
This agrees with the statement I made in the OP: "Spacetime is locally Minkowskian for a local stationary (accelerating) observer in Schwarzschiuld coordinates" and the short rigid falling elevator appears to be simply velocity length contracted to a local stationary accelerating Schwarzschild observer.
All that remains is to compute the velocity as measured by a local observer. The good news is that I worked this out once. The bad news, is I can't find out where. I did find http://arxiv.org/PS_cache/gr-qc/pdf/0111/0111073v1.pdf though, which has a formula.
From the reference you give, the velocity at r of the falling short elevator dropped from Schwarzschild coordinate R with an initial velocity of $v_0$, according to a stationary local Schwarzschild observer at r as the elevator passes is:

$$v_f = \sqrt{1-v_0} \sqrt{\frac{R/r-1}{R/rS-1}}$$
If you specify a sufficiently short elevator, then the radar methods will give the proper length, which won't change. Why make life difficult for yourself? It won't illuminate anything conceptually to use a long elevator - in fact, it'll obscure some of the actual physics.
I am not sure I can be convinced of that until someone actually finds a way to compute the length of a long falling elevator. Until then any conclusions about a long elevator is just conjecture.

Passionflower said:
In the semi-free falling case when a light beam leaves the bottom of the elevator at RBot to reach the top, the top R coordinate will decrease in the mean time. So we will have to take both the equation of motion for the top and the light beam to see where they meet. Then we have the new R coordinate, and from there we can do a light reflection in the other direction the same way but now for the top R coordinate Then we have the total coordinate time which can be converted to proper time, although this conversion may have a caveat.

Now the difficulty is that I do not know the equation of motion when we assume the elevator stays rigid. To keep things simple perhaps we can assume that the center of the elevator is free falling and both the top and bottom are adjusting to keep it rigid. Now Pervect gave a diff equation in the other topic related to the tidal effect that we perhaps can use as a first approximation for such an adjustment? The equation of motion of light is no problem.

Are you following or does the above approach make no sense to you?
I am following your approach, but I am not sure it is the simplest approach, mainly because of the caveats which can only be minimised for an infinitesimal elevator.

However, we now have equations for the distance a particle falls in a given time, elapsed proper time or coordinate time of a falling particle and the velocity of a particle dropped from a given height. From these we should be able to precisely locate the positions of a ball of unconnected particles even for a non infinitesimal ball of particles.

Yes we have some formulas but in my mind the fundamental question is still not answered and formulaically written down in e-ink. It is likely crystal clear in the minds of the experts who perhaps are baffled by my fruitless tinkering with irrelevant formulas, nevertheless I keep trying:

The fundamental question being, at least in my mind: what is the coordinate distance between two test particles with respect to a given R-coordinate and what is the coordinate distance of those same test particles if they are connected with an, as ridgidly as theoretically possible, rod.

So we can take a small numerical example (I like examples, that way I can verify if I am not just talking 'philosophy'):

In A Schwarzschild solution where Rs = 1 consider Clock A: RA = 20 and Clock B removed a coordinate distance 1 from A at RB = 21, we assume radial motion.

The free fall from infinity does not pose a problem in terms of discovering proper time if we switch things around and consider proper time in terms of 'proper time till ultimate doom', e.g. the arrival at the singularity. See the attached graph for a comparison between proper and coordinate time. As you can see coordinate time never gets beyond Rs while proper time gets all the way to the singularity.

Using (MTW 25.38):

$$-2/3\,{\it rs}\, \left( {\frac {r}{{\it rs}}} \right) ^{3/2}$$

We can also use coordinate time, but obviously we would never get to the singularity, by using:

$${\it rs}\, \left( -2/3\, \left( {\frac {r}{{\it rs}}} \right) ^{3/2}-2 \,\sqrt {{\frac {r}{{\it rs}}}}+\ln \left( \left( \sqrt {{\frac {r}{ {\it rs}}}}+1 \right) \left( \sqrt {{\frac {r}{{\it rs}}}}-1 \right) ^{-1} \right) \right)$$

The proper time to ultimate doom for both clocks is:

A: -59.62847939
B: -64.15605973

The number is negative because it is a 'time to' thing. So basically B has a longer life from our starting point than A.

Lets also check the proper distance between A and B at this starting point:

$$\sqrt {{\it ro}\, \left( {\it ro}-{\it rs} \right) }-\sqrt {{\it ri}\, \left( {\it ri}-{\it rs} \right) }+{\it rs}\,\ln \left( {\frac { \sqrt {{\it ro}}+\sqrt {{\it ro}-{\it rs}}}{\sqrt {{\it ri}}+\sqrt {{ \it ri}-{\it rs}}}} \right)$$

Where ri is RA and ro is RB we get:

Proper distance Between A and B at RA: 1.025325886

In other words, we can place a rod of a proper distance of 1.025325886 between R and R+1 meaning that the lower the coordinate value the more 'extra room is provided for' in this spacetime.

Now let's see what happens after 20 seconds of proper time for RA.

We use:

-59.62847939 - 20 = -39.62847939

Now we have to solve going from proper time to an R value.

We use (MTW 25.39a):

$$1.310370697\,\sqrt [3]{{\it rs}\,{\tau}^{2}}$$

RA20: 15.23114122

Now how do we go from here, I present three options, we can move B in a ratio of proper time, we can keep the R differential fixed or we can keep the proper distance fixed:

First we use some ratio of proper time:

Let's use:

RATime/RBTime * 20 Seconds

-59.62847939 / -64.15605973 * 20 = 18.58857282

So let's 'run' B for 18.58857282 seconds

Then RB20: 16.71722640

Assuming all this is correct what can we conclude:

The initial coordinate distance between A and B was 1 and the initial proper distance between A and B was 1.025325886

After a time shift of 20 seconds for A and (we assume) 18.58857282 seconds for B we get:

The coordinate distance is now: 1.48608518 and the proper distance is now: 1.534943739

This would indicate that due to the tidal effect a proper distance of 1.025325886 becomes 1.534943739 over a proper time for A of 20 seconds and a proper time for B of 18.58857282 seconds.

Now let's consider what happens if we instead try to keep the R distance fixed between the two measurements:

Then B20 becomes: 15.23114122 + 1 = 16.23114122
And the proper distance becomes: 1.033396851

Now how would that relate back to elapsed proper time for B, assuming this motion is inertial?

Finally we force the proper distance to stay fixed, then:

The coordinate value of B becomes:
B20: 20.59299168
However the proper time will change as B is not traveling inertially anymore.

So there are many approaches. But we need to find the right one.

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yuiop said:
I am not sure I can be convinced of that until someone actually finds a way to compute the length of a long falling elevator. Until then any conclusions about a long elevator is just conjecture.

I don't understand why you are resisting the obvious approach that I've mentioned several times : to find the length of a long elevator, you just use a chain of points all of which are stationary relaltive to each other and measure the distance along the chain.

This is basically inspired by "Born rigidity" (see the sci.physics.faq)
http://www.desy.de/user/projects/Physics/Relativity/SR/rigid_disk.html#star. In fact, you can take born-rigid motion of a ruler as a way to measure the length of a long elevator by using born-rigid rulers.

sci.physics.faq said:
I won't plunge right into Born's definition (as Pauli does). Instead I'll approach it by thinking atomistically. Imagine our solid as made up of a large number of atoms A1,. . ., An. Between any two nearby atoms Ai and Aj there is a "natural distance" dij, natural in the sense that if Ai and Aj are pushed together or pulled apart, stresses result, trying to restore the distance dij. Of course there are propagation delays, but if we start with the solid at rest in some inertial frame, and accelerate it gently, the resulting elastic waves in the solid should die out pretty quickly. Or we can pretend that exactly the right force is applied to each atom at all times, so that natural distances are preserved. Let the number of atoms tend to infinity (continuum approximation), let the stress/strain ratio tend to infinity, and apply forces gently enough so that the elastic waves can be ignored. In such a case we arrive at Born's definition. Born used coordinates, but I'll try for a coordinate-free rephrasing.

yuiop said:
I am following your approach, but I am not sure it is the simplest approach, mainly because of the caveats which can only be minimised for an infinitesimal elevator.
I
However, we now have equations for the distance a particle falls in a given time, elapsed proper time or coordinate time of a falling particle and the velocity of a particle dropped from a given height. From these we should be able to precisely locate the positions of a ball of unconnected particles even for a non infinitesimal ball of particles.

Yes, we should be able to do that, though the calculation is messy. It's also interesting to note that the differential equations of motion for the falling particles are formally the same as in the Newtonian case, it's just that the symbols r and tau stand for the schwarzschild r coordinate and proper time in the relativistic case, while we use radial distance and universal time for the Newtonian case. So as messy as they are, the results should be the same as for the Newtonian case.

Passionflower said:
The fundamental question being, at least in my mind: what is the coordinate distance between two test particles with respect to a given R-coordinate and what is the coordinate distance of those same test particles if they are connected with an, as ridgidly as theoretically possible, rod.

What do you mean by "coordinate distance"? If you like to calculate things, why not illustrate your concept on the surface of the Earth - it'll be a curved-space example.

So we can take a small numerical example (I like examples, that way I can verify if I am not just talking 'philosophy'):

In A Schwarzschild solution where Rs = 1 consider Clock A: RA = 20 and Clock B removed a coordinate distance 1 from A at RB = 21, we assume radial motion.

Is clock at supposed to be "at rest" at R=20 at the starting time? We'll skip over the question of the distance between clock A and clock B until you define what you mean by "distance". Though it looks like you're just subtracting the coordinates (?).

The free fall from infinity does not pose a problem in terms of discovering proper time if we switch things around and consider proper time in terms of 'proper time till ultimate doom', e.g. the arrival at the singularity. See the attached graph for a comparison between proper and coordinate time. As you can see coordinate time never gets beyond Rs while proper time gets all the way to the singularity.

Using (MTW 25.38):

$$-2/3\,{\it rs}\, \left( {\frac {r}{{\it rs}}} \right) ^{3/2}$$

If you use 25.38, then the clock will NOT be at rest at R=20. That equation will only work if the clocks are at rest at infinity.

We can also use coordinate time, but obviously we would never get to the singularity, by using:

$${\it rs}\, \left( -2/3\, \left( {\frac {r}{{\it rs}}} \right) ^{3/2}-2 \,\sqrt {{\frac {r}{{\it rs}}}}+\ln \left( \left( \sqrt {{\frac {r}{ {\it rs}}}}+1 \right) \left( \sqrt {{\frac {r}{{\it rs}}}}-1 \right) ^{-1} \right) \right)$$

The proper time to ultimate doom for both clocks is:

A: -59.62847939
B: -64.15605973

The number is negative because it is a 'time to' thing. So basically B has a longer life from our starting point than A.

Lets also check the proper distance between A and B at this starting point:

$$\sqrt {{\it ro}\, \left( {\it ro}-{\it rs} \right) }-\sqrt {{\it ri}\, \left( {\it ri}-{\it rs} \right) }+{\it rs}\,\ln \left( {\frac { \sqrt {{\it ro}}+\sqrt {{\it ro}-{\it rs}}}{\sqrt {{\it ri}}+\sqrt {{ \it ri}-{\it rs}}}} \right)$$

Where did this expression come from? It appeared out of the blue, and doesn't look familiar at all - I've got some serious doubts about its correctness...

pervect said:
What do you mean by "coordinate distance"?
Coordinate distance is simply the difference between two coordinate values.

pervect said:
If you like to calculate things, why not illustrate your concept on the surface of the Earth - it'll be a curved-space example.
This is about free falling clocks not about an experiment on Earth, the situations are not comparable. Furthermore the curvature in the vicinity of the Earth is too small, in fact there is currently no experiment possible to prove the correctness of the Schwarzschild solution.

pervect said:
Is clock at supposed to be "at rest" at R=20 at the starting time?
No, if you would have read correctly you would have seen I am using a clock falling fron infinity.
pervect said:
We'll skip over the question of the distance between clock A and clock B until you define what you mean by "distance". Though it looks like you're just subtracting the coordinates (?).
That is perhaps because you did not properly read what I was writing. It appears that you already made up your mind that 'I don't understand what the R value represents' and with this view you skimp over what I wrote and made your conclusions. At the same time you do not even seem to recognize the formula that integrates the R values to obtain the proper distance.

pervect said:
If you use 25.38, then the clock will NOT be at rest at R=20. That equation will only work if the clocks are at rest at infinity.
Yes you are absolutely correct, and since my clocks are falling from infinity it is an applicable formula.

pervect said:
Where did this expression come from? It appeared out of the blue, and doesn't look familiar at all - I've got some serious doubts about its correctness...
Which expression, you quote two expressions? One is similar to the one from MTW and the other you get when you integrate the R coordinate values to get the proper distance.

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Let me recommend something available online from Taylor, the "T" in MTW, on the topics of distances in GR>

http://www.eftaylor.com/pub/chapter2.pdf

"Curving"

I'll provide a short quote to hopefully generate interest.

Curving: 1 “Distances” Determine Geometry said:
Nothing is more distressing on first contact with the idea of curved spacetime
than the fear that every simple means of measurement has lost its
power in this unfamiliar context. One thinks of oneself as confronted with
the task of measuring the shape of a gigantic and fantastically sculptured
iceberg as one stands with a meterstick in a tossing rowboat on the surface
of a heaving ocean.

Were it the rowboat itself whose shape were to be measured, the procedure
would be simple enough (Figure 1). Draw it up on shore, turn it
upside down, and lightly drive in nails at strategic points here and there
on the surface. The measurement of distances from nail to nail would
record and reveal the shape of the surface. Using only the table of these
distances between each nail and other nearby nails, someone else can
reconstruct the shape of the rowboat. The precision of reproduction can be
made arbitrarily great by making the number of nails arbitrarily large.
It takes more daring to think of driving into the towering iceberg a large
number of pitons, the spikes used for rope climbing on ice. Yet here too the
geometry of the iceberg is described—and its shape made reproducible—
by measuring the distance between each piton and its neighbors.

But with all the daring in the world, how is one to drive a nail into spacetime
to mark a point? Happily, Nature provides its own way to localize a
point in spacetime, as Einstein was the first to emphasize. Characterize the
point by what happens there: firecracker, spark, or collision! Give a point
in spacetime the name event.

The T in MTW is Kip Thorne (Misner, Wheeler, and Thorne). Though black, my copy has not yet become a black hole; though a relativity theorist once argued to me that MWT disproves the priniciple of equivalence:

Drop the big black book. Doesn't it fall faster than everything else?

Passionflower said:
Lets also check the proper distance between A and B at this starting point:

$$\sqrt {{\it ro}\, \left( {\it ro}-{\it rs} \right) }-\sqrt {{\it ri}\, \left( {\it ri}-{\it rs} \right) }+{\it rs}\,\ln \left( {\frac { \sqrt {{\it ro}}+\sqrt {{\it ro}-{\it rs}}}{\sqrt {{\it ri}}+\sqrt {{ \it ri}-{\it rs}}}} \right)$$
pervect said:
Where did this expression come from? It appeared out of the blue, and doesn't look familiar at all - I've got some serious doubts about its correctness...
This expression is the integrated distance between the Shwarzschild radial coordinates ro and ri. As we understand it, it is equivalent to the distance obtained by laying infinitesimal rulers end to end and we have therefore been referring to it as the proper length of a stationary rod that is at rest in Schwarzschild coordinates and extends between those coordinates. The expression had already been mentioned of post #12 of this very thread https://www.physicsforums.com/showpost.php?p=2901038&postcount=12 and was recently brought up in posts #2 and #3 of the closely related open thread https://www.physicsforums.com/showthread.php?t=431712. It was also derived by DrGreg in post #15 of this old thread https://www.physicsforums.com/showthread.php?t=248015&page=2 and embellished upon by myself in posts #22 and #33 of the same old thread. If there is an error in the final form of the equation it is probably a mistake on my part. Would you care to check it for us?

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?

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Passionflower said:
Using (MTW 25.38):

$$-2/3\,{\it rs}\, \left( {\frac {r}{{\it rs}}} \right) ^{3/2}$$
Thanks for the expression Passionflower. If we define $\tau$ as the proper time remaining "until ultimate doom" then we have this equation:

$$\tau = -2/3\,{ r_s}\, \left( {\frac {r}{{ r_s}}} \right) ^{3/2}$$

and solve for r we get:

$$r = \left(\frac{9}{4} r_s \tau^2\right)^{1/3}$$

... which is numerically the same as the other equation you quoted:

$$1.310370697\,\sqrt [3]{{\it rs}\,{\tau}^{2}}$$
(Just in case anyone was wondering where the 1.310370697 factor came from. OK, maybe it was just me that was wondering)

pervect said:
What do you mean by "coordinate distance"?

Here is my way to to attempt to explain what Passionflower and myself are getting at by the term "coordinate distance" and maybe we can come to some sort of agreement in the context given below.

$$c^2d\tau^2 = (1-r_s/r) c^2 dt^2 - (1-r_s/r)^{-1} dr^2$$

where $cd\tau$ is the invariant timelike interval or proper time between two events that all observers can agree on.
Now reverse the signature to obtain the invariant spacelike interval or proper distance (dS) between two events defined as:

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

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

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

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.

Now if we have two events e1 and e2 with Schwarzschild coordinates (x1,t1) and (x2,t2) respectively, where t1 and t2 have the same time according to the SINGLE clock of the Schwarzschild observer at infinity, so that $\Delta t = t2-t1 = 0$, then the Schwarzschild coordinate distance between the two events should simply be $\Delta x = x2-x1$, even when x2-x1 is not infinitesimal, no? We do not expect the coordinate distance to be the same as the proper distance. When we integrate the differentials between the two events we are effectively transforming the coordinate distance to a proper distance.

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I really do not understand what the fuss here is about coordinate distance.

A distance between coordinates is pretty unambiguous if you ask me.

Say we have two coordinate values: R=2 and R=3

The coordinate distance is simply 3 - 2 = 1

For a proper, e.g. physical distance, we obviously need a metric. In static spacetimes this is simple. If we want to have a physical distance for the Schwarzschild metric then in our case we have to use:

$$\int _{{2}}^{{3}}\!{\frac {1}{\sqrt {1-{\frac {{\it rs}}{r}} }}}\,{dr}$$

Integrating this gives us:

$$\sqrt {9-3\,{\it rs}}-\sqrt {4-2\,{\it rs}}+{\it rs}\,\ln \left( { \frac {\sqrt {3}+\sqrt {3-{\it rs}}}{\sqrt {2}+\sqrt {2-{\it rs}}}} \right)$$

So if we assume Rs is 1, then the proper distance between R=2 and R=3 is: 1.300118429

In the graph below observe the increasing 'extra room' for distance for increasingly lower coordinate values RX and RX+1 for RS=1. All the way up to the Schwarzschild horizon. Beyond the horizon we can also calculate the distance from the horizon all the way up to, but not including, R=0. The same applies for area and volume.

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Passionflower said:
I really do not understand what the fuss here is about coordinate distance.

A distance between coordinates is pretty unambiguous if you ask me.

Say we have two coordinate values: R=2 and R=3

The coordinate distance is simply 3 - 2 = 1
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.

For a proper, e.g. physical distance, we need a metric. In static spacetimes this is simple. If we want to have a physical distance for the Schwarzschild metric we have to use:

$$\int _{{2}}^{{3}}\!{\frac {1}{\sqrt {1-{\frac {{\it rs}}{r}} }}}\,{dr}$$

Integrating this gives us:

$$\sqrt {9-3\,{\it rs}}-\sqrt {4-2\,{\it rs}}+{\it rs}\,\ln \left( { \frac {\sqrt {3}+\sqrt {3-{\it rs}}}{\sqrt {2}+\sqrt {2-{\it rs}}}} \right)$$

So if we assume Rs is 1, then the proper distance between R=2 and R=3 is: 1.300118429
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. We know that for an object falling from infinity the local velocity is $\sqrt{(r_s/r)}$ so the velocity length contraction factor is $\sqrt{(1-r_s/r)}$ and the total length contraction factor is $(1-r_s/r)$. 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. Hopefully someone here can confirm or deny this.

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yuiop said:
Again, I think this is fine if the measured object is not moving with respect the Schwarzschild coordinates.
I am trying yo understand why you think this is a problem if the object moves, we can just 'freeze the image', e.g. keep coordinate time fixed and calculate. No?

yuiop said:
We know that for an object falling from infinity the local velocity is $\sqrt{(r_s/r)}$ so the velocity length contraction factor is $\sqrt{(1-r_s/r)}$ and the total length contraction factor is $(1-r_s/r)$.
Why do you assume this is different from for instance an example in the Rindler metric (Yes, I know this is a curved spacetime situation, but I am just looking at Lorentz contraction here), I mean in the Rindler case the proper distance remains the same with increasing coordinate velocities, and that makes sense because it is a proper distance. The object in question after all goes with a proper velocity of zero. GR does not do away with this principle of relativity of motion.

Also when you talk about 'gravitational length contraction' it might be useful if you explain exactly what you mean by that. In the graph above I showed that for decreasing equal coordinate distances more additional proper distance is created, e.g. there is more proper distance between (R=3, R=2) than (R=4, R=3). The same happens for volumes, e.g. there is more space between (R=3, R=2) than (R=4, R=3) if we adjust the volumes by the volume obtained from an Euclidean based calculation.

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Passionflower said:
I am trying yo understand why you think this is a problem if the object moves, we can just 'freeze the image', e.g. keep coordinate time fixed and calculate. No?
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 using the equation you were using, you are in fact calculating the proper length of a stationary rod that spans those two coordinates. Even though one end of the moving rod and the stationary rod are both at R1 at time T and the other end of the moving rod and the stationary rod happen to coincide at coordinate R2 at time T, it does not mean the proper length of the stationary rod and the proper length of the moving rod are the same (even though they have the same coordinate length). As Pervect mentions, to a local observer, local measurements are Minkowskian and the moving ruler going past him is length contracted relative to the ruler he has in his hand if they both have the same proper length or if they both appear to have the same length as measured by the stationary local observer, then the proper length of the falling ruler (as measured by an observer riding on the ruler) is longer than the proper length of the ruler that is not falling (as measured by the stationary local Schwarzschild observer).

Consider a purely SR scenario. A moving rod is passing through a barn and to an observer at rest with the barn, the rod and the barn appear to have "the same length" when the "image is frozen", but we know that because the rod is moving relative to the barn observer, the rod has a longer proper length than the barn's proper length. The barn observer's measurement of the moving rod is in fact a coordinate measurement. An observer riding with the moving rod measures the rod's proper length. I won't go on. I am fairly sure you are familiar with the details and resolution of the barn and pole paradox. All I am doing is upending the barn and dropping the pole through it.

Why do you assume this is different from for instance an example in the Rindler metric (Yes, I know this is a curved spacetime situation, but I am just looking at Lorentz contraction here), I mean in the Rindler case the proper distance remains the same with increasing coordinate velocities, and that makes sense because it is a proper distance. The object in question after all goes with a proper velocity of zero.
Yes, I agree the proper length of the falling rod remains constant. It is the coordinate length of the falling rod that changes as it falls and to the Schwarzschild observer at infinity the length contraction is a factor of both the gravitational and velocity related time dilation factors. To a local stationary observer further down that measures the rod as it goes past, the rod only appears to be length contracted by the velocity time dilation factor that is a function of its local velocity. To the local observer, for a sufficiently local region, spacetime appears to be flat or Minkowskian and the calculations are those of SR. The falling rod and the local observer, when they are approximately and momentarily co-located, are subject to the same gravitational time dilation factor and so the gravitational time dilation factor is not needed for local calculations.

GR does not do away with this principle of relativity of motion.
I have not done away with it either. In fact I have introduced the fact into our equations, that when an object has motion relative to the observer, it appears to have a coordinate length that is length contracted relative to the proper length.

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yuiop said:
Here is my way to to attempt to explain what Passionflower and myself are getting at by the term "coordinate distance" and maybe we can come to some sort of agreement in the context given below.

$$c^2d\tau^2 = (1-r_s/r) c^2 dt^2 - (1-r_s/r) dr^2$$

The "abbreviated" Schwarzschild metric is:$$ds^2 = (1-r_s/r) c^2 dt^2 - \frac{dr^2}{1-r_s/r}$$ (Rindler (11.13))
$$dS^2 = -(1-r_s/r) c^2 dt^2 + (1-r_s/r)^{-1} dr^2$$

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

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".

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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

$$g_{r'r'} = g_{rr} \frac{\partial r}{\partial r'} \, \frac{\partial r}{\partial r'} + g_{rt} \frac{\partial r}{\partial r'} \, \frac{\partial t}{\partial r'} + g_{tt} \frac{\partial t}{\partial r'} \, \frac{\partial t}{\partial r'}$$

$$g_{r't'} = g_{rr} \, \frac{\partial r}{\partial r'} \, \frac{\partial r}{\partial t'} + g_{rt} \frac{\partial r}{\partial r'} \, \frac{\partial t}{\partial t'} + g_{tt} \frac{\partial t}{\partial r'} \, \frac{\partial t}{\partial t'}$$

$$g_{t't'} = g_{rr} \frac{\partial r}{\partial t'} \, \frac{\partial r}{\partial t'} + g_{rt} \frac{\partial r}{\partial t'} \, \frac{\partial t}{\partial t'} + g_{tt} \frac{\partial t}{\partial t'} \, \frac{\partial t}{\partial t'}$$

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.
$$g_{r'r'} = g_{rr} \alpha^2$$

$$g_{t't'} = g_{tt} \beta^2$$

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.

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Passionflower said:
Also when you talk about 'gravitational length contraction' it might be useful if you explain exactly what you mean by that. In the graph above I showed that for decreasing equal coordinate distances more additional proper distance is created, e.g. there is more proper distance between (R=3, R=2) than (R=4, R=3). The same happens for volumes, e.g. there is more space between (R=3, R=2) than (R=4, R=3) if we adjust the volumes by the volume obtained from an Euclidean based calculation.

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(1/2) = 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.

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