# Proper distance in GR

by pervect
Tags: proper distance
Emeritus
P: 7,620
 Quote by yuiop I am looking for a practical operational method to measure proper distance in GR to try and get an intuitive feel for the concept. One difficulty is that using a rod as a ruler is problematic in GR because even in free fall the ruler is subject to tidal forces and its proper length is being physically changed making it useless as a measuring device. Would the following proposed method work for measuring proper distance in free fall? 1) Attach a master clock to the centre of gravity of the falling object. 2) Attach further clocks above and below the master clock at intervals defined in (3). 3) Arrange the clocks so that the radar distance between any two clocks is the same and the clocks are spaced close enough to each other so that the radar distance measured from either end of the unit gap is approximately the same to an agreed accuracy. 4) Attach a mechanism to each clock that adjusts its spatial separation from its neighbouring clocks so that the unit gap length as defined in (3) is actively maintained to provide continuous active calibration of the ruler. In the above set up only the master clock will be inertial in the sense that it experiences no proper acceleration. Since all the secondary clocks experience proper acceleration and since the proper acceleration is a function of the proper distance from the master clock it would be interesting to find out is the proper distance can be defined in terms of proper acceleration. If the above method does define a mechanism to measure proper distance in the free falling frame, the change in proper length of a natural free falling object due to tidal forces could in principle be measured by such a device attached to the object.
The method you describe for distance is due to Born, it's the concept of using Born rigid motion to define length, though it's one I also like and have espoused in the past.

A more detailed calculation confirms to my satisfaction that the fermi-normal distance will give the same result as the measurement of distance via a Born rigid ruler.

(I felt I needed to confirm that the timelike worldlines of the accelerating observers at constant fermi-normal distance were orthogonal to the space-like geodesics along which the distance was measured. This will be true if the metric in these coordinates is diagonal away from the origin, which appears to be the case. Using the same series approach but including higher order terms, I convinced myself that metric was in fact diagonal.)

 Secondly, would anyone here agree that the "Fermi normal distance" described by Pervect in this thread: http://www.physicsforums.com/showthread.php?t=435999 is in fact a proper distance in the free falling frame? (Note: I do not think that Pervect has claimed the Fermi normal distance is the proper distance in the free falling frame, but I suspect it might be.)
It is a proper distance in the sense that it's a distance measured along a specific curve. Passionflower's distance will also be a proper distance - if he ever gets around to specifying the curve along which it's measured, that is :-)

However, I don't believe that passionflower's distance will be equivalent to the distance measured by dropping a Born rigid ruler.
 P: 3,967 Just a quick note to acknowledge that Passionflower pointed out a major mistake in my calculations for the motion of a falling body with an initial velocity at infinity in post #69. I have now edited that post and hopefully the equations are now correct. Thanks for checking Passionflower. It is good to know that someone is checking the posted equations so that they might be a useful and accurate reference to someone in the future .
P: 3,967
 Quote by pervect It is a proper distance in the sense that it's a distance measured along a specific curve. Passionflower's distance will also be a proper distance - if he ever gets around to specifying the curve along which it's measured, that is :-) However, I don't believe that passionflower's distance will be equivalent to the distance measured by dropping a Born rigid ruler.
As far as I can tell, Pf's distance is the distance measured by a series of free falling observers each with their own clock and very short Born rigid ruler. The observers would have to synchronise clocks at infinity and then jump at regular intervals of say one second. At the time the leading observer arrives at the event horizon, each observer notes their spatial separation from their nearest neighbours at that time (according to their own local clock) and the distance to the event horizon according to a given observer at "that time" is the sum of all the individual measurements. The observers have to be very close together and the Born rigid rulers very short. Unlike Fermi normal observers and clocks, the Pf free fall observers are not physically connected in any way and do not experience proper acceleration. They will notice that their separation from their immediate neighbours is continually changing according to measurements using their Born rigid rulers or radar measurements. The Pf distance should also agree with the Schwarzschild coordinate distance and with what is measured by a free falling observer using a rolling wheel meter like the odometer of a car, as also pointed out by Pf. I am pretty sure the Pf measurement is closely related to the PG coordinate system. It would be interesting to check that out.
 P: 1,555 You are welcome yuiop, I definitely can use some practice in checking mistakes in formulas as the formulas I write myself are often full of mistakes. I attempted to generalize the formulas to obtain the proper velocity of a free falling observer wrt a local stationary observer, in the attempt to generalize the, what Pervect calls, 'the passionflower distance'. Then perhaps we can do the same thing but then for the 'Fermi distance'. Apparently the velocity function can be parametrized, if I am not mistaken such thing was developed by Hartle (or perhaps he copied it before, I don't know that). Basically the proper velocity wrt a local stationary observer of a radially free falling observer can be described as: $$v=\sqrt {{\frac {{A}^{2}-g_{{{\it tt}}}}{{A}^{2}}}}$$ where: $$g_{{{\it tt}}}=1-2\,{\frac {m}{r}}$$ In the case the free fall is from infinity A simply becomes 1. In the case the free fall has an initial velocity then A is calculated as: $$A={\frac {1}{\sqrt {1-{v}^{2}}}}$$ Now, if we assume the formula below as given by yuiop is both correct and not an approximation: $$\frac{dr}{dtau} = \sqrt{\frac{2m}{r} - \frac{2m}{R}}$$ then there is a catch. For A smaller than 1 the free fall starts from a given height. I was trying to obtain the correct formula in order to plug in the right A in such a case and here is where I got some surprises (I assume for the greater minds on the forum this is yet another demonstration of my lack of understanding). I did not succeed in expressing this in terms of r only. Initially I was trying to reason that if in some way I could 'subtract' the escape velocity at the given R and convert that into A I would get the expected results. But that did not work, it turns out that A is no longer constant during the free fall from a given height, we could express it in the following way to get the desired results, but it is ugly: $$A = -{\frac {\sqrt { \left( -rR+2\,mR-2\,rm \right) R \left( -r+2\,m \right) }}{-rR+2\,mR-2\,rm}}$$ This 'monstrosity' is not even real valued! Now the interesting question is why, assuming again the formula based on this is exact, does the parameter depend on r? Perhaps I made a mistake or is perhaps the formula to obtain the proper velocity from a given R only an approximation? For instance it is rather tempting to calculate A for a drop at h independently of r by using: $$A=\sqrt {1-2\,{\frac {m}{h}}}$$ The result is close to yuiop's formula but not identical. The 'shape' of the slabs look rather similar.
 Sci Advisor P: 8,470 Is it true that each hovering observer's local surface of simultaneity in their local inertial rest frame is parallel to a line of constant Schwarzschild time and varying radius? If so, maybe Passionflower's distance would be the proper distance along a single spacelike curve where the tangent at each point along the curve is parallel to the local surface of simultaneity of a freefalling observer at that point?
P: 1,555
 Quote by yuiop Mathpages gives a parameter K defined as $$K = \sqrt{1-V^2}$$ or alternatively as: $$K = 1/\sqrt{1-2m/R}$$ It is obvious that the Hartle parameter A is related to the mathpages parameter by A = 1/K. This suggests you should end up with: $$A = \sqrt{1-2m/R}$$ for the constant parameter in terms of the apogee height R.
Right, I think I did that, but then the results do not seem identical to the formula you quoted (see the graph). Could you confirm? It looks like it is off by some linear factor. I probably made some mistake.

Hopefully we can work out the kinks because then we have a generic velocity formula for a free falling observer. Then it is simply a matter of finding the Lorentz factor based on the obtained velocity v(r) and divide (because it is length contraction! ) the integrand by the Lorentz factor to get the correct integrand to obtain a proper distance in all free falling radial scenarios (except for acceleration that is, but I think it is better to get the kinks out free falling motion first). By the way, the parametric approach has the advantage that we can extend things to a solution with a rotating black hole by applying some adjustments. As far as I am concerned that would then cover most areas where the notion of distance makes some sense, as the concept of proper (e.g. physical) distance in non-stationary spacetimes do not make much sense to me (except perhaps in things like FLRW models where distances are 'pumped up' in time). To me, but that is likely because I am simple minded, the generic distance discussion is rather academic for arbitrary non-stationary spacetimes as there is really not much to calculate.

By the way once this is sorted out it would be really interesting to lay out (in a separate topic) all the Doppler shifts (including a decomposition into kinematical and gravitational components) for all radial motion wrt a stationary local observer, an observer from infinity and of course wrt other free falling observers. As far as I know, no such generic formula is available.
P: 3,967
 Quote by Passionflower Apparently the velocity function can be parametrized, if I am not mistaken such thing was developed by Hartle (or perhaps he copied it before, I don't know that). Basically the proper velocity wrt a local stationary observer of a radially free falling observer can be described as: $$v=\sqrt {{\frac {{A}^{2}-g_{{{\it tt}}}}{{A}^{2}}}}$$ where: $$g_{{{\it tt}}}=1-2\,{\frac {m}{r}}$$
O.K. I have figured it now. The equation given by Hartle is not proper velocity (dr/ddtau), but the coordinate velocity (dr'/dt') according to a local stationary observer using his local stationary rulers and clocks. Proper velocity is the velocity using the proper time of the falling clock which is a different thing.

Starting with the Schwarzschild falling coordinate velocity I gave in #69:

$$\frac{dr}{dt} = (1-2m/r) \sqrt{\frac{2m}{r}(1-V^2) +V^2}$$

and using dr'/dr = 1/sqrt(1-2m/r) and dt/dt' = 1/sqrt(1-2m/r) to relate local measurements of the stationary observer to the observer at infinity such that dr/dt = dr'/dt' (1-2m/r), the local coordinate velocity is:

$$\frac{dr'}{dt'} = \sqrt{\frac{2m}{r}(1-V^2) +V^2}$$

This can be written using A = 1/sqrt(1-v^2) as:

$$\frac{dr'}{dt'} = \sqrt{\frac{2m/r +V^2A^2}{A^2} }$$

$$\Rightarrow \frac{dr'}{dt'} = \sqrt{\frac{2m/r + A^2 - A^2 +V^2A^2}{A^2} }$$

(because A^2-A^2 =0)

$$\Rightarrow \frac{dr'}{dt'} = \sqrt{\frac{2m/r + A^2 - A^2(1-V^2)}{A^2} }$$

$$\Rightarrow \frac{dr'}{dt'} = \sqrt{\frac{2m/r + A^2 - 1 }{A^2}$$

$$\Rightarrow \frac{dr'}{dt'} = \sqrt{\frac{ A^2 - (1 -2m/r) }{A^2}$$

$$\Rightarrow \frac{dr'}{dt'} = \sqrt{\frac{ A^2 - g_{\it tt} }{A^2}$$
P: 1,555
 Quote by yuiop O.K. I have figured it now. The equation given by Hartle is not proper velocity (dr/ddtau), but the coordinate velocity (dr'/dt') according to a local stationary observer using his local stationary rulers and clocks.
Just to verify we are on the same page, you do not mean coordinate velocity in terms of Schw. coordinates right? Because clearly a local stationary observer would not measure such coordinates with local clocks and rulers. Also the free falling observer would measure the same velocity right, since this is simply the principle of relativity at work, right (e.g. time dilation, length contraction, locally Minkowskian)? But if so, then what is proper velocity in this context if you call this one coordinate velocity? Celerity?

Since:

$${dt \over d\tau} = {A \over g_{tt} }$$

It should be staightforward to convert this coordinate velocity to proper velocity.

By the way it would be helpful is someone could attach the relevant page from Hartle, unfortunately I do not have it: http://books.google.com/books?id=azZ...ed=0CC0Q6AEwAA
P: 3,967
 Quote by Passionflower Just to verify we are on the same page, you do not mean coordinate velocity in terms of Schw. coordinates right? Because clearly a local stationary observer would not measure such coordinates with local clocks and rulers. Also the free falling observer would measure the same velocity right, since this is simply the principle of relativity at work, right (e.g. time dilation, length contraction)? But if so, then what is proper velocity in this context if you call this one coordinate velocity? Celerity?
I was just trying to make clear the distinction between proper velocity and coordinate velocity. Proper velocity (or celerity) is coordinate distance (which depends on the observer) divided by the proper time of the moving object. In post #69 I gave the proper velocity of the falling object according to the Schwarschild observer at infinity as:

$$\frac{dr}{dtau} = \sqrt{\frac{2m}{r} + \frac{ V^2}{(1-V^2)}}$$

where tau is the proper time of the falling clock. This can be converted to the proper velocity as measured by a local observer at coordinate radius r by using dr'/dr = sqrt(1-2m/r) so that:

$$\frac{dr'}{dtau} = \frac{1}{\sqrt{1-2m/r}} \sqrt{\frac{2m}{r} + \frac{ V^2}{(1-V^2)}}$$

and this can be simplified using A = 1/sqrt(1-V^2) and $g_{tt} = (1-2m/r)$ to:

$$\frac{dr'}{dtau} = \sqrt{\frac{A^2 - g_{tt}}{g_{tt}}}$$

The equation for the local velocity of the falling object given by Hartle is obviously not the proper velocity.

 Since: $${dt \over d\tau} = {A \over g_{tt} }$$ It should be staightforward to convert this coordinate to proper velocity.
Yes we can do it that way too. Starting with the Hartle equation for local velocity:

$$\frac{dr'}{dt'} =\sqrt {{\frac {{A}^{2}-g_{{{\it tt}}}}{{A}^{2}}}}$$

the local proper velocity is:

$$\frac{dr'}{dtau} = \frac{dr'}{dt'}*\frac{dt'}{dt}*\frac{dt}{dtau} = \frac{A}{\sqrt{g_{tt}}} \sqrt {\frac {{A}^{2}-g_{tt}}{A^2}} = \sqrt{\frac{A^2 - g_{tt}}{g_{tt}}}$$
P: 1,555
 Quote by yuiop I was just trying to make clear the distinction between proper velocity and coordinate velocity. Proper velocity (or celerity) is coordinate distance (which depends on the observer) divided by the proper time of the moving object. In post #69 I gave the proper velocity of the falling object according to the Schwarschild observer at infinity as:
Ok, it looks we are on the same page.

After some tinkering, the simplest generic formula I came up with is not the 'Hartle' approach but this:

$$v_{(r)} = \sqrt {2m \left( {r}^{-1}-{{\it r_0}}^{-1}-{\frac { \left( -r+1 \right) {{\it v_0}}^{2}}{r}} \right) }$$

I worked out a position-velocity equivalent, and it turned out it is much simpler to express an initial velocity as a (virtual) position than vice versa.

Lorentz factoring the equation gives:

$${1 \over \gamma} =\sqrt {1-2\,m \left( {r}^{-1}-{r_{{0}}}^{-1}-{\frac { \left( -r+1 \right) {v_{{0}}}^{2}}{r}} \right) }$$

Multiplying 1/gamma by the integrand used by a stationary observer to obtain the distance for an arbitrary radial free falling observer gives:

$$\int _{{\it r1}}^{{\it r2}}\!\sqrt {1-2\,m \left( {r}^{-1}-{{\it r_0}}^ {-1}-{\frac { \left( -r+1 \right) {{\it v_0}}^{2}}{r}} \right) }{\frac {1}{\sqrt {1-2\,{\frac {m}{r}}}}}{dr}$$

A few plots:
P: 3,967
 Quote by Passionflower Ok, it looks we are on the same page. After some tinkering, the simplest generic formula I came up with is not the 'Hartle' approach but this: $$v_{(r)} = \sqrt {2m \left( {r}^{-1}-{{\it r_0}}^{-1}-{\frac { \left( -r+1 \right) {{\it v_0}}^{2}}{r}} \right) }$$
Before I check this out in detail, can you confirm that to use your formula, that if r0 is not infinite then v0 must be set to zero and if v0 is not zero then r0 has to be set to infinite?
P: 1,555
 Quote by yuiop Before I check this out in detail, can you confirm that to use your formula, that if r0 is not infinite then v0 must be set to zero and if v0 is not zero then r0 has to be set to infinite?
Yes, as far as I understand both parts 'on' would lead to something unphysical.
P: 3,967
 Quote by Passionflower After some tinkering, the simplest generic formula I came up with is not the 'Hartle' approach but this: $$v_{(r)} = \sqrt {2m \left( {r}^{-1}-{{\it r_0}}^{-1}-{\frac { \left( -r+1 \right) {{\it v_0}}^{2}}{r}} \right) }$$
If I set v0 = 0 then the above equation reduces to:

$$v_{(r)} = \sqrt {2m \left( {r}^{-1}-{{\it r_0}}^{-1}\right) }$$

which agrees with the mathpages equation I gave in #69 for the proper velocity dr/dtau:

 Quote by yuiop ... the proper velocity of a free falling particle dropped from R in terms of the proper time of the particle is: $$\frac{dr}{dtau} = \sqrt{\frac{2m}{r} - \frac{2m}{R}}$$
Now if I set r0 = infinite in your equation, it reduces to:

$$\frac{dr}{dt} = \sqrt{\frac{2m}{r}(1-v_0^2) +2mv_0^2}$$

which does not agree with either the mathpages coordinate velocity or proper velocity:

 Quote by yuiop The coordinate velocity of a free falling particle at r, with initial velocity of V at infinity is: $$\frac{dr}{dt} = (1-2m/r) \sqrt{\frac{2m}{r}(1-V^2) +V^2}$$ ... the proper velocity of a free falling particle in terms of the proper time of the particle is: $$\frac{dr}{dtau} = \sqrt{\frac{2m}{r} + \frac{ V^2}{(1-V^2)}}$$
although it is similar to the coordinate velocity. More tinkering required I think

You did not specify if $v_{(r)}$ is intended to be a proper or coordinate velocity or if the observer that makes the measurement is local or at infinity.

[EDIT] Actually looking back I am guilty of the same thing in #69 and did not make it clear that all measurements in that post are according to the Schwarzschild observer at infinity.
P: 1,555
 Quote by yuiop Now if I set r0 = infinite in your equation, it reduces to: $$\frac{dr}{dt} = \sqrt{\frac{2m}{r}(1-v_0^2) +2mv_0^2}$$ which does not agree with either the mathpages coordinate velocity or proper velocity: although it is similar to the coordinate velocity. More tinkering required I think
There are different kind of velocities, pure coordinate velocity for an observer at infinity will obviously be zero at the EH.

 Quote by yuiop You did not specify if is intended to be a proper or coordinate velocity or if the observer that makes the measurement is local or at infinity.
The velocity this formula is supposed to track is the velocity as measured by a local stationary observer. I am not using the observer at infinity in any way, this observer is by the way physically speaking rather uninteresting.

Perhaps that is the source of some of the confusion we had communicating this. At all times I talked about the velocity as measured by a local stationary observer.

So far I do not see anything wrong with the formula, and the plots look right as well.

Perhaps you could present an alternative and then we can plot those out as well. I would not have been wrong the first time. :)
Emeritus
P: 7,620
 Quote by yuiop As far as I can tell, Pf's distance is the distance measured by a series of free falling observers each with their own clock and very short Born rigid ruler. The observers would have to synchronise clocks at infinity and then jump at regular intervals of say one second. At the time the leading observer arrives at the event horizon, each observer notes their spatial separation from their nearest neighbours at that time (according to their own local clock) and the distance to the event horizon according to a given observer at "that time" is the sum of all the individual measurements. The observers have to be very close together and the Born rigid rulers very short. Unlike Fermi normal observers and clocks, the Pf free fall observers are not physically connected in any way and do not experience proper acceleration. They will notice that their separation from their immediate neighbours is continually changing according to measurements using their Born rigid rulers or radar measurements. The Pf distance should also agree with the Schwarzschild coordinate distance and with what is measured by a free falling observer using a rolling wheel meter like the odometer of a car, as also pointed out by Pf. I am pretty sure the Pf measurement is closely related to the PG coordinate system. It would be interesting to check that out.
You've gotten a key point - because PF's ensemble of observers doesn't maintain a constant distance from each other, they don't define a Born rigid distance.

The motion of the observers can't be ignored and is important to specify, because different observers have different notions of space and time.

As far as using small, born-rigid rulers goes - if you are in an area of space-time small enough that the effects of curvature are small, radars and rulers agree. This is why the SI folks were able to switch from a physical, rigid rod, to a radar-like method of counting the wavelengths of a particular frequency of a cesium standard without a major upheaval. When you consider how atoms maintain their distance, it's not surprising, the forces between nearby atoms are based on the same electromagnetic fundamentals that the radar is based on.

One of the advantages of the radar concept for measuring distance is that it makes the implied concept of simultaneity more manifest than using rigid bars.

Let one wolrdline hold the radar. The other worldline reflects the radar signal. Then the midpoint between the time of signal emission and its reception on the radar world-line is simultaneous with the reflection of the radar signal on the reflection worldline.

This defines a short segment of a curve on the radar world-line, that propagates in a spatial direction. It's an alternate way of defining "orthogonal" without the math, though the 4-vector approach remains more convenient in my opinion. And if you imagine repeating the process, by putting a bunch of radars on a bunch of worldlines, and adjusting the timing so that the short line-segments all connect, you'll generate the entire curve, the curve of simultaneity, the one that PF's distance is being measured along.
 P: 6 I want to know that is it possible that we will get all our lost things when the big CRUNCH occur?
P: 3,967
 Quote by Passionflower The velocity this formula is supposed to track is the velocity as measured by a local stationary observer. I am not using the observer at infinity in any way, this observer is by the way physically speaking rather uninteresting.
In that case you formula still does not reduce to the correct equations for the local velocity when r0 = infinite or v0 = 0. These are the correct equations:

The local velocity of a falling particle with initial velocity v0 at infinity, according to an observer at r, was derived in #79 and is:

$$\frac{dr'}{dt'} = \sqrt{\frac{2m}{r}(1-v0^2) +v0^2}$$

The local velocity of a falling particle with apogee r0 according to an observer at r is:

$$\frac{dr'}{dt'} = \sqrt{\frac{2m(r^{-1} - r0^{-1})}{(1-2m/r0)}}$$

 Perhaps that is the source of some of the confusion we had communicating this. At all times I talked about the velocity as measured by a local stationary observer. So far I do not see anything wrong with the formula, and the plots look right as well. Perhaps you could present an alternative and then we can plot those out as well. I would not have been wrong the first time. :)
A general formula that combines both the above equations should be something like (if I have got it right):

$$\frac{dr'}{dt'} = \sqrt{\frac{2m(r^{-1}(1-v0^2) - r0^{-1})}{(1-2m/r0)}+v0^2}$$

Just to be clear, I am not talking about proper velocity and I assume you are not either.
P: 3,967
 Quote by Roshanjha14 I want to know that is it possible that we will get all our lost things when the big CRUNCH occur?
Yes, all your mysteriously lost odd socks will be right there next to you, but they will still be rather difficult to find because everyone elses missing odd socks and the rest of the universe will be right there next to you too.

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