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The length of a falling elevator

  1. Sep 24, 2010 #1
    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 any one of the questions, that would be a significant start. :smile:
     
    Last edited by a moderator: Apr 25, 2017
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  3. Sep 24, 2010 #2

    pervect

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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.
     
  4. Sep 24, 2010 #3
    Hi Pervect,

    I realise 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)?
     
  5. Sep 25, 2010 #4

    pervect

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    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?
     
  6. Sep 25, 2010 #5
    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:

    Schwarzschild Radius: R0 = 2
    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:

    [tex]\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
    [/tex]

    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:

    [tex]2\,x-6+4\,\ln \left( x-2 \right) [/tex]

    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:

    [tex]\sqrt{1-{2 \over r}}[/tex]

    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?
     
    Last edited: Sep 26, 2010
  7. Sep 25, 2010 #6
    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.

    Your comments are appreciated.
    Regards.
     
    Last edited: Sep 26, 2010
  8. Sep 25, 2010 #7
    Does this imply a limitation of freefall coordinate velocity acquired due to acceleration, dependant on the large mass but independant of the distance fallen??
     
  9. Sep 25, 2010 #8
    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.
     
    Last edited: Sep 25, 2010
  10. Sep 25, 2010 #9
    OK good analogy. To make my question more specific:
    Freefalling from infinity towards a massive blackhole 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
     
  11. Sep 25, 2010 #10

    pervect

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    We're getting into definitional issues again, I think. Lets 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 travelled "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.
     
  12. Sep 25, 2010 #11
    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 fundemental 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" [itex]\Delta r = r2-r1[/itex] where r1 and r2 are Schwarzschild spatial radial coordinates?
     
  13. Sep 25, 2010 #12
    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.

    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:

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

    That gives us:

    [tex]
    \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)
    [/tex]

    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?
     

    Attached Files:

    Last edited: Sep 26, 2010
  14. Sep 26, 2010 #13
    Hi.
    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.
     
    Last edited: Sep 26, 2010
  15. Sep 26, 2010 #14
    Agree.

    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
     
  16. Sep 26, 2010 #15
    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?
     
    Last edited: Sep 26, 2010
  17. Sep 26, 2010 #16

    pervect

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

    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.

    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.

    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.
     
  18. Sep 27, 2010 #17

    pervect

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    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

    [tex]
    \frac{dr}{d\tau} = - \sqrt{\frac{2M}{R} - \frac{2M}{r}}
    [/tex]

    [tex]
    \frac{dt}{d\tau} = \frac{ \sqrt{ 1 - \frac{2M}{R} } } {1 - \frac{2M}{r}}
    [/tex]

    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 [itex]\tau[/itex]+[itex]\epsilon[/itex] to the first order the particle's position will be

    [tex]
    r = r + \epsilon \frac{dr}{d\tau}
    [/tex]

    [tex]
    t = t + \epsilon \frac{dt}{d\tau}
    [/tex]

    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.

    [tex]
    g_{tt}\, p\, \frac{dt}{d\tau} + g_{rr} \, q \, \frac{dr}{d\tau} = 0
    [/tex]

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

    [tex]
    g_{tt}\, p^2 + g_{rr} \, q^2 = 1
    [/tex]

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

    [tex]
    g_{tt} = -1 + \frac{2M}{r}
    [/tex]

    [tex]
    g_{rr} = \frac{1}{1 - \frac{2M}{r}}
    [/tex]

    Then the coordinates of our following particle should be:

    [tex]
    t = t + \epsilon*p
    [/tex]
    [tex]
    r = r + \epsilon*q
    [/tex]

    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
    [tex]
    p = \frac{\sqrt{\frac{2M}{R} - \frac{2M}{r}}}{1 - \frac{2M}{r}}
    [/tex]

    [tex]
    q = \sqrt{1-\frac{2M}{R}}
    [/tex]

    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.
     
    Last edited: Sep 27, 2010
  19. Sep 27, 2010 #18
    Do you assume dr/dt= 0 at R?

    Then isn't the correct formula:

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

    And

    [tex]
    \frac{dt}{d\tau} = {r\sqrt {{\frac {R-2\,M}{R}}} \over r-2\,M }
    [/tex]

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

    I missed this on initial viewing of your posting.

    I am not sure what this is: constant energy at infinity?
    Can you explain?
     
    Last edited: Sep 27, 2010
  20. Sep 27, 2010 #19

    pervect

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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

    [tex]
    d \tau = \frac{dr} { \sqrt{\frac{2M}{r} - \frac{2M}{R}} }
    [/tex]

    and also on this page
    [tex]
    R = \frac{2M}{1-E^2}
    [/tex]

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

    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.
     
    Last edited: Sep 27, 2010
  21. Sep 27, 2010 #20
    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.


    From the reference you give, the velocity at r of the falling short elevator dropped from Schwarschild coordinate R with an initial velocity of [itex]v_0[/itex], according to a stationary local Schwarzschild observer at r as the elevator passes is:

    [tex]v_f = \sqrt{1-v_0} \sqrt{\frac{R/r-1}{R/rS-1}} [/tex]



    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 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.
     
  22. Sep 28, 2010 #21
    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):

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

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

    [tex]
    {\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)
    [/tex]

    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:

    [tex]
    \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)
    [/tex]

    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 lets 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):

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

    So the answer is:
    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?
    Answer: 17.28894242 seconds

    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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    Last edited: Sep 28, 2010
  23. Sep 28, 2010 #22

    pervect

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    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.

    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.
     
  24. Sep 28, 2010 #23

    pervect

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    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.

    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 (?).

    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.

    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...
     
  25. Sep 28, 2010 #24
    Coordinate distance is simply the difference between two coordinate values.

    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.

    No, if you would have read correctly you would have seen I am using a clock falling fron infinity.


    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.

    Yes you are absolutely correct, and since my clocks are falling from infinity it is an applicable formula.

    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.
     
    Last edited: Sep 28, 2010
  26. Sep 28, 2010 #25

    pervect

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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.

     
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