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Co-accelerating spaceships (Bell's Spaceship Paradox), how does this occur in SR?

  1. Feb 13, 2006 #1
    I'll describe the problem in my own words.

    The co-accelerating ships problem:

    -There are two identical spaceships which are initially at rest in "lab-frame".

    -These two ships perform IDENTICAL acceleration procedure.

    -The ships are launched simultaneously in the lab-frame.

    http://www.saunalahti.fi/~anshyy/PhysicsForums/Simultaneity03.jpg" [Broken]
    Red ones are the space ships. Right one is the "front ship". Blue line is an observer who stays at rest, added in there just for convenience. The acceleration events are instantaneous here, but you can imagine any sort of real-world acceleration instead)

    Now, since the accelerations are identical in lab-frame, it should hold true that the distance between the ships does NOT change during the acceleration; As measured by the lab-frame, the distance between ships before and after the acceleration is exactly the same.

    Obviously in the mechanics of SR, the acceleration procedures are NOT identical from the point of view of the ships. The distance between the ships should stretch in their own POV:

    From the "POV" of FRONT SHIP:
    Immediately after acceleration, the front ship exists in such an inertial frame that the rear ship must not have been launched yet. If you assert it has, then the front ship will receive information about the launch of the other ship at speeds slower than C

    From the "POV" of the REAR SHIP:
    Vice versa happens. Immediately after the acceleration, the rear ship exists in such an inertial frame, that the front ship must have been launched much earlier than the rear ship. If you assert it wasn't launched earlier, then the rear ship will receive information about the launch of the front ship at speeds faster than C.

    POV is used here to refer to how things "are" in the inertial coordination system of an observer, according to Lorentz-transformation. POV is NEVER used in the meaning of what the observer can SEE; This talk concerns Lorentz-contraction.

    In other words, the front ship will accelerate away from the other ship while it is still sitting on the launch pad. Here's what it all looks like from the inertial coordination system that the ships will end up to after acceleration.

    http://www.saunalahti.fi/anshyy/PhysicsForums/Simultaneity04.jpg" [Broken]
    Black lines are planes of simultaneity

    Even though the ships did go through the same acceleration procedure, after the fact these procedures exist in different moments in time.

    We should arrive at the same conclusion even if there is a steel rod between the ships.

    And that's not all. Since the ships are performing an identical acceleration, they should stay in the same inertial coordination system at all times. In other words, their notion of simultaneity should be identical at all times. In other words the ships (& the rod) should keep their length from their own perspective, and contract from the perspective of the blue observer.

    So it seems that SR very concretely requires that from the POV of the front ship, the rear ship must still be on launch pad after the launch, AND it must be in the same inertial coordination system at all times (i.e. NOT on the launch pad).

    There is odd asymmetry in Lorentz-contraction. It occurs to external objects when you switch inertial frames. But at the same time, it doesn't seem to occur if it is the external objects that are switching frames. And if the observer has any volume, he should stretch by the same mechanic that causes contraction (At least if he is being accelerated from the front and from the rear simultaneously).

    When I first asked about this, it was noted that there is no standard way to construct a coordinate system where non-inertial observer is at rest.

    However, if we place the launch pads at far enough distances from each others, and/or use rapid enough acceleration, it is trivial to show that AFTER the acceleration the front ship can exist in such an inertial frame, that the rear ship must not have been launched yet, IF it is true that light approaches the front ship in this new inertial coordination system at speed C.

    I was also informed of a FAQ-page regarding this problem. I didn't understand the explanation, if any was even offered:

    http://math.ucr.edu/home/baez/physics/Relativity/SR/spaceship_puzzle.html" [Broken]

    They first give instructions to draw similar spacetime diagrams as I have offered above, but with non-instantaneous acceleration. Then they note:

    It doesn't make any difference who thinks the accelerations are constant and who doesn't. All that matters is that the acceleration procedures themselves are identical. Surely one acceleration procedure looks the same when performed in any location of the lab-frame.

    Then they say:
    What does this mean? Just pick the curve to be something else? Just decide that of the two identical accelerations, the other one is not identical?

    What is the actual explanation?
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Feb 13, 2006 #2


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    It depends on what you mean by "immediately". There will not, in fact, be any sudden jump in simultaneity, but there will be a gradual shift in simultaneity as the ship accelerates.

    The "point of view" of an accelerating observer needs to be defined, like any other coordinate system. A good choice is to use the "point of view" of an instantaneously co-moving observer.

    This choice of POV is covered in textbooks (specifically MTW's Gravitation). The resulting expression is that the "POV" coordinates (which act fine locally and become non-unique too far away from the spaceship) are translated into inertial coordinates by the following expressions.

    As I mentioned in another post
    t = (1/g + \chi) \mathrm{sinh}(g \tau)
    x = (1/g + \chi) \mathrm{cosh}(g \tau)

    (t,x) are the inertial coordinates
    [itex](\tau,\chi)[/itex] are the "POV" coordinates.

    Plot the lines of simultaneity ([itex]\tau = constant[/itex])

    Let the acceleration, g, be 1 for simplicity.

    Then at [itex]\tau=0[/itex], the lines of simultaneity are given by the expressions

    [itex] t = (1+\chi)[/itex]
    [itex] x = 0 [/itex]

    Plot this parametrically - you'll see a horizontal line, x=0, as you'd expect.

    Now let's evaluate the expression at [itex]\tau=.1[/itex]

    You get approximately

    [itex] t = (1+\chi)*1.005[/itex]
    [itex] x = (1 + \chi)*.1[/itex]

    Plot these parametrically - you'll see a slightly tilted horizontal line. This line starts at the location of the spaceship at t=.1, and has a slope of .1

    Eventually these coordinate lines cross (the lines for tau=0 and tau=.1), this happens at x=0. (But the ship is at x=1).

    See the attached plot. The single curved line is the plot of the ships position (it starts at x=1, t=0). The various straight lines are lines of simultaneity at tau = .1, .2, .5, and 1

    I've talked at length about the significance of the coordinate lines intersecting previously - basically, multiple coordinates get assigned to the same point. This is rather ill-behaved, but perhaps not fatal. I usually don't use coordinates in this region, just because they act strangely, so I may not be aware of all the pitfalls of such usage. You are the person who wants to assign the ontological status of "reality" to coordinates, whereas I regard coordinates as being observer dependent, and thus in a sense not really "real". You are also the person who wants to use the coordinates beyond the point where they overlap - I'm the person who is leery of such usage. So when you complain about how weird they act, I tend to think "Then don't do it!"

    Attached Files:

    Last edited: Feb 13, 2006
  4. Feb 13, 2006 #3


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    The coordinate system of an accelerated observer is not an inertial coordinate system. Only when the coordinates of the same inertial observer are used consistently is the coordinate system inertial. The coordinate system of an accelerated observer is decidely non-inertial.

    It is true that coordinates assigned to a forward-runing clock can run backwards when the clock is beyond the "Rindler horizon". This is just the region where I don't use the coordinates anymore because a single point has a multiple coordinates, thus the mapping of points to coordinates is not 1:1, but into (or was it onto? - never can keep that straight, but it's not 1:1 which must be both).

    THis is the region where the coordinate lines cross in my previous post, at x=0.

    This region (x=0) is also an event horizon - light signals emitted from behind the rindler horiozon (i.e. x<0) will never reach the spaceship!

    If you look at the equation for accelrated motion I gave, you can see why. The signal of a lightbeam is given by x=t. The accelerated spaceship with an acceleration of unity has coordinates x=cosh(tau), t=sinh(tau). This means that x is always greater than t, because
    [tex]\mathrm{cosh} \tau = (e^{\tau}+e^{-\tau})/2 \,\, \mathrm{sinh} \tau = (e^{\tau} - e^{-\tau})/2[/tex]

    and thus cosh(tau) > sinh(tau).

    Therfore there is no solution for x=t, i.e. for cosh(tau) = sinh(tau), i.e. for the light beam reaching the spaceship.

    The Faq is trying to tell you that while the accelerations of the two spaceship appear to be the same in the lab frame, the accelerations of the two spaceships do not appear to be the same in the non-inertial coordinate system of the spaceships themselves.

    The "upper" spaceship will appear to have clocks that tick faster than the lower spaceship, due to "gravitational time dilation". Gravitational time dilation occurs only in non-inertial frames, but the accelrating spaceships coordinate systems are just such frames.

    Both spaceships will agree about this, in the accelerated frame higher objects have clocks that tick faster than lower objects. Because of this clock rate difference, coordinate accelerations will not be the same between the two spaceships.

    In the problem as specified, the front spaceship will always have a greater coordinate accleration than the rear spaceship.

    This means that any string connected betweent the two spaceships will break as the spaceships pull apart.
    Last edited: Feb 13, 2006
  5. Feb 13, 2006 #4
    I gave this one some thought before. I found the Usenet Physics FAQ explanation to be overly complex for my tastes. At the bottom of this thread is a layman's explanation.

    Also consider The Pinwheel Paradox, which resolves to Bell's Spaceship Paradox for any two adjacent floors of the pinwheel. (You'll have to wade through the conversation but it's not long.) The symmetry of the pinwheel helps intuit both Bell's paradox and the concept of excess radius in general relativity.
  6. Feb 13, 2006 #5


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    Sorry for the large number of responses, but there is one more fact I forgot to mention.

    The set of coordinates ([itex]\tau,\chi[/itex]) which I will from now on call "accelerated coordinates" does not cover all of space-time.

    If you refer to the equations I posted earlier which convert accelerated coordinates to the corresponding non-accelerated coordinates, no value of ([itex]\tau,\chi[/itex]) maps to (t=0,x=10), as

    t^2 - x^2 = (1+[itex]\chi[/itex])^2

    Thus t^2 - x^2 must always be positive. Thus "accelerated coordinates" may not exist for a given point (t,x) specified in unaccelerated coordinates.

    (I already mentioned that when they do exist, accelerated coordinates are not necessarily unique, but I'll mention it again).

    This is yet another reason to view and treat accelerated coordinates cautiously.
  7. Feb 14, 2006 #6
  8. Feb 14, 2006 #7
    Seems like you've been thinking about the exact same problems that I have :)
    I was thinking about the pinwheel paradox at some point too, but I figured it out when I realized that I was doing actually the same mistake what you were doing in the beginning of that thread; considering that from your perspective the next dude is aged more, and from HIS perspective the next one is aged more, etc...

    I figured it out when I realized that this is in fact the same thing as considering two observers who are simply approaching each others without any acceleration. Let's say you both started moving from "lab frame" at the same instant. When your own clock is showing, say 100 seconds, the clock of the other is showing like 200 seconds (from your inertial frame). You cannot now just hop into the perspective of the other observer, whose clock is showing 200 seconds, and assert that at the same moment your clock must show 400 seconds.

    And after you've passed each others the same thing happens. The clock of the other observer is always showing less than your clock, but you can't just jump immediately from the perspective of one observer to another, otherwise you end up zig-zagging back in time.

    So with the pinwheel problem too is simply about the plain old failure of simultaneity.

    Hmmm, in fact, let's hope they never invent immediate quantum teleportation, because if you had to teleport from spaceship to earthly McDonald's for lunch break, then when you teleported back to ship, it could be again the beginning of your lunch break, and you'd be in an infinite lunch-break loop, and get really fat. ;)

    Have you by any chance also solved the problem regarding two spinning wheels? If you have two wheels spinning on the same axis to separate directions, they both should regard themselves as larger than the other wheel. This is not the same as two trains shrinking and both regarding themselves longer than the other because of failure of simultaneity, since the shrinking of the wheels occurs at direction other than the motion. I can't see how this is solved at all.
  9. Feb 14, 2006 #8


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    As far as the rotating disks go, I favor the explanation by Tartaglia


    which is very interesting, but somewhat heavy reading.

    In a nutshell, if you put a chalk mark around the disk and walk around the cirumference, you may wind up at the same location as the chalkmark, and thus you think you've travelled around the circumference. But when you look at the situation more closely, you realize that no matter how slowly you walk, your time will not agree with the time of a clock left on the chalkmark. The sign of the discrpenancy will depend on whether you walked around the disk in the same direction as the rotation, or in the opposite direction.

    If you have three observers, one stationary at the chalkmark, one who walks around in the same direction as the rotation, and one who walks around in the direction opposite to the rotation, when they all get back together, all of their watches will read different times.

    While you may tend to want to ignore this and say that you are concrened with length and not time, it bites you in the end, because time and space "mix together" in relativity. One man's time is another man's space. One ultimately gets hopelessly muddled when one thinks of a circle around a rotating disk as a closed geometric object. The explanation is that it's not really closed, and it turns out that the defintion of circumference is ambiguous.
  10. Feb 15, 2006 #9


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    Maybe it´s important to point out again that the shrinking occurs only in the direction of the motion: The circumference shrinks, the radius stays constant -> circumference/diameter != PI -> Spacetime is not flat -> better use GR.
    And for your model of acceleration slowing processes down: The dilation is then proportional to the gravitational field, not to the gravitational potential as observed and described by GR. This model leads to wrong predictions and you better not use to get a better understanding of reality.
  11. Feb 15, 2006 #10
    Yeah, it was probably important to point it out, although I don't quite grasp it yet how the circumference can shrink without the whole wheel shrinking. I have kind of thought of it as if the radius becomes curved in the 3D-space, which would also mean that the object has in fact shrunk. But I'll try to think about it and start another thread if I get hopelessly lost :)

    Thank you pervect for your points as well.

    Hmmm, I seem to have missed your point. The higher the gravitational field, the lower the gravitational potential, the higher the time dilation?

    In simple words, the stronger the acceleration effect is being felt by an observer (be it gravitational or inertial), the stronger the time dilation he suffers as observed by a non-accelerating observer. Yes?
  12. Feb 15, 2006 #11


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    No. You may float perfectly weightless at the center of the earth, but your clock would tick slower than a reference clock eg at the north pole. That´s contrary to an effect proportional to the gravitational field strength. It´s the difference in gravitational potential that matters, not field strength.
  13. Feb 15, 2006 #12
    I don’t understand why so may smart people can make something simple so complex they get it wrong.

    Why would you think:
    Is it because you bought the false argument:
    But that is NOT what was given in your OP.
    You gave at t = 0 both frames for ship 1 and ship 2 start accelerating together, that can only mean there respective frames are equivalent! Also known as, the SAME FRAME, until one of them changes acceleration – what could be simpler.
    Let’s just make your set up a little clearer:

    Use 3 very large ROCKETS 1, 2, & 3 with ladders on their sides facing each other in a triangle. Rocket 1 has red spaceship #1 on step 3. Rocket 2 has red spaceship #2 on step 5 and Rocket 3 has a Blue observer on step 1.
    (Note: your diagram clearly shows these three to be at fixed different altitudes at t = 0 and all times prior so they cannot be “launched” from a common altitude launch pad.)
    Rockets provide all the power needed to keep the ships observer and frames (ladders) all accelerating the same. They may as well be attached to each other.

    Now the group of three have a non-powered ladder gaining on them though the center of the triangle moving at speed V. On this ladder is passengers 1 & 2 on steps 3 & 5 respectively and anther blue observer on step 1.

    Now at time = 0 the three rockets reach speed V just as the powered ladder at fixed speed V is level with the three ladders on the rockets. Passengers 1 and 2 step across into ships 1 & 2 as Rocket 3 runs out of fuel and stops accelerating. The two blue observer continue together at the same speed. Strings tied between various steps between ladders 1, 2, & 3, all strings between ladder 3 connecting to 1 or 2 break at t = 0.

    Now we can build the equivalent with out the expensive rockets. Three ladders on earth surface with #3 supported over a very deep hole. At t=0 the support holding ladder #3 is removed – – etc. etc.

    The two red lines in your OP graph are the passengers, the blue line is the observer on the ladder of fixed speed V. Every thing else is relative.

    No matter how long ladders 1 & 2 continue to accelerate, OR sit there under on the surface near a deep hole, the strings tied between them will never break I don’t care from who’s reference they are observed. The length of the strings and distances between the steps may well change based on the view of different observers, but those distance and length changes will be identical.

    Complex explanations & math that ‘prove’ differently, just means to me that an error was made, not some profound explanation or paradox.
    Last edited: Feb 15, 2006
  14. Feb 16, 2006 #13
    Oh I see.
    I presume there is such experimental data?
  15. Feb 16, 2006 #14
    Very interesting assertion.

    In fact, in order for the hole scenario to be completely equivalent to your initial description of ladders in space, the 3rd ladder should not be initially supported above the well, but rather thrown up from the hole so that at the topmost point of its trajectory it would just meet the other ladders, and the passengers would step on the ladders resting on earth (that way the 3rd ladder is non-accelerating at all times).

    The reason I bought pervect's argument is that I remember hearing that clocks placed "higher up" in the direction of acceleration will run faster, but I must say I haven't quite grasped this idea, and now that I'm thinking about it, it does seem a bit cumbersome idea.

    Like, let's say we have a completely uniform gravitational field in some room, and then we have a little rocket toy that is blasting at such strength that it will levitate absolutely stationary in this gravitational field.

    Now it would seem that we can place this toy in any place of the uniform field, and it will stay stationary in relative to the room.

    But GR claims that if we have two toys, the upper toy will tick faster than the lower toy, and so it will also accelerate more strongly.

    I would find it hard to believe, that if I have one toy levitating in the field, then just by placing another toy somewhere below it, the upper toy would suddenly start to gain distance on the lower toy.

    The description of all this in Feynman's "Six not-so-easy pieces" is quite superficial;
    "Suppose we put a clock "A" at the "front" end of a ship, and we put another identical clock "B" at the "tail". If we compare these two clocks when the ship is accelerating, the clock at the head seems to run faster relative to the one at the tail. To see that, imagine that the front clock emits a flash of light each second, and that you are sitting at the tail comparing the arrival of the light flashes with the ticks of clock B.

    The first flash travels the distance L1, and the second flash travels the shorter distance L2. It is a shorter distance because the ship is accelerating and has a higher speed at the time of the second flash.

    Higher speed compared to what? Definitely not compared to you, since you are traveling inside the ship. While a non-accelerating observer would see it like he describes, what does this have to do with people inside the ship exactly?

    He continues;
    The same thing will also happen for all the later flashes. So if you were sitting in the tail you would conclude that clock A was running faster than clock B.
    Another odd thing, how does the way you see the clock A running correlate with the way it IS running? He doesn't mention a failure of simultaneity anywhere (which I suspect must play a part here), so it would seem like he is just describing how each signal is arriving "faster" than the previous one, but all these signals would still have begun their journey one second apart, and so the clocks would in actuality tick at the same rate, as observed inside the ship.

    So, some help here, anyone? If you want to explain it with math, please include common sense explanation too.

    Can you Randall offer any alternative explanation to the original problem?

    Yeah, it must be said that I also find complex math explanations to be dangerously deceptive. And I am now looking for the internal structure of GR rather than fundamental nature of nature. The more I learn about the theory of relativity, the less I consider it to describe anything fundamental. I think accepting its macro-level descriptions as the fundamental level of nature is just insult to intellect.

    It's like the assertion that an electron is a particle and a wave. Please. No such thing as "a particle and a wave" can be fundamental. I don't know why such a thing is even being implied by smart people. Particles and waves are semantical concepts, neither of which fits to describe an electron. That doesn't mean electron is a particle and a wave any more than a swimming cat is a mammal and a fish.

    Obviously what we understand as an electron and its behaviour is a manifestation of some fundamental mechanism that we don't know of yet. Look at the behaviour, look at the data, and figure out what could cause such a thing to exist. Don't stop at the notion that it simply is "a particle and a wave".
  16. Feb 23, 2006 #15

    The above message about the toy rockets blasting in an uniform gravitational field or in an accelerating room... What's going on?
  17. Feb 23, 2006 #16


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    I think you (ANssih) had the right idea before you let RandallB confuse you.

    You intended spaceships A and B to stay a constant distance apart in the inertial frame, yes? This implies that they do not stay a constant distance apart in their own coordinate systems.

    It's always possible that I misunderstood you, but I don't think that's the case here.

    If I appear to be mostly ignoring RandallB, that's because I am mostly ignoring him. Life's too short for me to spend a lot of time talking to someone (RandallB) who doesn't listen.

    For a numerical treatment of the problem, I will refer you to some previous posts:
    (BTW, I highly recommend actually sitting down and doing the math, it will teach you things that endless discussions just can't.)


    is one example of a coordinate system that an accelerated observer can use.


    discuss at length another coordinate system accelerted observers can use, based on 'radar coordinates'. The radar coordinates are a bit distorted because they assume that the speed of light is constant in the accelerated frame, when the speed of light actually depends on distance. But you may be able to "come to grips" with these coordinates numerically a bit easier than the other ones I gave, because a derivation is presented.

    For a derivation of the first set of coordinates, I will refer you to MTW's textbook, "Gravitation". Unfortunately, parts of it may be a bit advanced.
    Last edited: Feb 23, 2006
  18. Feb 23, 2006 #17
    Well, that was the case with the original problem. I haven't plotted with the math you gave yet, partly because I'm being hit with a flu :) And partly because it seems pretty evident that the clocks towards which we are accelerating should advance faster in time if I'm just thinking about how the simultaneity planes are tilting (regardless of which direction the clocks are moving). So that just goes to show that the notion of simultaneity is not the same for two observers who are accelerating "uniformly" in a lab frame?

    But then the question is, is it not possible to have a room with uniform acceleration (let's say, inside a spaceship), and then have a little rocket accelerating inside the room, blasting with such strength that it could sit completely stationary inside the room. And do so ANYWHERE inside the room?
  19. Feb 23, 2006 #18
    No the thrown ladder your referring to is equally to the non-powered ladder at a fixed speed the catches up to the rockets just as they accelerate up to and past that fixed speed. Both cases feel like the same free fall in comparison to the rockets. And now kicking out the support on the third ladder will duplicate that third rocket running out of fuel, joining the “fixed speed” ladder in a common free fall, now in a gravity field.
    No, GR doesn’t say that. No flaw in GR equivalence, – no paradox here.
    Common mistake, your thinking of earth’s gravity field, not the one you defined “a completely uniform gravitational field”.
    We don’t have one of those here on earth’s spherical mass! To duplicate that, you need to use a wall of mass with a uniform density/area built as a disc, edges going to an effective infinity for your experiment. Then you do have the up and down uniform gravitational field that does not change vertically, so no gravitational red shift time dilation. Too many people incorrectly expect that earth style difference in space too. There have just built their gravitational force source incorrectly to be “Equivalent” to the in space rocket experiment reality. (This is the source of the upper rocket somehow gaining a “higher speed” due to time dilation to break the string.)

    Don’t forget the detail: The ships are accelerating at the same start time, but start at different altitudes. That’s why I put the toy ships on different positions of the ladders of large rockets. That way we can see the rockets using a common launch pad & time. Hence the altitude (distance) between them will always be the same from either ship’s (Or rocket carrying the ship) view. Even as they appear to change clock speed and lengths from other reference frame views. No string will be breaking here.

    Little details like these are what can get you to an incorrect conclusion or seeing a paradox when there is none. When pervect makes a mistake he has a tendency to shoot the messenger or some excuse rather than admit it. A mistake yelled loader is still a mistake. You just need to give him a little extra space, while you work out the details yourself.
  20. Feb 23, 2006 #19


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    If it's the one that comes with a very drippy nose, it's making it's way around the world. I am just getting over it here, everyone else in the house has had it too, and I was talking to someone in Czechlosovokia just the other day who had the same thing.

    Yep, exactly.

    Sure. However, if you mount an accelerometer on the "stationary" rocket, the accelerometer reading on that "stationary" rocket will decrease as the altitude of the rocket increases. It will be constant for any given "stationary" rocket, because that rocket will maintain a constant altitude - however, different rockets will have different constant accelerations required to "hold station" which depends on where they are located.

    We just got through a long derivation in another thread for a more complicated version of the problem you are talking about.


    In this thread it is derived that if the equations of motion for an accelerated obserer are

    x = cosh(g*tau)/g
    t = sinh(g*tau)/g

    (note- this is correct, my post #3 has a typo that's too old for me to correct that interchanges x and t)

    then the equations of motion for a "stationary" observer with respect to x and t can be most simply written as

    x' = cosh(g' * tau) / g'
    t' = sinh(g' * tau ) / g'

    where g' is not equal to g. (There is an alternate form of the equations of motion derived in that thread, but later on it is pointed out that they only look different than the above, that they are mathematically equivalent).

    The description of these curves is that they are all hyperoblae which share the same "hyperbolic point", which gives rise to the name "hyperbolic motion" for the motion of an accelerated observer.

    Just in case you didn't know, cosh(x)= (e^x + e^-x)/2
    while sinh(x) = (e^x - e^-x)/2, these are the hyperbolic trig functions.

    What this means is that the "stationary" rocket will have a lower proper acceleration depending on its height, where the proper acceleration is the aaccleration read by an accelerometer mounted on the hovering rocket. Mathematically the proper acceleration is derived by taking the magnitude of the 4-acceleration, which is just the deriviative of the 4-velocity with respect to proper time.

    (I don't know if you know much about 4-vectors. They are used througout the above thread, though, and are an essential tool of special relativity.)

    The higher the "stationary" rocket, the lower the (constant) reading on its accelerometer. If the "stationary" rocket is below the reference rocket, it actually has to accelerate harder. This acceleration increases towards infinity as the stationary rocket nears the hyperbolic point located at the origin of the coordinate system, a distance c/g "below" the reference rocket. This point also marks the plane of the "Rindler horizon", which I think I've talked about before (?).

    ps - Anssih, if you see some point made by RandallB that you want me to address, please quote it specifically and point it out to me - there's no guarantee that I've read it.
    Last edited: Feb 24, 2006
  21. Feb 23, 2006 #20
    You framed the problem well by putting the room in a spaceship. Therefore the apparent gravity throughout the room is uniform and any solution that calls for a different acceleration to “hold station” between different heights in the room is obviously wrong. Obvious common sense will make that clear.

    But what of that room on earth you wish was equivalent. The shape of the gravity field your using on earth, is not equivalent as there is a difference from top to bottom.

    To build an equivalent field see:http://www.mathpages.com/home/kmath530/kmath530.htm" [Broken]
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