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A question about a thought experiment in space

  1. Mar 12, 2007 #1
    Attached to this post is a thought experiment in space. It is only one page (38kb) in microsoft Word. At the end of the page is my question, and I hope some of you may help me about it. Thank you in advance.
     

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  3. Mar 12, 2007 #2

    russ_watters

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    All that thought experiment shows is whether the ship is at rest wrt the two buoys. How would you know if the buoys are at an absolute state of rest (assuming such a state existed)?
     
    Last edited: Mar 12, 2007
  4. Mar 12, 2007 #3
    I am not understanding your above reply. Indeed, the experiment (Attached to the thread) does not mention any buoy at all. The experiment only considers one spaceship (ABC), with three apparatus and one observer inside it. Sorry, but I think there has been some misunderstanding. Can you please look again at the experiment? Any way, thanks for your attention.
     
  5. Mar 12, 2007 #4

    Mentz114

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    An observer in the spaceship will only ever see T_ab = T_cb. The velocity of light will always be same inside the ship irrespective of any motion.

    That is the first principle of special relativity and appears to be confirmed by experiment.

    Oh, I should also add that there is no such thing as a state of absolue rest.
     
  6. Mar 12, 2007 #5

    The experiment does not deny that the velocity of the light pulse is the same and constant inside the spaceship. Just because of that, if the spaceship is moving to the right, the pulse from C will strike B before the pulse from A does it, and Tc<Ta. In the same way, if the spaceship is moving to the left the observer will find Tc>Ta. This seems to be obvious. See for example, figure 3-1, at page 63, of the book SPACETIME PHYSICS, of Professors Taylor and Wheeler, edited by W. H. FREEMAN AND COMPANY (1997). It is also recognized by Professor Paul A Tipler, in his book "Physics for Scientists and Engineers", Fourth Edition, Volume 3. And finally it is just the old story of Einstein´s long train, whose front and back ends were striken by two lightnings. It becomes obvious that Ta = Tc will only happen if the spaceship is not moving. But not moving with respect to what? The only possible answer is that it is not moving with respect to spacetime. In other words. It must be in absolute rest in space-time. Also the experiment itself seems to be one capable of distinguishing physically if a body is moving at v=constant or if it is at rest.
     
  7. Mar 12, 2007 #6

    Mentz114

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    No, you're wrong and so are your sources ( unless you're misinterpreting them).

    I don't want to argue about this - there is no absolute rest frame.

    Imagine an expanding sphere. Every point on the surface has the same relationship to other points so you cannot distinguish betwen them.
    How can you define an absolute frame in this case except arbitrarly, which means it is not absolute.

    There is no absolute rest frame.

    I'm outa here.
     
  8. Mar 12, 2007 #7

    russ_watters

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    I'm calling the two objects on the ends "buoys" for simplicity - that is essentially what they are (signaling devices for determining your position/motion).
    Why is that the only possible answer? You are measuring the speed with respect to the two objects at the ends. So you are only saying the other object is not moving with respect to them. In order to know it is not moving wrt the other two objects, you would need to know that they are stationary.

    You're going to need to figure this out quickly - I don't want to seem short here, but this is a pretty simple concept and we don't entertain crackpottery here.

    Edit: BTW, there is no need to do a thought experiment here. We already have a system that does this - it is called GPS. But instead of each GPS receiver interrogating the satellites, the satellites are synchronized by a ground station and constantly send out a time signal. The receiver gets the time signal and calculates the location of the receiver wrt the satellites.
     
    Last edited: Mar 12, 2007
  9. Mar 12, 2007 #8

    JesseM

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    Marlos, you are ignoring the fact that different reference frames disagree about whether the pulse from A was emitted "at the same instant" as the pulse from C, and that this disagreement about simultaneity ensures that every frame agrees both pulses reached B at the same moment and that both pulses were traveling at the same speed, regardless of the speed they see the ship moving (for example, a frame that observe the ship moving to the right will observe that A emitted its pulse before C, while a frame that observes the ship moving to the left will observe that C emitted its pulse before A). Are you familiar with the concept of the "relativity of simultaneity"?
     
  10. Mar 13, 2007 #9
    Mr. Russ Watters. 1. The experiment is not measuring any speed among the objects in the spaceship. All of them, A and C (the emitters), and B (the receiver), are fixed inside the spaceship and their relative speed is none. 2. The experiment, due to its purpose, says nothing about other objects. 3. The name of this forum contains two times the world HELP. And all I asked for was help to clear my doubts, specially the help of people like you, who certainly has a strong knowledge on Physics. 3. What I expected then, was that people like you, would, without any intellectual prejudice, make an attentive appreciation of the experiment, and with reasonable arguments to accept or reject it, in adherence to the purposes and rules of this phorum. Instead, I received from you the the stinking smell of dogmatism, and the sick desire of imposing humiliation (English is not my native language, but I discovered in the Dictionary the meaning of the word Crackpottery).I am not a specialist on Relativity. I am only trying to understand it, asking help to clear my doubts.
     
  11. Mar 13, 2007 #10
    OK, Jesse. You are right about how other observers will see the events inside the spaceship. But remember that the experiment, to fulfil its purpose, need not to consider other observers. Just look at each possibility the experiment generates, and try to judge if the correspondent conclusions of the observer are right or wrong. Then you can decide about its validity. Thanks.
     
  12. Mar 13, 2007 #11

    JesseM

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    Well, if the observer concludes that in his own frame, the two signals were emitted at the same time and traveled at the same speed, then he's right. But if he concludes that his point of view is "right" in some absolute sense, while other frames are "wrong" in an absolute sense, then you need to explain why you think the point of view of other frames is not equally valid.

    Also, I don't understand why you think it matters whether an observer is "inside the spaceship" or not, all that matters is the observer's speed relative to the ship. If an observer is driving a car at a constant speed through the ship, then her rest frame will be different than the ship's frame, and in her rest frame she will conclude that the two light signals from A and C were emitted at slightly different times.
     
  13. Mar 13, 2007 #12

    russ_watters

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    Ahh, I misunderstood that part. Sorry. (a better drawing would help)

    Ok, so this just becomes a Michelson-Morley type experiment hoping to show an anisotropy of the speed of light? If the objects are all at rest wrt each other, then there is nothing happening here, not even a relativity of simultenaity issue. The objects A and C can be triggered either by object B or be synchronized separately and constantly fire time-coded signals at object B. Since the speed of light is constant regardless of the motion (but not acceleration) of the ship, the signals always travel the distance in the same time and the operator of the ship would always think he was stationary.

    You may be envisioning a situation such as a boat traveling against a current or even ultrasonic sensors with sound travleing with/against the wind. Such setups would show motion wrt to the fluid medium because things like the speed of sound are not constant across different reference frames.
     
    Last edited: Mar 13, 2007
  14. Mar 13, 2007 #13
    Thanks for your attention, and I would say that now you got the real conception of the experiment. Lets put it in a little diferent way. The sapaceship is ABC, as before. B is in the center and the emitters A and C are at the opposite ends of the spaceship. The movement of ABC, if any, can only be along the X axis. A and C emit a pulse to B, simultaneously. If the spaceship is eventually moving to the right, the pulse from C will travel a certain distance Dc<(CB=AB), to reach B (because B is aproaching the pulse, due to his movement to the right, with the spaceship, just in the direction of point C, from where the pulse departed). On the other hand, the pulse coming from A, will have to travel a bigger distance Da>(AB=CB) to reach B (because B is moving away from the pulse, due to his movement to the right, with the spaceship). Being Da>Db, and because c is constant, Ta=Da/c > Tc=Dc/c.

    Now, if the ship is eventually moving to the left, receiver B will report Ta<Tc. This seems clear to me, and is exactly what happens in the story of the Einstein`s long train, where an observer, in the middle of the train, must see the light coming from the lightning that striked the front end of the train, before he sees the light coming from the back end of the train,etc, etc. In consequence of the above, my reasoning is that, if the receiver B eventually reports Ta=Tc, the spaceship cannot be moving to the right, neither to the left, and consequently it is not moving, and it should be, I suppose, in a kind of Absolute Reference Frame. And that any spaceship where Ta=Tc, would be in the same situation.

    Additionally, as you know, it is said and accepted that “there is no physical experiment through which one can say if a body is moving uniformly or if it is at rest”. Accordingly to the above reasonings, this experiment can do it (indeed, if the observer finds Ta<>Tc, the body (spaceship) is moving. If Ta=Tc, the body is not moving).

    Finally I should tell that I cannot really see where the above reasoning hurts the Theory of Relativity, just because they contain nothing against the two basic principle of that so solid theory: 1. The velocity of light is the same for all observers in inertial frames; 2. The laws of physics remain the same for all those observers.

    Thanks again for being tolerant with those reasonings which may very well prove to be only silliness, fruit of my lack of better knowledge. That is my big doubt.
     
  15. Mar 14, 2007 #14

    JesseM

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    But there is no absolute truth about whether the spaceship is moving to the right in relativity, it just depends on what frame you choose. If the light signals from A and C are emitted simultaneously in the ship's rest frame, then obviously in this frame the ship is not moving at all, so the light signals reach B at the same time. And if this is viewed by some other frame where the ship is moving to the right, then in this frame the signals from A and C are not emitted simultaneously, instead the signal from A is emitted prior to the signal from C, in such a way that both signals reach B at the same time. So this example doesn't allow you to judge that one frame's view is "more correct" than any other.

    You could also imagine what would happen if the signals from A and C were emitted simultaneously in a frame where the ship was moving to the right. In this case, the signal from C would reach B before the signal from A. But this doesn't indicate that the ship is moving in an absolute sense...in the rest frame of the ship, it's easy to explain the same observations, since in this frame the signal from C was emitted before the signal from A, so of course the signal from C reached B first. Again, this example doesn't lead you to prefer one frame's perspective over another in any absolute sense.
     
  16. Mar 14, 2007 #15

    russ_watters

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    What you have wrong is thinking that the distances are different when the spaceship is moving. If the apparatus is physically fixed, it is easy to see that the distances are fixed. You can measure it with a tape measure! Since you are saying that the distances depend on the speed of the ship, that's a violation of #2.

    Since the pieces of the apparatus are fixed with respect to each other, if the time for the signal to transit varies, the speed at which they transit (the speed of light) must very as well. So that would violate #1.

    Btw, those are reversed... #2 is the first postulate and predates Einstein's version of relativity. How you said it wasn't even true for Galileo's version. It would probably help for this problem to get them in order since you need to start with what you know: that due to the principle of relativity, the length of your apparatus is fixed in your reference frame. This was the starting point for the MM experiment and the reason they postulated that the speed of light was frame dependent.

    It may also help you to apply some arbitrary math here. How would you actually do the calculations? What does the speed/distance/time equation look like? Since the two pieces of information you measure are the distance between the parts of the apparatus and the timing of the signals, it is s=d/t....making s (which is the speed of light) vary.
     
    Last edited: Mar 14, 2007
  17. Mar 14, 2007 #16
    Jesse. I think that we are discussing exactly the core of the experiment, and what you say is in perfect adherence to the Theory. I accept that the spaceship is at rest in its reference frame. What I am doing, which seems to be somewhat “not accepted”, is to forget that convention of reference frames, and imagining the spaceship on space, alone, full closed, with no windows, and imagining a way for the observer in it to discover if the spaceship, with himself and his apparatus, is moving uniformly or not. For that, the observer only can and needs to cope with the spaceship itself, and its apparatus A, B and C. He even don´t need to know about the concept of reference frames. Also, I am treating the problem only using classical kinematics, with the velocity of light constant. Looking at the problem that way, the experiment seems to be valid, because if the spaceship is moving to the left or to right, B wil register Ta<>Tc. And this happens because the velocity of the pulse inside de spaceship is independent of the eventual real movement of the spaceship itself in space.

    To make it clearer maybe I will need to make a diagram and demonstrative yet simple kinematic calculus, but I am afraid the diagram, being a little complicated, will be deformed because I have to do it in Microsoft Word, accordingly to the rules of the forum.
     
  18. Mar 14, 2007 #17
    Mr Russ. I think that we are discussing exactly the core of the experiment, and what you say is in perfect adherence to the Theory. What I am doing, which seems to be somewhat “not accepted”, is to forget that convention of reference frames, and imagining the spaceship on space, alone, full closed, with no windows, and imagining a way for the observer in it to discover if the spaceship, with himself and his apparatus, is moving uniformly or not. For that, the observer only can, and only needs to cope with the spaceship itself, and its apparatus A, B and C. He even don´t need to know about the concept of reference frames. Also, I am treating the problem only using classical kinematics, with the velocity of light constant. Looking at the problem that way, the experiment seems to be valid, because if the spaceship is moving to the left or to right, B wil register Ta<>Tc. And this happens because the velocity of the pulse inside de spaceship is independent of the eventual real movement of the spaceship itself in space.

    To make it clearer maybe I will need to make a diagram and demonstrative yet simple kinematic calculus, but I am afraid the diagram, being a little complicated, will be deformed because I have to do it in Microsoft Word, accordingly to the rules of the forum.
     
  19. Mar 14, 2007 #18

    JesseM

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    But how? Do you understand that if the observer sets the clocks at A and C so that they are synchronized in his rest frame, then if the two signals are emitted at the same time according to those clocks, the signals will both reach B at the same moment? Do you understand that all frames would agree about this prediction, regardless of whether the ship is moving in that frame or not? Remember, if the ship sets the clocks so they are synchronized in the ship's rest frame, then another frame will see the clocks as out-of-sync, and thus see the signals emitted at different times, in just the right way so that both signals will be predicted to arrive at B at the same moment given the assumption that both signals travel at c in this other frame.
    Yes, in a frame where the ship is moving, the two signals will take a different amount of time between being emitted and reaching B, due to the motion of the ship; this is just as true in relativity as it is in classical kinematics. But this will be exactly compensated for by the fact that the different signals were emitted at different moments in this frame, so both signals are predicted to reach B at the same moment (in all frames). Again, this is assuming that the observer on the ship had clocks at the front and back which were synchronized in his rest frame, and that A and C emitted their signals when the clock at each end showed the same reading.

    If you don't want the observer to pick the moment the signals are sent using clocks which have been synchronized according to the Einstein synchronization procedure (where two clocks are judged to be 'synchronized' in their mutual rest frame if they both show the same reading at the moment they are hit by light from a source turned on at the midpoint of the two clocks), then how do you want the observer on the ship to pick the moment the two signals are sent, exactly?
     
  20. Mar 14, 2007 #19
    Mr Russ. I think that we are discussing exactly the core of the experiment, and what you say is in perfect adherence to the Theory. What I am doing, which seems to be somewhat “not accepted”, is to forget that convention of reference frames, and imagining the spaceship on space, alone, full closed, with no windows, and imagining a way for the observer in it to discover if the spaceship, with himself and his apparatus, is moving uniformly or not. For that, the observer only can, and only needs to cope with the spaceship itself, and its apparatus A, B and C. He even don´t need to know about the concept of reference frames. Also, I am treating the problem only using classical kinematics, with the velocity of light constant. Looking at the problem that way, the experiment seems to be valid, because if the spaceship is moving to the left or to right, B wil register Ta<>Tc. And this happens because the velocity of the pulse inside de spaceship is independent of the eventual real movement of the spaceship itself in space.

    To make it clearer maybe I will need to make a diagram and demonstrative yet simple kinematic calculus, but I am afraid the diagram, being a little complicated, will be deformed because I have to do it in Microsoft Word, accordingly to the rules of the forum.
     
  21. Mar 14, 2007 #20

    JesseM

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    You say that you want to "forget that convention of reference frames", but your experiment depends on the idea that the signals from A and C are emitted "at the same time"...so unless you specify which frame's definition of simultaneity you're using, or else specify a physical procedure for A and C to decide when to emit their signals (like my suggestion earlier of having clocks at each end of the ship which have been synchronizing using the Einstein synchronization convention, which involves setting each clock based on the moment they're hit by light from a source at their midpoint), then your experiment is simply ill-defined.

    So again, please answer the question from my last post: how do you want the observer on the ship to pick when each of the two signal is sent?
     
    Last edited: Mar 14, 2007
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