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A thought experiment

  1. Jan 1, 2007 #1
    I stand shoulder to shoulder with a group of people, we all face the same direction. We all have synchronized watches and nobody is moving.

    We face a second group of people, facing us. They all have synchronized watches and nobody is moving.

    Now suppose that my group feels like they are in a gravitational field that runs along the group from left to right. Pretend somehow the other group does not.

    Given that the only difference between the two groups is the presence of a gravitational field for one group, how will their watches compare?
     
  2. jcsd
  3. Jan 1, 2007 #2
    The group in the gravitational field cannot have synchronized watches, assuming each member is at a different potential.
     
  4. Jan 1, 2007 #3
    That's very interesting. Where can I learn what you just stated?
     
  5. Jan 1, 2007 #4
    Look up tests of general relativity, particularly the ones with atomic clocks at different heights (eg. aboard planes and GPS satellites).
     
  6. Jan 1, 2007 #5

    pervect

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    For another source, look at "gravitational time dilation", http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/gratim.html

    As is predicted by GR, and confirmed by experiments (for instance the Harvard tower experiment section in the above URL, aka the Rebka-Pound experiment), clocks at different heights run at different rates.

    Thefore if you synchoronize two clocks at different heights, they don't stay synchronized - because one clock, the higher clock, "ticks faster".

    It doesn't matter whether the "different heights" occur because the clocks are in an actual gravitational field, or in the pseudo-gravitational field of an accelerating rocket, the same effecthappens.

    To really describe what we mean by "tick faster" and "tick slower" demands that we describe how we compare the clocks. The answer in this case is fairly simple - light signals always have a constant round-trip propagation delay between the two points in question (the high clock and the low clock) as measured by either observer. (The observers won't agree on the propagation delay, but both observers will agree that the round trip propagation delay is constant).

    If we assume that the propagation delay in each direction is also constant (given that we observe that the round trip propagation delay is constant) we can simply use light signals to compare the rates of the two clocks, taking into account some fixed propagation delay for the signal. Doing this, we find that the two clocks do not tick at the same rate.
     
  7. Jan 1, 2007 #6
    Here
    Chris, do you know of a more up to date version?
     
  8. Jan 2, 2007 #7
    Its easy to see why an observer located at the position of the higher clock will conclude that the lower clock ticks slower in the case of an accelerating rocket - successive pulses transmitted at one second intervals by the lower clock (in the rocket tail) will travel an additional distance to reach the front because the higher clock is moving away from the source so each light pulse must travel further - likewise for the lower clock - pulses transmitted by the higher clock will arrive sooner at the lower clock because each pulse travels a shorter distance (the lower clock is rising to meet the pulses) - but in a uniform G field the changing distance explanation isn't applicable - so what physical principle is at work? If the pulses emitted by the lower clock actually take longer to arrive at the upper clock in a uniform G field we have a violation of the tenants of SR. And if they don't what is the physics? The experiments support Einstein's equivalence - so is the effective distance dependent upon the way it is measured relative to the gravitational potentials?
     
  9. Jan 2, 2007 #8

    Hurkyl

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    Gravitational time dilation, of course.

    The result of this experiment is almost literally the definition of what it means for (one particular component of) the curvature of space-time to be nonzero.
     
  10. Jan 2, 2007 #9

    Chris Hillman

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  11. Jan 3, 2007 #10
    Thanks for your response Hurkyl. What puzzles me is that the mechanism that leads to time dilation in the Schwarzschild solution is not the same as the mechanism that determines the frequency shift for the accelerating rocket.
    In the case of the rocket, since acceleration per se does not alter the rate of an ideal clock, it would seem that the frequency shift is observational - so that a later comparison of the two clocks would show they have run at the same rate during the entire acceleration phase even though the lower clock has observed the upper to run at a faster rate and the upper clock has observed the lower to run at a slower rate. In the case of different gravitational potentials, a later comparison of the clocks would show that the lower clock has logged less time
     
  12. Jan 3, 2007 #11
  13. Jan 3, 2007 #12

    pervect

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    I don't see why or how you could think this.

    An example might help. Suppose we have a rigid (Born rigid) rod, and we acclerate the tail end of the rod at 1g for a proper time of one minute (i.e. a time of one minute as measured by a local clock at the tail end of the rod) and then we stop accelerating the rod.

    The top end of the rod will, in order to satisfy the rigidity conditions, acclerate for a longer time at a lower acceleration (in terms of proper time and proper acceleration). For example, for some particular length of the rod, the top end of the rod will accelerate at a proper accleration of 1/2 g for a proper time of two minutes.

    A rough space-time diagram of the situation can be seen at

    http://en.wikipedia.org/wiki/Image:Bell_observers_experiment2.png

    The lower magnitude of the acceleration of the second spaceship isn't really drawn correctly on this diagram though, I'm recylcling this diagram from a different application, namely Bell's spaceship paradox, which is closely related.


    Putting it another way: if the spaceships acclerate at the same rate, a string connecting them will break (they won't maintain the same distance).

    If you demand that the spaceships maintain the same separation (born rigid motion) the lead spaceship has to accelerate at a lower rate so that it doesn't break the string.


    You might also try the article by H Nikolic aka Demystifier at http://arxiv.org/abs/physics/9810017
     
    Last edited: Jan 3, 2007
  14. Jan 3, 2007 #13

    Hurkyl

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    Actually, "uniformly accelerating rocket" and "tower in uniform gravitional field" are geometrically identical -- they are the same situation represented in different coordinate charts.


    I assume you are doing an SR-style analysis in an inertial reference frame? You are right that acceleration doesn't affect the (coordinate) rate of a clock. But (coordinate) speed does: the lower clock has been travelling with a greater speed than the upper clock throughout the entire journey.

    A quick heuristic way to see this is that, as measured by this inertial frame, the rocket is length-contracting as it speeds up. That can only happen if the front of the rocket moves more slowly than the tail of the rocket. (As pervect has pointed out)

    Since the lower clock has been moving faster, it's been running more slowly than the upper clock, and happily we have agreement with the gravitational case:
     
  15. Jan 4, 2007 #14
    With due respect, I would at this point side with the Cern theoretic group in concluding that the string would not break. But, as Bell points out, those who first reach this conclusion frequently change their mind after further consideration. The analysis depends upon the reality of a physical length contraction in the comoving frame to explain the time difference between the clocks - I have a problem with that - but perhaps the light will be better in the morning.

    Regards

    Yogi
     
  16. Jan 4, 2007 #15

    Hurkyl

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    Minor correction -- the analysis assumes that for every comoving inertial frame, at the instant1 when the frame is comoving with (some part of) the rocket, the length1 of the rocket is exactly its rest length.

    1: as measured in said comoving frame
     
  17. Jan 4, 2007 #16
    There is no length contraction in the comoving frame.
     
  18. Jan 4, 2007 #17
    No that would mean someone traveling with the rocket could physically see the rocket change length. That can’t happen, the only place that sees the length of the rocket change is some other reference frame not comoving with the accelerating frame of the rocket both the front, back and those inside the rocket all moving in a common frame.
    The key that even Bell himself missed here is IMO really quite simple in principle:
    How do you figure out when the front of the rocket starts to move, so that is starts at the same time as the back of the rocket? And no, that does not happen as a result of there being attached to each other! The structure that separates the two points is not rigid enough to serve as a true “ridged” connection, any normal structure will compress at these forces, so each point must be accelerated independently and at an appropriate simultaneous coordination.
    If done correctly the simple part is recognizing that any outside observer in a frame other then the rocket ship should be able to make two observations:
    First, the distance of space between point A and B. This will always decrease for other frames and by the appropriate gamma function.
    Second the length of the string attached to points A and B. Again this will decrease for other frames and by the appropriate gamma function.
    Now by what logic could anyone rationally explain how the gamma function that predicts the change in distance and length would not be identical for both!
    IMO Bell got this wrong, and if he were still available I’d ask him to provide the formula to detail the differences between the functions to describe length vs. distance. I think asking him that would have been be enough for him to be convinced the string would not break as well.

    The hard part is seeing at what times the two ends should start, there is no reason to expect the start time to be coordinated by the rocket ship clock. The relativistic problem of simultaneity is still real and we should not expect the front and back clocks to remain synchronized once acceleration starts.
    So how do we establish the correct start time for each end?
    We will have to find a proper “preferred frame”!
    This will NOT violate the SR rule against there being A Preferred Frame. Because once we establish such a “local preferred frame” to get our ship started by, we would not be able to coordinate that frame with a similarly defined “local preferred frame” for some rocket ship some 10 billion light years away. It would be nice if we could, but Hubble expansion demands that the two “local preferred frames” cannot be a common frame. Therefore, the idea of no universal preferred reference frame is still preserved as correct
     
  19. Jan 4, 2007 #18


    Here is a mathematical solution to Bell's paradox:

    http://www.ph.utexas.edu/~gleeson/NotesChapter13.pdf
     
  20. Jan 4, 2007 #19

    pervect

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  21. Jan 4, 2007 #20

    pervect

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    With all due respect, we have only hearsay evidence that the Cern group ever came to such a conclusion. We certainly have no published papers by the Cern group which claim the string does not break. We do, however, have numerous published papers which show that the string does break.

    The moral of the story is (or should be) that people do put more care into their published papers than an informal discussion over lunch.

    There really isn't any question that according to SR, the string breaks. It's easier to see this if one is capable of following the arguments - but the proof isn't hard, if you can solve a triangle, you can show that the string breaks. My own modest effort is online in the Wikipedia article on the topic (and the talk pages have an endless argument with Rod Ball over this very issue.)
     
    Last edited: Jan 4, 2007
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