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Near-instant Acceleration / Clocks

  1. Apr 8, 2015 #1
    If I have a "train" at the "platform" and wish to send the train accelerating to near-C velocity in near-zero time from my frame.

    I presume that there is no theoretical limit on the acceleration (as far as SR goes).

    I presume that the clocks throughout the train will all appear barely to have moved in my frame after acceleration.

    If this all has no issues, my question would be:

    What do observers on the train make of the great desynchronization which has taken place between the clocks on board.

    Thank you.
     
  2. jcsd
  3. Apr 8, 2015 #2

    PeterDonis

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    Correct, in the sense that kinematically, any finite value of proper acceleration is possible, no matter how large.

    However, SR does impose a finite limit on the tensile strength of materials (basically that the speed of sound in the material cannot exceed the speed of light). So there is a limit on what acceleration you can impose on an object of finite size without breaking its structure--basically, the length of the object cannot exceed ##c^2 / a##, where ##a## is the acceleration. If the object is longer than this, internal forces would have to propagate faster than light to hold it together.

    (I should note that this is not, strictly speaking, an issue for your problem, since you can always in principle make the object's length as short as required. But I think it's worth mentioning anyway.)

    By "in my frame", do you mean in the frame in which the train is initially at rest (before the acceleration)? If so, then yes, your presumption is correct; but your assumption about the implications is not. See below.

    You are assuming that there is a "great desynchronization" from the standpoint of observers on the train. Rather than assuming, you should calculate.
     
  4. Apr 8, 2015 #3

    DaveC426913

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    This is an engineering limitation, though, not a physical limit, yes? For example, if the object were entirely ferro-magnetic, each atom would be accelerated directly, bypassing the need for excessive structural cohesion.
     
  5. Apr 8, 2015 #4

    PeterDonis

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    Not really. Even if you accelerate every atom directly, if the distance between atoms is greater than ##c^2 / a##, the atom in back won't be able to keep up with the atom in front (it would have to move faster than light to do so). In other words, the atom in back has to be closer to the atom in front than the atom in front's Rindler horizon is.
     
  6. Apr 9, 2015 #5

    A.T.

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    If the train keeps its proper length, then it contracts to almost zero length in the platform frame. That definitely requires some movement of the clock at the ends of the train.
     
  7. Apr 9, 2015 #6
    I am deriving this from two facts.
    * from the platform frame, the clocks at the ends remain greatly in synch
    * when two frames are moving at high speed wrt each other, there is great difference in perceived simultaneity - thus if platform observer determines the clocks are still close to synch, then the riders must find that the clocks are greatly out of synch.
    There are a number of factors, and the devil will be in the details, I guess.
    * at very high V, train length in platform frame approaches zero
    * if acceleration is defined to finish at both ends at about the same time in the platform frame, this implies simultaneity is different on the train ("both ends won't complete acceleration at the same time")
    * if acceleration is defined to finish at both ends at about the same time in the train frame, , this implies simultaneity is different on the platform ("both ends won't complete acceleration at the same time")
     
  8. Apr 9, 2015 #7

    Nugatory

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    That's why you have to calculate it.
     
  9. Apr 9, 2015 #8

    PeterDonis

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    Yes. One key detail is that you have not specified which of these two cases...

    ...you are talking about.

    (Another detail is that you are assuming that there is a single "train frame" after the acceleration, when in fact there may not be.)
     
  10. Apr 9, 2015 #9
    Actually my greatest interest is *after* a brief acceleration phase - when near-C is maintained. I presume that we're talking about one train frame in that case.

    I had intended to say that if possible it would be technically feasible to get both ends of a rigid train to stop accelerating near-instantaneously in the platform frame after a brief great acceleration. If that were possible, this is what generates the paradoxes. Perhaps that means it is not possible.
     
  11. Apr 9, 2015 #10

    PeterDonis

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    But what happens after that depends on how you specify what happens during that phase.

    Again, rather than presuming, you should calculate.

    Sure, this is possible (and it corresponds, I take it, to the first of the two possibilities you listed, which I quoted in my previous post). Then what you've specified is a Bell Spaceship Paradox scenario: once the brief acceleration has finished, in the rest frame of either end of the train (they are not the same frame), the train is stretched, and the accelerations of the two ends of the train were out of sync. In the rear end's rest frame, the front end did its acceleration much earlier than the rear end, so the train was stretched; and in the front end's rest frame, the rear end did its acceleration much later than the front end, so the train was stretched.
     
  12. Apr 9, 2015 #11
    "I had intended to say that if possible it would be technically feasible to get both ends of a rigid train to stop accelerating near-instantaneously in the platform frame after a brief great acceleration."

    Not so rigid after all, then, this stretched out train.

    It sounds like the case in which the train remains rigid cannot be the same as the case in which both ends attain target velocity near instantaneously in the platform frame.
     
  13. Apr 9, 2015 #12

    PeterDonis

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    Correct.
     
  14. Apr 9, 2015 #13

    Nugatory

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    "Rigid" means that all the parts of a body maintain a constant distance from one another. "Maintaining a constant distance" means that when one part of the body (for example, the nose) is at a particular location, then another part (for example, the tail) is at the same time at a location that is that constant distance away. And of course that phrase "at the same time" is your warning to watch out for the relativity of simultaneity - if it's "at the same time" in one frame, it won't be in another.

    There are no truly rigid bodies when we take relativistic effects and acceleration into account. The closest we can come is the concept of "Born rigid motion" (google for it).
     
  15. Apr 10, 2015 #14

    I can picture this if the accelerating force is pulling the object, it can be "ripped" apart. But then I can't picture it happening (tearing apart) if the accelerating force is pushing the object. What am I missing? All I can imagine happening is the rear pushing through the front unimpeded.
     
  16. Apr 10, 2015 #15

    PeterDonis

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    The case we're talking about doesn't really correspond to either "pulling" or "pushing". Each individual atom has an external force applied to it; the force is not applied to just one part of the object and then transmitted to the rest of the object by inter-atomic forces within the object.
     
  17. Apr 10, 2015 #16

    stevendaryl

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    You're right. If you accelerate a rocket by pulling on the front with a proper acceleration [itex]a[/itex], then the rocket will fall apart if

    [itex]L > \frac{c^2}{a}[/itex]

    If you accelerate a rocket by pushing on the rear with proper acceleration [itex]a[/itex], then there is no limit on the length of the rocket. However, the front of the rocket will have a proper acceleration that decreases with the length of the rocket:

    [itex]a_{front} = a_{rear} \frac{1}{1 + \frac{a_{rear} L}{c^2}}[/itex]
     
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