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On Length contraction and FTL

  1. Sep 14, 2008 #1
    before I start, I should say that I'm a layman and as such will probably say something which sounds quite stupid to the rest of you. I will clarify anything that needs clarifying

    Ok, so we have

    A - a 2 light year long garage with two doors that can swing open and closed
    B - a 4 light year long pole with engines on every inch of it, so that it can decelerate (near) equally along every part of itself to an arbitrary degree < c.

    I have two scenarios:

    Scenario 1: The FTL Message


    with respect to the garage

    with respect to the pole

    The pole is traveling at a constant velocity of 0.88c and as such undergoes length contraction. It is approx 47% of its rest length with respect to the garage.

    It is programmed to calculate ahead of time to decelerate along its entire length when (WRT the garage) the pole's end point has reached the garrage middle.

    When the pole's end reaches the mid point of the garage (WRT the garage), somebody places a message on it's end, the pole then decelerates at 0.99..c

    WRT: Garage

    WRT: Pole

    Now, as far as I understand it. At the point of placing the message from the point of view of the pole, it's end hasn't actually crossed the threshold of the garage and is still outside but the message from the garage mid-point is on it.

    With respect to the garage, the pole is fully inside at the time the message is placed. Both the pole and the garage can agree at what time the message is placed.

    WRT: Both

    has the message traveled FTL?

    Scenario 2: The status of the door in objective reality


    with respect to the garage

    The pole is once again traveling at a constant velocity of 0.88c and as such undergoes a length contraction and is approx 47% of its rest length with respect to the garage.

    It is again programmed ahead of time to decelerate along its entire length (near) simultaneously with respect to itself when its mid point reaches the mid point of the garage (from the frame of the garage).


    With respect to the garage, the pole is fully inside. The garage doors swing shut


    now one of two things happens:


    a. the pole stops inside the garage, expands along its length and breaks through the garage door(s)
    b. the pole stops, but hasn't actually crossed one or more of the garage doors so the garage door(s) crash down onto the pole

    (it seems embeded images don't work on this forum)

    have at it... :)

    Attached Files:

    • 10.gif
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  2. jcsd
  3. Sep 14, 2008 #2

    Doc Al

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    The "message" was placed on the left end of the pole when the left end of the pole crossed the midpoint of the garage, right? At that point, everyone agrees that the left end of the pole is in the middle of the garage, but they disagree as to whether the right end of the pole is inside or outside the garage at that time.


    Why does the left end of the pole end up outside the garage?
  4. Sep 14, 2008 #3

    this is not my understanding of it (though I'm probably wrong).

    I thought the pole expands in the opposite direction from it's direction of velocity (as its decelerating) with reference to another object, and contracts in the direction of velocity as it's accelerating, thus the left part of it sticks out
  5. Sep 14, 2008 #4

    Doc Al

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    What exactly is not your understanding of it? Didn't you arrange that the message be placed on the left end of the pole when that end passed the middle of the garage?

    Your diagram shows the left end of the pole--with message attached--way outside and to the left of the garage.
    The pole "expands" because, from the view of the garage, different parts deaccelerate at different times. The back end (left end) deaccelerates first, followed by the other sections. Assuming that the pole can deaccelerate in such a way as to preserve its proper length without being ripped apart, the left end of the pole would still be moving towards the right.

    Regardless of the details of the "expansion", why would you think that the left end of the pole--or any other section--would travel faster than light?
  6. Sep 14, 2008 #5


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    The Lorentz contraction formula is only meant to tell you the length of an inertial rigid object as seen in different inertial frames, it doesn't really tell you how an object will change length when it accelerates, that depends on the details of what forces are applied to different sections of the object, and in general it can't remain truly rigid when it accelerates, sections of it will be physically compressed or stretched in the frame of an inertial observer temporarily at rest relative to that section. Once external forces stop being applied it may return to its original rest length and behave as an inertial rigid object again, but as with a spring that has been stretched or compressed and returns to its own relaxed length, this will take some time, it won't happen instantaneously as soon as the external forces are turned off (and until it does return to a relaxed rest length there will still be internal forces due to the physical stretching/compression of different sections). Certainly regardless of what forces are acting on different sections of the pole at different times, every section must move slower than light at every moment in every inertial frame.
    Last edited: Sep 14, 2008
  7. Sep 22, 2008 #6
    sorry to ressurect this thread

    ok, perhaps I haven't made myself clear (or I'm understanding this wrong)

    There exists a moment where somebody in the garage (from his/her point of view) can put two messages A-B (two to avoid ambiguities on the pole's areas of expansion) on each end of the pole as it is going through the garage's mid point

    To the pole, either message A or message B will be outside of the garage light years from the it's mid point as soon as it is placed. Also, to the pole one of the messages has never been inside the garage

    I suppose this all comes down to:

    a: whether the pole can expand to it's original length FTL as it decelerates along it's body (near) simultaneously

    b: how exactly the pole views one of the messages being placed upon itself and from where, when that part of the pole is never inside the garage when the message is placed
  8. Sep 22, 2008 #7

    Doc Al

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    Yes, according to garage frame observers, messages can be simultaneously attached to each end of the pole while the pole is entirely in the garage. Of course, from the pole frame a message was placed on the front end long before the second one was placed on the back end.
    According to pole observers, when the message is placed on the front of the pole the back of the pole is still way outside of the garage (but the second message has not yet been placed on that end).
    No, everyone agrees that when the message was placed on the front end, the front end was inside the garage; similarly, everyone agrees that when the message was placed on the back end, the back end was inside the garage.
    Last edited: Sep 22, 2008
  9. Sep 22, 2008 #8

    I understand what you are saying, the messages placed from the garages point of view are not simultanous (like the doors) from the point of view of the pole

    this however, does not deal with the fundamental two questions:

    * the pole's (near simultaneous) deceleration, and the speed of the expansion. How fast does it expand?

    * Both garage doors close whilst the pole is inside (POV garage). The pole then decererates along it's body. To the pole, the garage doors should crash down on the part of it inside the garage, to the garage the pole should expand and poke through the door(s)

    which one happens?
  10. Sep 22, 2008 #9

    Doc Al

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    "Near simultaneous" according to whom?
    What do you mean by "then"? Does each part of the rod decelerate simultaneously according to the the garage? (I don't think you mean that.)
    It's still not clear to me exactly how this deceleration is to take place.

    But I again ask, regardless of the details of the expansion, are you thinking that somehow some part of the rod moves faster than light speed?
  11. Sep 22, 2008 #10
    But they cannot say where and when message A is as they are attaching message B. Even in the simple case of not changing the speed of the pole -- That is the point of simultaneity. Both frames claim the other has the shorter standard for measure, and because of that that other frame has “Slow Clocks” that are terribly out of sync with each other.
    Only the correct “Prefered Frame” can correctly resolve the questions you are asking.

    So can you establish which frame is the more preferred?
    Ref: SR Simultaneity; No you cannot observe or define any one frame as preferred.

    Therefore when you change the relative speed between the barn and pole, you cannot claim to know which or is getting longer (or shorter) the barn or the pole.
    At least not until you can show the ablity to define a preferred frame.
  12. Sep 22, 2008 #11
    to whomever (one or the other) wants to see it decelerate along it's body simultenously

    it is

    if the rod were to decelerate from a single point on it's body, it's decleration would travel down it's light cone and it would expand in sections (to my knowledge) at the speed of light

    the deceleration at this point is inconsequential. It can be programmed any which way before the experiment starts.

    I understand it is impossible for it to decelerate simultaneously along it's body in both of the reference frames

    I'm thinking (possibly incorrectly) the following facts.

    Lets go purely from the POV of the garage:

    * When they are both at rest the pole is 4 light years long and the garage is 2 light years long
    * Every part of the pole can decelerate and come to rest in say 15 seconds
    * If the pole undergoes this deceleration whilst in the garage it will expand to it's rest length in 15 seconds
    * It will have increased in size from 1.88 light years to 4 light years in 15 seconds (FTL)
  13. Sep 22, 2008 #12
    indeed, in my hasty diagrams above both the garage and pole become shorter simultaneously to each other

    my issue is this, if we make the poll twice as long at rest. When the message is placed on the pole whilst it is within garage (POV garage) and the pole decelerates to rest, the message is sticking out of the back of the garage and never entered

    also, there's the issue of the "crash down" Vs. the "poke through". If both doors are shut and the pole decelerates whilst "inside" the garage (POV garage) - one or the other has to happen to the doors
  14. Sep 22, 2008 #13

    Doc Al

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    It makes a difference.
    Realize that if each segment of the rod deaccelerated uniformly with respect to the garage, then its length would not change. (Of course, good luck arranging to exert the required forces, which will destroy the pole.)

    What I suspect that you want to do is to decelerate the rod in such a way that no permanent or excessive stresses build up that would crush the rod. As viewed from the garage, that would mean that the back end of the rod decelerates first, while the front end keeps going, thus stretching the rod out as it slows down. (Of course, such a gentle deceleration may take well over your allotted 15 seconds.)

    Again, regardless of how you do it, no part of the rod is going to end up moving at light speed.
    Not sure what you're saying here. Do you mean that if a part of the rod where to decelerate, then that deceleration would spread through the rod at light speed? No it wouldn't! Mechanical stress of one part of the rod exerting a force on another would spread much slower than light speed (closer to the speed of sound). (Please re-read JesseM's post.)
    OK, just for fun, let's say that every part of the pole decelerates at the same time (as seen in the garage) and comes to rest in 15 seconds. (Pretend the pole is made of some incredible springy material that squeezes but doesn't break.) As I said before, that means that somehow, by applying external forces throughout the body of the pole, you have stopped the pole in its "shortened" state--it's only 1.88 light years long and under tremendous stress.
    No, all you did was stop it. Now, if you release those external forces, the pole will expand (again, because it's made of special super springy material), eventually settling on its normal unstressed length of 4 light years. (And busting through the garage doors.)
    No, the expansion will take who knows how long, but nothing's going to move at light speed.

    No. We agreed before that the message would be placed on the back end of the pole when the pole was in the garage. Everyone agrees about that.
    If, as the pole decelerates, internal stresses end up pushing the back end of the pole out of the garage (let's assume we can do that without destroying the pole) then it must "poke through" the closed doors.
  15. Sep 22, 2008 #14
    Don't change the lengths now - at rest keep both the Pole and the Barn at 2LY and don't worry about deceleration.
    The question is when they are moving WRT each other do you think you can identify which one is actually shorter than the other without identifying a Preferred Frame?

    If so you needed to detail exactly how – because none of what you are talking about ("crash down" or “poke through") can make any sense unless you do.

    My guess is you need to focus on building a more complete understanding of the point Einstein makes on simultaneity with SR, to understand the paradox.
  16. Sep 22, 2008 #15
    hey spike:

    Please don't be apologetic for being ignorant, everone is (no one knows everything).
    On the contrary, it's wise to ask questions, stupid not to.

    This is the old 'pole-barn' paradox.
    For something closer to reality, let's say the the building is 20' long and the pole is 40'. Let's land the pole at Edwards AFB on one of those dry lake beds.
    Next we truck it to the building and compare lengths. Will the pole fit completely inside the building?
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