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Barn-Paradox Type Question

  1. Dec 10, 2014 #1
    1. The problem statement, all variables and given/known data
    Two rockets, A and B each of proper length Lo, approach each other along the x axis with speed v. their noses cross at Ta = Tb = 0. A man with a gun at the tail of rocket A is aiming perpendicular to the relative motion, and shoots at the very moment he see's the nose of A cross the tail of B. What happens ? Does he, or does he not, hit B ? (knowing that in A's reference he missed because B is contracted, and in B he hits because A is contracted)

    The problem also states that you can neglect the time of travel of the bullet, and the time of communication between the nose and the tail of A.

    2. Relevant equations


    3. The attempt at a solution

    Ok so I got this question in my relativity test yesterday as a variation of the barn-pole paradox, and I guess I got thinking a bit too far but to me it seems as though you cannot neglect the time of communication between the nose and the tail of A. (Although you can neglect the time of travel of the bullet since this could all be happening in 1D). My reasoning being that it actually changes the outcome of if the bullet hits or not.

    As I read the problem, there is clearly a causal relationship between the events 1: ''Tail B crosses nose of A'' and 2: ''A shoots'' and these events are related by a time-like space, ie there exists no frame of reference where event 2 happens after event 1. Therefore, B cannot observe A shoot before his tail crosses A's nose.

    This is the point where I started thinking that the time of communication is crucially important since it leads to different outcomes, and A will hit or not hit B will depend on the speed of B, since A will miss B if the light from A's tail arrives before B's nose, and he will hit if B's nose arrives before the light from A's tail. The critical speed being at 1/sqroot(2) (at this speed, B's nose arrives exactly ay the same time as the light).

    Finally, the other problem I have thinking about this is that for A, if he misses, he will shoot in front of B whereas for B, if A misses he will have shot behind B. Should I have a problem with this ? Can I still give a coherent answer to this question while neglecting the time of communication ?
     
    Last edited: Dec 10, 2014
  2. jcsd
  3. Dec 10, 2014 #2

    Doc Al

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    Staff: Mentor

    The key word here is "see", so the travel time of the signal to reach the tail of A is critical.

    That's OK, since you can have the rockets pass arbitrarily close.

    Not a good assumption.

    You need to figure out, for each frame, how much time does it take the light signal to reach the tail of A and compare that to the time it takes for the rockets to clear each other.

    I don't think so. Not in a relativity problem!
     
  4. Dec 10, 2014 #3
    At Ta = 0, where is the tail of B's rocket located as reckoned from A's frame of reference?
    As reckoned from A's frame of reference, at what time does the tail of B's rocket pass the nose of A's rocket?
    As reckoned from A's frame of reference, at what time does an observer in the tail of A's rocket see the tail of B's rocket pass the nose of A's rocket?
    What are the coordinates of this event, as reckoned from A's frame of reference?
    What are the coordinates of this event, as reckoned from B's frame of reference?

    Chet
     
  5. Dec 10, 2014 #4
    Taking the nose of A to be X = 0, X+ pointing in front of A towards B
    Now answering the questions in order:


    at Ta = 0 ==) Xtb = L/Ω (Can't find symbol for lambda on my keyboard so this will have to do)

    T1 = (Lo/Ω)/V

    T2 = T1 + Lo/c

    (Xta = -Lo, T2)

    X'ta = Ω (-Lo - V((L/Ω)/V + Lo/c)) = -Ω(Lo + (Lo/Ω + BLo)
    T'2 = Ω ((L/Ωv) + Lo/c) - (-Lo)(V/cˆ2) = Ω(L/ΩV + Lo/c + LoV/cˆ2)

    This is hard to do on a computer. Did I get it right ?
     
  6. Dec 10, 2014 #5
    Although I know I could get an answer just with the Lorentz transformation equations, I'm trying to get an answer reasoning through the process of what is going on (which was after all the question in the exam: explain your reasoning for why A hit/missed B).

    I'm thinking then that the answer is dependent on the speed of B. If the information gets to the tail of A before B's nose, then he shoots in front of B and misses. If B's nose gets there first, then he hits (Obviously he can't 'miss' by shooting behind B, since the information travelling at c will always arrive before B's tail).

    But if I consider that the info gets there instantaneously, this is tantamount to saying that the two events ''A shoots'' and ''B's tail crosses A's nose'' happen instantaneously for A. But then in this case B will see A shoot first and then his tail crossing A's nose, meaning there is no causal relationship between the two events (unless I'm wrong here), which seems to be a contradiction with what the problem clearly states.

    The exact phrasing (translated from french in the problem is: ''When the observer in rocket A sees that the nose of his rocket coincides with the tail of rocket B, he shoots his canon''.

    and

    ''you can neglect the time of flight of the bullet, as well as the time of communication between the nose and the tail of rocket A''

    Is there a clear contradiction between those two phrases then ? I'm maybe planning to go see the teacher about it, but I don't want to be making a distinction about something that maybe doesn't really make a difference.

    Thanks !
     
  7. Dec 10, 2014 #6
    You analyzed this very well. But there is an error in the last step. B is going in the minus x direction, so there should be a + sign in front of the v instead of a - sign. What do you get then?

    Incidentally, once you see the answer, consider what the result would have been if the transmission of information from the tip to the tail of A were indeed instantaneous.

    Chet
     
    Last edited: Dec 10, 2014
  8. Dec 10, 2014 #7
    Yes. That's what I get.

    Chet

    PS, I accidentally deleted your previous post. Sorry. The quote in this thread captures most of it.
     
    Last edited: Dec 10, 2014
  9. Dec 11, 2014 #8
    The interesting thing about these results is that, if the communication time is treated as zero, the cannon ball misses rocket B ahead of the rocket at ##L_0(1-\gamma)## (as reckoned from B's frame of reference). However, if the communication time is taken into account, enough extra time is added for the cannon ball to always hit rocket B somewhere between its nose and tail, depending on the relative speed of the rockets.

    Chet
     
  10. Dec 11, 2014 #9

    Doc Al

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    Just to add to what Chestermiller already explained...

    Good. I think it's quite useful to think in terms of what's going on--in terms of moving clocks and measuring rods--in addition to solving it more formally using the LT.

    Here's how I would work it out. First according to A. Measuring time from the moment that the nose of A crosses the tail of B, how long does it take for the light signal to reach the man with the gun? That time is ##L_0/c##. How far does the tail of B get in that time? Measured from the starting point (the nose of A), it goes a distance ##(v/c)L_0##. So clearly the tail of B has not passed the tail of A. Where's the nose of B at that time? Just a distance ##L_0/\gamma## ahead of the tail, so a distance of ##(v/c)L_0 + L_0/\gamma = L_0 (v/c + 1/\gamma)## from the nose of A. You should be able to show that that distance is always greater than ##L_0## for any nonzero speed, and thus the bullet always hits rocket B.

    Then do a similar exercise according to the B frame. You'd better get the same result, since the collision of bullet and rocket B is an event that must happen in any frame.

    I agree. It makes no sense to have "instantaneous signaling" in a relativity problem.

    Yes there is. Ask your teacher what he or she meant by that last bit.
     
  11. Dec 11, 2014 #10
    Hey, thanks guys for replying to my problem.

    Yeah so I redid the problem and come at the same conclusion that if time of communication is taken into account then A will always hit B. I dunno why but the first night when I posted here I had calculated that A would hit if the relative speed was under a certain speed, and would hit if it was over that speed. I tried refinding it to see my initial mistake but it is in a the garbage by now.

    It was part of what got me confused because in the case A misses, I had to have A's shot pass behind B instead of in front, and it didn't feel right. (Although 'feelings' aren't that useful in relativity so I kinda assumed with was true even though reallly counter-intuitive)


    It was in my finals exam so I'll be sure to ask when I go see my correction. Thanks again for the help everyone !
     
  12. Dec 11, 2014 #11
    I think I initially made the same algebraic mistake and came to the same conclusion. I forgot to multiply one of the terms in parenthesis by gamma (when I applied the distributive law), and two of the terms involving L canceled with each other. This left a third term which allowed B to be hit if beta was less than 1/sqrt(2) and caused B to be missed in the rear if beta was greater than1/sqrt(2).

    Chet
     
  13. Dec 11, 2014 #12

    PeroK

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    It's more obvious in B's frame:

    When the signal is sent, the nose of B is already past the tail of A. And, as the tail of B can't outrun the light signal, it's going to be a hit!
     
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