Where are the bullets?

Discussion in 'Physics & Math' started by fess, Feb 29, 2020.

  1. fess Registered Senior Member

    Messages:
    97
    A train is traveling at great speed on a huge circular track. An observer in the middle of the car sees 2 men standing at either end of the car shoot at each other simultaneously. The bullets travel at exactly the same speed, hit each other in the exact middle of the car and fall straight down

    Another observer watches from a station as the train goes by and because of the movement of the train and distance the light has to travel, sees gunman A fire first. That gunman's bullet travels father than the gunmans B bullet before they meet and fall to the floor, not in the middle of the car.

    The train continues its journey and eventually returns to the station and stops. Can both observers agree on what happened? Where will the bullets be?
     
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  3. Halc Registered Senior Member

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    They're in a rotating reference frame, so both bullets will exhibit different levels of centrifugal force. If the train moves at sufficient speed to generate reasonable G forces going around that circle, the bullets will miss each other. This is Newtonian physics and requires nothing near light speed.

    Assuming the train is on a straight track, the bullets meet and fall in the middle of the car. The circular track seems to detract from the point you're trying to make. The gunmen shoot simultaneously in the frame of the train, something that isn't necessary to specify in Newtonian physics.

    The bullets meeting an falling at the X drawn in the middle of the car is objective fact. The observer on the platform (or anywhere else) witnesses that no matter where he stands.
    If he sees the one shot firing first, maybe it's because he's closer to one shooter than the other, or maybe he sees the guy in the rear firing first because he in fact fires first in that frame. You haven't specified where the observer is standing, so maybe the train passed by a minute before the duel takes place. I don't know. Best place for him to stand is right where the bullets meet, so the train is centered on him at that moment, not when the guns are fired.
    What we do know is that the bullets fall to the floor in the center of the car and move along thereafter at the speed of the train.

    The circular track makes the bullets miss each other. They don't have to meet again. Each observer can email what they saw and compare.
    They agree the bullet falls in the middle of the car. The observer mid-car says the gunmen shot simultaneously, and the platform guy might say he saw either one shoot first depending on who was closer, but if he takes the extra distance for light travel into account, he'll say the rear gunman shot first.
     
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  5. fess Registered Senior Member

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    Sorry, I only used a’circular’ track to demonstrate that the train would eventually return to the station where the stationary observer saw the 2 men fire non-simultaneously.
    I don’t understand why both observers would see the bullets meet at the center of the car if the stationary observer saw 1 gunman fire first He would need to perceive the 2nd bullet fly faster than the 1st
     
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  7. Halc Registered Senior Member

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    Of course the bullets are different speeds. Suppose the bullet speed is +/-vb and the train speed is vt. The bullet from the guy in the rear is vt + vb* relative to the platform guy, and vt - vb from the guy in front, who is shooting backwards.

    * Not true at relativistic speeds. The addition of relativistic velocity is more like
    v = (v1 + v2) / (1 + (v1 v2/c*c))
     
    Last edited: Feb 29, 2020
  8. Janus58 Valued Senior Member

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    2,185
    I assume when you say, "The bullets travel at exactly the same speed", you mean relative to the train as measured from the train.
    However, This does not equate to them traveling at the same speed with respect to the trains as measured from the tracks. Velocities don't add the same way in Relativity as they do with according to Newton.
    The Relativistic rule is W= (v+u)/(1+uv/c^2)
    Here we will assign v as the velocity of the train relative to the tracks, and u the velocity of the bullet as measured from the train (positive for the bullet fired in the direction of the train's motion and negative for the other bullet.)

    So let's say that v is 0.25c and u is 0.5c ( this makes thing more noticeable).
    Thus Bullet 1, fired in the direction the train is moving at, as measured from the tracks, (0.25c+0.5c)/(1+0.5c(0.25c)/c^2 =0.6666...c, relative to the tracks, and the difference between this and the train is 0.41666...c
    Bullet 2, fired in the opposite direction, travels at (0.25c-.05c)/(1-.25c(0.5c)?c^2) = - 0.2857c relative to the tracks, a difference of 0.5357c from the train. In other words, According to the track observer, the bullets have different speeds with respect to the train. One bullet is fired before the other, but has lower speed difference between itself and the train than the other bullet does, and thus both bullets still meet at the center of the train.
     
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  9. James R Just this guy, you know? Staff Member

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    35,101
    In relativity, all observers always agree on what happens, regardless of their reference frame. Events in spacetime are events in spacetime. If one of those events happens to be two bullets colliding and falling to the floor in the centre of the train, then all observers will agree that's what happens, in every frame. Different observers might disagree about where that event happened and when it happened, since different observers will use their own rulers and clocks to measure those things. But they can never disagree about what happened.
     
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