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Special Relativity: Bullets on a train

  1. Sep 30, 2013 #1
    1. The problem statement, all variables and given/known data

    A train moves at speed v. Bullets are successively fired at speed u (relative to the train) from the back of the train to the front. A new bullet is fired at the instant (as measured in the train frame) the previous bullet hits the front. In the frame of the ground, what fraction of the way along the train is a given bullet, at the instant (as measured in the ground frame) the next bullet is fired? What is the maximum number of bullets that are in flight at a given instant, in the ground frame?

    2. Relevant equations

    Δt = γ(Δt' + vΔx'/c^2)
    u = (u'+v)/(1+u'v/c^2)

    3. The attempt at a solution

    In the ground frame, the time elapsed between the bullet being fired and hitting the front is given by the Lorentz transformation:

    Δt = γ(Δt' + vΔx'/c^2)

    If the train has proper length l', then in the train frame Δt' = l'/u' and Δx'=l', so this gives:

    Δt = γ(l'/u' + vl'/c^2)

    So in the ground frame, the next bullet is fired at Δt. The distance travelled by a bullet in this time in the ground frame is therefore uΔt.

    Using u = (u'+v)/(1+u'v/c^2) and Δt = γ(l'/u' + vl'/c^2)

    Distance = (u'+v)/(1+u'v/c^2)*γ(l'/u' + vl'/c^2)

    Now when I multiply this out I just get a huge mess that doesn't simplify into anything useful, so I feel like I'm doing this completely wrong. Could anyone give me any guidance?

    Thanks
     
  2. jcsd
  3. Sep 30, 2013 #2

    mfb

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    2016 Award

    Staff: Mentor

    The difference between "time between shots in the train frame" and "time between shots in the ground frame" is just γΔt'. There is no additional term in the transformation, as the shooter does not move in the frame of the train - just time dilation.
    You'll need the speed of the bullet in the ground frame, afterwards you can calculate everything in the ground frame and you don't have to care about relativity any more.
     
  4. Oct 1, 2013 #3
    It's my understanding that Δx' is the distance between events in the train frame which is surely l', meaning the transformation is not simply γΔt'?
     
  5. Oct 1, 2013 #4

    mfb

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    Staff: Mentor

    The shots always happen in the same position in the frame of the train.
     
  6. Oct 1, 2013 #5
    Ok I see that this is true if the two events are one shot being fired and another being fired. So I now have:

    Δt = γΔt' = γ[itex]\frac{l'}{u'}[/itex]

    The distance travelled by a bullet in the ground frame is therefore:

    uΔt = uγ[itex]\frac{l'}{u'}[/itex]

    Using the velocity addition formula this gives

    uΔt = [itex]\frac{v+u'}{1+\frac{uv}{c^2}}\frac{\gamma l'}{u'}[/itex]

    Now in the ground frame the train also travels vΔt, so subtracting this from uΔt will give us the actual distance of the bullet travelled in the ground frame. This gives:

    [itex]\gamma^2 l\frac{v+u'}{u'+\frac{vu'^2}{c^2}} - \gamma \frac{vl'}{u'}[/itex]

    [itex]\gamma^2 l(\frac{v+u'}{u'+\frac{vu'^2}{c^2}} - \frac{v}{u'})[/itex]

    Which is much larger than l when v tends to c, which cannot be correct so I must be missing some other step.

    Edit: Just realised I actually needed to simplify some more. And the [itex]\gamma^2[/itex] cancels out to give:

    [itex] distance = \frac{u'^2}{u'^2+\frac{vu'^3}{c^2}}l[/itex]

    Which is always less than l, for all c and u.
     
    Last edited: Oct 1, 2013
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