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Bullet-hole paradox

  1. Aug 14, 2011 #1
    Bullet-hole "paradox"

    This is a variant of the so-called "bullet-hole paradox" (the guns can be lasers and the train a spaceship):

    Two guns are mounted a distance of 40 ns apart (SR unit system) on the embankment beside a railroad track. The barrels of the guns project toward the track so that they almost touch a speeding train as it passes by. The train moves at 3/5 the speed of light with respect to the ground. Suppose the two guns fire simultaneously (in the ground frame) as the train passes leaving two bullet holes in the train.

    The Lorentz transformations show that in the train frame, the space-time coordinate separation between the events of the guns firing is 50 ns apart spatially and 30 ns in time (with the forward gun firing before the rear gun). In this frame the distance between the guns is seen as 32 ns.

    My questions are these: (1) In the train frame how far apart do the bullet holes apear to be?; (2) If the train is stopped after the firing of the guns, how far apart are the holes measured to be?

    As the train is a physical object I expect the answer in both cases to 40 ns. Can anyone justify this or contradict it?
     
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  3. Aug 14, 2011 #2

    xts

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    Re: Bullet-hole "paradox"

    68 ns - 50 ns for spatial diff plus 18=3/5*delay (if your calculations are correct, I haven't checked them)

    Still 68 ns. From the train conductor's perspective nothing changes.

    Wrong! The train, as passing the guerilla forces, was in motion, and thus got dilated. So bullets hit "compressed" train 40 ns apart, but after its "decompression" the holes are 68 ns apart.
     
    Last edited: Aug 14, 2011
  4. Aug 14, 2011 #3

    clem

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    Re: Bullet-hole "paradox"

    I calculate that the holes will be inflicted 40 ns apart in the moving train, simultaneously in the track rest frame.
    This means that they will be 40\gamma=50 ns apart in the rest frame of the train.

    I don't see why you draw a distinction between the "train frame" and the stopped train.
     
  5. Aug 15, 2011 #4

    xts

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    Re: Bullet-hole "paradox"

    Sorry - they'll be 50ns, not 68 - you've already calculated the spatial distance between events, not just between guns.

    Did I? Just contrary. I told you there is no distinction. As an answer to your question (2)
     
  6. Aug 15, 2011 #5

    jtbell

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    Re: Bullet-hole "paradox"

    In the ground frame, the guns are 40 ns (distance) apart. The Lorentz factor is [itex]\gamma = 1 / \sqrt{1 - 0.6^2} = 1.25[/itex]. The guns fire simultaneously in the ground frame. Call the gun that is closer to the front of the train F, and the other one R.

    In the train frame, the guns are moving towards the rear at speed 0.6c, with gun R "in the lead". The distance between them is contracted to 40 / 1.25 = 32 ns. They do not fire simultaneously. Gun F fires first, then gun R. The time difference is [itex]\gamma v \Delta x / c^2[/itex], where [itex]\Delta x[/itex] is the distance between the guns in the ground frame. 1.25 * 0.6 * 40 = 30 ns. During this time, gun R travels a distance of 0.6 * 30 = 18 ns. So the distance between the two gunshots in the train frame, and therefore the distance between the holes in the side of the train, is 32 + 18 = 50 ns.
     
    Last edited: Aug 15, 2011
  7. Aug 15, 2011 #6
    Re: Bullet-hole "paradox"

    My reply to XTS and CLEM remarks failed to get posted last night.

    However, at this point, we all seem to be in agreement with the following:
    (1) In the ground frame the guns are 40 ns apart and the holes in the train are 40 ns apart.
    (2) In the train frame the guns appear to be 32 ns apart. Because gun F is seen to fire 30 ns before gun R, the distance between the holes turns out to be 50 ns.

    The seeming "paradox" arises because of the disparity between the 32 ns measured gun spacing and the 50 ns spacing of the holes in the train frame. My basic question is not about this, but asks: If the train is stopped so that it is now stationary in the ground frame, how far apart are the holes then measured to be? 40 or 50 ns?

    The stopping of the train involves deceleration, so that there is an assumption that while the train is undergoing this non-inertial process the spacing of the holes is unaffected. But SR does not posit that real physical objects are modified by their motions; i.e., by being stretched or compressed like rubber bands, but only that the results of measuring space-time coordinates depends on the state of motion of the coordinate syatems involved due to the constancy of the speed of light.

    So, is it 40 or 50 ns -- and why?
     
  8. Aug 15, 2011 #7

    pervect

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    Re: Bullet-hole "paradox"

    If the train is Born rigid, the holes will still be 50 ns apart when it stops.
     
  9. Aug 15, 2011 #8

    jtbell

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    Re: Bullet-hole "paradox"

    Your question could also be asked about the length of the train as a whole. Suppose a passenger measures it to be 100m long while it is in motion, by walking from end to end with a measuring tape. Why should it be any other length when at rest, assuming that it was brought to a stop in a way that doesn't over-stress it and cause the frame to crumple?
     
  10. Aug 15, 2011 #9

    clem

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    Re: Bullet-hole "paradox"

    My statement "I don't see why you draw a distinction between the "train frame" and the stopped train." was for NSEFF.

    Final question: How can be so much discussion about such a simple problem?
     
  11. Aug 15, 2011 #10
    Re: Bullet-hole "paradox"

    PERVECT:
    As I remarked, Special Relativity does not posit conditions on physical objects, such as rigidity, for the equations to apply. Contraction arises merely due to postulated constancy of the speed of light in all inertial frames, and how the length of an object must rationally be defined.

    JTBELL:
    Your passenger is measuring the length of the train essentially at rest (if he walks slowly, << c, when making his measurement). You recognize that this is not the canonical way of defining length though?
    I agree that a 100 m train (as measured at rest in the ground frame) will look shorter if it flies by at high speed and, if subsequently brought to rest, will still again 100 m long. However, my question (40 or 50 ns) relates to a physical object which has been modified (the two bullet holes) while flying by -- and seemingly, there are two measurements, both in rest frames, disagreeing as to the spacing of the bullet holes.

    CLEM:
    There is no disagreement about the simple problem part -- the arithmetic describing what each observer sees.
    But no one seems to have an answer to my fundamental question about how far apart the bullet holes are after the train is brought to rest.
     
  12. Aug 15, 2011 #11

    jtbell

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    Re: Bullet-hole "paradox"

    The canonical way to measure the length of an object is to put a meter stick (or measuring tape) up next to the object, at rest with respect to the object. Laying out a measuring tape along the floor of the train from one end to the other fits this description. Of course the passenger has to walk from one end of the train to the other in order to set this up, but the measuring tape is stationary with respect to the train during the actual measurement (reading off the markings at the two ends).
     
  13. Aug 15, 2011 #12
    Re: Bullet-hole "paradox"

    JTBELL:
    The esssential canonical feature I was refering to was that the position readings at the two ends of the object had to be established simultaneously.
     
  14. Aug 15, 2011 #13

    jtbell

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    Re: Bullet-hole "paradox"

    Simultaneity is a potential problem only if the measuring rod/tape and the object being measured are moving with respect to each other.
     
  15. Aug 15, 2011 #14

    clem

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    Re: Bullet-hole "paradox"

    50 ns, as I answered last night.
     
  16. Aug 15, 2011 #15
    Re: Bullet-hole "paradox"

    CLEM:
    So your contention is that the simultaneously fired bullets did not enter the train 40 ns apart spatially as it appeared they did in the ground frame? That this ground frame viewpoint is erroneous? What is your argument for this? This is principally where my hangup is with the question I posed.
     
  17. Aug 16, 2011 #16

    pervect

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    Re: Bullet-hole "paradox"

    I believe you already mentioned that in the ground plane , the interval was 40 ns purely spatial, but in the train plane, the interval was 50 ns spatial and 30ns timelike, which is the same lorentz interval (50^2 - 30^2 = 40^2).

    Now all you have to do is believe your calculations. YOu seem to be taking a step backwards in your remarks above...

    The other part you might want to calculate in more detail is the Born rigid motion. But basically it behaves pretty much like you'd expect, if the bullets were 50 ns apart in the train frame spatially, they stay 50 ns apart if the motion is Born-rigid.

    There's not a lot out there on it, you might try http://www.mathpages.com/home/kmath422/kmath422.htm
     
  18. Aug 18, 2011 #17
    Re: Bullet-hole "paradox"

    Time to close this thread out. The following is the explaination I was looking for:

    The answer must be 50 ns just as measured in the train frame.

    This is because as the train slows, observers on the ground will see the train lengthen as the Lorentz contraction effect decreases. The bullet holes being part of the train will also appear to spread apart to ground frame observers. When the train finally comes to rest, meter sticks on the ground will finally agree with those on the train, and both will agree that the holes are 50 ns apart.

    In the train frame, there can be no physical changes inside the train as it slows down. Train observers always see the bullet holes to be 50 ns apart, no matter how they are moving. For if observers on the train saw the bullet holes (originally measured to be 50 ns apart when the train was moving at its original speed) move closer together as the train slowed to a stop -- how could they explain this? That the front bullet hole moved back a few seats?

    Only the 50 ns answer makes any logical sense.

    The ground observers originally see the bullet holes as being 40 ns apart, the same spacing as the two guns. As the train slows down, the gun spacing stays at 40 ns, but the distance between the bullet holes, which are in the moving train, expands in proportion to the length of the train as it expands. The bullet holes are always next to the same seats on the train, it is just that the distance between them expands just like every other distance measured between objects on the train as the train slows down.

    PERVECT: Born rigid motion is new to me. I'll check out your reference. Thanks.
     
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