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Bullet Paradox

  1. Jan 12, 2006 #1
    I posted this paradox in another forum, but no one could figure it out. I got a few 'figure it out for your self''s but that does not satisfy me. Mabye there is someone in this forum that can find a solution.

    http://bulletparadox.bravehost.com/index.html
     
  2. jcsd
  3. Jan 12, 2006 #2

    pervect

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    It appears to be the usual problem, due to the relativity of simultaneity.

    Events that are simultaneous in one frame (i.e. the experiment frame) are not in general simultaneous in another frame (i.e. the bullet frame).

    In particular, the bullet being fired from A and the light beam being fired from C occur at the same time in the experiment frame. They do not, however, occur at the same time in the bullet frame.
     
  4. Jan 12, 2006 #3
    When the bullet is fired it is in the same reference frame as the experiment. It accelerates very quickly, granted (its a futuristic bullet), but it still is at rest the instant the light is at C. The question, I suppose, is what happens during the acceleration. However, other problems arise if you are to say that the light distance does not 'contract' at the same rate as the experiment apparatus. What do you think?
     
    Last edited: Jan 12, 2006
  5. Jan 12, 2006 #4

    JesseM

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    You don't have to worry about acceleration, you could just imagine the bullet coming in at constant velocity from far away and passing by A at the precise moment the light beam reached that point.

    Anyway, the easy way to figure this out is just to use the Lorentz transformation to find the coordinates of different events first in the apparatus frame, then in the bullet frame. In the apparatus frame, we can set the coordinates of the events as:

    light hits Y, splits into two beams: x=0, y=0, t=0
    left beam reaches U: x=-0.5, y=0, t=0.5
    right beam reaches V: x=0.5, y=0, t=0.5
    left beam reaches A as bullet passes A (assume distance from U to A is 0.2 light seconds): x=-0.5, y=0.2, t=0.7
    right beam reaches C: x=0.5, y=0.2, t=0.7
    right beam reaches B: x=-0.1, y=0.2, t=1.3
    bullet reaches B: x=-0.1, y=0.2, t=1.5

    ...with x and y in units of light-seconds, and t in units of seconds.

    Now, the Lorentz transformation is:
    [tex]x' = \gamma (x - vt)[/tex]
    [tex]y' = y[/tex]
    [tex]t' = \gamma (t - vx/c^2)[/tex]
    with [tex]\gamma = 1/\sqrt{1 - v^2/c^2}[/tex], so with v=0.5c for the bullet frame, [tex]\gamma[/tex] is around 1.1547.

    So, the coordinates of all these events in the bullet frame are:

    light hits Y, splits into two beams: x'=0, y'=0, t'=0
    left beam reaches U: x'=-0.866, y'=0, t'=0.866
    right beam reaches V: x'=0.289, y'=0, t'=0.289
    left beam reaches A as bullet passes A: x'=-0.981, y'=0.2, t'=1.097
    right beam reaches C: x'=0.173, y'=0.2, t'=0.520
    right beam reaches B: x'=-0.866, y'=0.2, t'=1.559
    bullet reaches B: x'=-0.981, y'=0.2, t'=1.790

    You can check my math to see if I made any mistakes, but these numbers indicate that the beam reaches the mirror before the bullet in both frames. In the bullet's frame, you can see that the bullet passes A about 0.577 seconds after the right beam hits mirror C, so the right beam has a head start towards B in this frame. The "paradox" arises from forgetting the different definitions of simultaneity in the two frames, and wrongly supposing that if the beam leaves C at the same moment the bullet passes A in the apparatus frame, then this would be true in the bullet frame as well. Once you notice that the bullet passes A later than the light beam bounces off C in the bullet's own frame, there is no paradox, and both frames agree the light reaches B before the bullet.
     
  6. Jan 12, 2006 #5
    If we are to imagine the bullet coming in at a constant velocity from far away, as you say, then you are right that these events will have a different order from the bullets point of view. This is because the time taken by the light as it travels from Y to U and from Y to V is not the same in each reference frame.

    But, as the experiement is set up, the Y to U and Y to V travel time ( for the light ) is equal for both reference frames. And this is exactly why the math doesn't work out.

    So what effect does the acceleration of the bullet have to fix this? How does the light pulse move relative C (still in the bullet reference frame) as the bullet is accellerating.

    I have been thinking about this problem for quite a while now, and I know that the only solution is in the acceleration of the bullet. However anything I can think of to correct it just causes a paradox in another way.
     
  7. Jan 12, 2006 #6

    JesseM

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    No, that isn't part of the experimental setup, it's just an assumption made in the author's analysis of what will happen in this situation. But if the distance from Y to U is equal to the distance from Y to V, and if both beams are emitted from Y at the same moment, it's simply impossible that the travel time from Y to U could be equal to the travel time from Y to V in both frames, according to a valid relativistic analysis of the experimental setup. If you look at the frame where U and V are at rest, then since they are both equal distances from Y, if light travels at the same speed in all directions it must take the same amount of time to reach U and V in this frame; but in the frame where U and V are moving, V is moving towards the point where the light beams were emitted and U is moving away from that point, so if light travels at the same speed in all directions in this frame, the light obviously has to hit V before it hits U. The only way the two beams could hit U and V at the same time in this frame would be if the light beam heading for V moved more slowly than the beam heading for U, which would be a violation of one of the two basic postulates of relativity.

    Perhaps part of the problem is that we are meaning different things by "reference frames"? Maybe you are imagining the "bullet's frame" is one where U and V are initially at rest, but then when the light hits A all the mirrors accelerate to 0.5c relative to the bullet. But the rules of special relativity, like the rule that light travels at c in all frames, only apply to inertial frames, so if the bullet accelerates relative to the apparatus (which is moving inertially), you can't assume these rules still apply in the non-inertial coordinate system where the bullet is at rest throughout the experiment (I think you'd have to introduce a constant gravitational field to understand things from the perspective of this non-inertial coordinate system--see here for an explanation of how to analyze the twin paradox from the point of view of a non-inertial coordinate system). When I talk about the two frames in the problem, I'm talking about the inertial frame where the apparatus is at rest, and the inertial frame where the apparatus is moving to the left at 0.5c throughout the experiment.
     
    Last edited: Jan 12, 2006
  8. Jan 13, 2006 #7

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    Yes- the two events are simultaneous in the inertial frame of the experiment, but are not simultaneous in the inertial frame of the bullet travelling at .5c.

    Using a non-inertial frame is possible but not recommended for the beginning student (just as it is not recommended for beginners in Newtonian mechanics). A frame in which the bullet accelerates is such a non-inertial frame.

    The advanced student can deal with the fact that the acceleration of the bullet causes the clock at "V" to appear to tick more quickly than the clock at U during the time in which the bullet is accelerating, creating a time offset that remains after the acceleration stops. (The accelerating coordinate system gives the clock at V the appearance of gravitational time dilation - but in this case, time is accelerated rather than dilated. The amount of time acceleration is proportional to the acceleration times the distance. The formal proof of this involves using the Rindler metric, which is a standard way of describing the local coordinate system of an accelerated observer.

    Howver, the advanced student would not be likely to make the error of assuming that two events that were simultaneous in one frame were simultaneous in all frames in the first place.

    On the whole I'd recommend that the problem be analyzed in inertial frame, which makes the failure of the web-page analysis in the bullet frame easiest to understand (one does not need to understand what a metric is to follow the analysis when its done in this manner).

    The bottom line is that the web page analysis in the bullet frame is correct, but the analysis of the bullet frame is incorrect (if it were done correctly, it would agree with the first analysis).
     
    Last edited: Jan 13, 2006
  9. Jan 13, 2006 #8
    The bullet does not begin to accelerate until after the pulse of light is past U and V. This confuses me?.? If the bullet's acceleration causes time acceleration differences at U and V, for some reason I cant see how that can matter if the light pulses were already reflected before the time difference is created.
     
    Last edited: Jan 13, 2006
  10. Jan 13, 2006 #9

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    The simplest answer is probably this. In the moving frame, we know that the two events were not simultaneous, and in the stationary frame, we know that the two events were simultaneous.

    As the bullet accelerates, it is changing its definition of simultaneity from that of the stationary object to that of a moving object by virtue of its acceleration.

    This makes distant clocks "above" the bullet appear to tick more quickly, and distant clocks "below" the bullet to tick more slowly.

    This effect is interpreted as being due to "gravitational time dilation" when the accelerating observer applies the principle of equivalence (the idea that gravity is indistinguishable from inertial forces). In order for the accelerating observer to believe he is stationary, he must believe that all of space-time is experiencing a gravitatataional field.

    This field starts instantly when he turns on his rockets (according to his current defintion of "instantly") - there is no speed-of-light delay - and stops "instantly" (again, according to his current defintion of simultaneity) when he stops his rockets.

    On the whole, I would still strongly recommend understanding the problem in terms of non-accelerated coordinate systems first - it is too easy to get confused if one attempts to "jump" right into understanding accelerated coordinate systems.
     
  11. Jan 13, 2006 #10

    JesseM

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    I don't know how to analyze things from the point of view of a non-inertial frame, but I have read that the speed of light is not necessarily constant in such frames, so this might be part of the answer--you can't assume the light beam will travel from C to B at constant speed, or that its speed at any given moment will be twice the speed of the bullet at the same moment.
     
  12. Jan 13, 2006 #11

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    Jesse: The easiest way to actually analyze what's going on in the accelerated frame is to use the Rindler metric

    [tex]
    d\tau^2 = (1+g x)^2 dt^2 - dx^2 - dy^2 - dz^2
    [/tex]

    Here g is the acceleration, assumed to be in the 'x' direction, and c is assumed to be equal to 1.

    You can see that this metric basically includes a height-variable time dilation factor (1+gx) with the other metric coefficients being normal (unity).

    The mechanics are fairly similar to the way one deals with the Schwarzschild metric of a black hole, it's just that the metric here is different.

    The biggest thing to watch out for is that the coordinate system of an accelerated observer is only defined locally - if you get too far away from the observer, coordinates become ill-behaved in that they specify multiple locations in space. MTW's gravitation still has the best treatement that I know of for accelerated observers, along with the detailed derivations and the appropriate cautions.
     
  13. Jan 13, 2006 #12
  14. Jan 13, 2006 #13

    JesseM

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    Again, that statement is about whether acceleration affects the speed of a clock as seen in an inertial reference frame. It is true that a clock's instantaneous rate of ticking depends only on its instantaneous speed in an inertial reference frame, but you can't generalize such rules to non-inertial frames. Most any statement you see about SR is only meant to apply to inertial reference frames, this is why it's better not to confuse yourself with non-inertial frames unless you're very well-versed in how they work.
     
  15. Jan 13, 2006 #14

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    There are two cases: when the clock is accelerating, and when you are accelerating (i.e. when you are in a non-inertial frame).

    There is never an effect when the clock accelerates.

    Furthermore, if you are at the same location as the clock, there is no effect when you accelerate, either. This is what DK talks about when he talks about a momentarily co-moving inertial frame (the author of the FAQ you cite).

    However, when the clock is NOT located at the same spot as you are, and you accelerate (you are in a non-inertial frame), there IS an effect on the clock that depends on its position.

    This is unfortunately not explained very clearly in the FAQ, but it's referenced indirectly in the section on "What about the equivalence principal".

    I'll see if I can come up with some on-line references which explain this more fully, right now I have to run.

    [add]
    http://panda.unm.edu/courses/finley/p570/handouts/accelobserv.pdf

    goes through the math in more detail, but it may be a difficult read
     
    Last edited: Jan 14, 2006
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