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I think I remember hearing about this years ago, but I cant remember what was said about it. If a bullet is fired into the front of a speeding train, then the bullet has to change direction, which means it will momentarily stop. If the bullet stops, then so must the train, how can this be possible.

Is this xeno/zenos paradox?? or how is it connected to it

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I think I remember hearing about this years ago' date=' but I cant remember what was said about it. If a bullet is fired into the front of a speeding train, then the bullet has to change direction, which means it will momentarily stop. If the bullet stops, then so must the train, how can this be possible.

Is this xeno/zenos paradox?? or how is it connected to it[/quote']

Why must the train stop? metal can bend and the bullet deforms. the train is also not completely rigid. in the frame of the train, its not moving anyway.

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I think I remember hearing about this years ago' date=' but I cant remember what was said about it. If a bullet is fired into the front of a speeding train, then the bullet has to change direction, which means it will momentarily stop. If the bullet stops, then so must the train, how can this be possible.

[/quote']

The bullet stops in which frame? If you're talking about the rest frame of the train then there is no big deal here because the train is stopped the entire time. If you're talking about the rest frame of the Earth then the bullet never stops, and neither does the train.

Is this xeno/zenos paradox?? or how is it connected to it

No. Zeno of Elea has 4 paradoxes of motion attributed to him (by Aristotle), but this isn't one of them.

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I am talking about the rest frame of the earth. How wouldnt the bullet stop?. i can understand that the train neednt stop as the metal can be deforemed. but in all frames, the bullet stops doesnt it?

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Ack, I was too sloppy. I should have said that the bullet doesn't turn around in the Earth frame only if the recoil speed of the bullet is less than the speed of the train. In that case the bullet will be seen by the Earth based observer to collide with the wall and then continue moving forward, but slower. At no time will the speed of the bullet be zero in the Earth frame. So in that case, it is not true that the bullet turns around in all frames.

But now let's look at what happens when the recoil speed of the bullet is greater than that of the train.

Let [imath]v_B[/imath] be the initial speed of the bullet as measured from the train, [imath]v_T[/imath] be the (constant) speed of the train as measured from the Earth, and let [imath]v_B>v_T[/imath]. Further, let the bullet hit the front wall of the train at time [imath]t=0[/imath] and let the deformation and recoil take place from [imath]t=0[/imath] to [imath]t=T[/imath]. Let the acceleration of the bullet be constant during the entire collision-deformation-recoil process (this is for simplicity). We can write down the following function for the velocity of the bullet with respect to the train. (Sorry for how it looks; I tried to define a piecewise function as an array, but the system said I had a syntax error in my LaTeX. Couldn't find it though.).

[imath]v_{BT}(t)=v_B[/imath] for [imath]t \leq 0[/imath]

[imath]v_{BT}(t)=v_B-\frac{2v_B}{T}t[/imath] for [imath]0 \leq t \leq T[/imath]

[imath]v_{BT}(t)=-v_B[/imath] for [imath]t \geq T[/imath]

Note that I assumed that this collision is perfectly elastic in the frame of the train.

Now the question is, at what time [imath]t_{stop,BT}[math] is the speed zero? It can only be zero during the deceleration, so we'll set that piece of [imath]v_{BT}[/imath] equal to zero. Solving the equation yields [imath]t_{stop,BT}=\frac{T}{2}[/imath], which is exactly the midpoint of the deceleration process (makes sense).

Now let's transform to the Earth frame with the Galilean transformation. If we call the velocity of the bullet as measured from the Earth frame [imath]v_{BE}[/imath] and the velocity of the train as measured from the Earth frame [imath]v_{TE}[/imath] then the Galilean transformation reads as follows: [imath]v_{BE}=v_{BT}+v_{TE}[/imath], which leads to:

[imath]v_{BE}(t)=v_B+v_T[/imath] for [imath]t \leq 0[/imath]

[imath]v_{BE}(t)=v_B+v_T-\frac{2v_B}{T}t[/imath] for [imath]0 \leq t \leq T[/imath]

[imath]v_{BE}(t)=-v_B+v_T[/imath] for [imath]t \geq T[/imath]

Now let's ask the same question that we asked for the train's point of view: When is the speed of the bullet equal to zero in the Earth frame? Set the middle piece of [imath]v_{BE}(t)[/imath] equal to zero and solve. You will get [imath]t_{stop,BE}=\frac{v_B+v_T}{2v_B}T[/imath].

This resolves the apparent paradox. Yes, the bullet turns around in the Earth frame, provided that it recoils with a speed greater than that of the train. But it doesn't turn around at the same time in both frames. When the bullet turns around in the Earth frame, is moving with respect to the train, and thus there is no reason to suppose that the train must be stopped at that time.

Now if you want to see this done with the Lorentz transformation, you're going to have to start paying buddy.

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Of course there is no problem if time is discreet.

There will be a planck-length moment where everything has stopped.

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I think I remember hearing about this years ago' date=' but I cant remember what was said about it. If a bullet is fired into the front of a speeding train, then the bullet has to change direction, which means it will momentarily stop. If the bullet stops, then so must the train, how can this be possible.

Is this xeno/zenos paradox?? or how is it connected to it[/quote']

It's related to the paradoxes (paradoxi?) in that it deals, or doesn't, with infinitesimals. And we've discussed it, or something close, before.

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