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Hello, I've just registered to ask this question. This is something that's been bothering me for some time. I have a non-elastic collision of 2 masses in 1-dimension observed by two different frames of reference, and the total momentum of the system is conserved for one of them but not for the other. I'm probably making a mistake but I don't know where.

Here's a picture that shows what's going on in both frames. I represented the frames of reference S and S' as dots for simplicity.

The problem is this: according to the [itex]S[/itex], the two objects should stick together and stop after the collision, as predicted by both the classical and the relativistic definition of momentum. But the problem is, when you carry out the calculations for the velocities in the [itex]S'[/itex] frame, the sum of the momenta of the objects doesn't come out to be equal to the momentum of the combined object.

Okay, so the definition of the relativistic momentum is this: [itex]\overrightarrow{P}=\frac{m \overrightarrow{v}}{\sqrt{1-\frac{v^2}{c^2}}}[/itex]

In the first frame, both classical and relativistic definitions agree that the momentum of the object [itex]M_3[/itex] (let's call it [itex]P_3[/itex]) should be 0, because:

[itex]m u + m (-u) = 0[/itex]

and it doesn't matter if you use the relativistic definition because the factor [itex]\gamma = \frac{1}{\sqrt{1-\frac{u^2}{c^2}}}[/itex] is the same both objects, and it gives the same result if you multiply both sides by [itex]\gamma[/itex]. So we know that the objects will stop after colliding.

Now let's consider the event in the reference frame of [itex]S'[/itex]. Using the Lorentz transformation, the relative velocities of the objects [itex]M_1[/itex] and [itex]M_2[/itex] with respect to [itex]S'[/itex] are found to be:

[itex]u_1' = \frac{u - v}{1 - \frac{u v}{c^2}}[/itex] and [itex]u_2' = \frac{-u - v}{1 + \frac{u v}{c^2}}[/itex]

So the momentum of the first object [itex]P_1'[/itex] should be equal to:

[itex]\frac{m u_1'}{\sqrt{1-\frac{u_1'^2}{c^2}}}[/itex]

To simplify it a little before substituting for [itex]u_1'[/itex], let's divide both the numerator and the denominator with [itex]|u_1|[/itex]:

[itex]P_1 = \frac{m \frac{u_1'}{|u_1'|}}{\sqrt{\frac{1}{u_1'^2}-\frac{1}{c^2}}}[/itex]

We know that [itex]\frac{u_1'}{|u_1'|}[/itex] is equal to the sign of [itex]u_1[/itex] which is equal to the sign of [itex]u - v[/itex], since the denominator is always positive. Now let's write [itex]u_1'[/itex] in terms of [itex]u[/itex] and [itex]v[/itex].

[itex]P_1' = \frac{m sgn(u-v)}{\sqrt{\frac{1}{\big(\frac{u - v}{1 - \frac{u v}{c^2}}\big)^2}-\frac{1}{c^2}}} = \frac{m sgn(u-v)}{\sqrt{\frac{1-\frac{2 u v}{c^2}+\frac{(u v)^2}{c^4}}{u^2-2 u v+v^2}-\frac{1}{c^2}}} = \frac{m sgn(u-v)}{\sqrt{\frac{c^2-2 u v+\frac{(u v)^2}{c^2}-(u^2-2 u v+v^2)}{c^2 (u^2-2 u v+v^2)}}}[/itex]

[itex] = \frac{m sgn(u-v)}{\sqrt{\frac{c^2-u^2-v^2+\frac{(u v)^2}{c^2}}{c^2 (u-v)^2}}} = \frac{m sgn(u-v) |u-v|}{\sqrt{1-\frac{u^2}{c^2}-\frac{v^2}{c^2}+\frac{(u v)^2}{c^4}}} = \frac{m (u-v)}{\sqrt{1-\frac{u^2}{c^2}-\frac{v^2}{c^2}(1-\frac{u^2}{c^2})}}[/itex]

[itex]= \frac{m (u-v)}{\sqrt{(1-\frac{u^2}{c^2}) (1-\frac{v^2}{c^2})}}[/itex]

Okay, that was a long calculation but now it's in a form that's easier to read. We know that the original equation we found for [itex]P_1'[/itex] was the same as the one for [itex]P_2'[/itex], except for the negated initial velocity [itex]-u[/itex]. So if we plug in [itex]-u[/itex] for [itex]u[/itex] in the equation we just derived for [itex]P_1'[/itex], we should get the equation for [itex]P_2'[/itex]:

[itex]P_2' = \frac{m (-u-v)}{\sqrt{(1-\frac{u^2}{c^2}) (1-\frac{v^2}{c^2})}}[/itex]

Now let's find the momentum of [itex]M_3[/itex]. We know that it is at rest for [itex]S[/itex], so for [itex]S'[/itex] it must be moving at [itex]v[/itex] along with [itex]S[/itex]:

[itex]P_3' = \gamma (2m) (-v) = \frac{-2 m v}{\sqrt{1-\frac{v^2}{c^2}}}[/itex]

If the momentum is conserved, [itex]P_1' + P_2'[/itex] should be equal to [itex]P_3'[/itex].

[itex]P_1' + P_2' = \frac{m (u-v)}{\sqrt{(1-\frac{u^2}{c^2}) (1-\frac{v^2}{c^2})}} + \frac{m (-u-v)}{\sqrt{(1-\frac{u^2}{c^2}) (1-\frac{v^2}{c^2})}} = \frac{-2m v}{\sqrt{(1-\frac{u^2}{c^2}) (1-\frac{v^2}{c^2})}}[/itex]

and

[itex]P_3' = \gamma (2m) (-v) = \frac{-2 m v}{\sqrt{1-\frac{v^2}{c^2}}}[/itex]

Here [itex]P_3'[/itex] is missing the factor [itex]\frac{1}{\sqrt{1-\frac{u^2}{c^2}}}[/itex] that [itex]P_1' + P_2'[/itex] has so they can't be equal.

I appreciate it if you bothered to read this far. I don't understand what I may have done wrong. I've already asked a few people but couldn't get an answer. If you have an idea, please drop a comment below. Thank you.

Here's a picture that shows what's going on in both frames. I represented the frames of reference S and S' as dots for simplicity.

The problem is this: according to the [itex]S[/itex], the two objects should stick together and stop after the collision, as predicted by both the classical and the relativistic definition of momentum. But the problem is, when you carry out the calculations for the velocities in the [itex]S'[/itex] frame, the sum of the momenta of the objects doesn't come out to be equal to the momentum of the combined object.

Okay, so the definition of the relativistic momentum is this: [itex]\overrightarrow{P}=\frac{m \overrightarrow{v}}{\sqrt{1-\frac{v^2}{c^2}}}[/itex]

In the first frame, both classical and relativistic definitions agree that the momentum of the object [itex]M_3[/itex] (let's call it [itex]P_3[/itex]) should be 0, because:

[itex]m u + m (-u) = 0[/itex]

and it doesn't matter if you use the relativistic definition because the factor [itex]\gamma = \frac{1}{\sqrt{1-\frac{u^2}{c^2}}}[/itex] is the same both objects, and it gives the same result if you multiply both sides by [itex]\gamma[/itex]. So we know that the objects will stop after colliding.

Now let's consider the event in the reference frame of [itex]S'[/itex]. Using the Lorentz transformation, the relative velocities of the objects [itex]M_1[/itex] and [itex]M_2[/itex] with respect to [itex]S'[/itex] are found to be:

[itex]u_1' = \frac{u - v}{1 - \frac{u v}{c^2}}[/itex] and [itex]u_2' = \frac{-u - v}{1 + \frac{u v}{c^2}}[/itex]

So the momentum of the first object [itex]P_1'[/itex] should be equal to:

[itex]\frac{m u_1'}{\sqrt{1-\frac{u_1'^2}{c^2}}}[/itex]

To simplify it a little before substituting for [itex]u_1'[/itex], let's divide both the numerator and the denominator with [itex]|u_1|[/itex]:

[itex]P_1 = \frac{m \frac{u_1'}{|u_1'|}}{\sqrt{\frac{1}{u_1'^2}-\frac{1}{c^2}}}[/itex]

We know that [itex]\frac{u_1'}{|u_1'|}[/itex] is equal to the sign of [itex]u_1[/itex] which is equal to the sign of [itex]u - v[/itex], since the denominator is always positive. Now let's write [itex]u_1'[/itex] in terms of [itex]u[/itex] and [itex]v[/itex].

[itex]P_1' = \frac{m sgn(u-v)}{\sqrt{\frac{1}{\big(\frac{u - v}{1 - \frac{u v}{c^2}}\big)^2}-\frac{1}{c^2}}} = \frac{m sgn(u-v)}{\sqrt{\frac{1-\frac{2 u v}{c^2}+\frac{(u v)^2}{c^4}}{u^2-2 u v+v^2}-\frac{1}{c^2}}} = \frac{m sgn(u-v)}{\sqrt{\frac{c^2-2 u v+\frac{(u v)^2}{c^2}-(u^2-2 u v+v^2)}{c^2 (u^2-2 u v+v^2)}}}[/itex]

[itex] = \frac{m sgn(u-v)}{\sqrt{\frac{c^2-u^2-v^2+\frac{(u v)^2}{c^2}}{c^2 (u-v)^2}}} = \frac{m sgn(u-v) |u-v|}{\sqrt{1-\frac{u^2}{c^2}-\frac{v^2}{c^2}+\frac{(u v)^2}{c^4}}} = \frac{m (u-v)}{\sqrt{1-\frac{u^2}{c^2}-\frac{v^2}{c^2}(1-\frac{u^2}{c^2})}}[/itex]

[itex]= \frac{m (u-v)}{\sqrt{(1-\frac{u^2}{c^2}) (1-\frac{v^2}{c^2})}}[/itex]

Okay, that was a long calculation but now it's in a form that's easier to read. We know that the original equation we found for [itex]P_1'[/itex] was the same as the one for [itex]P_2'[/itex], except for the negated initial velocity [itex]-u[/itex]. So if we plug in [itex]-u[/itex] for [itex]u[/itex] in the equation we just derived for [itex]P_1'[/itex], we should get the equation for [itex]P_2'[/itex]:

[itex]P_2' = \frac{m (-u-v)}{\sqrt{(1-\frac{u^2}{c^2}) (1-\frac{v^2}{c^2})}}[/itex]

Now let's find the momentum of [itex]M_3[/itex]. We know that it is at rest for [itex]S[/itex], so for [itex]S'[/itex] it must be moving at [itex]v[/itex] along with [itex]S[/itex]:

[itex]P_3' = \gamma (2m) (-v) = \frac{-2 m v}{\sqrt{1-\frac{v^2}{c^2}}}[/itex]

If the momentum is conserved, [itex]P_1' + P_2'[/itex] should be equal to [itex]P_3'[/itex].

[itex]P_1' + P_2' = \frac{m (u-v)}{\sqrt{(1-\frac{u^2}{c^2}) (1-\frac{v^2}{c^2})}} + \frac{m (-u-v)}{\sqrt{(1-\frac{u^2}{c^2}) (1-\frac{v^2}{c^2})}} = \frac{-2m v}{\sqrt{(1-\frac{u^2}{c^2}) (1-\frac{v^2}{c^2})}}[/itex]

and

[itex]P_3' = \gamma (2m) (-v) = \frac{-2 m v}{\sqrt{1-\frac{v^2}{c^2}}}[/itex]

Here [itex]P_3'[/itex] is missing the factor [itex]\frac{1}{\sqrt{1-\frac{u^2}{c^2}}}[/itex] that [itex]P_1' + P_2'[/itex] has so they can't be equal.

I appreciate it if you bothered to read this far. I don't understand what I may have done wrong. I've already asked a few people but couldn't get an answer. If you have an idea, please drop a comment below. Thank you.

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