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MARKUS KLUTE: Welcome back
to 8.20, Special Relativity.

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So in this section,
we want to look again

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at collisions and study
momentum conservation.

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And so we have
this scenario here

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in which we have two balls
colliding with velocities uA

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and uB, mass mA and mB.

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And after the collisions
there, the mass

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is changed to mC and mD.

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And the velocities
are uC and uD.

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Momentum conservation
tells you that the product

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of the mass and velocities
and the sum of the two

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particles before and after
the collisions is the same.

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But now what happens if
you boost the system?

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If we look at the very
same system with the

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boosted reference frame, and we
can just simplify the case here

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by just considering
the x direction.

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So the question is if
momentum is conserved

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in a frame s, like the one we're
looking here in this picture,

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is the momentum also conserved
in a moving reference

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frame with relative velocity v?

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And so you can show this quite
easily, that this is actually

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not the case, right?

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So you write down
the velocities.

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The momentum equation
is the same as before

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with boosted velocities.

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And you find that
this is not the case.

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You can easily show
this by, for example,

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setting the right part
of this equation to 0,

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and see whether or
not this equation will

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hold true in general.

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And it doesn't.

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But in the last section, we
introduced the new concept

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of proper velocity.

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So how about redefining
momentum through proper velocity

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and just saying mA times
proper velocity A plus mB times

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proper velocity B is equal to
mC times proper velocity C,

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and so on, and see whether
or not we can learn something

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from this equation?

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So why don't we write down
this very same equation

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and with proper velocity
and see whether or not

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it's invariant in the
Lorentz transformation?

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OK.

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Again, a good moment
to stop the video

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and just work out the math
by yourself, on your own.

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So I did this here.

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And I'm just doing this
for the x component.

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So we have our proper
velocity vector,

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which is gamma times C, gamma
times uX, gamma times uY,

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and sometimes uZ.

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And so the x component
is the first component.

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And in the boosted
reference frame,

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the proper velocity of the x
component in our boosted frame

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is gamma times the proper
velocity of the particle

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A, first component, minus
beta, upper velocity

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of A0's component.

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And then you can
write your equation.

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And this should be our momentum
energy, momentum conservation

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equation.

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OK.

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And the Lorentz transformation?

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We find this one here.

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And so all we need
to do now in order

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to show that this is always
true, or true in general,

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is to reassign, relabel,
reorder the individual terms.

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And I did this here
to make this visible.

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You see everything behind, in
this bracket behind the beta,

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is equal to the equation we
had before with the minus sign

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in between, which means it's 0.

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And everything we see on the
top of the first component,

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it also be 0.

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That's for the boosted
reference frame.

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So we see here, and it just
shows this for the x component.

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You can show this
for all components

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that momentum is conserved.