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[CLICKING]
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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.