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So I would now like
to consider systems

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in which the total momentum
of the system is zero.

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And show that L, the
angular momentum,

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is independent of the
choice of point A. So let's

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consider this.

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Let's draw a picture
here and let's

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make our system a bunch of
discrete particles, particle

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one particle two.

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And let's call this
the j-th particle.

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And let's just choose
a point A right here,

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and let's choose a
point B right there.

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And I'll show that L
for this system about A

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is going to be equal to
L of the system about B.

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In order to make an
angular momentum diagram

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we have the momentum
Pj of this particle.

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And likewise, for all
the other particles.

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And I'll draw a
vector R A,j Likewise,

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for B I'll draw
the vector R B,j.

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And I'm also going to
draw a vector from B

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to A. I'll make that capital.

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This vector is a constant
because the points A

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and B, here, A and
B are fixed points.

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Now let's calculate
the angular momentum

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about A for the system.

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It's the sum of j goes from 1 to
n of the vector R A,j cross Pj.

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And I can use the angular
momentum of the system about B

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is j goes from 1 to
n of R B,j cross Pj.

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Now I'll use the
vector triangle,

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that the vector from B to j
is equal to the vector from B

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to A plus the
vector from A to j.

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So we have that R B,j is equal
to the vector from B to A plus

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the vector from A to j.

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And I'll substitute that into
our expression for the angular

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moment about B. So the angular
momentum about B is the sum j

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goes from 1 to n of R
B,A plus R A,j cross Pj.

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Now the vector
product distributes

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over vector addition.

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So this is two sums.

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So let's write them both out.

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J goes from 1 to n of R B,A
cross Pj plus the sum j goes

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from 1 to n of R A,j cross Pj.

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Now already you're seeing
that this term is the angular

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momentum of the system
about A. This term

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is the interesting one.

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R P,A is a constant vector.

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No matter which
particle I choose,

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one, two, the j-th particle,
this vector is always the same.

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So I can pull it out of the sum
and I get R B,A cross product

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the sum of j goes
from 1 to and of Pj.

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Now we've already
recognized this term

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as the angular momentum
of the system about A.

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And this is just the total
momentum of the system.

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So we have equal to R B,A
cross the total momentum

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of the system plus the angular
momentum of the system about A.

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And that's the angular
momentum of the system about B.

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So this is our general result
about how angular momentum

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differ between two points.

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But now, if P system equals to
zero, the first term is zero,

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then L of the system
about B is equal to the L

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of the system about A.
We've proved our proposition

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that if the total
momentum of the system

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is zero then the
angular momentum doesn't

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depend on the point we choose.

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Now what's interesting here is
that the reference frame moving

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with the center of
mass, by definition,

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has P in that frame,
let's call it the system.

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And this is the cm frame
is zero by definition.

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So this is one more reason why
the center of mass reference

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frame is an important
reference frame.

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Because in the center
of mass reference frame,

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the total momentum, by
definition, is zero.

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And therefore, the
angular momentum

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is independent of
any point that you

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choose in that reference frame.

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And so we can say
that the system, when

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we talk about a system's
angular momentum,

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we're referring to the
angular momentum in the center

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of mass reference frame.