Conservation of angular momentum

[Anonymous No.2576890]

I remember an argument Feynman used to show that the conservation of angular momentum implies the conservation of linear momentum. I've never seen this anywhere else, though, so is it possible that he was mistaken? The proof goes something like this:

Suppose we have a system of particles. Conservation of L doesn't depend on where the we choose the axis, so we can choose an axis that is very far from the all the particles. Then, all the particles will have the same position vector r relative to this axis, which doesn't vary with time. Then
<span class="math"> \displaystyle\ \frac{d}{dt} \bold{L_{TOT}} = \frac{d}{dt} \sum \bold {L_i} = \frac{d}{dt} \sum \bold{r} \times \bold{p_i} = \bold{r} \times \frac{d}{dt} \sum \bold{p_i} = \bold{0}[/spoiler]
implies
<span class="math">\displaystyle\ \frac{d}{dt} \sum \bold{p_i} = \bold{0}[/spoiler].
The only flaw with this argument that I can see is that r will be infinite, but I don't see how that invalidates the proof. Any insight?

[Anonymous No.2576908]

Umm, does this work?
<span class="math">\displaystyle\ \frac{d}{dt} \boldsymbol{L_{TOT}} = \frac{d}{dt} \sum \boldsymbol {L_i} = \frac{d}{dt} \sum \boldsymbol{r} \times \boldsymbol{p_i} = \boldsymbol{r} \times \frac{d}{dt} \sum \boldsymbol{p_i} = \boldsymbol{0}[/spoiler]
implies
<span class="math">\displaystyle\ \frac{d}{dt} \sum \boldsymbol{p_i} = \boldsymbol{0}[/spoiler]

[Anonymous No.2576921]

He's definitely not mistaken.

[Anonymous No.2576922]

\bf gives you bold and works on 4chan.

[Anonymous No.2576930]

<span class="math">\displaystyle\ \frac{d}{dt} \mathbf{L_{TOT}} = \frac{d}{dt} \sum \mathbf {L_i} = \frac{d}{dt} \sum \mathbf{r} \times \mathbf{p_i} = \mathbf{r} \times \frac{d}{dt} \sum \mathbf{p_i} = \mathbf{0}[/spoiler]
implies
<span class="math">\displaystyle\ \frac{d}{dt} \sum \mathbf{p_i} = \mathbf{0}[/spoiler]

[Anonymous No.2576939]

>>2576921
We'll, this would be a pretty important result, so I would have expected it to be mentioned in most (or at least one) mechanics book.

[Anonymous No.2576946]

>>2576890
Nope

Spatial translation symmetery -> conservation of linear mometum

Spatial rotation symmetery -> conservation of angular mometum

http://en.wikipedia.org/wiki/Noether's_theorem
http://en.wikipedia.org/wiki/Symmetry_in_physics

You can have systems with conservation of angular momtum and no conservation of linear mometum, and visa versa

[Anonymous No.2576949]

>>2576890
>doesn't know how to derive conservation laws

[Anonymous No.2576948]

>>2576946
Yeah, I was thinking along these line too, but the proof seems strong to me...

[Anonymous No.2576955]

>>2576948
That proff is shit, and it only applicable under certain circumstances. It does not work in general.

[Anonymous No.2576954]

OOOHHH wait I think I figured it out: the fact that, if the conservation of linear momentum works w/ respect to one axis, it's true for all axis, depends on the fact that the center of mass has a constant velocity, which is equivalent to conservation of momentum.

[Anonymous No.2576990]

>>2576946
Yes, but a key point is that you can change the axis. If you're allowed to change the axis, you can generate a translation using a composition of rotations.

[Anonymous No.2577021]

>>2576930
You can't just take r out of the derivative like that; it's a function of time. What you need is something like this:
<div class="math">\frac{d}{dt} \mathbf{L_{TOT}} (axis~1) - \frac{d}{dt} \mathbf{L_{TOT}} (axis~1) = 0</div>
<div class="math">\frac{d}{dt} \sum (\mathbf{r_i} - \mathbf{c_1}) \times \mathbf{p_i} - \frac{d}{dt} \sum (\mathbf{r_i} - \mathbf{c_2}) \times \mathbf{p_i} = 0</div>
<div class="math">\frac{d}{dt} \sum (\mathbf{c_2} - \mathbf{c_1}) \times \mathbf{p_i} = 0</div>
<div class="math">(\mathbf{c_2} - \mathbf{c_1}) \times \frac{d}{dt} \sum \mathbf{p_i} = 0</div>
and since we can choose <span class="math">\mathbf{c_2} - \mathbf{c_1}[/spoiler] arbitrarily:
<div class="math">\frac{d}{dt} \sum \mathbf{p_i} = 0</div>

>>2576954
You are correct.

[Anonymous No.2577031]

>>2577021
First equation should have been
<div class="math">\frac{d}{dt} \mathbf{L_{TOT}} (axis~1) - \frac{d}{dt} \mathbf{L_{TOT}} (axis~2) = 0</div>
(second term is angular momentum about the point <span class="math">\mathbf{c_2}[/spoiler])