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I was watching a video of a Prof Feynman Cornell lecture from the 60s. He was explaining angular momentum in terms of area traced out by a mass in orbit, and correctly noted that his explanation held true for a mass travelling in a straight line at constant velocity. It got me thinking about how the conservation of angular momentum must be related closely to the conservation of linear momentum. The challenge is that their units vary - $\rm kg~ m^2/ s$ against $\rm kg~ m/s\;.$

I only have an undergrad degree in Mech Engineering. So my maths is not too bad, but not up to the standard of many reading this question. Anyway...can it be easily shown, somehow that angular momentum translates to linear momentum as R increases to infinity?

Appreciate any help.

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  • $\begingroup$ good question; could you add a link to the video? "can it be easily shown, somehow that angular momentum translates to linear momentum as R increases to infinity" - is this what Feynman said in the lecture? $\endgroup$ – Mozibur Ullah Sep 9 '16 at 6:08
  • $\begingroup$ Hi. If you Google "Feynman Messenger Lectures" you get a Cornell site. It is the third video on the list titled Great Conservation Principles. He does not say those exact words. Also if you Google " relationship of angular momentum to linear momentum" one of the first hits is a Stanford site. Here it says Linear momentum can be considered a renormalised case of angular momentum when r reaches infinity. Or something like that. I can't cut and paste too well from my phone sorry. Thanks everyone for showing interest in my question. $\endgroup$ – Simon Jeffery Sep 11 '16 at 0:57
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Given that conservation of linear momentum is tied to invariance of a system under translations, that conservation of angular momentum is tied to invariance of system under rotations, and that these two symmetries are independent of each other, then I would say no, linear momentum can not be considered a subset of angular momentum.

The special case of a particle moving at constant speed in a straight line is one in which the system is both translationally symmetric and rotationally symmetric, and so both linear and angular momentum of the particle are conserved.

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  • $\begingroup$ It's easy to break translational symmetry while preserving rotational symmetry (about a particular point). But how do you break rotational symmetry while preserving translational symmetry? $\endgroup$ – Rococo Sep 9 '16 at 4:29
  • $\begingroup$ @Rococo. Well that's an interesting question that I'd have to think about. Off the top of my head, thinking classically, I could imagine a spinning disk with equal and opposite forces applied to opposite sides of the disk, preserving linear momentum (due to the zero net outside force) while applying a non-zero torque. Now, this argument isn't at the level of Lagrangian mechanics, and after all these are constraint forces (and maybe even non-holonomic), so I'm not sure how far you can push all of this. I can't remember if you can even talk about symmetry in this kind of situation... $\endgroup$ – march Sep 9 '16 at 4:37
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    $\begingroup$ I think Rococo already knows this, but one possible answer is a flat torus, i.e. a square with opposite edges identified. The square picks out a preferred orientation but there's still translational symmetry. $\endgroup$ – knzhou Sep 9 '16 at 6:20
  • $\begingroup$ @knzhou oh, that's interesting. And you flatter me- I suspected that there was an answer like this, but I've spent almost no time thinking about physics on curved surfaces. $\endgroup$ – Rococo Sep 10 '16 at 22:58

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