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I'm confused on the concept of angular momentum a little bit. I understand that it is a vector which is equal to the cross product of the position and the linear momentum. But in the case of a two body system (e.g. star and planet or classical hydrogen atom) where both bodies orbit around the centre of mass of the system. My question is, in this case, is the angular momentum a property of the whole system or a property of each body?

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  • $\begingroup$ An angular momentum can be assigned to each of these bodies. But the values depends on the frame of reference, since it's comprised of momentum and position. Adding these in a vectorial way then gives you the total angular momentum of the system. Note that there can be even an angular momentum for free particles if one chooses a frame of reference, whose origin doesn't coincide at any point with the trajectory of this point mass. The crucial thing is the conservation of angular momentum in every frame. $\endgroup$ – DomDoe Apr 25 '18 at 10:11
  • $\begingroup$ It depends on what you want to do in the math. So I would say that the answer is 'yes.' If I randomly want $x$ to be 5, I can say $x = 5$. It's just a matter of what you want to define (and what you define is based on what you think is important) $\endgroup$ – DWade64 Apr 26 '18 at 12:10
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It is a property of each body and a property of the whole system, because it is an additive property: the kind of property whose total comes from the sum of the parts – like internal energy, kinetic energy, momentum, mass; and unlike temperature, internal pressure, stress.

In classical mechanics, rotational momentum is an additive property of any distribution of mass. Suppose you have mass within a volume element $\mathrm{d}\tau$, with density $\rho$. To define the rotational momentum of this mass element you need to (1) choose a frame of reference, so that you can meaningfully say what the position $\boldsymbol{r}$ and the velocity $\boldsymbol{v}$ of that mass is, and therefore calculate its momentum $\rho\boldsymbol{v}$; (2) choose a point $\boldsymbol{o}$ (at rest in that reference frame) as origin.

Then the rotational momentum $\boldsymbol{l}$ of that mass element in that reference frame with respect to that origin is $$\boldsymbol{l} := (\boldsymbol{r}- \boldsymbol{o}) \land \boldsymbol{v}\rho.$$

To say that rotational momentum is additive, it means that if you have a distribution of mass $\rho(\boldsymbol{r})$ within a volume $V$, the total rotational momentum is the integral of the rotational momenta of the mass elements: $$\boldsymbol{L}_V := \int_V \boldsymbol{l}(\boldsymbol{r})\, \mathrm{d}\tau = \int_V (\boldsymbol{r}- \boldsymbol{o}) \land \boldsymbol{v}(\boldsymbol{r})\,\rho(\boldsymbol{r})\,\mathrm{d}\tau.$$

When you apply this formula to two or more point masses, it says that each mass has its own rotational momentum, and the two masses together also have a total rotational momentum, which is the sum of those of the two masses. So it is a property of the whole system and of each body.

If you're asking "where" the total rotational momentum resides, or "to which point it is attached", the answer is that it isn't really attached to any point; although it can be useful to think of it as attached to the centre of mass of the system. This fact is deeply connected with the Euclidean geometry of space in classical mechanics. In fact, the property of being additive makes no sense for vectors in general relativity: there, it's a problem to sum vectors located at different points, because you need to "move" them, keeping each parallel to itself, to a single point first. But space-time is not Euclidean in general relativity, and the result depends on the trajectories along which you "move" the vectors. So in general relativity it's less meaningful to speak of the total rotational momentum of the whole system (although you can use alternative concepts in special circumstances).

A thorough discussion is given for example in:

  • C. A. Truesdell: A First Course in Rational Continuum Mechanics. Vol. 1: General Concepts (2nd ed., Academic Press 1991), § I.8,

and, accompanied by historical remarks, in:

  • C. A. Truesdell: Essays in the History of Mechanics (Springer 1968), ch. V: Whence the Law of Moment of Momentum?.
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