# Conservation of angular momentum and spin

I have a point mass particle r from a axis of rotation however the molecule is two atoms spinning about it's own center of mass If I wanted to calculate total angular momentum of my system should I not include the spin momentum as well

While I understand that they cancel to zero can't they interact with translational momentum to alter the balance?

• The orbital angular momentum(about origin) has the direction $r$x$mv$ out of the screen. The spin angular momentum has direction into the screen from the diagram. They subtract. In effect, the total angular momentum is given by the angular momentum of each atom about their centre of mass + the angular momentum of centre of mass about the origin(vector addition of course) – Prasad Mani Oct 9 '16 at 10:08
• @CountTo10 Yes your correct but if the rotation was counter clockwise the RH rule would make them additive ? Is that why these degrees of freedom aren't counted- if we were to look at a many particle system this motion is completely random. So there will be as many clockwise as c-clockwise rotations- on average they cancel out . but cant they still interact with translational momentum ? – Tom Chester Oct 9 '16 at 10:13
• Internal degrees of freedom are counted. i dont know what your question exactly is. but if it is a rigid body, then such pairs of atoms cannot exist – Prasad Mani Oct 9 '16 at 10:27
• @PrasadMani yes I wanted to know if these internal degrees of freedom can interact with translational ? As I could see such interactions affecting the individual balance of translational and internal while conserving the momenta of the system as a whole . – Tom Chester Oct 9 '16 at 10:37

In any system one can compute the orbital angular momentum by computing $$\mathbf{L} = \sum_n \mathbf{r}_n \times \mathbf{p}_n$$ for all the particles/point masses in the system. Here $\mathbf{p}_n$ represents their momentum vectors and $\mathbf{r}_n$ are their position vectors relative to some common origin. This means that in general the orbital angular momentum depends on the definition of the origin.
Spin is associated with an internal degree of freedom of the particles. For fermions, for instance, one can have $\mathbf{S}=\pm \hbar/2 \hat{z}$ (assuming we measure it along the $z$-direction). It is important to note that spin does not correspond to a physical rotation of the particle or system of particles. It is an intrinsic property of the particles.
The total angular momentum is now the sum of the orbital angular momentum and the spin angular momentum $$\mathbf{J} = \mathbf{L} + \mathbf{S} .$$ Conservation of angular momentum applies to the total angular momentum and not to the orbital angular momentum or the spin angular momentum separately. Hope this clarifies the matter and answers you question.