Is it possible to write the relative angular momentum of three particles as a vector product of single position vector and single position momentum? I think that relative angular momentum can be defined as total angular momentum minus centre of mass angular momentum. To simplify let's assume equal masses. For instance for two particles system the position vector is simply relative coordinate and the momentum vector is the conjugate of the relative coordinate:

$$ \mathbf L_r = \mathbf L_{tot} - \mathbf L_{cm}=\mathbf r_1 \times \mathbf p_1+\mathbf r_2 \times\mathbf p_2 - \frac{1}{2}(\mathbf r_1 + \mathbf r_2) \times(\mathbf p_1+\mathbf p_2)= \frac{1}{2}( \mathbf r_1 -\mathbf r_2) \times (\mathbf p_1 - \mathbf p_2) $$

I originally came up to the question thinking about QM but a classical argument shouldn't be much different.

  • $\begingroup$ What do you find for three particles? $\endgroup$ Apr 16, 2023 at 0:10
  • $\begingroup$ You know how to express $\vec L_\text{tot}$ and $\vec L_\text{cm}$ in terms of the positions and momenta, consequently you can write $\vec L_\text{r}=\vec L_\text{tot}-\vec L_\text{cm}$ in terms of the positions and momenta, for any number of particles. I'm not sure what kind of answer you are looking for, can you clarify? $\endgroup$
    – Puk
    Apr 16, 2023 at 3:14
  • $\begingroup$ @Puk I want to write it as a product of a single position vector times a single momentum vector, like in last parte of the formula for two particles $\endgroup$
    – Mattia
    Apr 16, 2023 at 9:14
  • $\begingroup$ @CosmasZachos What do you mean? $\endgroup$
    – Mattia
    Apr 16, 2023 at 9:15
  • $\begingroup$ You wrote a formula for two particles. Did you apply it to three? $\endgroup$ Apr 16, 2023 at 11:46

1 Answer 1


You got your own answer in your comment: the relative coordinate part only simplifies for two components. Expand all phase-space variables into an average piece and an excursion from it, $$ \bar {\mathbf r}\equiv {1\over n}\sum_{i=1}^n {\mathbf r_i}, \qquad \leadsto \qquad {\mathbf r_i} \equiv \bar {\mathbf r} + \tilde {\mathbf r_i} \implies \sum_{i=1}^n \tilde {\mathbf r_i}=0, \\ \bar {\mathbf p}\equiv {1\over n}\sum_{i=1}^n {\mathbf p_i}, \qquad \leadsto \qquad {\mathbf p_i} \equiv \bar {\mathbf p} + \tilde {\mathbf r_i} \implies \sum_{i=1}^n \tilde {\mathbf p_i}=0. $$ so that $$ \mathbf L = \sum_{i=1}^n \mathbf r_i \times \mathbf p_i= {n}~\bar {\mathbf r} \times \bar {\mathbf p}+ \sum_{i=1}^n \tilde {\mathbf r_i} \times \tilde {\mathbf p_i}, $$ a celebrated expression: no cross terms. Both terms in the sum of the second, excursion, term are identical, so they collapse to your simple product; but from 3 on, they do not; why should they? There are more independent relative/excursion coordinates.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.