I am confused by two different definitions of the angular momentum of a particle P about a moving point Q, with point O as the origin of the inertial frame. I learned the first definition from my sophomore dynamics course, and now I am taking analytical dynamics and the textbook gave a different definition.

The first definition looks like this: enter image description here

$$ \vec{h}_{P/Q} = \vec{r}_{P/Q} \times m\vec{v}_{P/O} $$

Where h_P/Q is the angular momentum of particle P w.r.t point Q; r_P/Q is the relative position of P w.r.t Q, v_P/O is the inertial velocity of particle P w.r.t a fixed point O.

The second definition looks like this: enter image description here

$$ \vec{h}_{P/Q} = \vec{r}_{P/Q} \times m\vec{v}_{P/Q} $$

Basically the velocities are taken w.r.t different points.

Which one is the correct one?

  • $\begingroup$ Your equations are formatted in a way that's hard to understand or recognize. Can you add some explanation in words? $\endgroup$
    – RC_23
    May 25 at 20:29
  • $\begingroup$ What is "velocity with respect to a fixed point"? Velocity is independent of a reference point (assuming we do not change frame) $\endgroup$
    – DanDan0101
    May 25 at 20:39
  • $\begingroup$ @DanDan0101 velocity w.r.t a fixed point is the velocity relative to point O, which is the origin point of the inertial frame. In the second equation, velocity relative to the moving point Q is used, but it is still the velocity as seen from the inertial frame. $\endgroup$
    – Eric CZH
    May 25 at 20:42
  • $\begingroup$ I edited your question to write the equations using MathJax/Latex that is used on this site. It is pretty easy to learn. Feel free to make a change if it does not reflect your intent. $\endgroup$
    – RC_23
    May 25 at 20:44
  • $\begingroup$ @RC_23 Thank you for helping me with the formatting! $\endgroup$
    – Eric CZH
    May 25 at 20:46

2 Answers 2


If the angular momentum is calculated with respect to Q, both the position vector and the velocity vector must related to Q. So, the second expression is correct.

If the momentum is calculated with respect to the origin of the inertial frame: $\mathbf L = \mathbf r_{OP} \times \mathbf v_{OP} = (\mathbf r_{OQ} + \mathbf r_{QP}) \times \mathbf v_{OP} $

  • $\begingroup$ Thank you for clarifying this for me! $\endgroup$
    – Eric CZH
    May 25 at 20:46

Oh I think it is just a matter of conventions. And it is more commonly accepted to take the velocity relative to point O, as I just saw from an MIT course on engineering dynamics and my sophomore textbook on dynamics. Taking the velocity relative to Q would give a different set of equations; but there should not be much problems if one stays consistent.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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