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When defining angular momentum or rather calculating angular momentum what is the difference in the use of the terms "with respect to" , "about a point" or "in the frame of" ?

Are the angular momentum about the centre of mass and in the frame of the centre of mass two different physical quantities?

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  • $\begingroup$ They're quite equivalent, yes. $\endgroup$ – FGSUZ Jan 11 at 21:27
  • $\begingroup$ Then what does this statement signify ; the angular momentum of a body about any point in the centre of mass frame same. $\endgroup$ – Harsh Somani Jan 11 at 21:33
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You recognize that angular momentum is a vector which varies by location. Quite similar to velocity which differs by location. But also just like all other vectors, the orientation of the reference frame changes the values of the vector (components).

Here is my interpretation of the following expressions:

  • "with respect to" => The is the location where angular momentum is measured about.
  • "about a point" => Same as above, but more common. "Angular momentum about the center of mass" or "Angular momentum about the pivot" are pretty clear of what it means. If you split the body into many infinitesimal particles, each located at $\boldsymbol{r}_i$ with respect to the reference point, then the angular momentum at the reference point is $$ \boldsymbol{L} = \sum_i \boldsymbol{r}_i \times ( m_i \boldsymbol{v}_i ) = \mathrm{I}_{com} \boldsymbol{\omega} + \boldsymbol{r}_{com} \times m \boldsymbol{v}_{com} $$
  • "in the frame of" => in the coordinate frame of. This designates the orientation of the components of the vector. "In the frame of the center of mass" is a little ambiguous because the COM is a point which has no orientation. But "In the frame of body ..." is clear that angular momentum is described in a local coordinate system. When this is omitted the world coordinate system is assumed.
  • "of" => This should tell of us which body angular momentum we are dealing with.
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  • $\begingroup$ So I get it about a point and in the frame of are two different terms but why so while writing the torque equality about an accelerated point I have to take care for a pseudo force acting on the system. I mean I am writing it in the ground frame about an accelerated point so why is there a pseudo force involved $\endgroup$ – Harsh Somani Jan 12 at 9:15
  • $\begingroup$ @HarshSomani - the equations of motion in an inertial frame should not involve pseudo forces. Do you have a reference to what you are talking about? $\endgroup$ – ja72 Jan 12 at 18:27
  • $\begingroup$ for instance a rod kept free on a friction less table is hit on one of its ends by a particle travelling perpendicular to it. Can I conserve the angular momentum of the system (the particle and the rod) about any point on the rod without considering a pseudo force? $\endgroup$ – Harsh Somani Jan 13 at 10:21
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The angular momentum in a given inertial reference frame is not a unique quantity but is given with respect to some origin $O$:

$\vec{L}_O = \vec{r} \times \vec{p}$

There are two different ways of describing the situation which affect the value of $\vec{L}_O$: we can change our arbitrary origin, or we can 'boost' to another inertial frame which is going at constant speed $\vec{u}_b$ with respect to the lab frame. Each of these has a different effect:

Changing Origin

Changing origin shifts every position vector by a constant $\vec{r}_O$ but leaves all momenta unchanged. Each particle's angular momentum changes by $$\vec{L}_O \mapsto \vec{L} = \vec{L}_O + \vec{r}_O\times\vec{p}$$ If you are in the centre of mass frame where total momentum vanishes ($\vec{P}=\sum \vec{p}_i = 0$) then these contributions cancel and the angular momentum is the same about any origin.

Boosts

At least at the instant of the boost, the positions are left unchanged and the momenta all shifted by a constant $m\vec{v}_b$. Each angular momentum changes like so:

$$ \vec{L} \mapsto \vec{L}' = \vec{L} + \vec{r}\times m\vec{v}_b$$

like for the change of origin these variations do not typically cancel but we can find a condition for them to. We want the change in $L$ to be zero:

$$\Delta L = \left(\sum_i m_i\vec{r}_i\right)\times \vec{v}_b = 0 \implies \sum_i m_i\vec{r}_i = 0$$

the last condition is that the centre of mass is chosen as the origin.

Conclusion

When talking about the angular momentum of a body or system you need to give your choice of origin and a choice of inertial frame. There are simple equations for transforming $\vec{L}$ when you change either one. The phrase "with respect to" or "about the point" almost always refer to the choice of origin whereas anything using the word "frame" may be safely assumed to be talking about boosts. The angular momentum is possibly best looked at in the "zero momentum frame" which is the only frame where the choice of origin is not important, and if one chooses to put the origin at the centre of mass of the system then the choice of inertial reference frame also ceases to affect the value.

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