Undergrad Axial angular momentum calculation

[cianfa72]

TL;DR

About the definition of axial angular momentum and its calculation using moment of inertia tensor

Hi, I'd ask for clarification about some topics on angular momentum calculation.

Consider a system of particles moving w.r.t. a given inertial frame $\mathcal A$. Picked a point $P$ in $\mathcal A$ one defines the angular momentum vector $\vec L$ w.r.t. the reference point/pole P.

Furthermore one defines the axial angular momentum $L_a$ as the component along $a$ of the system's angular momentum $\vec L$ calculated w.r.t. a pole $P$ on the axis $a$. Such a notion is well defined since it doesn't depend on the particular pole $P$ chosen on the $a$ axis to do the calculation.

Next the notion of system's moment of inertia tensor $I$ is introduced that allows to calculate the axial angular momentum $L_a$ for the system supposed to rotate "as a rigid body" about any given $a$ axis with angular velocity $\omega$.

 

[cianfa72]

Btw, from a mathematical viewpoint, angular momentum is a covector, i.e. a linear map from $\mathbb R^3$ as vector space to the field $\mathbb R$. Therefore the moment of inertia tensor $\mathbf I$ (shorted as inertia tensor) is actually a (0,2)-rank tensor that "eats" a vector (angular velocity $\vec \omega$) and returns a covector (angular momentum $\mathbf L$).

Then when it comes to matrix representation, fixed a basis in $\mathbb R^3$ and the associated dual basis in $\mathbb R^{3*}$, the inertia tensor is represented in such a basis by a symmetric matrix $[{I}]$.

Should the moment of inertia $\mathbf I$ be associated in some way with the system's CoM?

 

[A.T.]

Should the moment of inertia $\mathbf I$ be associated in some way with the system's CoM?

it's often useful to express $\mathbf I$ wrt to the CoM, but you can use any other point.

 

[cianfa72]

it's often useful to express $\mathbf I$ wrt to the CoM, but you can use any other point.

Ok, however I'm confused from the following.

Let's say the origin $O$ of reference frame coincides with the rigid body's CoM. Suppose the body isn't symmetric about the frame $z$-axis.

Then we ask for the angular momentum $\mathbf L$ w.r.t the pole $O$ of the body when rotating with constant angular velocity $\vec \omega$ along the $z$-axis. Such an angular momentum vector $\mathbf L$ (or better covector) isn't directed along $z$ but basically rotates about it.

On the other hand, by definition of inertia tensor $\mathbf I$, we have $$\mathbf L = \mathbf I \vec \omega$$ Does this mean that inertia tensor is (or can be) function of the body's "absolute position" w.r.t. the given reference frame?

 

[A.T.]

Ok, however I'm confused from the following.

I'm not sure If I understand what you are confused about, but this might help:
https://en.wikipedia.org/wiki/Parallel_axis_theorem

 

[TSny]

Ok, however I'm confused from the following.

Let's say the origin $O$ of reference frame coincides with the rigid body's CoM. Suppose the body isn't symmetric about the frame $z$-axis.

Then we ask for the angular momentum $\mathbf L$ w.r.t the pole $O$ of the body when rotating with constant angular velocity $\vec \omega$ along the $z$-axis. Such an angular momentum vector $\mathbf L$ (or better covector) isn't directed along $z$ but basically rotates about it.

On the other hand, by definition of inertia tensor $\mathbf I$, we have $$\mathbf L = \mathbf I \vec \omega$$ Does this mean that inertia tensor is (or can be) function of the body's "absolute position" w.r.t. the given reference frame?

For a given coordinate system, the components $I_{xx}, I_{xy},$ etc. of the inertia tensor depend on the orientation of the rigid body relative to the coordinate axes. So, if the body is rotating such that the mass distribution of the body changes relative to the coordinate axes, at least some of the tensor components will change with time.

However, if you choose axes that are fixed to the body (so the axes rotate with the body), then the components of the inertia tensor relative to these axes will be time independent.

For example, consider a "tilted dumbbell" rotating about the z-axis with constant angular velocity $\boldsymbol{\omega}$ as shown:

$\boldsymbol L$ is the angular momentum of the dumbbell and $\boldsymbol L$ precesses around the z-axis. If the xyz axes are fixed in the lab, then all the inertia tensor components relative to these axes will change with time (except for $I_{zz}$).

But, if we fix the blue coordinate axes to be the dumbbell so that the coordinate system rotates around the z-axis with the dumbbell, all of the inertia tensor components relative to this body-fixed coordinate system will be time independent.

 

[cianfa72]

But, if we fix the blue coordinate axes to be the dumbbell so that the coordinate system rotates around the z-axis with the dumbbell

I don't understand what you mean by "fix the blue coordinates axes to be the dumbbell" ?

What I got is that the "new" coordinate system is supposed to rotate around the z-axis "anchored" with the dumbbell.

all of the inertia tensor components relative to this body-fixed coordinate system will be time independent.

Yes, definitely.

 

[cianfa72]

So, fix a cartesian coordinate system in the lab with origin in the dumbbell's CoM. Use $(\theta,\varphi)$ spherical coordinates to describe the orientation of the dumbbell (system) relative to the coordinate axes. Then the inertia tensor $\mathbf I$ will depend on $(\theta, \varphi)$, hence its matrix representative in the given coordinate system will have entries $I_{i,j}$ function of $(\theta, \varphi)$ as well.

In the end, let me say, the inertia tensor behaves like a (0,2)-rank tensor field since it depends on the system configuration space point (charted for instance by $(\theta, \varphi)$ spherical coordinates).

 

[TSny]

I don't understand what you mean by "fix the blue coordinates axes to be the dumbbell" ?

What I got is that the "new" coordinate system is supposed to rotate around the z-axis "anchored" with the dumbbell.

Yes, it is common in the study of rigid body dynamics to use "body-fixed" axes. These axes rotate with the body. So, the components of the body's inertia tensor relative to these axes do not change with time.

In the example I used, you could choose the blue coordinate system to rotate with the dumbbell as body-fixed axes. Often, however, the orientation of the body fixed axes is chosen to make the off-diagonal components of the inertia tensor equal to zero. This is always possible. For the dumbbell, this occurs if you choose one of the body fixed axes to be along the axis of the dumbbell.

 

[cianfa72]

As general question: does the notion of inertia tensor also extend to non-rigid systems?

 

[ergospherical]

It might make sense to think in those terms if the deformation timescale is quite long compared to the timescale of the dynamics. Maybe, to give an arbitrary example, you are studying the rotational dynamics of a galactic disk, whose density profile is changing slowly with time. Otherwise, you are moving into the realm of continuum or fluid mechanics.

 

[TSny]

As general question: does the notion of inertia tensor also extend to non-rigid systems?

For any system of particles at some instant, you could calculate $I_{xx}, I_{xy},$ etc for some coordinate system. But, I don't know of any use for doing this unless the system is a rigid body. I'm only familiar with using the inertia tensor for rigid bodies.

 

[cianfa72]

For any system of particles at some instant, you could calculate $I_{xx}, I_{xy},$ etc for some coordinate system.

Ok yes, one can calculate such matrix entries at some time $t$ in a given coordinate system using the definition of inertia tensor.

However it is useless since it is no longer true that the angular momentum $\mathbf L$ w.r.t. the system's CoM of the system rotating with angular velocity $\vec \omega$ about the CoM at time $t$ is given by $$\mathbf L = \mathbf I \vec \omega$$

But, I don't know of any use for doing this unless the system is a rigid body. I'm only familiar with using the inertia tensor for rigid bodies.

Ok, definitely.

 

[wrobel]

from a mathematical viewpoint, angular momentum is a covector

That's wrong. Angular momentum is a vector. It is a pseudovector to be precise. Just by definition: $\sum_k m_k\boldsymbol r_k\times\boldsymbol {\dot r}_k$.

 

[cianfa72]

That's wrong. Angular momentum is a vector. It is a pseudovector to be precise. Just by definition: $\sum_k m_k\boldsymbol r_k\times\boldsymbol {\dot r}_k$.

I was taking the more advanced viewpoint in which the momentum $\boldsymbol p$ is actually a co-vector (see for instance MTW chapter 2.5).

 

[weirdoguy]

I was taking the more advanced viewpoint in which the momentum p is actually a co-vector

It is a covector, but that does not mean that angular momentum also is.

 

[cianfa72]

Better, I'd say the angular momentum is a bi-vector -- https://en.wikipedia.org/wiki/Angular_momentum

 

[wrobel]

It is a covector,

It is a covector in the Lagrangian formalism but not in this context. Yes $\frac{\partial L}{\partial \dot x}$ is a covector on the configuration manifold. In the Newtonian formalism the mass is a scalar; the velocity is a vector thus $\boldsymbol p=m\boldsymbol v$ is a vector in $\mathbb{R}^3$

 

[Sagittarius A-Star]

Better, I'd say the angular momentum is a bi-vector -- https://en.wikipedia.org/wiki/Angular_momentum

Yes. In this case, the cross product must be replaced by the wedge product.
$\mathbf L = \mathbf r \wedge \mathbf p$

 

[A.T.]

Yes. In this case, the cross product must be replaced by the wedge product.
$\mathbf L = \mathbf r \wedge \mathbf p$

Using the wedge product (bi-vector) instead of the cross product (pseudo-vector) avoids the issues pseudo-vectors have under mirror transformations (or system handedness changes). Cross-products introduce asymmetry into physically symerical situations, but are easier to represent and visualize than wedge products.

Here a previous thread with a nice video on this:

Undergrad Post in thread 'How deep does it need to go for a physics theory to be consistent?'

Mar 15, 2025

I would rather prefer answer on the metter...

You asked what the point of pseudovectors is. I gave you a practical example from physics, where it matters if something is a vector or pseudovector.

You can find more examples here:
https://en.wikipedia.org/wiki/Pseudovector

Here, when you mirror $\vec{I}$, then $\vec{B}$ is not just mirrored, but mirrored and negated. So $\vec{I}$ and $\vec{B}$ behave differently under a mirror transformation.

This anti-symmetric behavior might seem somewhat weird. But it becomes clearer, when you consider that in physics...

• A.T.

• 

 

[wrobel]

Using the wedge product (bi-vector) instead of the cross product (pseudo-vector) avoids the issues pseudo-vectors

Please bring a concrete example of a problem from classical mechanics (we are discussing classical mechanics in this thread are not we?) that provides the issues you are speaking about.

 

[A.T.]

Please bring a concrete example of a problem from classical mechanics (we are discussing classical mechanics in this thread are not we?) that provides the issues you are speaking about.

https://en.wikipedia.org/wiki/Pseudovector

 

[wrobel]

Thanks a lot.

https://en.wikipedia.org/wiki/Pseudovector

citation from this wiki article:

certain discontinuous rigid transformations such as reflections

A discontinuous linear operator in the finite dimensional space. That's nice indeed.
Besides this ignorance I do not see any issues with use of pseudo tensors.
The example from the article with angular momentum is irrelevant as well. Both sides of the equation $\boldsymbol {\dot L}=\boldsymbol\tau$ contain pseudo tensors of the same type so that the equation is invariant.

Moreover all the formulas are invariant. For example the distribution of velocities in a rigid body is
$$\boldsymbol v_B=\boldsymbol v_A+\boldsymbol\omega\times \boldsymbol{AB}.$$
Here the expression $\boldsymbol\omega\times \boldsymbol{AB}$ is a vector not a pseudo vector.

UPD:

Using the wedge product (bi-vector) instead of the cross product (pseudo-vector) avoids the issues pseudo-vectors have under mirror transformations (or system handedness changes). Cross-products introduce asymmetry into physically symerical situations,

No they do not. Because if it were so, then mechanical equations that contain cross-products would be incorrect in physically symmetrical situations, but it is not so.

 

[cianfa72]

As far as I can tell, the notion of pseudo-vector arises from its definition as cross product of two vectors. We all know how components of a "normal" vector change w.r.t. a change of basis. Namely, when the basis vectors flip, the vector's components flip their sign as well. This assume the passive transformation point of view: the vector itself stays fixed, what changes are the vector's components.

The key point is how the angular momentum is actually defined. In principle, upon a change of basis, one must recompute the components of the vectors entering the cross product and recompute it. What does one discover ? That upon a flip of the basis vectors the components of the cross product (calculated in that basis) do not flip their sign.

In this sense the angular momentum -- by its very definition -- is actually a pseudo-vector. Then one can take the more general/powerful definition as bi-vector by using the wedge/external product.

 

[wrobel]

Then one can take the more general/powerful definition as bi-vector by using the wedge/external product.

OK, solve the following problem without pseudovectors and show me a gain from the bi-vector's language usage.

A coin of radius $r$ rolls about a cone without slipping. At a moment $t=t_0$ the value of acceleration of the coin's lowest point $A$ is given: $a=|\boldsymbol a_A|$. Find the velocity of the point that lies on the coin's rim at the horizontal diameter.

 

[cianfa72]

OK, solve the following problem without pseudovectors and show me a gain from the bi-vector's language usage.

In the context of Newtonian physics I believe there is no gain from the bi-vector language/definition.

Suppose to extend to an euclidean space of dimension 4 or higher. In that case, the pseudo-vector definition can no longer be used, and the bi-vector definition becomes mandatory.

 

[A.T.]

Cross-products introduce asymmetry into physically symmetrical situations ...

... if it were so, then mechanical equations that contain cross-products would be incorrect in physically symmetrical situations...

Maybe I should have been more clear here: Cross-products introduce asymmetry into the analysis of physically symmetrical situations. The asymmetry introduced by the cross-product is on purely conventional (as in: per convention) level, as I already explained in the previous thread that I linked above:

It's also important to note, that the actual observable physics is perfectly symmetric between the two mirrored cases above. If you shoot electrons with symmetric initial velocities through each $\vec{B}$ field , their deflections and thus paths will be symmetric. It's only that not directly observable quantity $\vec{B}$, whose mathematical representation was chosen by a non-symmetric convention, that unsurprisingly breaks the symmetry. But only in our diagrams and calculations.

... show me a gain from the bi-vector's language usage....

If there was a net gain, we would be using the wedge-product instead of the cross-product. The only gain is conceptual, in avoiding asymmetric vectors in analyses of physically symmetric situations and avoiding arbitrary rules like the right-hand-rule. But using the cross-product has also its positives, as I've also already explained:

... it's simpler to visualize/draw arrows, than parallelograms with orientation arrows around their boundary (see diagram below). You can also apply the math you already know from other vector quantities,

 

[cianfa72]

Maybe I should have been more clear here: Cross-products introduce asymmetry into the analysis of physically symmetrical situations. The asymmetry introduced by the cross-product is on purely conventional (as in: per convention) level, as I already explained in the previous thread that I linked above:

Taking the active transformation viewpoint, suppose it sends the basis vectors to their flipped versions. In the the given basis the active transformation is given by the following matrix $A = -I$ where $I$ is the 3x3 Identity matrix.

Now take the vectors $\vec v$ and $\vec p$ and compute the cross product of their (active) transformed versions $\vec v' = A \vec v$ and $\vec p' = A \vec p$, i.e. $Av \times Ap = \vec v \times \vec p$.

As you can see this is not the transformed version of the cross product $\vec v \times \vec p$ understood as a "normal" vector, since the latter is $- (\vec v \times \vec p)$.

 

[wrobel]

Suppose to extend to an euclidean space of dimension 4 or higher. In that case, the pseudo-vector definition can no longer be used

Definition of what? If you want to say that pseudovectors arise only as the cross product in 3D then that is a mistake: there is a broad theory of pseudotensors in arbitrary manifolds. Study the Hodge star operation, study what the tensor density is. In particular, there is no problem to generalize the concept of a cross product to any Riemannian manifold of odd dimension.

 

[cianfa72]

Definition of what? If you want to say that pseudovectors arise only as the cross product in 3D then that is a mistake: there is a broad theory of pseudotensors in arbitrary manifolds.

Yes, of course you can define a pseudovector also in 4D or higher dimensions. The point is, however, that it is no longer true that exists a pseudovector that is the dual of a given bivector (just to begin with the dimensions no longer match).