Is angular momentum taken about a point or an axis?

[etotheipi]

One part of König's theorem states that $\vec{L} = \vec{L}_{\text{COM}} + \vec{L}^{'}$. The term $\vec{L}^{'}$ simply refers to the angular momentum wrt. the centre of mass. This is just a point, and doesn't have an axis implicitly associated with it (we have infinitely many choices!).

The most general definition of angular momentum of a rigid body, $$\vec{L} = \sum_{i} m_{i} \vec{r}_{i} \times \vec{v}_{i}$$ likewise doesn't require an implicit axis in order for it to be computed, only an origin point.

I wondered then whether we take angular momentum about a point, or an axis? The same could be said about the moment of inertia. We often speak of the moment of inertia about an axis, however why can't we just speak about the moment of inertia about a point?

 

[PeroK]

One part of König's theorem states that $\vec{L} = \vec{L}_{\text{COM}} + \vec{L}^{'}$. The term $\vec{L}^{'}$ simply refers to the angular momentum wrt. the centre of mass. This is just a point, and doesn't have an axis implicitly associated with it (we have infinitely many choices!).

The most general definition of angular momentum of a rigid body, $$\vec{L} = \sum_{i} m_{i} \vec{r}_{i} \times \vec{v}_{i}$$ likewise doesn't require an implicit axis in order for it to be computed, only an origin point.

I wondered then whether we take angular momentum about a point, or an axis? The same could be said about the moment of inertia. We often speak of the moment of inertia about an axis, however why can't we just speak about the moment of inertia about a point?

You perhaps should be able to work some of this out for yourself. AM is a vector (pseudo-vector to be precise), is defined about a point and has three components.

If a rigid body is rotating there is an axis of rotation (why?). The AM about any point on this axis is the same (why?), so you can talk about the AM about that axis.

The moment of inertia is actually a tensor. See, for example:

http://farside.ph.utexas.edu/teaching/336k/Newtonhtml/node64.html

 

[etotheipi]

The AM about any point on this axis is the same (why?), so you can talk about the AM about that axis.

If I spin a ball anticlockwise on a massless string in a horizontal circle, then the ball has some angular momentum computed from the point where my hand is. It should point straight upward.

I define the axis of rotation to be perpendicular to the plane of the rotations and passing through my hand.

Then I choose a point 1 metre above my hand, on the axis of rotation, and try to compute $\vec{L}$ again. Except now the position vector from my newly chosen origin is angled downward, whilst the velocity vector is still in the plane of the rotations. So wouldn't the angular momentum vector point off in a slightly different direction?

Apologies if I'm being daft, I'm slightly out of it at the moment!

 

[PeroK]

If I spin a ball anticlockwise on a massless string in a horizontal circle, then the ball has some angular momentum computed from the point where my hand is. It should point straight upward.

I define the axis of rotation to be perpendicular to the plane of the rotations and passing through my hand.

Then I choose a point 1 metre above my hand, on the axis of rotation, and try to compute $\vec{L}$ again. Except now the position vector from my newly chosen origin is angled downward, whilst the velocity vector is still in the plane of the rotations. So wouldn't the angular momentum vector point off in a slightly different direction?

Apologies if I'm being daft, I'm slightly out of it at the moment!

I don't understand this. Without loss of generality we can take the axis of rotation to be the x-axis. This tells you something about the motion of any point in the body. Then you use the properties of the cross product.

 

[etotheipi]

I don't understand this. Without loss of generality we can take the axis of rotation to be the x-axis. This tells you something about the motion of any point in the body. Then you use the properties of the cross product.

I tried drawing a diagram; if we compute $\vec{L}$ from different points along the x-axis, the magnitude is invariant but the direction appears to change, since the direction of $\hat{n}$ from the cross product changes if the position vector $\vec{r}$ from different points along the same axis also varies.

 

[PeroK]

I tried drawing a diagram; if we compute $\vec{L}$ from different points along the x-axis, the magnitude is invariant but the direction appears to change, since the direction of $\hat{n}$ from the cross product changes if the position vector $\vec{r}$ from different points along the same axis also varies.

Do the calculation of the cross product about the origin. Point at $\vec r = (x, y, z)$, $\vec v = (0, v_y, v_z)$.

 

[etotheipi]

Do the calculation of the cross product about the origin. Point at $\vec r = (x, y, z)$, $\vec v = (0, v_y, v_z)$.

$\vec{L} = m\begin{pmatrix}x\\y\\z\end{pmatrix}\times\begin{pmatrix}0\\v_y\\v_z\end{pmatrix} = \begin{pmatrix}yv_z - zv_y\\-xv_z\\xv_y\end{pmatrix}$

If we choose a point along the axis at the centre of rotation (i.e. $x=0$) then $\vec{L}$ only has an $x$ component, as expected. Though not for non-zero $x$.

Though it seems then the magnitude is also not invariant!

 

[PeroK]

$\vec{L} = m\begin{pmatrix}x\\y\\z\end{pmatrix}\times\begin{pmatrix}0\\v_y\\v_z\end{pmatrix} = \begin{pmatrix}yv_z - zv_y\\-xv_z\\xv_y\end{pmatrix}$

If we choose a point along the axis at the centre of rotation (i.e. $x=0$) then $\vec{L}$ only has an $x$ component, as expected. Though not for non-zero $x$.

Though it seems then the magnitude is also not invariant!

That's a good point. The AM in the direction of the axis of rotation is constant, but there are other components.

 

[kuruman]

Are you asking why $\vec r \times \vec p$ changes if you change $\vec r$? That is certainly the case for a point mass. For a non-rigid body, e.g. a gas, you can always calculate $\sum\vec r_i \times \vec p_i$ and call that the angular momentum of the gas. To calculate the angular momentum about an axis, you need an axis. For a rigid body (fixed relative distances) it is the locus of all points of mass $dm_i$ that are instantaneously at rest relative to each other while the remaining rotate in a circle about them. Calculating angular momentum about an axis makes sense for a rigid body because then you can use the idea of the moment of inertia tensor.

 

[etotheipi]

Are you asking why $\vec r \times \vec p$ changes if you change $\vec r$? That is certainly the case for a point mass. For a non-rigid body, e.g. a gas, you can always calculate $\sum\vec r_i \times \vec p_i$ and call that the angular momentum of the gas. To calculate the angular momentum about an axis, you need an axis. For a rigid body (fixed relative distances) it is the locus of all points of mass $dm_i$ that are instantaneously at rest relative to each other while the remaining rotate in a circle about them. Calculating angular momentum about an axis makes sense for a rigid body because then you can use the idea of the moment of inertia tensor.

The thing I'm struggling to understand is that I have only seen angular momentum mentioned relative to an axis (perhaps I haven't been looking hard enough before!).

However, the angular momentum appears to differ if we measure it at different points along the axis. So surely then there's no such thing as angular momentum about an axis, only about points?

 

[kuruman]

The thing I'm struggling to understand is that I have only seen angular momentum mentioned relative to an axis (perhaps I haven't been looking hard enough before!).

The definition $\vec L=\vec r \times \vec p$ shows dependence on the position vector $\vec r$ which depends on the choice of origin, a single point, not an axis with many points on it. You must have seen this definition before seeing $\vec L=I\vec\omega$ which involves an axis and requires a rigid body in order to be derived from the definition. It's in most intro physics textbooks.

 

[PeroK]

The thing I'm struggling to understand is that I have only seen angular momentum mentioned relative to an axis (perhaps I haven't been looking hard enough before!).

However, the angular momentum appears to differ if we measure it at different points along the axis. So surely then there's no such thing as angular momentum about an axis, only about points?

This is a good point. Suppose we have a rigid body rotating about a fixed axis of rotation. In that case, the component of AM in the direction of that axis is constant along that axis, whatever point is chosen. In the above example $L_x$ is constant for all points on the x-axis. Then, we can describe this as "the AM about that axis".

But, as you've shown, the other components of the AM are generally non-zero.

I must confess, I'd never noticed that subtlety before.

 

[wrobel]

Angular momentum about a point is a primary object. If an axis $\alpha$ passes through a point $O$ along a unit vector $\boldsymbol e$ then the momentum about this axis by definition is $L_\alpha:=(\boldsymbol L_O,\boldsymbol e)$. That is $L_\alpha$ is a projection of $\boldsymbol L_O$ on $\alpha$.

It is important: prove that $L_\alpha$ does not depend on what a point $O$ on the axis is chosen.

 

[etotheipi]

Angular momentum about a point is a primary object. If an axis $\alpha$ passes through a point $O$ along a unit vector $\boldsymbol e$ then the momentum about this axis by definition is $L_\alpha:=(\boldsymbol L_O,\boldsymbol e)$. That is $L_\alpha$ is a projection of $\boldsymbol L_O$ on $\alpha$.

It is important: prove that $L_\alpha$ does not depend on what a point $O$ on the axis is chosen.

Does the cross product we computed earlier prove the second part? The projection onto $\hat{x}$, $yv_z - zv_y$, is indeed independent of $x$.

 

[PeroK]

The thing I'm struggling to understand is that I have only seen angular momentum mentioned relative to an axis (perhaps I haven't been looking hard enough before!).

However, the angular momentum appears to differ if we measure it at different points along the axis. So surely then there's no such thing as angular momentum about an axis, only about points?

One more point. This is another example of where AM for motion of particles in a plane is relatively straightforward, but the full 3D AM of a rigid body is more complex.

In addition to centre of mass, you have concepts such as the radius of gyration and centre of percussion. I must admit I haven't looked at these in years, but I probably had the same sort of questions as you have when I first studied this.

As a textbook, I like Kleppner and Kolenkow, if you want to study this systematically. It's a good next step up from A-level material. I just like their style. The problems are algebraic and conceptual, rather than testing your calculator skills. The opposite of plug 'n' chug.

 

[etotheipi]

As a textbook, I like Kleppner and Kolenkow, if you want to study this systematically. It's a good next step up from A-level material. I just like their style. The problems are algebraic and conceptual, rather than testing your calculator skills. The opposite of plug 'n' chug.

Thanks for the suggestion, I'll look into how I might be able to get hold of a copy!

 

[wrobel]

Does the cross product we computed earlier prove the second part? The projection onto ^x\hat{x}, yvz−zvyyv_z - zv_y, is indeed independent of xx.

yes that's right

 

[vanhees71]

The angular momentum $\vec{L}$ is always defined referring to some fixed point in space, which you need to choose to define an inertial reference frame (which consists of the choice $O$ of this point and a Cartesian right-handed basis such that any point $P$ in space is defined by the position vectors $\overrightarrow{OP}=\vec{r}$). Then the angular momentum of an arbitrary set of point particles is defined as
$$\vec{L}=\sum_{i} m_i \vec{r}_i \times \dot{\vec{r}}_i=\sum_i \vec{r}_i \times \vec{p}_i.$$
Of course $\vec{L}$ changes under translations of the origin. Let the new origin $O'$ be given by $\vec{a}=\overrightarrow{OO'}$. Then $\vec{r}'=\overrightarrow{O'P} = \overrightarrow{O' O}+\overrightarrow{OP}=\vec{r}-\vec{a}$ and thus
$$\vec{L}'=\sum_i \vec{r}_i' \times \vec{p}_i' = \sum_i (\vec{r}_i-\vec{a}) \otimes \vec{p}_i = \vec{L}-\vec{a} \times \vec{P},$$
where $\vec{P}$ is the total momentum.