# Understanding the meaning of the directions of $\vec\omega$ and $\vec{L}$

I'm struggling with some basic intuition regarding the angular velocity $$\vec\omega$$ and angular momentum $$\vec{L}$$ vectors, for any arbitrary motion. Specifically, I'm trying to figure out what the idea is behind their directions. (The idea behind the magnitudes is more clear).

First I'll explain what (I think) I know, and please correct any mistakes.

We know that angular momentum is defined as $$\vec{L} = m\vec{r}×\vec{v}$$

And angular velocity is defined as $$\vec\omega = (\vec{r}\times\vec{v})/r^2$$

Regarding angular momentum: by the cross product, I understand that the angular momentum is normal to the plane spanned by $$\vec{r}$$ and $$\vec{v}$$. So my question is, what would be the intuition about the direction of this vector? Does it matter for good understanding, or is it kind of an arbitrary artifact of the cross-product?

About angular velocity: this one really gets me. So again, it's clear the angular velocity is perpendicular to both the position and the velocity vectors, and thus a normal to the plane spanned by them.

However - again, please explain the intuition for the direction of the angular velocity.

For circular motion on the xy plane, where the origin is the center of the circle, I understand. $$\vec\omega$$ will be parallel to the axis of rotation, which the particle rotates around.

However, this understanding seems to break down (?) for other cases. For example, in the case of circular motion on the xy plane - but choosing the origin not on the plane. Maybe in the center of the plane, but elevated some distance above the plane. In this case, $$\vec\omega$$ is no longer parallel to what seems to be the axis of rotation for the particle (which is still the Z axis). In fact, in this case, $$\vec\omega$$ changes its direction constantly.

So please explain the intuition, or idea, about the direction of $$\vec\omega$$ in the general case.

(Note - I have asked a similar question in the past few days, but it had to do with fixed-axis rotation only. This is asking about the general case).

• Commented Mar 29 at 17:46
• Commented Mar 29 at 21:44

It sounds like you got if figured out. The direction of $$\vec \omega$$ is always perpendicular to the plane defined by $$\vec r\times \vec v$$. Further, $$\vec \omega$$ is always parallel to the axis of rotation whenever $$\vec r$$ and $$\vec v$$ lie in the plane of rotation. This all stems from the use of the cross product in the definitions as you note in your post.

Now as far as intuition goes. Intuitively, defining the angular quantities in terms of cross products gives a sense of orientation to the motion, i.e. using the "right hand rule", one can define a sense of positive rotation or negative rotation respectively.

• By the way I just noticed something. According to the formulas above, both $\vec{L}$ and $\vec\omega$ are always perpendicular to the plane defined by $\vec{r}$x$\vec{v}$. So that means the angular velocity and angular momentum are always parallel, right? If so, than why did I read that they are not always parallel? Commented Mar 29 at 14:28
• @AvivCohn They are always parallel for point particles, however, for rigid bodies, we have that $\vec L=I\vec \omega$, where $I$ is the moment of inertia tensor, thus the angular momentum vector can be rotated away from the angular velocity or sense of rotation of the body. When this happens, a rigid body rotates in an "unbalanced manner", i.e. there is a precession of the angular momentum vector. This is the phenomenon behind an unbalanced or wobbly tire. Commented Mar 29 at 14:32
• @AvivCohn For most simple cases of rotation, even of rigid bodies, we do not need the full tensor treatment for $I$ ($I$ is diagonal). We can calculate it as a number like $mr^2$ for example, however, this is not always the case. Commented Mar 29 at 14:37
• I see, thanks. Regarding intuition: what I meant was kind of, what is the physical meaning of the direction of the angular momentum and angular velocity. And the difference between the two. Is there something of the sort? Commented Mar 29 at 14:45
• @AvivCohn, are you familiar with how we define normal vectors to patches of surfaces? Just like the direction of a normal vector tells us what plane the surface is tangent to at that point by being perpendicular to it, the $\vec\omega$ vector tells us what plane the rotation is happening in by being perpendicular to it. Commented Mar 29 at 16:29

There is direction vector (magnitude and direction) isn't fully meaningful, since the direction is ambiguous. We use:

$$\omega_i = \epsilon_{ijk}r_jv_k$$

which rotates like a vector and defines a direction ($$\uparrow$$),

but the physics quantity we looking at is really:

$$\Omega_{ij} = r_iv_j - r_jv_k$$

(with $$\omega_i = \frac 1 2 \epsilon_{ijk}\Omega_{jk}$$)

which is the antisymmetric part of a rank-2 Cartesian tensor. It clearly doesn't point in a direction, but does define a tensor alignment ($$\updownarrow$$), so we pick a convention to give it a sign. The alignment is of course perpendicular to the plane spanned by the vectors.

Since:

$$L_i = (mr^2)\omega_i$$

I'm not sure why $$\omega_i$$ is more difficult than $$L_i$$ in the context of the question.

Finally: These vectors are called axial-vectors. They rotate like vectors, but are even under coordinate inversion, so if you ever see something like:

$$\vec p \propto \vec L$$

it's either wrong or is the weak interaction, e.g.:

$$\vec p_{\beta^-} \propto \vec L_{^{60}Co}$$

• Often times the "axial vectors" are referred to as "pseudo vectors". Commented Mar 29 at 16:21
• @AlbertusMagnus Yes, they do that in particle physics; however, later I learned from math ppl that pseudo vector is preferred usage for vectors in an affine space; that it, they are defined relative to an origin that matters. So for instance $\vec r$ is affine while $\vec r_1 - \vec r_2$ is just a vector. So, uniform circular motion has constant $\vec L$ and $\vec \omega$ and zero $\vec{\tau}$ ....until you move the origin: then bam: they all oscillate, but in the end $d\vec L/dt = \vec{\tau}$ and so on.
– JEB
Commented Mar 29 at 19:41

And angular velocity is defined as $$\vec\omega = > (\vec{r}\times\vec{v})/r^2$$

No. Suppose the rotation of a rigid body as the Earth for example. There is only one angular velocity $$\vec\omega$$ for the body. But $$\vec{r}\times\vec{v}$$ is a different vector, with a different direction depending on the latitude of the point chosen.

$$\vec\omega$$ is defined to be a vector such that $$\vec{v} = \vec\omega \times \vec{r}$$. So, $$\vec{v}$$ is always orthogonal to $$\vec\omega$$ and $$\vec{r}$$, but $$\vec\omega$$ is not necessarily orthogonal to $$\vec{r}$$.

The definition of $$\vec \omega$$ is useful for a rigid body because it is the same for all points (for a given instant of time). By using the definition of cross product it can be shown that $$\vec L = m\vec r \times (\vec \omega \times \vec r)$$ leads to: $$\vec L = \vec\omega\int_V \rho(\vec r) r^2dV - \int_V \rho(\vec r) (\vec \omega \cdot \vec r) \vec r dV$$.

For some symmetries, (as the case of a disk, where $$\vec \omega \cdot \vec r$$ is always zero), the second term vanishes. And $$\vec L = I\vec\omega$$, where $$I = \int_V \rho(\vec r) r^2dV$$ is the moment of inertia.

• Thanks for answering. So where did $\vec\omega = (\vec{r} x \vec{v})/r^2$ come from? Is it necessarily correct only when we talk about a rigid body rotating about a fixed axis? Commented Mar 30 at 7:50
• For a point particle rotating around an axis. The second term of $\vec L$ vanishes. $\vec \omega = \frac{\vec L}{I}$. $I = mr^2$ and $\vec L = m\vec r \times \vec v$. Commented Mar 30 at 10:50