# Why are angles so weird?

I'm potentially asking quite a stupid set of questions here but I'm wondering if there is some overarching theory about why angles, and "rotational" quantities which deal with them, have such strange properties (to me - I might just have overthought this and confused myself).

Firstly, I understand that it is a useful convention to define angular velocity and acceleration to be normal to the plane of rotation; but wouldn't it be equally intuitive to define them "in the direction of the change"? That is, in 2D space, why wouldn't we define angular velocity to be in the $$\hat{\theta}$$ direction? After all, it always points anticlockwise when defined as $$\hat{\theta} = -\sin\theta\hat{i} + \cos\theta\hat{j}$$, so positive angular velocity would be in the anticlockwise direction and negative clockwise, so that holds up. Is the definition being perpendicular to the plane of motion because differentiating $$\omega\hat{\theta}$$ would then give us angular acceleration in the $$-\hat{r}$$ direction, or is there another reason?

Also, it confuses me that angles are considered to be a scalar quantity, but by this reasoning would we not consider them pseudovectors, too? In the same sense that position is a "displacement from the origin", some $$x\hat{i}+y\hat{j}$$, would we by using the same convention not say that an angle is a "rotational displacement from the origin", $$\theta\hat{k}$$?

Finally (and sort of unrelated), unlike every other scalar quantity I can think of, angles don't make any sense when multiplied together or divided by eachother. Is there a meaning to such an expression?

• One issue: you look down on a rotating object and note that it is rotating counter-clockwise. If you change position such that you are under that same object, you note that it is rotating clockwise. This leads to ambiguity that shouldn't exist for a vector quantity. Defining the rotation vector with the right hand rule and setting it normal to the plane of rotation avoids such ambiguity. Jun 14, 2023 at 20:21
• Ah that's also true, I've seen this explanation once but I guess I didn't get it as was looking at it wrong; my thinking was, if I look at a car from above and it's moving in a straight line in the x-direction, looking at it from below I will define that as the -x direction, but that's not a problem, so why is this? It's kind of a weird transformation since we are inverting two of the three dimensions. Jun 14, 2023 at 20:49
• It is often more useful to think about the plane of rotation instead of the axis (generally, cross products are about planes [more precisely 2-forms] not vectors). Angular velocity defines the plane of rotation (it is pointing along the plane's normal). So the angular velocity gives you the thing that remains constant (the plane) not the thing that changes or the direction of change (any point in the plane changes, all points change in different directions by different amounts) Jun 15, 2023 at 3:47
• More on angular velocity and vectors. Jun 15, 2023 at 5:33
• Check out this pedagogical paper: arxiv.org/abs/2207.03560 Jun 17, 2023 at 17:18

Your concerns are valid, and for that reason, there is also an alternative way to view angular velocities, angular momentums, etc.: They are not vectors, but bivectors. A bivector is essentially an oriented plane segment, where "oriented" can be interpreted either as specifying a top and bottom of the plane, or as specifying a direction of rotation within the plane. The bivector describing the angular velocity of a rotating disc, for instance, is a plane segment parallel to the disc, with area proportional to the absolute value of angular velocity, and with its direction corresponding to the direction of the disc's rotation.

The reason why this isn't widely taught is that it requires quite a bit of heavy math in the form of the exterior product: bivectors can be obtained by taking the exterior product $$v\wedge w$$ of two vectors $$v$$ and $$w$$. But it turns out that the exterior product of two 3d vectors behaves very similarly to the cross product $$v\times w$$ of the two vectors. In fact, it is so similar that if we replace every bivector with a vector in such a way that exterior products are replaced by cross products, essentially nothing changes on the algebra side, so we can keep working in the familiar framework of vectors instead of bivectors without losing any utility. So instead of introducing bivectors, angular velocities and similar objects are often described using cross products. And those happen to be perpendicular to the plane of rotation.

• The cross product vector also tells you what direction to look at the rotation so it appears to rotate clockwise. Without a perspective there is no such thing as clockwise. Jun 15, 2023 at 13:35
• Angles are in a different category than angular velocity or angular momentum. The latter are vectors (well, 2-forms or bivectors isomorph to vectors, to be precise), while the former are not vectors (the sum of finite rotations is not commutative in 3D). Jun 16, 2023 at 14:45
• Here's an excellent pedagogical treatment of how to use bivectors to represent rotations, which you should find accessible: arxiv.org/abs/2207.03560 Jun 17, 2023 at 17:17
• Thank you, this is excellent! Jun 26, 2023 at 16:08

Comparing cross products 3D space and 2D space is fraught with peril, at the level of your question. So, let's stick to 3D.

(1) Would be a mess, and has been answered.

(2) Note that:

$$\vec a \cdot \vec b = |\vec a||\vec b|\cos{\theta_{ab}}$$

The L.H.S is a manifest scalar, as are the magnitudes of the vectors on the R.H.S. Thus: $$\cos{\theta_{ab}}$$ is a scalar, which should convince you that the angle is also scalar.

Regarding any "pseudo" nature, the above formula transforms under parity as:

$$(-\vec a) \cdot (-\vec b) = |-\vec a||-\vec b|\cos{\theta_{ab}}$$

so the cosine term doesn't change sign, but cosine is even, so the angle can.

Try the cross product's magnitude:

$$|\vec a \times \vec b| = |\vec a||\vec b|\sin{\theta_{ab}}$$

which transforms as:

$$|(-\vec a) \times (-\vec b)| = |-\vec a||-\vec b|\sin{\theta_{ab}}$$

so:

$$P::\sin\theta_{ab} \rightarrow \sin\theta_{ab}$$

implies the angel's sign does not change. It's a scalar

(3) Multiplying angels absolutely make sense. Consider two orthogonal angle differentials on a unit sphere, $$d\theta$$ and $$d\phi$$ (polar and azimuth, respectively).

Now integrate their product (with Jacobian):

$$\int_{\theta=0}^\pi \int_0^{2\pi} \sin\theta (d\theta \, d\phi) =$$ $$\int_{\theta=0}^\pi (2\pi \sin\theta \, d\theta)$$ $$= 4\pi$$

Which is the solid angle subtended by a sphere.

The steradian is the SI unit of solid angle.

• Thank you! Your answer for (2) really helped me out. Thanks! Jun 14, 2023 at 20:33
• guess my conclusion is that angles are not that weird, and I should have majored in marine biology as I had originally planned. Jun 14, 2023 at 20:35
• @krill there is alway architecture. Art Vandelay's firm loves to hire marine biologists. Or maybe the Yankees?
– JEB
Jun 14, 2023 at 20:39
• @krill So in (3) I multiplied orthogonal angels. Multiplying like-angels may not make sense.. I can't think of anything. But then do we multiply scalars like temperature? Pressure? The only case for angles is on a parameterized polar plot, what is the rate of change of the radial change vs. angle: $d^2r/d\theta^2$, which in SI would be meters per rad-squared...not very convincing.
– JEB
Jun 14, 2023 at 20:49
• (1) You can not [ & need not ] @ the Answerer when Commenting to that Answer. My Current Comment will always be notified to Answerer , automatically , & to @krill , because I included you. (2) I am not sure why Marine Biology & Architecture & ETC are involved here , I guess it is some In-Joke !
– Prem
Jun 15, 2023 at 13:42
1. We'd like the angular velocity vector to be constant as a particle traverses its way around a circle, since a particle undergoing uniform circular motion should have uniform angular velocity. In other words, the direction of $$\mathbf{\omega}$$ shouldn't be a function of $$\theta$$ at all, whereas it is in your definition.

2. You can certainly make this definition but it's not how one would usually think about angles geometrically. For example, your pseudovector wouldn't be rotationally invariant since by rotating through $$2\pi$$ you'll pick up an extra factor of $$2\pi \hat{k}.$$ Also, it seems more confusing to do trigonometry with pseudovectors than with scalars.

3. Why wouldn't a product or quotient of angles make sense? For example, $$\frac{\theta}{2\pi}$$ represents the fraction of a circle spanned by an arc subtending angle $$\theta$$, and higher powers of angles are everywhere in physics, like in the exact expression for the period of a simple pendulum.

• That makes sense! I agree with (3) - the ratio of two angles makes sense, but what would it mean physically to multiply two angles? What even is a angle squared? That is, we can interpret length*length as area, but there doesn't seem to be an analogous way of thinking about multiplying two angles? Jun 14, 2023 at 19:08
• In the limit of small angles $\theta \approx \sin{\theta}$ so a (radian) angle can be viewed simply as a ratio of lengths. Now consider a particle on top of a sphere of radius $R$ that rolls through an angle $\theta$. The horizontal distance it travels is $R\sin\theta \approx R\theta,$ but then the vertical distance it travels is approximately $R\theta \sin (\theta/2) \approx R \theta^2/2.$ In this case, the product of angles corresponds simply to a product of ratios between lengths, which is intuitive because angle is fundamentally just a ratio of lengths $\theta=\frac{s}{r}.$ Jun 14, 2023 at 19:19
• Can add to 1: the related conservation of angular momentum makes this physically tangible too, since it simulatenously expresses the invariance of the plane of motion and of the angular momentum magnitude combined. A spinning wheel in the lack of friction will conserve both speed and the direction of its axis too, as often demonstrated via a gyroscope
– Amit
Jun 14, 2023 at 20:20
• @krill multiplication of angles does make sense when you think of an object like a sphere, taking the product of two angles gives the solid angle subtended by those two angles, you can read about solid angles in steradian. Jun 14, 2023 at 21:28
• They don't have to be orthogonal, or one should say you can always find an orthogonal basis for your set-up, but it was more to illustrate the point that products of angles can have real use in the world. Jun 14, 2023 at 21:31

I think it might also help if you change your point of view from angles to rotations. Angles are a way to parameterize rotations.

(1) A rotation happens in some plane, and it just so happens that in 3D space there is a pseudo vector orthogonal to every plane (in 4D, for example, there would be a pseudo-plane orthogonal to every plane). This is why we can use pseudo vector of rotational velocity to characterize a rotation in 3D space. (see "Dual tensor" if you want to learn more about it)

(2) Even in 3D, angles of rotation can't be considered vectors or pseudo-vectors because they don't add up with one another as vectors do. Only if you consider rotations in the same plane you get $$\theta_{1+2}\hat{k} = \theta_1\hat{k} + \theta_2\hat{k}$$. In general case, if you consider two consecutive rotations done in different planes this kind of expression doesn't hold. However, if you look at angles of infinitesimal rotations, they would, up to the 1st order, add up in a proper "pseudo-vectorish" way, so angular velocity, being a fraction of infinitesimal change in angle over infinitesimal time turns out to be a proper pseudo vector.

(3) Rotations can be "multiplied", that is define $$(R_2R_1)(\vec{x}) \equiv R_2(R_1(\vec{x}))$$. The way this translates to transfomration of angles of individual rotations is more complicated. Note also that in general case $$R_2R_1 \neq R_1R_2$$