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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?

Qmechanic
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krill
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    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. – David White Jun 14 '23 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. – krill Jun 14 '23 at 20:49
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    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) – tobi_s Jun 15 '23 at 03:47
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    More on angular velocity and vectors. – Qmechanic Jun 15 '23 at 05:33
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    Check out this pedagogical paper: https://arxiv.org/abs/2207.03560 – tparker Jun 17 '23 at 17:18

4 Answers4

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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.

Vercassivelaunos
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  • 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. – candied_orange Jun 15 '23 at 13:35
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    Hold your right hand just under your chin. Rotate that hand so that your thumb points down. Notice when you close your fingers they curl clock wise. You're looking in the same direction your thumb is pointing. Now hold your hand over your head, point your thumb at your head and close your fingers the same way. Now they're curling counter clockwise. Clockwise only exists with a perspective. The cross product vector defines a scalar and that perspective. It's not simply orthogonal to the plane of rotation. The direction maters. – candied_orange Jun 15 '23 at 19:14
  • 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). – GiorgioP-DoomsdayClockIsAt-90 Jun 16 '23 at 14:45
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    Here's an excellent pedagogical treatment of how to use bivectors to represent rotations, which you should find accessible: https://arxiv.org/abs/2207.03560 – tparker Jun 17 '23 at 17:17
  • Thank you, this is excellent! – krill Jun 26 '23 at 16:08
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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.

Michael Hardy
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JEB
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  • Thank you! Your answer for (2) really helped me out. Thanks! – krill Jun 14 '23 at 20:33
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    guess my conclusion is that angles are not that weird, and I should have majored in marine biology as I had originally planned. – krill Jun 14 '23 at 20:35
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    @krill there is alway architecture. Art Vandelay's firm loves to hire marine biologists. Or maybe the Yankees? – JEB Jun 14 '23 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 '23 at 20:49
  • Yeah so I think it only works if you add that constraint? because also if we add angles, we preserve the property that angle mod(2$\pi$) = angle. This cancels in a way that makes sense for the ratio. If we multiply two non-orthogonal angles, we are doing something a bit weird, and the result would look something like $(\alpha + a2\pi)(\beta +b2\pi)=(\alpha\beta + 2\pi(a\alpha + b\beta) + 4ab\pi^2)$ which, whatever that is, doesn't look like it's going to be an angle... at least i think that's kind of an essential property of an angle? – krill Jun 14 '23 at 20:57
  • [https://stackexchange.com/users/3114321/jeb] I think this modulo property also holds for solid angles, since $4\pi$ steradians covers all space! – krill Jun 14 '23 at 21:38
  • sorry I can't seem to @ you – krill Jun 14 '23 at 21:41
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    (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 '23 at 13:42
  • ah thanks! sorry, didn't realise :) – krill Jun 15 '23 at 13:57
  • +1. Love the "fraught with peril" ツ – John Doty Jun 15 '23 at 14:24
  • @JohnDoty that's code for "No Clifford Algebra today" – JEB Jun 15 '23 at 16:03
  • I find the notation $\displaystyle \int_{\theta=0}^\pi \int_0^{2\pi} \sin\theta (d\theta , d\phi)$ odd and potentially confusing. If this is supposed to be an iterated integral, then either it is $$ \int_{\theta,=,0}^\pi \left( \int_{\varphi,=,0}^{2\pi} \sin\theta, d\varphi\right) , d\theta$$ or it is $$\displaystyle \int_{\varphi,=,0}^{2\pi} \left( \int_{\theta,=,0}^\pi \sin\theta, d\theta\right) , d\varphi.$$ On the other hand, if it's supposed to be an integral with respect to area in the plane, [character limit approaching] $\ldots\qquad$ – Michael Hardy Jun 15 '23 at 18:30
  • $\ldots,$ On the other hand, if it's supposed to be an integral with respect to area in the plane, then it's either $$ \iint\limits_{(\theta,\varphi),\in,[0,\pi],\times , [0,2\pi]} \sin\theta , d(\theta,\varphi)$$ or it's $$ \iint\limits_{(\varphi,\theta),\in,[0,2\pi],\times , [0,\pi]} \sin\theta , d(\varphi,\theta).$$ These two are really to different ways to say the same thing, and one cannot always say that about the iterated one-dimensional integrals in the previous comment. $\qquad$ – Michael Hardy Jun 15 '23 at 18:34
  • A parenthetical comment: Above I wrote "one cannot always say that about the iterated one-dimensional integrals in the previous comment." The reason is that things like the following can happen: $$ \int_0^1 \left( \int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2}, dy \right) , dx = +\frac\pi2, $$ but $$ \int_0^1 \left( \int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2} , dx \right) , dy = -\frac\pi2. $$ This is possible because of the lack of absolute convergence, i.e. $$ \iint\limits_{[0,1],\times,[0,1]} \left| \frac{x^2-y^2}{(x^2+y^2)^2}\right| , d(x,y) = +\infty. $$ – Michael Hardy Jun 15 '23 at 18:38
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  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.

Roger Yang
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  • 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? – krill Jun 14 '23 at 19:08
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    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 $\theta=\frac{s}{r}.$ – Roger Yang Jun 14 '23 at 19:19
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    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 '23 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. – Triatticus Jun 14 '23 at 21:28
  • @Triatticus yeah that makes sense, but only if they are orthogonal! Otherwise, modulo property breaks down, doesn't it? I elaborated on that in the comments on the answer below. – krill Jun 14 '23 at 21:30
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    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. – Triatticus Jun 14 '23 at 21:31
  • @krill: An angle squared is a solid angle, which comes up occasionally as the inner surface of a sphere, or piece thereof. – Joshua Jun 15 '23 at 04:19
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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 $R_2R_1 \neq R_1R_2$

xaxa
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