]> On the Dimension of Angles

# On the Dimension of Angles

I've discussed elsewhere my scepticism about the orthodox treatment of angles as dimensionless quantities, so here I turn my attention to what it means, for various physical quantities, if we don't. I'll use the radian as unit for these purposes, since that's what SI takes as unit of angle, while treating it as dimensionless, but note also that I consider the turn to be at least as compelling a choice of unit for angles. In the notation of dimensions, I'll use M for mass, L for length, T for time and A for angle.

## Rotational mechanics

Start with a torque: this is normally introduced as a force, along a line, times the perpendicular distance from the line along which it acts to some axis about which we want to measure the torque. However, that contains a cross product, and I have (as we'll see below) suspicions that cross product entails units of angle. So, instead, look at a torque's other two characterisations. It is an amount of work done per unit angle that a system turns about some axis – as such, it has dimensions M.L.L/T/T/A. It is also the rate of change, over time, of the angular momentum of the system, implying that angular momentum's dimensions are time × torque or M.L.L/T/A, momentum times lengh over angle.

Note that this per angle factor in angular momentum aligns also with regarding Planck's constant as being of this kind. In Planck's original formulation, it was energy divided by frequency; frequency is cycles per unit time, and using turn as the model of cycle makes this energy times time per turn. Of course, time per turn is also called period. Orthodoxy, lacking the units of angle in either, then has Dirac's constant be the result of dividing Planck's by 2.π – which we can write as ℏ = h/2/π or as ℏ/radian = h/turn, so if we define Planck's constant to be this quantity, with a per angle factor in its dimension, we unify the two constants into one.

Note that length per radian is a characteristic hint that lengths measured along a circumference are being measured in terms of a radius. To review my suspicions about the cross product entailing units of angle, notice that we can write torque as force times length per angle and angular momentum as momentum times length per angle; in both cases, the length per angle is showing up in place of a cross product with dispacement in the formula for the relevant quantity. The dimension of displacement is of course length, so the cross product appears to bring in a factor of per radian.

In the following, I'll need squared distances: these come from a metric g, a linear map from displacements to gradients, with gradients themselves being linear maps from displacements to scalars; for displacements p, q the scalar g(p, q) is the inner product or dot product of the two displacements, symmetric between p and q and equal to half the difference between the squared length of p+q and the sum of squared lengths of p and q separately; p and q are perpendicular precisely if g(p, q) = 0, i.e. p+q is the hypotenuse of a right-angle triangle with p and q as sides.

Next, let us consider the other representation of angular momentum, as a moment of inertia times an angular velocity. The angular velocity w is a rate of rotation, angle turned per time taken to do so, with dimensions A/T; it is also a vector along the axis of rotation, so that scaling it by moment of inertia gives angular momentum, and the inner product of that with half the angular velocity gives the kinetic energy of the rotation. The velocity of any given part, whose location is at displacement r from some point on the axis of rotation, of a body rotating in this manner is then v = cross(w, r) and needs a per radian to turn that L.A/T into L/T. Its acceleration is likewise a = cross(w, v) / radian (and this is also true if v includes a component of motion in addition to the cross(w, r) due to rotation – this changes cross(w, v) by, in the rotating frame of reference, the Coriolis term).

This, in turn, makes the moment of inertia M.L.L/A/A, mass times the square of length per angle. Multiplying that by half g(w, w) gives the rotational kinetic energy, which we can also express as a sum over parts of the system of half mass times the square of v = cross(w, r) / radian, so the moment of inertia sums mass times g(cross(w, r), cross(w, r))/g(w, w) and divides by the square of radian. If we take a unit vector u in the direction of w, g(cross(w, r), cross(w, r))/g(w, w) is reduced to g(cross(u, r), cross(u, r)), the squared length of cross(u, r).

Since cross product maps its two vectors to one perpendicular to each of them, of length equal to the area of the parallelogram that has them as sides, and u is perpendicular to cross(u, r), cross(cross(u, r), u) has the same length as cross(u, r), thanks to the length of u being 1; and cross(cross(u, r), u) = r −g(u, r).u is the component of r perpendicular to u (since g(u) maps it to zero and adding a multiple of u to it gets you r). So the length of cross(u, r) is in fact the distance of r from the axis.

## The cross product and rotations

So this examination of torque, angular momentum, moment of inertia and rotational motion fits with the cross product having dimension inverse angle. The cross product is a feature of three-dimensional space (in other dimensions it can't be coaxed into giving you a vector), linked to the metric (a.k.a. dot product). From our ({gradients}| g |{displacements), linear between dual spaces, we can obtain μ = √det(g) which encodes the measure of volume: it consumes the displacements along the three edges out of a corner of a parallelpiped and delivers the volume of that solid. Shearing the parallelpiped doesn't change its volume, so adding a linear combination of two of those edge displacements to the third doesn't change the value, implying that μ is antisymmetric in its three parameters. There's no obvious reason for either g or μ to implicate any units of angle.

Together with the metric, the measure also delivers the cross product as cross(x, y) = (x·μ·y)/g – so what is the significance of the per radian we're finding wherever cross is used ? One obvious option is that where physics is using cross, it has some physical reason to insert a per radian; but since the radian is a purely mathematical unit, it is possible that it lurks in the mathematics, not the physical application, of the cross product. In particular, the fact that rotational motion (essentially a purely abstract description of the relationship between two frames of reference) needs to divide cross products by radian to get the right dimensions is a hint that the latter is what matters.

So let's look at rotation. If we have a position expressed as a displacement r from some point on an axis, with unit vector u along that axis, we have a vector cross(r, u) perpendicular to the axis and r; and cross(u, cross(r, u)) = r −u.g(r, u) is then the component of r perpendicular to u, as noted above. We can write our position r in the form u.g(r, u) +(r −u.g(r, u)) in which the first term is in the direction of u, hence unaffected by rotation about the axis parallel to u, while the second is perpendicular to u (as g(u) maps it to g(u, r) −g(u, u).g(r, u); and g is symmetric, with g(u, u) = 1), so rotation through angle a about our axis produces Cos(a) of it plus Sin(a) of some vector of the same length perpendicular to u; this latter vector is ±cross(r, u). Taking the ± as + we get the direction clockwise about u as seen looking along it, but the orthodox direction of positive angles is anticlockwise, so I'll take − as the sign and rewrite −cross(r, u) as +cross(u, r).

So rotating r through angle a about u maps u.g(r, u) to itself and r −u.g(r, u) = cross(cross(u, r), u) to Cos(a).cross(cross(u, r), u) +Sin(a).cross(u, r), so it maps r to the sum of this last and u.g(r, u). Let W be the magnitude of our angular velocity w, so w = W.u, and a point initially at r shall, when rotating at angular velocity w about the axis, be mapped at time t to rot(W.t.u, r) = u.g(r, u) +Cos(W.t).cross(cross(u, r), u) +Sin(W.t).cross(u, r). The derivatives of this are:

• radian.drot(W.t.u, r)/dt = −W.Sin(W.t).cross(cross(u, r), u) +W.Cos(W.t).cross(u, r)