Angles

Where two lines meet, there is an angle. We can add angles together, we can multiply them by numbers (yielding angles), we can divide one angle by another to get a number; in all this, angles are just like lengths. We can measure the sizes of angles relative to one another but, until we chose an angle to use as unit, angles aren't numbers. A quarter turn isn't π/2 or, indeed, 1/4; it's a quarter turn; and it is a right angle; but it isn't a number. Here turn serves as a unit of angle entrenched in the English language: it's the angle through which one has to turn to get back to where one started. A quarter of that angle is the right angle one gets between perpendicular lines.

Angles also have direction: so there are two quarter turns – one clockwise, the other anticlockwise. I'll distinguish these as the right angle and the left angle. For compatibility with the notion of turn right or turn left, taken in context of some orthodox descriptions, presume that a right angle is a quarter turn clockwise while a left angle is a quarter turn anticlockwise; and describe the left angle as positive, right angle as negative; however, all that matters is that left and right be in opposite senses and that one is negative, the other positive.

The SI unit of angle is the radian; 2.π radians = one turn. I'll come back to why later. The right (or left) angle, already mentioned, is another candidate for use as a unit of angle; as is the about-face or half turn. I can't say I'm much interested in the degree, though – it's a hang-over from the Babylonian approximation to the year as 360 days, albeit 360 is a highly factorisable number (it has 5 as a factor once, 3 twice and 2 thrice). There's also a unit called the grad, equal to turn / 400, which I believe originated in military usage (among gunners); this subdivides the right angle into 100 equal parts, making it marginally tidier than the degree; but I do not find it a compelling choice for unit of angle – at best, it is the centi-unit when the chosen unit is the quarter turn.

Rotation through a whole turn gets you back to where you were when the turn started. In two dimensions, a half turn negates both co-ordinates; repeating it to make a whole turn illustrates (−1)**2 = 1; the quarter turn will in due course illustrate what a square-root of −1 looks like. It is worth noting that one needs at least two dimensions to have angles; with more dimensions, more complex options enter the picture, but our discussion of any given angle can, at least, be reduced by projection onto the (at most) two dimensions spanned by the lines forming the angle. In one dimension, we still – sort of – have angles; but only half and whole turns: as before, the half turn takes the rôle of −1, we see its square is 1 and might pause to dream of the quarter turn as its square root.

One way or another, there are good reasons in geometry and linear algebra for wanting to measure angles in turns or nice easy fractions of the turn. So why have the learned bods of the international institutes chosen the radian, an irrational fraction of the turn, as the SI unit ?

To answer that, it will be necessary to discuss some trigonometry; which shall reinforce the case for the turn but equip us to state the case for the radian.

A little trigonometry

Figure: two lines come out of a right angle, one going forward the other upwards; the front of the former is joined to the top of the latter by a line, called the hypotenuse; this is longer than either of the original two lines. At the front of the triangle, the hypotenuse meets the forward line; the angle between them is labelled a; the sides are also labelled with h on the hypotenuse, h.Sin(a) on the upright and h.Cos(a) on the forward edge. The labels are the size of the angle and lengths of the sides.

By such a diagram or otherwise, the functions Sin (short for sine and pronounced like sign) and Cos (for cosine) are defined; each takes an angle as input and produces a ratio (a pure number) as output; a is an angle, Sin(a) is the length ratio of the upright to the hypotenuse, as Cos(a) is for forward to hypotenuse. [The ratio of upright to forward is called the tangent, written Tan(a) and equal to Sin(a) / Cos(a). But it and the rest of that family are peripheral to the present discussion.] We can use Sin and Cos to express Pythagoras' theorem as:

• Sin(a).Sin(a)+Cos(a).Cos(a)=1 for every angle a.

The diagram only really gives us Sin(a) and Cos(a) for a between zero and a quarter turn; and, at that, interprets a as positive, so we'd better call that quarter turn the left angle (positive by the conventions chosen above). However, various structural truths about Sin and Cos emerge, in particular it is possible to compute the Sin and Cos of a+b and a−b from the Sin and Cos of a and b, at least wherever a+b or a−b was an angle for which Sin and Cos are defined; by accepting the answers this gives where Sin and Cos aren't defined, we can extend the two functions to arbitrarily large angle. Here are the relevant formulae:

• Sin(a+b) = Sin(a).Cos(b) +Cos(a).Sin(b)
• Cos(a+b) = Cos(a).Cos(b) −Sin(a).Sin(b)
• Sin(a−b) = Sin(a).Cos(b) −Cos(a).Sin(b)
• Cos(a−b) = Cos(a).Cos(b) +Sin(a).Sin(b)

For the subtraction formulae with a = b, we promptly obtain Sin(zero) = 0 and, via Pythagoras' theorem, Cos(zero) = 1. Applying the same formulae with a = zero instead, we obtain

• Sin(−b) = Sin(zero−b) = 0.Cos(b) −1.Sin(b) = −Sin(b)
• Cos(−b) = Cos(zero−b) = 1.Cos(b) +0.Sin(b) = Cos(b)

i.e. Sin is an odd function (negating its input negates its output) and Cos is an even function (negating input doesn't change output). These, in turn, turn the formulae for a−b into special cases of the formulae for a+b, by simply replacing b with −b. (They also imply that Tan = Sin/Cos is odd.)

Just as the original triangle only established Sin and Cos for angles between zero and a quarter turn, it only strictly lets us assert Pythagoras' theorem for those angles. However, by establishing that the above sum and difference formulae yield results which satisfy Pythagoras' theorem for a+b and a−b when it holds for a and b, we can induce the theorem's validity for arbitrary angles, just as we can infer the Sin and Cos of arbitrary angle using the above formulae. From Sin being odd and Cos even, we can immediately infer that Pythagoras' theorem holds for −b whenever it holds for b, saving us the need to examine the a−b case. So, given Pythagoras' theorem for angles a and b:

Sin(a+b).Sin(a+b) +Cos(a+b).Cos(a+b)

= (Sin(a).Cos(b) +Cos(a).Sin(b)).(Sin(a).Cos(b) +Cos(a).Sin(b)) + (Cos(a).Cos(b) −Sin(a).Sin(b)).(Cos(a).Cos(b) −Sin(a).Sin(b))

= Sin(a).Sin(a).Cos(b).Cos(b) +2.Sin(a).Cos(a).Sin(b).Cos(b) +Cos(a).Cos(a).Sin(b).Sin(b) +Cos(a).Cos(a).Cos(b).Cos(b) −2.Sin(a).Cos(a).Sin(b).Cos(b) +Sin(a).Sin(a).Sin(b).Sin(b)

= (Sin(a).Sin(a) +Cos(a).Cos(a)).Cos(b).Cos(b) +(Cos(a).Cos(a) +Sin(a).Sin(a)).Sin(b).Sin(b)

= Sin(b).Sin(b) +Cos(b).Cos(b) = 1

by applying Pythagoras first to a and then to b.

Furthermore, we can (by adding and subtracting suitable formulae above) infer the product formulae:

• 2.Sin(a).Cos(b) = Sin(a+b) +Sin(a−b)
• 2.Cos(a).Sin(b) = Sin(a+b) −Sin(a−b)
• 2.Cos(a).Cos(b) = Cos(a−b) +Cos(a+b)
• 2.Sin(a).Sin(b) = Cos(a−b) −Cos(a+b)

I'll use left as a positive quarter turn and right a negative one. In our original triangle, the angle at the top of the hypotenuse is just left−a (because the sum of angles in the triangle is a half turn, 2.left, and the right angle accounts for left of that, leaving left for the other two). This is equal to a precisely when a is left/2. Consequently the two sides other than the hypotenuse are equal; so Sin(a).Sin(a) = Cos(a).Cos(a) and their sum is 1, so each is 1/2, whence Sin(left/2) = 1/√2 = Cos(left/2). Thus

• Sin(left) = Sin(left/2).Cos(left/2) +Cos(left/2).Sin(left/2) = 1
• Cos(left) = Cos(left/2).Cos(left/2) −Sin(left/2).Sin(left/2) = 0
• Sin(turn/2) = Sin(left).Cos(left) +Cos(left).Sin(left) = 0
• Cos(turn/2) = Cos(left).Cos(left) −Sin(left).Sin(left) = −1
• Sin(turn) = Sin(turn/2).Cos(turn/2) +Cos(turn/2).Sin(turn/2) = 0
• Cos(turn) = Cos(turn/2).Cos(turn/2) −Sin(turn/2).Sin(turn/2) = 1

and we can use the fact that Sin is odd and Cos is even to infer the corresponding results for multiples of the right angle (i.e. negative multiples of left).

It follows immediately that

• Sin(a +turn) = Sin(a).1 +Cos(a).0 = Sin(a)
• Cos(a +turn) = Cos(a).1 −Sin(a).0 = Cos(a)
• Sin(a +turn/2) = Sin(a).(−1) +Cos(a).0 = −Sin(a)
• Cos(a +turn/2) = Cos(a).(−1) −Sin(a).0 = −Cos(a)
• Sin(a +left) = Sin(a).0 +Cos(a).1 = Cos(a)
• Cos(a +left) = Cos(a).0 −Sin(a).1 = −Sin(a)
• Sin(a +right) = Sin(a).0 +Cos(a).(−1) = −Cos(a)
• Cos(a +right) = Cos(a).0 −Sin(a).(−1) = Sin(a)

These tell us that the two sinusoids are periodic (they repeat themselves after a turn), advancing either by a half turn negates it, and each may be obtained by advancing or backing up the other by a quarter turn. We should also note one further important structural fact (for symmetry, expressed here in two ways):

• Sin(left −a) = 1.Cos(a) −0.Sin(a) = Cos(a),
• Cos(left −a) = 0.Cos(a) +1.Sin(a) = Sin(a).

which you can actually read off from the original triangle which I used to define Sin and Cos, by looking at the angle at the top of the hypotenuse, which is left−a. We can use the addition formulae to compute Sin and Cos of any multiple of an angle, as polynomials in the Sin and Cos of the original angle. In particular, this lets us express the (known) Sin and Cos of the turn as polynomials in the Sin and Cos of the result of dividing the turn by any positive integer; by solving the resulting polynomial equations, we can (with suitable care) infer the Sin and Cos of such fractions of the turn; by scaling those fractions up, we are thus able to compute the Sin and Cos of any rational (i.e. integer divided by positive integer) multiple of the turn. Since the relevant polynomials all have integer coefficients and the Sin and Cos of the (quarter, half and whole) turn are integers, we also infer that the Sin and Cos of any rational multiple of the turn is an algebraic number (i.e. one that's a solution to some polynomial equation whose coefficients are all integers). This could sanely be construed as a powerful argument in favour of using the turn (or a rational multiple of it) as unit of angle.

That pretty much re-iterates the virtues of the turn and its fractions as units of angle; but it also sets the scene for the perspective that leads SI to measure angles in radians. The attentive will notice that, when a is small (by comparison to a quarter turn or other angle of roughly its size) Sin(a) is approximately proportional to a; the very careful will see that, in fact, Sin(a) is approximately 2.π.a/turn = a/radian for such small a. However, rather than approaching this approximately, I'll derive the constant of proportionality as a side-effect of …

Differentiating Sin and Cos

Draw a circle of radius R: use its centre as the origin for cartesian co-ordinates, [x, y]. Each position on the circle may be characterised by the radial line from the origin to it; which, in turn, may be uniquely identified by the angle between it and a chosen co-ordinate direction. Indeed, if this angle is a, measured from the positive x-axis in the sense which has the positive y-axis's angle equal the positive quarter turn (i.e. left), the position on the circle may readily be shown to be [x, y] = R.[Cos(a), Sin(a)]. Draw a tangent to the circle at our position at angle a. This tangent is at right angles to the radius, so is parallel to the unit vector [−Sin(a), Cos(a)].

Now, for a given angular velocity w (whose units are angle/time), we can consider an object moving around the circle so that its angle at time t is just w.t: its position is then [x, y] = R.[Cos(w.t), Sin(w.t)] and its velocity is V.[−Sin(w.t), Cos(w.t)] for some speed V (whose units are length/time). It is easy enough to verify that the absolute value of V is the speed of our moving object, which is just the circumference divided by the period. The circumference is 2.π.R, the period is turn/w, so we obtain abs(V) = 2.π.R.w/turn = R.w/radian. As to the sign of V, observe that (when w is positive) y is increasing exactly when x is positive while x is increasing exactly when y is negative: whence we find V = R.w/radian.

Thus (cancelling out the common factor of R), [Cos(w.t), Sin(w.t)] ←t has derivative [−Sin(w.t), Cos(w.t)].w/radian ←t, from which we may infer that the derivative of Sin is Sin' = Cos/radian and that of Cos is Cos' = −Sin/radian. These may be re-written as:

• Sin'(a) = Sin(a+left)/radian
• Cos'(a) = Cos(a+left)/radian

Thus differentiation of the sinusoids is add a quarter turn to the input, divide the output by radian. SI uses the radian as its unit of angle because one can read this as saying that differentiation wants to measure angles in radians; yet, even differentiation also uses the quarter turn.

Rotating vectors

For the object going round a circular path, above, let its position be p; so p = R.[Cos(w.t), Sin(w.t)] for some angular velocity, w, and the object's velocity is R.[−Sin(w.t), Cos(w.t)].w/radian. Its accelleration is then −R.[Cos(w.t), Sin(w.t)].w.w/radian/radian, so ddp/dt/dt = −p.(w/radian).(w/radian). Thus, again, the radian shows up as the convenient unit of angle for use when looking at the (very common) case of the simple harmonic oscillator, a system whose dynamics are described by the second derivative of a quantity being equal to a negative multiple of the quantity itself. This special case likely contributed heavily to SI's choice in the matter.

Of course, in these terms, radian wants to be an imaginary unit (as time is, in relation to spatial distance; the speed of light is more compellingly a square root of −1 than the radian is a dimensionless unit); then the negation involved in the above gets swallowed by radian. This also fits nicely with the fact that 2.π is the period along the imaginary axis of the exponential function on the complex plane.

Fourier's complication

Now it's possible to take (pretty much) any function (U: f |V) with U and V both real vector spaces, and apply a transformation, F, to it, defined (using i as a square root of −1) by:

• F(U:f|V) = (U: T.integral(U: exp(i.Q.w·x).f(P.x) ←R.x |V) ←S.w :dual(V))

for any choice we like of scalings P, Q, R, S and T. (The Fourier transform uses R = 2.π, Q = −1, P = S = T = 1; its inverse uses P = Q = R = S = T = 1.) It is easy enough to show that replacing S with 1 and Q with Q/S doesn't change F; nor does replacing R with 1, P with P/R and Q with Q/R; nor does replacing R with 1 and T with T/power(dim, R) where dim is the dimension of V. If we apply this transformation twice we get

• F&on;F = scale.(: (U: f(−x) ←x |V) ←(U:f|V) :)

for some scale computable from P, Q, R, S and T. Aside from the scaling, F&on;F is just (: f&on;negate ←f :), with negate = (: −x ←x :). Clearly repeating this yields the identity as F&on;F&on;F&on;F. Thus, give or take a scaling, F is a fourth root of the identity; if we chose P, Q, R, S and T suitably, so as to get scale = ±1, it'll deserve to be regarded as having unit size. The choice P = Q = S = 1 with R = √(2.π) yields scale = 1; but so does the choice Q = 2.π, P = R = S = T = 1. While the former is most useful in justifying the Fourier transform, it is hard to motivate other than as the way to achieve unit size; the latter, on the other hand, is easily enough motivated.

Chosing Q = 2.π effectively choses exp(2.π.i.w·x) as the canonical sinusoid being used by Fourier's analysis. The inner product w·x of the inputs (to the original function, f, and its transform, F(f)) is the natural scalar to obtain from a member of V and a member of dual(V); while the function exp(2.π.i.t) ←t is a sinusoid with period 1. Thus, in effect, the Fourier transform begs us to use period = 1 sinusoids rather than period = 2.π.i ones; in effect, it's asking us to use the turn as our unit of angle, in preference to the radian.

This preference becomes particularly stark in the case of the real transform, where no complex numbers are involved at all. For this we define

• C = (: ({reals}: integral(: Cos(k·x.turn).f(x) ←x |V) ←k |dual(V)) ←({reals}:f|V) :)
• S = (: ({reals}: integral(: Sin(k·x.turn).f(x) ←x |V) ←k |dual(V)) ←({reals}:f|V) :)

and obtain f = C(C(f)) +S(S(f)) cleanly without any factors of 2.π showing up at all; whereas the radian-based equivalents (using cos(k·x), i.e. Cos(k·x.radian), rather than Cos(k·x.turn); and likewise for Sin) are doomed to a scattering of factors of 2.π.

The two-dimensional solid angle

While we're at it, note that SI also defines the steradian, a unit of solid angle. Just as the whole of a circle subtends an angle of one turn about its centre, so equally a sphere's surface subtends a solid angle of one whole shell (for want of a better word) about its centre. The idea behind solid angle is that, for example, if a light source is radiating energy out in all directions equally, any object receives a share of that energy in proportion to the fraction of a whole shell that's covered by the radial projection of the given object onto the chosen shell. Thus, if a light-bulb hangs in the mouth of a cave, half the light from the bulb goes into the cave and half of it goes to the outside world; when projected onto a sphere about the bulb, the mouth of the cave appears as a great circle of the sphere, dividing it into equal parts, one of which faces into the cave, the other outwards.

Since we live in a Minkowskian universe, it might also make sense to consider what analogue of fraction of a whole spherical shell can be made intelligible for the metric of space-time, whose spheres are hyperboloids (of one sheet if of space-like radius, of two sheets if of time-like radius) and hence, in particular, infinite.

Astronomers use a measure of solid angle called the square degree; this is the solid angle subtended by a square on the sphere whose sides subtend one degree. For any tiny enough angle k, a similar square with sides subtending k will have area equal to the square of radius times k/radian; dividing this by the square of radius we get the number of steradians it corresponds to, which is just the square of k/radian. Thus the steradian is, in a meaningful sense, just the square of the radian. One full shell, 4.π sr, is thus 2.turn.radian or turn.turn/π; in square degrees, that comes to 129600/π or just under 41253.

Angle as scaled fraction of the whole

Now, the total area of a sphere's surface is 4.π times the square of the sphere's radius; this is the product of the sphere's circumference and diameter. Crucially, it grows in proportion to the square of the sphere's radius.

In the two-dimensional world, the one-dimensional circle's circumference grows proportional to the radius of the circle; the fraction of the circle covered by the projection, radially onto it, of any given object is just the angle (measured in turns) subtended by that object as seen from the circle's centre. The length of the relevant piece of the circumference is then just the angle multiplied by the radius times 2.π/turn. The radian = turn/(2.π) equips us to restate that as: the length of the piece of circumference is just the angle, measured in radians, times the radius.

Likewise, in three dimensions, a two-dimensional portion of a sphere's surface, covered by the projection, radially onto it, of some given object, has area equal to the solid angle it subtends, times the square of the sphere's radius, times 4.π/shell. Just as measuring an angle in radians gave us what to multiply the radius by to get the length of a piece of circumference, we can introduce a solid angle steradian = shell/(4.π) and, by measuring solid angles in this unit, get the scalar by which to multiply the square of radius to obtain the area of a portion of our sphere that subtends the given solid angle.

The question I want to address is: can we derive the solid angle from the angle (e.g. angle.angle), is it a separate and independent entity, or is it in fact properly dimensionless ? SI, in chosing to define the steradian, declares solid angle to be an independent kind of dimension, separate from the angle. This begs the question of what must happen at each higher dimension; do we get separate and independent n-dimensional angles for each positive integer n ?

Varying dimension

An immediate thought from that is to follow down in dimension; at dimension 1, the 1-sphere's perimeter comprises two points, independent of radius, and a zero-angle counts how many of those two points it embraces; it's clearly just a number. At dimension zero, the 0-sphere is a point with no boundary (or, if you will, with a boundary made of 0 pieces, each of which is the −1 simplex and varies in size proportional to 1/radius), so begs no notion of angle. Having no geometric intuition about – and, in particular, no notion of angle for – negative dimensional analogues, it doesn't make sense to chase below dimension zero. It is, however, noteworthy that this strange domain gives zero area and volume to negative even dimension and steadily growing (albeit alternating in sign) volume and area for negative odd dimension: even dimension is special.

Going up in dimension, we'd get the 4-sphere, of total area 2.π.π times the cube of radius, requiring a unit of 4-angle equal to 1/(2.π.π) of a 4-shell; the 5-sphere with area 8.π.π/3 times the fourth power of radius, requring a unit of 5-angle equal to 3/(8.π.π) of a 5-shell; and so on. Taking A as mapping from dimensions to the surface measure of the unit sphere at each, we obtain A(n+2) = 2.π.A(n)/n for each positive dimension, n. Note that A(dim) < 1 for dim > 18, so the required units of angle for dimension greater than 18 would exceed the whole shell at such dimensions. The two-step nature of the iteration might point to the right unit of angle at each dimension being a suitable power of solid angle times, for even dimensions, plain old angle; if solid angle was honestly dimensionless, we'd thereby escape from needing endlessly many new forms of angle unit.

A minor curiosity

The ancient HAKMEM document contains the following fascinating information:

PROBLEM 45 (Gosper):

• Take a unit step at some heading (angle).
• Double the angle, step again.
• Redouble, step, etc.
• For what initial heading angles is your locus bounded?

PARTIAL ANSWER (Schroeppel, Gosper):

When the initial angle is a rational multiple of [a half turn], it seems that your locus is bounded (in fact, eventually periodic) iff the denominator contains as a factor the square of an odd prime other than 1093 and 3511, which must occur at least cubed. (This is related to the fact that 1093 and 3511 are the only known primes satisfying

• power(2, P) = 2 mod P.P)

But a denominator of 171 = 9 * 19 never loops, probably because 9 divides phi(19). Similarly for 9009 and 2525. Can someone construct an irrational multiple of [a half turn] with a bounded locus? Do such angles form a set of measure zero in the reals, even though the measure in the rationals is about .155? About .155 = the fraction of rationals with denominators containing odd primes squared = 1 − product(: 1 − 1/P/(P +1) ←P :{odd primes}). This product = .84533064 ± a smidgen, and is not, alas, sqrt(pi/2) ARCERF(1/4) = .84534756. This errs by 16 times the correction factor one expects for 1093 and 3511, and is not even salvaged by the hypothesis that all primes > a million satisfy the congruence. It might, however, be salvaged by quantities like 171.

If, in fact, all solution angles are rational multiples of a half turn, this could be construed as geometry favouring the turn, again …

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Written by Eddy.