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; its name comes from the angle
an upright

line makes with the horizontal.

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 (gunners measuring elevation); 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 power(2, −1) = 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.

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
(including, for completeness, the formulae for tangents, which are easily
derived from those for Sin and Cos – and, satisfyingly, only involve
Tan):

- Sin(a+b) = Sin(a).Cos(b) +Cos(a).Sin(b)
- Cos(a+b) = Cos(a).Cos(b) −Sin(a).Sin(b)
- Tan(a+b) = (Tan(a) +Tan(b)) / (1 −Tan(a).Tan(b))
- Sin(a−b) = Sin(a).Cos(b) −Cos(a).Sin(b)
- Cos(a−b) = Cos(a).Cos(b) +Sin(a).Sin(b)
- Tan(a−b) = (Tan(a) −Tan(b)) / (1 +Tan(a).Tan(b))

The addition formulae can (for instance) be obtained by using the given definition to infer the co-ordinates of the linear map which implements a rotation through an angle a; (: [x.Cos(a) −y.Sin(a), x.Sin(a) +y.Cos(a)] ← [x,y] :); applying this for angles a and b we can compose the linear maps, by matrix multiplication, to obtain an implementation of rotation through angle a+b, entirely in terms of the cosine and sine of a and b; we can equally apply the same reasoning for a+b as we did for a and b to obtain rotation through a+b in terms of the cosine and sine of a+b itself; comparing this with the composite yields the first two formulae given above.

Either by similar reasoning or by substituting c=a+b, d=a, c−d=b and subsequent re-naming, one can obtain the difference formulae. (Alternatively, consider a rectangle, tilted relative to one's preferred co-ordinates, and its bounding box with respect to those co-ordinates; the angles between a diagonal of the rectangle, an edge of it and an edge of the bounding box can then give us two formulae for the edges of the bounding box, in terms of the length of the rectangle's diagonal, that give one the same equations.)

The formulae for Tan can be obtained by dividing the result for Sin by the result for Cos and dividing both numerator and denominator of the result by Cos(a).Cos(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 …

As it happens, there's only a very few rational multiples of the turn that have a rational as even one of their Sin, Cos or Tan: most pythagorean triangles have angles (aside from the right angle) that aren't rational multiples of the turn. In fact, Tan(turn/8) = 1 and Cos(turn/6) = 1/2 suffice to imply all of the other cases, given that Cos(turn/4 −a) = Sin(a) from the definitions, taken together with the usual addition and subtraction formulae.

This result was proved by Hadwinger, based on Scherrer's proof that no square or cubic lattice contains a regular polygon with five sides or more than six sides.

For the polygon with more than six sides, if we translate each edge of the polygon to put an end of it at one vertex of the polygon, we necessarily get a polygon with the same number of edges, centred on that vertex, whose vertices are closer together than those of the original. If every vertex of the original polygon was on our lattice, each edge had whole number co-ordinates, so the new polygon's vertices are also on the lattice; but it's smaller. Since we can repeat this process with each smaller polygon we obtain, this gives us an endless supply of ever smaller polygons, always shrinking by the same factor that's less than one. That must ultimately give us a polygon whose sides are shorter than the separation of nearest neighbours in our lattice; this necessarily doesn't have all vertices on the lattice, hence the original polygon can't have had all vertices on the lattice.

For the pentagon, pick a vertex: translate a copy of each of the two edges neither opposite the vertex or adjacent to it, each along the edge joining one end of it to the vertex, so that it ends up inside the pentagon with what was its nearer vertex moved to the chosen vertex. Now translate a copy of each edge out of the chosen vertex along the other, so as to bring its other end to the position of the other vertex of the line we just translated along it. This has repositioned these four edges of the original pentagon to make each a diagonal of a smaller pentagon, sharing the chosen vertex. Again, by repeating this process, we can make arbitrarily small pentagons, that thus can't all lie on any given integer lattice, but would if the original pentagon's vertices all did, hence the original pentagon's vertices didn't all lie on the lattice.

Now, if we have some angle with rational Cos, the angle-multiplying formulae let us express the Cos of any multiple of that angle as a polynomial function of the give nrational; the result is thus also rational, so every multiple of our angle with rational Cos also has rational Cos. If it's a rational angle, it's turn.p/q for some whole numbers p, q; by eliminating common factors, we can ensure that p and q are coprime; and, since our trigonometric functions have turn as their period, we can reduce p mod q. As it's coprime to q, the successive multiples of p, mod q, deliver all the distinct numbers mod q. So the points {[Cos(i.turn.p/q), Sin(i.turn.p/q)]: i in q} are equally {[Cos(i.turn/q), Sin(i.turn/q)]: i in q}, the vertices of the regular polygon with q verties equally spaced round the unit circle. Because each Cos(i.turn/q) is a rational for each i in q, we can put each of these q rationals into coprime form as a ratio, compute the lowest common multiple of the denominators and scale up the figure to get a regular q-gon whose vertices's first components are all whole numbers. By the same process as above, if q is 5 or greater than six, this gives us a shrinking sequence of q-gons whose vertices must also have whole first components; since that's not possible, q must in fact be 1, 2, 3, 4 or 6. Rational multiples of the turn that have rational cosine are necessarily multiples of turn/6 or turn/4.

The case for rational Sin is fiddlier (since only the *odd* multiples
of the angle have their Sin as polynomials in the original rational Sin; the
even multiples need a factor of Cos, which isn't given to be rational) but can
trivially be obtained by observing that Cos(turn/4 −a) = Sin(a), so if
Sin(a) is rational and a is a rational multiple of the turn, then turn/4
−a must be some multiple of turn/4 or turn/6.

For Tan, the ratio of Sin and Cos, the multiples of an angle give values
that can be written as ratios of polynomials in the base angle's tangent; so,
again, if this is rational, so are the tangents of all multiples of the angle,
except for any that may be infinite (an honorary rational

as 1/0 is a
ratio of naturals). So a rational angle with rational tangent gives rise to a
polygon of equally-spaced vertices round the unit circle for which each ray from
the origin through one of these vertices does pass through at least some
positions in a two-dimensional integer lattice. It's not clear to me, however,
where the reasoning goes from there, as we're not obviously assured that there's
a radius at which all of these rays pass through integer co-ordinates.

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 characterized 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.

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 acceleration 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.

This leads to the functions (for which, to match orthodoxy, I use lower-case names, in contrast to Capitalised Sin and Cos) of sin = (: Sin(t.radian) ←t :) and cos = (: Cos(t.radian) ←t :), which map dimensionless inputs to dimensionless outputs. The factor of radian in their input ensures that sin' = cos and cos' = −sin. With these functions, we get exp(i.t) = cos(t) +i.sin(t) and can express cos and sin in terms of exp; this also leads to sin and cos having nice elegant Taylor series expansions. In this form, with radian as implicit unit of angle, cos and sin are thus nicely related to exp, easy to compute and have each other (give or take a sign but no scaling) as derivative. This is the other big argument in favour of the radian as unit of angle.

The flip side of these nice properties of sin and cos is that we often need to represent rational multiples of the turn as angles; when we do, we have to represent them as rational multiples of the irrational number π as input to sin and cos, or as offset to an existing input. (For contrast, few situations naturally call for rational multiples of the radian, although our use of radian as unit may sometimes lead to considering them.) In particular, if you want to re-write their derivatives in the offset form I gave above, you get sin'(t) = sin(t+π/2) and cos'(t) = cos(t+π/2). Having π sneak in to the input all over the place is less unwelcome than having factors of π show up (as they would using turn as unit) in the derivatives and in the Taylor series, but it's still far from welcome.

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∘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∘F is just (: f∘negate ←f :), with negate = (: −x ←x :). Clearly repeating this yields the identity as F∘F∘F∘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.

Choosing Q = 2.π effectively chooses 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.

None the less, even Fourier isn't entirely ambiguous on this: when it comes to considering the transform applied to derivatives of a function, the radian-based form of the transform gives a power of the input times the transform of the original function; the turn-based version, however, complicates this answer with some factors of 2.π; likewise, when transforming a function scaled by a power of its input, the radian-based transform gives a derivative of the transform of the original function, while the turn-based one again drags in some factors of 2.π. As ever, when derivatives are involved, the radian wins; but the turn wins in most other cases.

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.

Where two lines meet in an angle, if we draw a circle about the point where
they meet, small enough that both lines cut it, we can measure the length of the
arc of the circle between where the two lines cut the circle. (Measure the
arc's length along the circumference, not the length of the chord that takes a
short-cut via the interior of the circle.) Dividing the length of the arc by
the radius of the circle gives you the size of your angle, measured in
radians. Equally, if you start with an arc of a circle and connect its
end-points to the centre by straight lines, the angle these make at the centre
is, in radians, the result of dividing the length of the arc by the radius of
the circle. Analogously, the steradian is defined by: if we connect the
boundary of a region on the surface of a sphere to the centre, by straight lines
forming a (not necessarily circular) conical

surface, the solid angle at
the apex of the cone

, measured in steradians, is just the area of the
region divided by the square of the radius.

If we look at a distant object, the lines from our eyes to the apparent
boundary of the object form just such a (not necessarily circular) cone

;
we can intersect this with any sphere between us and the object, centred on our
eyes, and measure the solid angle by looking at the portion of the sphere
contained within the cone. That portion is what's covered by projecting the
object radially, towards the sphere's centre, onto the sphere. The object is
said to subtend

the resulting solid angle at

our eyes. Likewise,
in two dimensions, if we project a figure radially onto a circle, to cover a
portion of its circumference, we obtain the angle that the figure subtends
at

the centre of the circle.

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 square(radius.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.

Strictly speaking, the appropriate notion of a square

on the surface of a sphere requires us to use, as its straight

edges,
arcs of great circles (these are the circles in which the sphere meets planes
through its centre; the sense in which they are straight

is that the
shortest path, in the surface of the sphere, between two points on it, is always
an arc of a great circle); and the figure is square

if the four angles in
which the edges meet are equal – in which case they are (at least a
little) *bigger than* quarter turns, thanks to the sphere's
curvature. For small enough angles, this makes no practical difference, so we
can properly define the square of an angle

a by: find some number k large
enough that for squares whose sides subtend a/k at a sphere's centre, there is
negligible difference in area between a Euclidean square in a plane tangent to
the sphere at the plane's centre and a square in the surface of the sphere; now
measure the solid angle such a square subtends at the centre and multiply it by
k.k; this is the square of a. The important thing is that all sufficiently
large k give the same answer (give or take negligible differences); or, to put
it formally, the solid angle this gives you as a×a tends to a definite
limit as 1/k tends to zero. If the sphere's radius is R, the tiny squares have
side R.a/k/radian and area R.R.(a/radian).(a/radian)/k/k, hence subtend solid
angle steradian.(a/radian).(a/radian)/k/k, whence the square

of a is
a×a = steradian.(a/radian).(a/radian). In particular, substituting a =
radian, we get radian×radian = steradian. Technically, this × isn't
scalar multiplication, but we can now see that it behaves about as much like
scalar multiplication as the multiplication of lengths of sides of a square, so
it makes sense to denote it the same way and just write a.a for the square of
the angle a. Which is just a long-winded way of formalizing what I said
previously.

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.

From the above, we have solid angle as square of plane angle, with shell =
2.turn.radian. SI, in choosing to define the steradian, declares solid angle to
be an independent kind of dimension, separate from the angle. This shall oblige
it to give independent n-dimensional angles

for each positive integer n,
where I am inclined to build such units up from the radian, with the whole
n-sphere

in n+1 dimensions (analogous to turn for n = 1 and shell for n = 2)
then being the surface area

(boundary measure)
of the sphere times the n-th power of
radian/radius.

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, requiring 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 (albeit this makes little
sense), we'd thereby escape from needing endlessly many new forms
of angle

unit. In any case, if a(n) is the n-dimensional angle

of
a full shell, we get a(n+2) = turn.radian.a(n)/n, with a(1) = 2, a(2) = turn (so
a(0) = 0 / radian, a(−1) = −2/turn/radian and stranger things
below). Going up, from 1, while avoiding factors of π, we get 2, turn,
2.turn.radian, turn.turn.radian/2, 2.turn.turn.radian.radian/3,
turn.turn.turn.radian.radian/8 and so on; aside from integer factors, we
alternate units of turn and radian as we go up in dimension. This encourages me
to embrace both units and accept their differences.

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

- 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

measurein 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 …

Although SI does specify the radian as unit of angle,
it has also opted to treat
angles (and solid angles) as dimensionless, hence to treat the radian (and thus
the steradian) as a synonym for the multiplicative identity, the number 1. From
the references cited in that decision, I infer
that this
document would tell me more about the reasons behind that decision, were I
willing to pay twenty dollars (before knowing whether it actually *does*
answer my questions) to read it. For now I shall make do with the abbreviated
reasoning given in the decision:

- that an angle is usually determined by computing a ratio of two lengths (likewise, a solid angle is computed from the ratio between an area and a squared length); and
- that (paraphrasing wildly) no scientific formalism treats them otherwise and is, thereby, clearer.

The latter is somewhat undermined by the decision itself mentioning counter-examples; I would also argue that the stray factors of 2.π needed when converting between frequencies and angular rates is a symptom of a need for this (and this is the source of the same factor between Dirac's and Planck's constants). In any case, the lack of a formalism that fully explores the ramifications of systematically taking units of angles seriously is not a reason to dismiss the possibility of trying to build such a formalism. Had such a formalism been attempted and found problematic, that would be a good argument for treating angles as scalars; but the fact that everyone so far has cut corners is not always a good reason to cut the same corners.

As it happens, this very subject is presently (in 2015) under review; the BIPM's consultative committee for units (CCU) has a working group on angles and dimensionless quantities who shall meet in late February 2015, with an eye to making proposals for a 2016 SI brochure revision.

As to the former reason, I find it unpersuasive. The length of an arc of a circle is indeed proportional to the angle the arc subtends; and the constant of proportionality is just the circle's radius per radian. This is, furthermore, a universal truth (at least for small enough circles, relative to the ambient curvature of the space in which the circles are studied), that is reflected in the differentiation of the trigonometric functions. All the same, when I compare this with the proportionality of mass of a piece of uniform wire and the length thereof, I find enough similarity – even though the constant of proportionality for angles is universal where that for the wire's mass is a mere physical property of the particular spool of wire – to be suspicious of conflating the angle with the ratio of lengths.

As explained above, there are valid grounds for debate as to which unit of angle to take as natural; and the cleanest resolution of this is to admit that both the turn and the radian are relevant and useful units; for which it is necessary to accept angles as distinct from pure scalars. The existence of a convenient isomorphism between a one-dimensional space and scalars can serve as a rationale for identifying the two, but when there's more than one such isomorphism – each convenient in its own ways, in which the other is inconvenient – it is better to acknowledge that the one-dimensional space is merely one-dimensional, rather than trying to conflate it with {scalars}. Each such isomorphism corresponds to a choice of unit in the one-dimensional space: so two units each with virtues, but each deficient where the other has virtues, make a case for distinguishing what those units measure from scalars. Angles form a one-dimensional space over scalars (that in many contexts warrants interpretation via equivalence modulo differences by whole numbers of turns), in which two values stand out as important enough to beg to be used as unit; but, as they are distinct (indeed, not rationally commensurable), each undermines the case for using the other as unit.

Furthermore, when we mutiply angles we do not get an angle – as argued above, we get a solid angle – whereas scaling an angle by a true number does indeed give an angle. There is no natural interpretation, in plane geometry, of a product of angles as an angle; or, indeed, of a mechanism for multiplying angles other than, in a higher dimension, to get a higher-dimensional angle. So treating angles as pure numbers is misguided.

Written by Eddy.