The angle that is really an area

A sine wave is a circle seen from the side: a point walks round the unit circle, its height is plotted against how far it has walked, and the plot is $\sin t$. The distance walked is the angle in radians, and radians are natural because they are the circle’s own ruler.

There is a second reading of the same number, and on the circle nobody needs it. The slice of the disc swept out by the radius as the point walks distance $t$ is a fraction $t/2\pi$ of the whole disc, whose area is $\pi$. So the swept area is exactly $t/2$. The angle is the arc length, and it is also twice the area swept, and on a circle those are the same number for a reason as simple as the formula for a sector.

The two readings stop agreeing the moment the circle is replaced by another curve. Choosing between them is the whole content of the hyperbolic functions.

Replacing the circle with a hyperbola #

The unit circle is the set of points with $x^2 + y^2 = 1$. Change one sign and the equation becomes $x^2 - y^2 = 1$: a hyperbola with two branches, of which the right-hand one passes through $(1, 0)$ exactly where the circle does, and bends away toward the two lines $y = \pm x$ instead of curling back.

Everything in the circular construction has an obvious analogue. Start at $(1, 0)$, move along the curve, and give each point a coordinate pair $(x, y)$ labelled by some measure of how far the point has gone. The functions $\cos$ and $\sin$ were the coordinates as functions of that measure; the new functions should be the same thing for the hyperbola. What is not obvious is which measure to use.

Arc length is the first thing to try, and it fails. The length of the hyperbola’s arc from $(1, 0)$ to a point is an integral that cannot be written in terms of any of the familiar functions — it is an elliptic integral, of the same kind as the perimeter of an ellipse — and the coordinates, written as functions of it, have no simple description at all. The rule that made radians natural on the circle produces nothing usable on the hyperbola.

Area works #

Try the other reading. Measure each point by twice the area swept by the segment from the origin as the point moves along the curve from $(1, 0)$: the region bounded by the two segments from the origin and the arc of the hyperbola between them.

The area can be computed. The region under the segment from the origin to $(x, y)$ is a triangle of area $xy/2$; subtract the area under the hyperbola between $x = 1$ and $x$, which is $\int_1^x \sqrt{s^2 - 1}\,ds$, and what remains is the swept region. The integral works out, and the swept area is

$$\frac12 \ln(x + y).$$

Call twice that area $u$. Then $x + y = e^u$. And since $(x + y)(x - y) = x^2 - y^2 = 1$, the other combination is $x - y = e^{-u}$. Adding and subtracting,

$$x = \frac{e^u + e^{-u}}{2} = \cosh u, \qquad y = \frac{e^u - e^{-u}}{2} = \sinh u.$$

The coordinates of the point that has swept area $u/2$ are the hyperbolic cosine and sine of $u$, and they are made of exponentials. The functions were defined by those formulas long before anybody drew the region, and the region explains why the names are right: they are to the hyperbola exactly what cosine and sine are to the circle, provided the parameter is read as an area.

The larger figure shows the difference that the construction does not hide. The circle’s point comes back: after area $\pi$ it has swept the whole disc and begins again, which is periodicity. The hyperbola’s point never comes back. It runs off along the branch, hugging the line $y = x$, and the area it has swept grows without limit while the point’s coordinates grow exponentially. Nothing on the hyperbola repeats, so nothing in $\cosh$ and $\sinh$ oscillates.

Equal slices, exponential steps #

The formula $x + y = e^u$ says something striking about points spaced evenly in area: they are not spaced evenly along the curve in any obvious sense, but they are spaced geometrically along the asymptote.

The reason is a linear map that plays the role rotation plays on the circle. In the coordinates $X = x + y$ and $Y = x - y$ — the directions of the two asymptotes — the hyperbola is $XY = 1$. The map that multiplies $X$ by $\lambda$ and divides $Y$ by $\lambda$ sends the hyperbola to itself, stretching along one asymptote and compressing along the other by the same factor. Its determinant is one, so it preserves area.

That map is called a squeeze, and it is to the hyperbola what a rotation is to the circle: it slides every point of the curve along the curve and carries every swept region to a region of the same area. Sliding by area $u/2$ therefore multiplies $X = x + y$ by a fixed factor, and doing it twice multiplies by that factor twice, so the factor is an exponential in $u$. That is where the $e$ comes from: not from any choice of base, but from area-preserving slides composing by multiplication while areas add.

The difference between the squeezes and the rotations is also the difference between the two families’ graphs. Rotating by $2\pi$ returns every point to where it started, so the rotations form a circle’s worth of maps, closed up on themselves, and every function built from them repeats. Squeezing never returns anything: the squeezes form a line’s worth of maps, running off to infinity in both directions, one for every real value of the area. A family of maps that closes up produces periodic functions and a family that does not produces monotone ones, and $\cos$ and $\cosh$ are the two cases of the same statement. The whole difference between oscillation and exponential growth is whether the group of area-preserving slides along the curve is compact.

The same squeeze runs through the theory of Pell’s equation, where the whole-number points on $x^2 - 2y^2 = 1$ are one solution slid repeatedly along the hyperbola by a fixed area-preserving map, equally spaced in exactly this parameter and exponentially spaced in ordinary coordinates.

A logarithm that was an area first #

The formula $u = \ln(x + y)$ says the hyperbolic parameter is a logarithm, and the history runs the other way round: the logarithm was recognised as an area under a hyperbola before anybody knew it was the inverse of an exponential.

In the asymptote coordinates the hyperbola is $XY = 1$, which is the graph of $Y = 1/X$. The area under that graph between $X = 1$ and $X = c$ is, by the calculus that came later, $\ln c$. But its key property was seen without calculus. In 1647 Grégoire de Saint-Vincent observed that stretching the region horizontally by a factor and compressing it vertically by the same factor — the squeeze again — carries the area under $1/X$ between $a$ and $b$ to the area between $\lambda a$ and $\lambda b$ without changing it. So the area from 1 to $ab$ splits into the area from 1 to $a$ plus the area from $a$ to $ab$, and the second equals the area from 1 to $b$. Areas under the hyperbola turn products into sums, which is exactly what a logarithm does, and his student Alphonse Antonio de Sarasa said so explicitly two years later.

The hyperbolic sector and the region under $1/X$ are related by that same squeeze-invariant geometry: the triangles at the two ends of the sector have equal areas, so the sector’s area equals the area under the curve between the corresponding values of $X$, scaled by the factor the coordinate change introduces. That is why the swept area is half a logarithm.

The inverse functions keep a trace of this in their names. The inverse of $\sinh$ is written $\operatorname{arsinh}$, and it is increasingly spelled that way rather than $\operatorname{arcsinh}$, because the prefix stands for area, not arc: $\operatorname{arsinh} y = \ln\big(y + \sqrt{y^2 + 1}\big)$ returns twice the area swept to reach height $y$. For the circle’s inverse functions the arc and the area readings agree, and the prefix arc was never wrong. For the hyperbola only the area reading is right, and the notation has been slowly catching up.

The identities, read off the squeeze #

Every circular identity has a hyperbolic twin with a sign changed, and the squeeze explains why without any calculation.

$\cosh^2 u - \sinh^2 u = 1$ is the statement that the point is on the hyperbola, as $\cos^2 t + \sin^2 t = 1$ is the statement that the point is on the circle.

The addition formulas come from composing slides. Sliding by $a$ and then by $b$ sweeps area $(a + b)/2$, so it is the slide by $a + b$. Written as matrices in $(x, y)$ coordinates, a slide by $u$ is

$$\begin{pmatrix} \cosh u & \sinh u \\ \sinh u & \cosh u \end{pmatrix},$$

and multiplying the matrices for $a$ and $b$ gives

$$\cosh(a + b) = \cosh a \cosh b + \sinh a \sinh b, \qquad \sinh(a + b) = \sinh a \cosh b + \cosh a \sinh b.$$

These are the circular formulas with one sign flipped, and the flipped sign is the one in the equation of the curve. On the circle, multiplying complex numbers adds angles because rotations compose; on the hyperbola, squeezes compose and areas add. The two families of identities are the same statement about two different groups of area-preserving maps.

The even and odd halves of one exponential #

Any function splits uniquely into an even part and an odd part. For $e^t$ the even part is $\cosh t$ and the odd part is $\sinh t$ — that is what the defining formulas say. The picture shows the split: the bowl and the S add, point by point, to the dotted exponential, and at negative $t$ they nearly cancel.

The circular functions are the same split taken along a different line. Euler’s formula $e^{it} = \cos t + i \sin t$ says $\cos$ and $\sin$ are the even and odd parts of the exponential evaluated on the imaginary axis. So $\cos t = \cosh(it)$ and $\sin t = -i\sinh(it)$, and the circle is the hyperbola seen at imaginary parameter. Turning $x^2 - y^2 = 1$ into $x^2 + y^2 = 1$ by replacing $y$ with $iy$ is the same substitution.

That is why the two families have identical shapes of formula, why the curve that is its own slope sits underneath both, and why the second derivative of $\cosh$ is $+\cosh$ where the second derivative of $\cos$ is $-\cos$. A quantity pulled back toward zero in proportion to its size goes round a circle; a quantity pushed away in proportion to its size runs out along a hyperbola.

The shape a chain hangs in #

The hyperbolic cosine was not invented for the hyperbola. It turned up first as the answer to a physical question, and the answer is worth seeing because it is so easily confused with something else.

Galileo suggested that a hanging chain forms a parabola, and for a shallow sag the two are almost indistinguishable. For a deep sag they are not: the chain’s sides are steeper near the posts and its bottom is flatter. In 1691 Leibniz, Huygens and Johann Bernoulli independently found the correct curve, the catenary, from the balance of forces on a short piece of chain. The horizontal tension is the same everywhere, the vertical tension grows with the weight of chain below, and that weight is proportional to arc length — so the slope’s rate of change is proportional to $\sqrt{1 + y'^2}$. The function whose derivative $\sinh$ satisfies exactly that equation is $\cosh$.

The catenary carries the hyperbolic functions in a second way that completes the picture. The arc length of $y = a\cosh(x/a)$ measured from its lowest point is $a\sinh(x/a)$. So on the hanging chain, $\cosh$ is the height and $\sinh$ is the distance along the chain — a curve on which the hyperbolic sine is an arc length, even though on the hyperbola it could not be.

What the shaded regions leave out #

The regions in the figures have their areas computed from the drawn polygons, and the agreement with $u/2$ to four decimal places is a check of the construction at the parameters drawn. It is not the derivation of $\tfrac12\ln(x + y)$, which needs the integral, and nothing in a picture shows that the area and the logarithm agree everywhere.

The figures also cannot show the failure of arc length. The claim that the hyperbola’s arc length is not an elementary function of its endpoint is a theorem about which integrals can be written in closed form, and a drawn arc has a length like any other. What makes area the right choice is invisible in both pictures: it is that area, unlike length, is preserved by the squeeze, and so the parameter it defines turns composition of slides into addition.

And the substitution $y \mapsto iy$ that turns the hyperbola into the circle has no picture at all. The waves figure shows the two families on real axes, where they look unrelated — one bounded and oscillating, one unbounded and monotone. That they are one function seen along two perpendicular lines through the complex plane is a statement about complex numbers, and the real graph is exactly the view in which it cannot be seen.

Where the area reading reaches further #

The hyperbolic parameter has physical meaning beyond chains, and the most striking example is special relativity. The inner product of space-time, $t^2 - x^2$, has a square that can be negative, and the transformations preserving it are exactly the squeezes above, acting on a time coordinate and a space coordinate. A change of reference frame moving at velocity $v$ is a squeeze by the parameter $u$ with $\tanh u = v$ — the rapidity. Rapidities add when frames are composed, as angles add when rotations are composed, while the velocities themselves combine by the hyperbolic tangent’s addition formula, which is why no sum of velocities below the speed of light ever reaches it.

The same functions give the geometry of the hyperbolic plane, where the right-triangle rule becomes $\cosh c = \cosh a \cosh b$, and the elliptic coordinates in which ellipses and hyperbolas with shared foci cross at right angles. In each case the parameter is the one the area defines, and in each case reading it as a length would give nothing.

Still open: whether e and π are related by anything #

The circle and the hyperbola are one curve at real and imaginary parameter, and the constants attached to them — $\pi$, the half-period of the circle’s functions, and $e$, the base of the hyperbola’s — are tied together by $e^{i\pi} = -1$. Both are known to be transcendental, satisfying no polynomial equation with whole-number coefficients; $\pi$’s transcendence is what settled the squaring of the circle.

What is not known is whether they are related in any other way. It is not known whether $e + \pi$ is irrational, nor whether $e\pi$ is. At least one of the two must be irrational — if both were rational, $e$ and $\pi$ would be the roots of a quadratic with rational coefficients, and they are not algebraic — but nobody can say which. A conjecture of Stephen Schanuel from the 1960s would imply that $e$ and $\pi$ are algebraically independent, satisfying no polynomial relation at all, and it remains unproved. The single identity connecting them is the one the circle and the hyperbola supply, and whether it is the only one is open.

The ruler the curve provides #

The radian was sold as the circle’s own ruler, and it turns out the circle has two rulers that happen to read the same. Arc length is the one that is easy to see and does not generalise. Swept area is the one that generalises, because it is preserved by the maps that move points along the curve — rotations on the circle, squeezes on the hyperbola.

Read that way, $\cosh$ and $\sinh$ are not an imitation of $\cos$ and $\sin$ with exponentials substituted in. They are the same construction on the other non-degenerate conic with a centre, with the parameter measured the only way the construction allows, and the exponential falls out of the area as the circle’s periodicity falls out of its closing up.