Orthogonal and Isogonal Trajectories

Pour iron filings on a sheet of paper, set a bar magnet underneath, and tap the paper. The filings snap into a pattern of curves that flow from the north pole to the south pole. Now imagine drawing, on top of that picture, a second family of curves — one curve through each point — each of which crosses every single magnetic field line at exactly a right angle. These second curves are a physical thing: they are the lines along which the magnetic potential is constant. Physicists call them equipotential curves. Mathematicians call them the orthogonal trajectories of the field lines.

The same thing happens with a pool of still water that you heat from the bottom. The temperature inside the water is a function of position: warmer near the heater, cooler near the top. If you draw the curves where the temperature is constant (the isotherms), and then draw the curves along which heat actually flows, the two families cross each other at right angles everywhere. One family is orthogonal to the other.

This is not an exotic coincidence. It is how electric fields relate to equipotentials, how fluid stream lines relate to potential lines, how the level curves of a mountain relate to the lines of steepest descent. Two families of curves, locked together by a right angle. And the tool that lets you compute one family from the other is a differential equation.

A specific family to stare at

Take the family of circles x^2 + y^2 = c^2 — all circles centred at the origin, one for each value of the constant c. Not one circle. The whole infinite collection: a tiny circle for c = 0.1, a bigger one for c = 1, a much bigger one for c = 5, and so on. Every point in the plane (except the origin) lies on exactly one of these circles.

Now suppose you want to draw a curve that, at every point it passes through, crosses the circle at that point at a right angle. What does such a curve look like? Stare at the picture for a moment.

At the point (1, 0), the circle through that point is x^2 + y^2 = 1. The tangent to the circle at (1, 0) is vertical. For your new curve to cross the circle at a right angle there, it must be horizontal at (1, 0) — it must point along the x-axis. The same reasoning at (3, 0), at (-2, 0), at any point on the x-axis: your new curve is horizontal there.

At the point (0, 2), the circle's tangent is horizontal. So your new curve must be vertical — pointing straight along the y-axis. The same at (0, -5), at (0, 1), and so on.

At the point (1, 1), the circle's tangent has slope -1. For perpendicularity, your new curve needs slope +1 there.

A pattern is forming. At every point (x, y), your new curve is pointing radially outward from the origin. So the new curves are simply the straight lines through the origin: y = mx for every slope m. This family is the orthogonal trajectories of the family of circles. And the picture of the two families together — circles and radial lines — is the one you've seen on every compass and every ripple in a pond.

The question is: how would you have found the second family if you had only been given the first one — without guessing by staring at the picture? That is the question this article answers.

From a family to a differential equation

The first idea is that any family of curves with a parameter c can be stripped of that parameter and turned into a differential equation — and the differential equation is what captures the slope of the family member at every point.

Take the circles x^2 + y^2 = c^2. Differentiate both sides with respect to x, treating y as a function of x:

Solve for \dfrac{dy}{dx}:

The constant c is gone. What remains is a rule that says: at every point (x, y) in the plane, the member of the family passing through that point has slope -\dfrac{x}{y}. At (1, 0) the slope is -\dfrac{1}{0}, which means vertical — matching what you saw above. At (0, 2) the slope is 0 — horizontal. At (1, 1) it is -1. Everything checks out.

This equation \dfrac{dy}{dx} = -\dfrac{x}{y} is the differential equation of the family. Every circle in the family satisfies it. The differential equation is what the family is, if you strip away the specific value of c.

The perpendicularity trick

Here is the one idea on which everything rests. If two curves cross at a point and are perpendicular there, the slopes of their tangent lines at that point multiply to -1:

This is a standard fact about perpendicular lines. So if the original family has slope m_1 at a point, the orthogonal trajectory through that point must have slope

Translated into differential equations: if the original family satisfies

then the orthogonal trajectories satisfy

In operational terms: write down the differential equation of the family, then replace \dfrac{dy}{dx} by -\dfrac{1}{\,dy/dx\,} = -\dfrac{dx}{dy}. Solve the new equation. Its solutions are the orthogonal trajectories.

Now watch it work.

Computing one from the definition

A second example, for the pattern

Two families of curves, perpendicular everywhere, both falling out of one differential equation rearrangement. That is the power of the method.

Isogonal trajectories

What if you do not want a right angle — what if you want every trajectory to cross every curve in the family at some other fixed angle, say 30^\circ? Those are called isogonal trajectories, from the Greek for "equal angles." Orthogonal is just the special case \alpha = 90^\circ.

The derivation uses a formula you already know from coordinate geometry. If two lines have slopes m_1 and m_2, the angle \alpha between them satisfies

Drop the absolute value for a moment (you can recover it by choosing the right branch) and solve for m_2 in terms of m_1:

So the recipe becomes: given the differential equation \dfrac{dy}{dx} = f(x, y) of the original family, the isogonal trajectories at angle \alpha satisfy

Set \alpha = 90^\circ and note that \tan(90^\circ) is infinite. Divide top and bottom by \tan\alpha and let \tan\alpha \to \infty: the f(x,y)/\tan\alpha and 1/\tan\alpha terms vanish, leaving \dfrac{dy}{dx} = \dfrac{1}{-f(x,y)} = -\dfrac{1}{f(x,y)} — exactly the orthogonal rule from before. The isogonal formula is a strict generalisation.

For a concrete taste: take the family y = mx (all lines through the origin). Its differential equation is \dfrac{dy}{dx} = \dfrac{y}{x}. The isogonal trajectories at angle \alpha satisfy

This is homogeneous of degree zero, and after the substitution y = vx and some patient algebra, the solution turns out to be a family of logarithmic spirals — curves of the form r = k e^{\theta \cot\alpha} in polar coordinates. Cut a spiral shell in half, look at the cross-section: a nautilus shell is a logarithmic spiral, and it crosses every radius through the centre at the same constant angle. Nature is running this construction for you.

Common confusions

A few things students reliably get wrong the first time they meet trajectories.

• "The orthogonal family is just the same equation with x and y swapped." Not in general. It happens to work for some simple families (like the circles-to-lines case, where the swap does give you lines) but in general you must write the differential equation first, make the \dfrac{dy}{dx} \to -\dfrac{dx}{dy} swap, and then solve. The swap is in the derivative, not in the coordinates.

• "You should eliminate the constant after the swap." No. Eliminate the parameter first, so that the differential equation depends only on x and y. The swap is a legal operation only on a differential equation that no longer mentions the parameter.

• "-\dfrac{1}{dy/dx} is undefined when the original slope is zero." It looks that way, but the underlying geometry is fine: if the original curve is horizontal at a point, the orthogonal trajectory is vertical there. In the differential equation this shows up as \dfrac{dx}{dy} = 0, which is perfectly valid. Just be willing to flip which variable you treat as independent when the formula threatens to divide by zero.

• "Orthogonal trajectories are unique." Orthogonal trajectories form their own family, with one curve through each point. Through any single point, though, there is exactly one trajectory — the same way one curve of the original family passes through each point. Two families, meshed together.

Going deeper

In polar coordinates

Sometimes the family is more natural to describe in polar coordinates (r, \theta) than in Cartesian coordinates. The perpendicularity condition transforms: the angle a curve makes with the radius vector at a point is captured by \tan\psi = r \dfrac{d\theta}{dr}. Two curves are orthogonal at a point when their \psi-angles differ by 90^\circ, which translates into

So the polar replacement rule is "substitute r \dfrac{d\theta}{dr} with -\dfrac{1}{r \, d\theta/dr}," which simplifies to the crisp rule: replace \dfrac{dr}{d\theta} with -r^2 \dfrac{d\theta}{dr}. Try it on the family r = 2a\cos\theta (circles through the origin with centres on the x-axis) and you will get another family of circles through the origin — this time with centres on the y-axis. The perpendicularity of the two families is a classical result and, once you compute it, obvious from the picture.

Why perpendicularity is the natural rule

In physics, two families of curves arise together all the time: level curves of a scalar field and flow lines of the gradient of that field. The gradient always points perpendicular to the level curves — that is not a choice, that is a theorem of multivariable calculus. So whenever you compute a scalar potential and its gradient, you automatically generate a pair of orthogonal families. Electric field lines and equipotentials. Temperature gradients and isotherms. Altitude contour lines on a map and the lines of steepest descent that a raindrop would follow. Orthogonal trajectories are how you compute one family from the other, in a regime where you can't see the scalar field itself.

Self-orthogonal families

Sometimes a family is its own orthogonal trajectories. The family of parabolas y^2 = 4a(x + a), where a is the parameter, has this property: its orthogonal trajectories form the same family with a different value of a. Check it by running the procedure — the differential equation you get after the swap has the same solution form as the original. Such families are called self-orthogonal, and they come up in the study of confocal conics (ellipses and hyperbolas sharing the same foci, where each ellipse is perpendicular to each hyperbola at every crossing).

Where this leads next

You now know how to take a family of curves, turn it into a differential equation, and extract the perpendicular family by one rearrangement. This is the first real application of differential equations — the first time solving an ODE tells you something geometric you did not already know.

• First Order - Variable Separable — the solving technique you used in both examples; every orthogonal trajectory problem boils down to a separable ODE (or an easy linear one).
• First Order - Reducible to Separable — for isogonal trajectories at non-right angles, the resulting ODE is often homogeneous in x and y; this article shows you the substitution that tames them.
• Applications of Differential Equations — other geometric and physical applications, from population growth to cooling to exponential decay.
• Differential Equations - Introduction — the general framework that orthogonal trajectories are a special case of.