Coordinate Systems in Physics

Try to describe the Earth's orbit around the Sun using ordinary (x, y, z) coordinates. You write down three numbers that are all changing at once, tangled together by x^2 + y^2 \approx \text{const}, and the "simple" fact that the planet stays a fixed distance from the Sun is buried inside a square root. Now switch to spherical coordinates, whose very first number r is the distance from the Sun. Suddenly the orbit is almost the single statement r \approx \text{const}. Nothing about the physics changed — only the language you chose to write it in.

That is the one idea of this page, and it is worth saying plainly before we drown in formulas: a coordinate system is a choice, and the smart choice is the one whose symmetry matches your problem. A ball on a string swinging in a circle wants polar coordinates. A current flowing down a long wire, or fluid down a pipe, wants cylindrical. A point charge, a star, a lone atom — anything that looks the same in every direction from a centre — wants spherical. Pick the coordinates that fit the symmetry and the mathematics collapses; pick the wrong ones and you fight algebra that need never have existed.

Below we meet the four workhorse systems, their conversion formulas, and the subtle, important idea of unit vectors that change as you move. That last point is where the deepest physics — and the classic mistakes — live. This page assumes you are comfortable with the mathematical toolkit and want to sharpen how you set problems up.

Cartesian: the flat grid you already know

Cartesian coordinates (x, y, z) label a point by three signed distances along three fixed, mutually perpendicular axes. Their great virtue is that the axes never move: the unit vectors \hat{\imath}, \hat{\jmath}, \hat{k} point in the same direction — along +x, +y, +z — no matter where you stand. That constancy is exactly why calculus is easy in Cartesian coordinates: when you differentiate a position vector, the unit vectors are constants and simply come along for the ride.

\mathbf{r} = x\,\hat{\imath} + y\,\hat{\jmath} + z\,\hat{k}, \qquad |\mathbf{r}| = \sqrt{x^2 + y^2 + z^2}.

The trouble is that Cartesian axes have no idea what your problem looks like. A circle, a sphere, a cylinder — none of them are natural on a square grid, and forcing them onto one produces square roots, awkward limits, and derivatives that hide the physics. When a problem has a centre or an axis of symmetry, we do better to build the coordinates around that symmetry.

Plane polar coordinates (r, \theta)

In two dimensions, polar coordinates replace (x, y) with a distance and a direction: r \ge 0, the straight-line distance from the origin, and \theta, the angle measured anticlockwise from the +x axis. Reading straight off a right triangle:

x = r\cos\theta, \qquad y = r\sin\theta,

and inverting to go back the other way,

r = \sqrt{x^2 + y^2}, \qquad \theta = \operatorname{atan2}(y, x).

Drag the slider under the figure. The point rides round a circle of fixed radius, and — this is the part to watch — the two little arrows at the point, \hat{r} and \hat\theta, rotate with it.

The unit vectors point "outward" and "anticlockwise" relative to where you are:

\hat{r} = \cos\theta\,\hat{\imath} + \sin\theta\,\hat{\jmath}, \qquad \hat{\theta} = -\sin\theta\,\hat{\imath} + \cos\theta\,\hat{\jmath}.

Because they carry \theta inside them, they are different vectors at different points. That single fact reshapes every polar derivative, as we will see.

Cylindrical coordinates (\rho, \varphi, z)

Take plane polar coordinates and stack them along a z axis: that is cylindrical coordinates. A point is given by \rho, its perpendicular distance from the z axis; \varphi, the azimuthal angle round that axis; and plain z, the height.

x = \rho\cos\varphi, \qquad y = \rho\sin\varphi, \qquad z = z, \rho = \sqrt{x^2 + y^2}, \qquad \varphi = \operatorname{atan2}(y, x), \qquad z = z.

This is the natural home of anything with a straight axis of symmetry: the magnetic field wrapping around a long current-carrying wire, water flowing down a pipe, a spinning shaft, the field of a charged rod. Ask "how far from the axis, how far around, how far along?" and the algebra of a would-be nightmare tidies itself into three nearly independent variables. The unit vectors are \hat\rho (away from the axis), \hat\varphi (around it) and the familiar constant \hat{k} (along it).

Because r is about to be needed for something else. In spherical coordinates r is the full distance from the origin (the length of the position vector), whereas the cylindrical \rho is only the distance from the z axis — the "shadow" of that length on the xy plane. Reserving \rho for the cylindrical radius keeps the two straight, and it makes the link between the systems clean: the cylindrical \rho equals the spherical r\sin\theta. Some books use s instead of \rho for the same reason.

Spherical coordinates (r, \theta, \varphi)

For anything organised around a single point — a planet orbiting a star, the electric field of a point charge, an atom's electron cloud — use spherical coordinates. A point is fixed by r, the distance from the origin; \theta, the polar angle measured down from the +z axis (so \theta = 0 is the north pole, \theta = \pi/2 the equator); and \varphi, the azimuthal angle round the z axis, just as in cylindrical. Reading off the geometry:

x = r\sin\theta\cos\varphi, \qquad y = r\sin\theta\sin\varphi, \qquad z = r\cos\theta,

and back again,

r = \sqrt{x^2 + y^2 + z^2}, \qquad \theta = \arccos\!\frac{z}{r} = \arccos\frac{z}{\sqrt{x^2+y^2+z^2}}, \qquad \varphi = \operatorname{atan2}(y, x).

The pattern to remember: z is the "up" component, r\cos\theta; the leftover length r\sin\theta is the radius of the horizontal circle at that height, and \cos\varphi, \sin\varphi share it out between x and y. The three unit vectors \hat{r}, \hat\theta, \hat\varphi point "outward", "southward along a meridian", and "eastward around" — and all three change from point to point.

The heart of it: unit vectors that move

Here is the idea that separates someone who has merely memorised the formulas from someone who can use them. In Cartesian coordinates the basis vectors are frozen, so a velocity is just \dot x\,\hat{\imath} + \dot y\,\hat{\jmath} — differentiate the components and stop. In polar coordinates you cannot stop there, because \hat{r} and \hat\theta themselves depend on \theta, which is changing in time. Differentiating \hat{r} = \cos\theta\,\hat{\imath} + \sin\theta\,\hat{\jmath} gives

\frac{d\hat{r}}{dt} = \dot\theta\,\hat\theta, \qquad \frac{d\hat\theta}{dt} = -\dot\theta\,\hat{r}.

Feed that into \mathbf{r} = r\,\hat{r} and the velocity and acceleration grow extra terms out of nowhere:

\mathbf{v} = \dot r\,\hat{r} + r\dot\theta\,\hat\theta, \qquad \mathbf{a} = \left(\ddot r - r\dot\theta^{\,2}\right)\hat{r} + \left(r\ddot\theta + 2\dot r\dot\theta\right)\hat\theta.

Those "extra" pieces are not bookkeeping — they are real physics. The -r\dot\theta^{\,2} is the centripetal acceleration that a satellite feels; the 2\dot r\dot\theta is the Coriolis term that swings the winds of a hurricane. They appear because the unit vectors turn as the particle moves. Miss that, and circular motion makes no sense at all.

Worked examples

Example 1 — Cartesian to polar, and back. A point sits at (x, y) = (-3, 3). Its radius is

r = \sqrt{(-3)^2 + 3^2} = \sqrt{18} = 3\sqrt{2} \approx 4.24.

The point is in the second quadrant (negative x, positive y), so the angle is \theta = \operatorname{atan2}(3, -3) = 135^\circ = 3\pi/4. Going back the other way confirms it: x = 3\sqrt{2}\cos 135^\circ = 3\sqrt{2}\left(-\tfrac{1}{\sqrt2}\right) = -3 and y = 3\sqrt{2}\sin 135^\circ = 3.

Example 2 — spherical to Cartesian. A point has r = 2, \theta = 60^\circ, \varphi = 90^\circ. Then

x = 2\sin 60^\circ\cos 90^\circ = 2\cdot\tfrac{\sqrt3}{2}\cdot 0 = 0, y = 2\sin 60^\circ\sin 90^\circ = 2\cdot\tfrac{\sqrt3}{2}\cdot 1 = \sqrt3 \approx 1.73, z = 2\cos 60^\circ = 2\cdot\tfrac12 = 1.

So (x, y, z) = (0,\ \sqrt3,\ 1). As a check, \sqrt{0 + 3 + 1} = 2 = r.

Example 3 — choosing the system. Which coordinates suit each problem, and why?

No — and believing they are is the single most common blunder in this whole subject. The Cartesian \hat{\imath} and \hat{\jmath} genuinely are the same arrow at every point in space, so their time-derivative is zero. But \hat{r} = \cos\theta\,\hat{\imath} + \sin\theta\,\hat{\jmath} carries \theta inside it: move to a point at a different angle and \hat{r} points somewhere new. So d\hat{r}/dt \ne 0 whenever \theta is changing.

This is precisely why the velocity in polar coordinates is not just \dot r\,\hat{r}. Treat the unit vectors as constants and you will "lose" the r\dot\theta\,\hat\theta term — and with it the entire physics of circular motion, the centripetal -r\dot\theta^2 and the Coriolis 2\dot r\dot\theta. The moving basis is a feature, not a nuisance: it is where the interesting forces come from. Never differentiate a polar or spherical position vector without differentiating its basis too.

In spherical coordinates, physicists and mathematicians swap two letters, and it causes real grief. In the physics convention used here (and in ISO 80000), \theta is the polar angle measured down from the z axis, and \varphi is the azimuthal angle round it. Many mathematics texts do the exact opposite: \theta for the azimuth and \varphi for the polar angle. Written as an ordered triple, physics says (r, \theta, \varphi) and maths often says (r, \varphi, \theta) for the same point.

Neither is wrong; they are just different dictionaries. Before you trust any spherical formula — a volume element, a gradient, a table of unit vectors — check which angle the author calls \theta. A sign or a \sin\leftrightarrow\cos swap that "makes no sense" is very often this convention clash in disguise. And while you are at it, prefer \operatorname{atan2}(y, x) over \arctan(y/x) for the azimuth: plain \arctan throws away the quadrant, mapping both (1, 1) and (-1, -1) to the same 45^\circ.