Generalized coordinates & transformations

1. Choose variables that belong to the motion

A pendulum's Cartesian coordinates are related by a fixed-length condition. Its angle, however, can be chosen directly. Once the angle is known, both Cartesian coordinates follow. The advantage is practical: we stop repeatedly enforcing a restriction that our coordinates could satisfy automatically.

Prerequisites: degrees of freedom and configuration space from Chapter 2; sine, cosine, and partial derivatives from the prerequisite guide.

Definition. Generalized coordinates $q_1,\ldots,q_n$ are independent parameters that locally specify a system's configuration, subject to the constraints already built into the coordinate description.

The word “generalized” does not mean a new kind of physical position. It means that we may use suitable variables other than Cartesian distances. Angles, distances along a track, extensions, and combinations of positions can all be useful.

The coordinates must be sufficient and locally independent. Using a pendulum's $x$ and $y$ as two unconstrained generalized coordinates would fail independence. Using only the angle for a bob on an extensible spring would fail sufficiency.

2. Transformation equations

Let $a$ label particles and $i$ label generalized coordinates. For $N$ particles with $n$ configuration freedoms, write

This is shorthand for three scalar transformations per particle: $x_a=x_a(q,t)$, $y_a=y_a(q,t)$, and $z_a=z_a(q,t)$. Knowing all the $q_i$ and the prescribed time determines every particle position.

The explicit $t$ allows moving geometry. It is distinct from the implicit time dependence of $q_i(t)$ along a motion. In a fixed-pivot pendulum, the transformation has no explicit time. In a prescribed moving-pivot pendulum, it generally does.

A fixed planar pendulum

Take $x$ to the right, $y$ upward, and measure $\theta$ from the downward vertical toward the right. For length $a$,

Substitute these into the constraint:

Every value of $\theta$ satisfies the fixed-length condition. It has been solved geometrically before the dynamics begins.

Polar coordinates for a free planar particle

Choose $q_1=r$ and $q_2=\theta$, where $\theta$ is measured from the positive $x$ axis. Then

Both $r$ and $\theta$ vary independently if no circle constraint is present. If the particle is confined to $r=a$, substitute that fixed value and keep only $\theta$. The same coordinate system can therefore describe an unconstrained planar particle or a constrained bead; the physical assumptions decide how many variables remain independent.

3. Meaning and dimensions of the notation

If $q_i$ is a length, $\dot q_i$ has units of speed. If $q_i$ is an angle, $\dot q_i$ has units of angular speed. Radians are dimensionless in dimensional analysis, but retaining “rad” helps identify the variable's meaning.

The dimensions of $\partial\mathbf r/\partial q_i$ are length divided by the dimensions of $q_i$. It need not be a unit vector. For the pendulum,

Its magnitude is $a$. A small angular change $\delta\theta$ produces an arc displacement of magnitude $a|\delta\theta|$. The factor $a$ is why this derivative is not simply the tangent unit vector.

The derivative $\partial\mathbf r_a/\partial q_i$ is a geometrical translator. It converts a change in your chosen coordinate into the physical displacement of particle $a$.

4. Local validity and coordinate choice

A useful coordinate system must distinguish nearby admissible configurations. For a circle, using $x$ alone works on an arc where a chosen sign of $y=\pm\sqrt{a^2-x^2}$ is understood. It does not distinguish upper and lower points globally. Near the leftmost or rightmost point, the description by a single branch of $y(x)$ also becomes inconvenient. The angle describes the circle more naturally, though its numerical value still requires a periodic convention.

Coordinate singularities do not necessarily indicate physical singularities. At $r=0$, all polar angles describe the same point. Cartesian coordinates remain regular there. A formula containing $1/r$ must therefore be used with attention to its coordinate domain.

There is no unique correct coordinate choice. If $q$ is a valid coordinate, a smooth invertible reparameterization may also be valid. For a pendulum, arc length $s=a\theta$ can replace angle. The final physical prediction must be the same after transforming back.

5. Applications

Example 1 · Cartesian position of a pendulum bob

For a pendulum of length $a=0.80\,\mathrm m$ at $\theta=30^\circ$ from the downward vertical, the Cartesian coordinates relative to the pivot follow directly from the transformation equations.

$x=0.80\sin30^\circ=0.40\,\mathrm m$ and $y=-0.80\cos30^\circ=-0.693\,\mathrm m$ to three significant figures. The negative $y$ puts the bob below the pivot. Also $\sqrt{x^2+y^2}=0.80\,\mathrm m$, so the result satisfies the geometry.

Example 2 · Parameterizing a helical wire

A fixed helical wire is described by $x=R\cos\phi$, $y=R\sin\phi$, $z=b\phi$, where $R>0$ and $b\ne0$ are constants. The parameter $\phi$ specifies both the position around the axis and the height along it.

Choose $q=\phi$, because one number fixes all three Cartesian coordinates. Differentiate each component:

Its squared magnitude is $R^2\sin^2\phi+R^2\cos^2\phi+b^2=R^2+b^2$. Thus a small change $d\phi$ corresponds to arc length $ds=\sqrt{R^2+b^2}|d\phi|$. Increasing $\phi$ by $2\pi$ raises $z$ by $2\pi b$, so these two values do not represent the same point, unlike a circle.

Example 3 · Coordinate transformations for a double pendulum

Take two rods of lengths $a,b$ in a vertical plane, with both angles $\theta_1,\theta_2$ measured from the downward vertical. The position of each bob is obtained by adding the relevant rod vectors.

The first bob has $x_1=a\sin\theta_1$, $y_1=-a\cos\theta_1$. The displacement from the first bob to the second is $(b\sin\theta_2,-b\cos\theta_2)$. Add the vectors:

Subtracting the first bob's position gives a vector of squared length $b^2$. Hence both rod constraints are built in. Notice that changing $\theta_1$ moves both bobs, whereas changing $\theta_2$ leaves the first bob fixed. Generalized coordinates need not correspond one-to-one with particles.

Example 4 · Pendulum with a translating pivot

Let the pivot of a fixed-length pendulum move as $(Ut,0)$, where $U$ is prescribed and constant. The bob moves both with the support and relative to it; the transformation must include both contributions.

The bob's position is $\mathbf r=(Ut+a\sin\theta,-a\cos\theta)$. Holding $\theta$ fixed while differentiating with respect to time gives $\partial\mathbf r/\partial t=(U,0)$. Holding time fixed while differentiating with respect to angle gives $\partial\mathbf r/\partial\theta=(a\cos\theta,a\sin\theta)$.

The first derivative describes transport by the support; the second describes motion relative to that support. Along a physical motion both effects contribute to velocity. Chapter 4 will combine them using the chain rule.

6. Coordinate conventions and local validity

Draw the angular convention before writing trigonometric transformations. The pendulum convention in this chapter differs from the polar angle convention; neither is wrong, but mixing them produces sign errors. Check the configuration at angle zero: our pendulum must point vertically downward.

Do not confuse $\partial\mathbf r/\partial q_i$ with $\partial q_i/\partial\mathbf r$, or with a force. It is a geometrical derivative. Do not silently set a prescribed support coordinate to zero after choosing a laboratory frame. Its contribution belongs in the transformation.

When a coordinate is only locally valid, state the branch. A square root with an unspecified sign does not uniquely locate a particle.

7. Practice problems

Attempt before opening Chapter 3 hints and solutions.

1. 3.1 Must a generalized coordinate have units of metres?

2. 3.2 If $q$ is an angle, what does $\dot q$ mean?

3. 3.3 Can one generalized coordinate move more than one particle?

4. 3.4 Are polar coordinates valid as unique coordinates at the origin?

5. 3.5 Explain the difference between explicit and implicit time dependence in $\mathbf r(q,t)$.

6. 3.6 Why are $x$ and $y$ not independent coordinates for a fixed-length planar pendulum?

7. 3.7 Interpret $\partial\mathbf r/\partial q_i$ and state its dimensions.

8. 3.8 Verify both fixed-length constraints using the double-pendulum transformations.

9. 3.9 For polar coordinates, calculate $\partial\mathbf r/\partial r$ and $\partial\mathbf r/\partial\theta$, and show that they are perpendicular.

10. 3.10 A pendulum has $a=1.2\,\mathrm m$ and $\theta=60^\circ$. Find its Cartesian position using this chapter's convention.

11. 3.11 For the helix, let $R=3\,\mathrm m$ and $b=4\,\mathrm m$. What arc distance corresponds to an increase $\Delta\phi=0.2$ rad?

12. 3.12 A bead lies on the parabola $y=cx^2$, with $c$ constant. Choose $q=x$ and calculate $\mathbf r(q)$ and $\partial\mathbf r/\partial q$.

13. 3.13 Define generalized coordinates and transformation equations, illustrating them for a pendulum, polar particle, and double pendulum.

14. 3.14 Discuss the advantages and limitations of generalized coordinates, including dimensions and local coordinate validity.

15. 3.15 For a pendulum, use $s=a\theta$ instead of $\theta$. Find $\partial\mathbf r/\partial s$ and interpret its length.

16. 3.16 On a helix with $b\ne0$, why can $\phi$ range over all real numbers rather than only one interval of length $2\pi$?

8. Coordinate choice and physical description

Different coordinate choices describe the same physical configuration. For a pendulum, angle and arc length are related by $s=a\theta$; neither changes the motion being described. What changes is the transformation from the chosen variable to the particle's position, and therefore the factors appearing in its derivatives.

A useful choice incorporates the constraints and distinguishes nearby configurations. Its range and any singular points must be specified. Once the transformations are known, differentiation converts coordinate changes into physical velocities and displacements. The response vector $\partial\mathbf r_a/\partial q_i$ supplies precisely this connection.