SICM Exercise 1.16

September 6, 2026

Solution to exercise 1.16 of Structure and Interpretation of Classical Mechanics by Gerald Jay Sussman and Jack Wisdom.

Central force motion

🛈 Note

Find Lagrangians for central force motion in three dimensions in rectangular coordinates and in spherical coordinates. First , find the Lagrangians analytically, then check the results with the computer by generalizing the programs that we have presented.

\[V(t; x, y, z) = U\left(\sqrt{x^2 + y^2 + z^2}\right)\]

\[T(t; x, y, z; v_x, v_y, v_z) = \frac{1}{2} m \left(v_x^2 + v_y^2 + v_z^2\right)\]

\[L = T - V = \frac{1}{2} m \left(v_x^2 + v_y^2 + v_z^2\right) - U\left(\sqrt{x^2 + y^2 + z^2}\right)\]

To get the Lagrangian in spherical coordinates, we have to write down the transformations from spherical to rectangular, namely \begin{equation*} x = r \sin \theta \cos \varphi \qquad y = r \sin \theta \sin \varphi \qquad z = r \cos \theta \end{equation*}

To get the velocity transformations we have to take the derivative of the position transformations.

\[v_x = Dx = \dot r \left(\sin \theta \cos \varphi\right) + r\left(\dot \theta \cos \theta \cos \varphi - \dot \varphi \sin \theta \sin \varphi\right)\]

\[v_y = Dy = \dot r \left(\sin \theta \sin \varphi\right) + r\left(\dot \theta \cos \theta \sin \varphi + \dot \varphi \sin \theta \cos \varphi\right)\]

\[v_z = Dz = \dot r \cos \theta - r \dot \theta \sin \theta\]

After taking the sum of squares of the velocities we can use the \(\sin^2 a + \cos^2 b = 1\) identitiy to simplify most expressions, the other terms cancel. The expression we are left with is

\[v_x^2 + v_y^2 + v_z^2 = \dot r^2 + r^2 \dot\theta^2 + r^2 \dot\varphi^2 \sin^2\theta\]

So the Lagrangian for the spherical system is

\[V(t; r, \theta, \varphi) = U(r)\]

\[T(t; r, \theta, \varphi) = \frac{1}{2} m \left(\dot r^2 + r^2 \dot\theta^2 + r^2 \dot\varphi^2 \sin^2\theta \right)\]

\[L = T - V = \frac{1}{2} m \left(\dot r^2 + r^2 \dot\theta^2 + r^2 \dot\varphi^2 \sin^2\theta \right) - U(r)\]

The following code implements this.

(define ((L-central-rectangular m U) local) 
  (let ((q (coordinate local)) 
	(v (velocity local))) 
    (- (* 1/2 m (square v)) 
       (U (sqrt (square q))))))

(define ((L-central-spherical m U) local) 
  (let ((q (coordinate local)) 
	(qdot (velocity local))) 
    (let ((r (ref q 0)) (theta (ref q 1)) (phi (ref q 2)) 
	  (rdot (ref qdot 0)) (thetadot (ref qdot 1)) (phidot (ref qdot 2))) 
      (- (* 1/2 m 
	    (+ (square rdot) 
	       (square (* r thetadot))
	       (square (* r phidot (sin theta)))))
	 (U r)))))

(define (s->r local) 
  (let ((spherical-tuple (coordinate local))) 
    (let ((r (ref spherical-tuple 0)) 
	  (theta (ref spherical-tuple 1))
	  (phi (ref spherical-tuple 2)))
      (let ((x (* r (sin theta) (cos phi))) 
	    (y (* r (sin theta) (sin phi)))
	    (z (* r (cos theta))))
	(up x y z)))))

(show-expression 
 (((Lagrange-equations 
    (L-central-rectangular 'm (literal-function 'U))) 
   (up (literal-function 'x) 
       (literal-function 'y)
       (literal-function 'z))) 
  't))

We get the same expressions as in the 2D example in the book (equations 1.62 and 1.63), but now we have a third equation for \(z\).

(show-expression 
 (((Lagrange-equations 
    (L-central-spherical 'm (literal-function 'U))) 
   (up (literal-function 'r) 
       (literal-function 'theta)
       (literal-function 'phi))) 
  't))

We get quite a complicated expression here. Later we will see that it is correct. Next, we will define a second Lagrangian for the spherical system that uses the coordinate transformation function.

(define (L-central-spherical-2 m U) 
  (compose (L-central-rectangular m U) (F->C s->r)))

Now we will take the difference of the two Lagrangians in spherical systems.

(show-expression 
 ((-
   ((Lagrange-equations 
     (L-central-spherical 'm (literal-function 'U))) 
    (up (literal-function 'r) 
	(literal-function 'theta)
	(literal-function 'phi)))
   ((Lagrange-equations 
     (L-central-spherical-2 'm (literal-function 'U))) 
    (up (literal-function 'r) 
	(literal-function 'theta)
	(literal-function 'phi))))
  't))
]=> (down 0 0 0)

As expected, the difference is zero, proving that the analytically derived Lagrangian is correct.

Tags:

Authors:

Dobi József