Category Archives: SICM

Ex 1.21 The dumbbell, 2.2.1

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Structure and Interpretation of Classical Mechanics

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b. Write the formal Lagrangian

\displaystyle{L(t;x_0, y_0, x_1, y_1, F; \dot x_0, \dot y_0, \dot x_1, \dot y_1, \dot F)}

such that Lagrange’s equations will yield the Newton’s equations you derived in part a.

~~~

[guess]


(define (U-constraint q0 q1 F l)
  (* (/ F (* 2 l))
     (- (square (- q1 q0))
        (square l))))

(define ((extract-particle pieces) local i)
  (let* ((indices (apply up (iota pieces (* i pieces))))
         (extract (lambda (tuple)
                    (vector-map (lambda (i)
                                  (ref tuple i))
                                indices))))
    (up (time local)
        (extract (coordinate local))
        (extract (velocity local)))))

(define q-rect
  (up (literal-function 'x_0)
      (literal-function 'y_0)
      (literal-function 'x_1)
      (literal-function 'y_1)
      (literal-function 'F)))

(show-expression q-rect)




(show-expression (q-rect 't))




(show-expression (Gamma q-rect))




(show-expression ((Gamma q-rect) 'w))




(show-expression ((Gamma q-rect) 't))




(show-expression (time ((Gamma q-rect) 't)))




(show-expression (coordinate ((Gamma q-rect) 't)))

[guess]

— based on /sicmutils/sicm-exercises

— Me@2021-04-27 05:03:59 PM

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2021.04.28 Wednesday (c) All rights reserved by ACHK

Ex 1.21 The dumbbell, 3

Posted on by

Structure and Interpretation of Classical Mechanics

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c. Make a change of coordinates to a coordinate system with center of mass coordinates x_{cm}, y_{cm}, angle \theta, distance between the particles c, and tension force F. Write the Lagrangian in these coordinates, and write the Lagrange equations.

~~~

[guess]

\displaystyle{ \begin{aligned} y_{cm} &= \frac{m_0 y_0 + m_1 y_1}{m_0 + m_1} \\ x_{cm} &= \frac{m_0 x_0 + m_1 x_1}{m_0 + m_1} \\ \end{aligned}}

\displaystyle{ \begin{aligned} y_{cm} &= \frac{m_0 y_0 + m_1 (y_0 + c(t) \sin \theta)}{m_0 + m_1} \\ x_{cm} &= \frac{m_0 x_0 + m_1 (x_0 + c(t) \cos \theta)}{m_0 + m_1} \\ \end{aligned}}

.

Let \displaystyle{M = m_0 + m_1}.

\displaystyle{ \begin{aligned} y_{cm} &= \frac{M y_0 + m_1 c(t) \sin \theta}{M} \\ x_{cm} &= \frac{M x_0 + m_1 c(t) \cos \theta}{M} \\ \end{aligned}}

\displaystyle{ \begin{aligned} y_0 &= y_{cm} - \frac{m_1}{M} c(t) \sin \theta \\ x_0 &= x_{cm} - \frac{m_1}{M} c(t) \cos \theta \\ \end{aligned}}

\displaystyle{ \begin{aligned} y_1 &= y_{cm} + \frac{m_0}{M} c(t) \sin \theta \\ x_1 &= x_{cm} + \frac{m_0}{M} c(t) \cos \theta \\ \end{aligned}}

.

The Lagrangian

\displaystyle{ \begin{aligned} L &= \frac{1}{2} m_0 (\dot x_0^2 + \dot y_0^2) + \frac{1}{2} m_1 (\dot x_1^2 + \dot y_1^2) + \lambda \left[ (x_1(t) - x_0(t))^2 + (y_1(t) - y_0(t))^2 - l^2 \right] \\ &= \frac{1}{2} m_0 (\dot x_0^2 + \dot y_0^2) + \frac{1}{2} m_1 (\dot x_1^2 + \dot y_1^2) - \frac{F}{2l} \left[ (c(t))^2 - l^2 \right] \\ \end{aligned}}

.

\displaystyle{ \begin{aligned} \dot y_0 &= \dot y_{cm} - \frac{m_1}{M} \dot c(t) \sin \theta - \frac{m_1}{M} c(t) \dot \theta \cos \theta\\ \dot x_0 &= \dot x_{cm} - \frac{m_1}{M} \dot c(t) \cos \theta - \frac{m_1}{M} c(t) \dot \theta \sin \theta \\ \end{aligned}}

…

[guess]

— Me@2021-04-17 05:40:46 PM

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2021.04.18 Sunday (c) All rights reserved by ACHK

Ex 1.21 The dumbbell, 2

Posted on by

Structure and Interpretation of Classical Mechanics

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Notice that these forces of constraint are proportional to the normal to the constraint surface at each instant and thus do no work for motions that obey the constraint.

b. Write the formal Lagrangian

\displaystyle{L(t;x_0, y_0, x_1, y_1, F; \dot x_0, \dot y_0, \dot x_1, \dot y_1, \dot F)}

such that Lagrange’s equations will yield the Newton’s equations you derived in part a.

~~~

[guess]

Eq (1.99):

\displaystyle{L(t;x,\lambda; \dot x, \dot \lambda) = \sum_\alpha \frac{1}{2} m_\alpha \dot{\mathbf{x}}_\alpha^2 - V(t,x) + \lambda \phi(t,x)}

\displaystyle{ \begin{aligned} L &= \frac{1}{2} m_0 (\dot x_0^2 + \dot y_0^2) + \frac{1}{2} m_1 (\dot x_1^2 + \dot y_1^2) + \lambda \left[ (x_1(t) - x_0(t))^2 + (y_1(t) - y_0(t))^2 - l^2 \right] \\ \end{aligned}}

.

The Lagrange equation:

\displaystyle{ \begin{aligned} D ( \partial_2 L \circ \Gamma[q]) - (\partial_1 L \circ \Gamma[q]) &= 0 \\ \end{aligned}}

\displaystyle{ \begin{aligned} \frac{d}{dt} \frac{\partial L}{\partial \dot x_0} - \frac{\partial L}{\partial x_0} &= 0 \\ \frac{d}{dt} \frac{\partial L}{\partial \dot x_1} - \frac{\partial L}{\partial x_1} &= 0 \\ \frac{d}{dt} \frac{\partial L}{\partial \dot y_0} - \frac{\partial L}{\partial y_0} &= 0 \\ \frac{d}{dt} \frac{\partial L}{\partial \dot y_1} - \frac{\partial L}{\partial y_1} &= 0 \\ \frac{d}{dt} \frac{\partial L}{\partial \dot \lambda} - \frac{\partial L}{\partial \lambda} &= 0 \\ \end{aligned}}

\displaystyle{ \begin{aligned} \frac{d}{dt} \left( m_0 \dot x_0 \right) + \left( \lambda 2 (x_1 - x_0)\right) &= 0 \\ \frac{d}{dt} \left( m_0 \dot x_1 \right) - \left( \lambda 2 (x_1 - x_0)\right) &= 0 \\ \frac{d}{dt} \left( m_0 \dot y_0 \right) + \left( \lambda 2 (y_1 - y_0)\right) &= 0 \\ \frac{d}{dt} \left( m_0 \dot y_1 \right) - \left( \lambda 2 (y_1 - y_0)\right) &= 0 \\ \frac{d}{dt} \left( 0 \right) - \left[ (x_1(t) - x_0(t))^2 + (y_1(t) - y_0(t))^2 - l^2 \right] &= 0 \\ \end{aligned}}

\displaystyle{ \begin{aligned} m_0 \ddot x_0 &= - 2 \lambda (x_1 - x_0) \\ m_0 \ddot x_1 &= 2 \lambda (x_1 - x_0) \\ m_0 \ddot y_0 &= - 2 \lambda (y_1 - y_0) \\ m_0 \ddot y_1 &= 2 \lambda (y_1 - y_0) \\ (x_1(t) - x_0(t))^2 + (y_1(t) - y_0(t))^2 - l^2 &= 0 \\ \end{aligned}}

.

Let \displaystyle{\lambda = - \frac{F}{2l}}:

\displaystyle{ \begin{aligned} m_0 \ddot x_0 &= F \frac{(x_1 - x_0)}{l} \\ m_0 \ddot x_1 &= -F \frac{(x_1 - x_0)}{l} \\ m_0 \ddot y_0 &= F \frac{(y_1 - y_0)}{l} \\ m_0 \ddot y_1 &= -F \frac{(y_1 - y_0)}{l} \\ \end{aligned}}

\displaystyle{ \begin{aligned} m_0 \ddot y_0 &= F \sin \theta \\ m_0 \ddot x_0 &= F \cos \theta \\ m_0 \ddot y_1 &= -F \sin \theta \\ m_0 \ddot x_1 &= -F \cos \theta \\ \end{aligned}}

[guess]

— Me@2021-04-09 11:18:07 PM

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2021.04.10 Saturday (c) All rights reserved by ACHK

Ex 1.21 The dumbbell

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…

Consider two massive particles in the plane constrained by a massless rigid rod to remain a distance l apart, as in figure 1.5.

…

a. Write Newton’s equations for the balance of forces for the four rectangular coordinates of the two particles, given that the scalar tension in the rod is F.

~~~

[guess]

\displaystyle{m_0 \ddot y_0 = F \sin \theta}

\displaystyle{m_0 \ddot x_0 = F \cos \theta}

\displaystyle{m_1 \ddot y_1 = - F \sin \theta}

\displaystyle{m_1 \ddot x_1 = - F \cos \theta}

[guess]

— Me@2021-04-04 03:51:05 PM

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2021.04.04 Sunday (c) All rights reserved by ACHK

Ex 1.20 Sliding pendulum

Posted on by

Structure and Interpretation of Classical Mechanics

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Consider a pendulum of length l attached to a support that is free to move horizontally, as shown in figure 1.4. Let the mass of the support be m_1 and the mass of the pendulum be m_2. Formulate a Lagrangian and derive Lagrange’s equations for this system.

~~~

[guess]


(define ((F->C F) local)
  (->local (time local)
           (F local)
           (+ (((partial 0) F) local)
              (* (((partial 1) F) local) 
                 (velocity local)))))

;

(define ((q->r l y1) local)
  (let ((q (coordinate local)))
    (let ((x1 (ref q 0))
          (theta (ref q 1)))
      (let ((x2 (+ x1 (* l (sin theta))))
            (y2 (- y1 (* l (cos theta)))))
        (up x1 y1 x2 y2)))))

(show-expression
  ((q->r 'l 'y_1) 
     (up 't
         (up 'x_1 'theta)
         (up 'xdot_1 'thetadot))))

;
 
(define (KE m vx vy)
  (* 1/2 m (+ (square vx) (square vy))))

(define ((T-rect m1 m2) local)
  (let ((q (coordinate local))
        (v (velocity local)))
    (let ((x1dot (ref v 0))
          (y1dot (ref v 1))
          (x2dot (ref v 2))
          (y2dot (ref v 3)))
      (+ (KE m1 x1dot y1dot)
         (KE m2 x2dot y2dot)))))

(show-expression
 ((T-rect 'm_1 'm_2)
    (up 't
        (up 'x_1 'y_1 'x_2 'y_2)
        (up 'xdot_1 'ydot_1 'xdot_2 'ydot_2))))
 
;

(define ((U-rect g m1 m2) local)
   (let* ((q (coordinate local))
     (y1 (ref q 1))
     (y2 (ref q 3)))
     (* g (+ (* m1 y1)
             (* m2 y2)))))

(show-expression
 ((U-rect 'g 'm_1 'm_2)
    (up 't
        (up 'x_1 'y_1 'x_2 'y_2)
        (up 'xdot_1 'ydot_1 'xdot_2 'ydot_2))))

;

(define (L-rect g m1 m2)
  (- (T-rect m1 m2) (U-rect g m1 m2)))

(show-expression
 ((L-rect 'g 'm_1 'm_2)
    (up 't
        (up 'x_1 'y_1 'x_2 'y_2)
        (up 'xdot_1 'ydot_1 'xdot_2 'ydot_2))))

;

(define (L-l l y1 g m_1 m_2)
  (compose
   (L-rect g m_1 m_2) (F->C (q->r l y1))))

(show-expression
 ((L-l 'l 'y_1 'g 'm_1 'm_2)
  (->local 't
	   (up 'x_1 'theta)
	   (up 'xdot_1 'thetadot))))

; 

(show-expression
 (((Lagrange-equations
    (L-l 'l 'y_1 'g 'm_1 'm_2))
   (up
    (literal-function 'x_1)
    (literal-function 'theta)))
  't))

;

[guess]

— Me@2021-03-26 08:22:01 PM

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2021.03.28 Sunday (c) All rights reserved by ACHK

Ex 1.19 A two-bar linkage, 2

Posted on by

Structure and Interpretation of Classical Mechanics

.

… Formulate a Lagrangian that describes the system and find the Lagrange equations of motion …

~~~

[guess]


(define ((U-rect g m1 m2 m3) local)
   (let* ((q (coordinate local))
 	 (y1 (ref q 1))
 	 (y2 (ref q 3))
 	 (y3 (ref q 5)))
     (* g (+ (* m1 y1)
	         (* m2 y2)
	         (* m3 y3)))))

;

(define (L-rect g m1 m2 m3)
  (- (T-rect m1 m2 m3) (U-rect g m1 m2 m3)))

(show-expression
 ((L-rect 'g 'm_1 'm_2 'm_3)
    (up 't
        (up 'x_1 'y_1 'x_2 'y_2 'x_3 'y_3)
        (up 'xdot_1 'ydot_1 'xdot_2 'ydot_2 'xdot_3 'ydot_3))))

;

(define (L-l l1 l2 g m_1 m_2 m_3)
  (compose
   (L-rect g m_1 m_2 m_3) (F->C (q->r l1 l2))))

(show-expression
 ((L-l 'l_1 'l_2 'g 'm_1 'm_2 'm_3)
  (->local 't
	       (up 'x_2 'y_2 'theta 'phi)
	       (up 'xdot_2 'ydot_2 'thetadot 'phidot))))

;

(show-expression
 (((Lagrange-equations
    (L-l 'l_1 'l_2 'g 'm_1 'm_2 'm_3))
   (up
    (literal-function 'x_2)
    (literal-function 'y_2)
    (literal-function 'theta)
    (literal-function 'phi)))
  't))

;

[guess]

— Me@2021-03-20 05:01:21 PM

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2021.03.20 Saturday (c) All rights reserved by ACHK

Ex 1.19 A two-bar linkage

Posted on by

Structure and Interpretation of Classical Mechanics

.

The two-bar linkage shown in figure 1.3 is constrained to move in the plane. It is composed of three small massive bodies interconnected by two massless rigid rods in a uniform gravitational field with vertical acceleration \displaystyle{g}. The rods are pinned to the central body by a hinge that allows the linkage to fold. The system is arranged so that the hinge is completely free: the members can go through all configurations without collision. Formulate a Lagrangian that describes the system and find the Lagrange equations of motion. Use the computer to do this, because the equations are rather big.

~~~

[guess]


(define ((F->C F) local)
  (->local (time local)
           (F local)
           (+ (((partial 0) F) local)
              (* (((partial 1) F) local) 
                 (velocity local)))))
 
;

(define ((q->r l1 l2) local)
  (let ((q (coordinate local)))
    (let ((x2 (ref q 0))
	      (y2 (ref q 1))
	      (theta (ref q 2)) 
          (phi (ref q 3)))
      (let ((x1 (+ x2 (* l1 (cos theta))))
	        (y1 (+ y2 (* l1 (sin theta))))
	        (x3 (+ x2 (* l2 (cos phi))))
	        (y3 (+ y2 (* l2 (sin phi)))))
        (up x1 y1 x2 y2 x3 y3)))))

;

(show-expression
  ((q->r 'l_1 'l_2) 
     (up 't
	     (up 'x_2 'y_2 'theta 'phi)
	     (up 'xdot_2 'ydot_2 'thetadot 'phidot))))

;

(define (KE m vx vy)
  (* 1/2 m (+ (square vx) (square vy))))

(define ((T-rect m1 m2 m3) local)
  (let ((q (coordinate local))
        (v (velocity local)))
    (let ((x1dot (ref v 0))
          (y1dot (ref v 1))
	      (x2dot (ref v 2))
	      (y2dot (ref v 3))
	      (x3dot (ref v 4))
	      (y3dot (ref v 5)))
      (+ (KE m1 x1dot y1dot)
  	     (KE m2 x2dot y2dot)
	     (KE m3 x3dot y3dot)))))

(show-expression
 ((T-rect 'm_1 'm_2 'm_3)
    (up 't
	    (up 'x_1 'y_1 'x_2 'y_2 'x_3 'y_3)
	    (up 'xdot_1 'ydot_1 'xdot_2 'ydot_2 'xdot_3 'ydot_3))))

;

[guess]

— Me@2021-03-12 05:37:27 PM

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2021.03.14 Sunday (c) All rights reserved by ACHK

Ex 1.18 Bead on a triaxial surface, 2

Posted on by

Structure and Interpretation of Classical Mechanics

.

A bead of mass m moves without friction on a triaxial ellipsoidal surface. In rectangular coordinates the surface satisfies

\displaystyle{\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1}

for some constants a, b, and c. Identify suitable generalized coordinates, formulate a Lagrangian, and find Lagrange’s equations.

~~~

[guess]

The generalized coordinates:

\displaystyle{\begin{aligned} x &= a \sin(\theta )\cos(\varphi),\\ y &= b \sin(\theta )\sin(\varphi),\\ z &= c \cos(\theta), \end{aligned}}

where

{\displaystyle 0\leq \theta \leq \pi ,\qquad 0\leq \varphi <2\pi .}

.

{\displaystyle {\begin{aligned} \dot x &= a \dot \theta \cos(\theta) \cos(\varphi) - a \dot \varphi\sin(\theta ) \sin(\varphi) \\ \dot y &= b \dot \theta \cos(\theta) \sin(\varphi) + b \dot \varphi \sin(\theta) \cos(\varphi) \\ \dot z &= - c \dot \theta \sin(\theta) \end{aligned}}}


(define ((F->C F) local)
  (->local (time local)
           (F local)
           (+ (((partial 0) F) local)
              (* (((partial 1) F) local) 
                 (velocity local)))))

;

(define ((e->r a b c) local)
  (let ((q (coordinate local)))
    (let ((theta (ref q 0)) 
          (phi (ref q 1)))
      (let ((x (* a (sin theta) (cos phi)))
	    (y (* b (sin theta) (sin phi)))
	    (z (* c (cos theta))))
        (up x y z)))))

;

(define ((L-rect m) local)
  (let ((q (coordinate local))
        (v (velocity local)))
    (* 1/2 m (square v))))

(define (L-e m a b c)
  (compose (L-rect m) (F->C (e->r a b c))))

;

(show-expression
  ((L-e 'm 'a 'b 'c)
   (->local 't
             (up 'theta 'phi) 
             (up 'thetadot 'phidot))))

;

(show-expression
 (((Lagrange-equations
    (L-e 'm 'a 'b 'c))
   (up (literal-function 'theta) (literal-function 'phi)))
  't))

[guess]

— Me@2021-03-01 06:22:50 PM

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2021.03.02 Tuesday (c) All rights reserved by ACHK

Ex 1.18 Bead on a triaxial surface

Posted on by

Structure and Interpretation of Classical Mechanics

.

A bead of mass m moves without friction on a triaxial ellipsoidal surface. In rectangular coordinates the surface satisfies

\displaystyle{\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1}

for some constants a, b, and c. Identify suitable generalized coordinates, formulate a Lagrangian, and find Lagrange’s equations.

~~~

[guess]

The generalized coordinates:

{\displaystyle {\begin{aligned}x&=a\sin(\theta )\cos(\varphi),\\y&=b\sin(\theta )\sin(\varphi),\\z&=c\cos(\theta),\end{aligned}}\,\!}

where

{\displaystyle 0\leq \theta \leq \pi ,\qquad 0\leq \varphi <2\pi .}

.

{\displaystyle {\begin{aligned} \dot x &= a \dot \theta \cos(\theta) \cos(\varphi) - a \dot \varphi\sin(\theta ) \sin(\varphi) \\ \dot y &= b \dot \theta \cos(\theta) \sin(\varphi) + b \dot \varphi \sin(\theta) \cos(\varphi) \\ \dot z &= - c \dot \theta \sin(\theta) \end{aligned}}}


(define theta (literal-function 'theta))

(define phi (literal-function 'phi))

; 

(define (x t) (* 'a (sin (theta t)) (cos (phi t))))

((D x) 't)

(show-expression ((D x) 't))

;

(define (y t) (* 'b (sin (theta t)) (sin (phi t))))

(show-expression (y 't))

((D y) 't)

(show-expression ((D y) 't))

[guess]

— Me@2021-02-16 07:20:25 AM

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2021.02.18 Thursday (c) All rights reserved by ACHK

Ex 1.17 Bead on a helical wire

Posted on by

Structure and Interpretation of Classical Mechanics

.

A bead of mass m is constrained to move on a frictionless helical wire. The helix is oriented so that its axis is horizontal. The diameter of the helix is d and its pitch (turns per unit length) is h. The system is in a uniform gravitational field with vertical acceleration g. Formulate a Lagrangian that describes the system and find the Lagrange equations of motion.

~~~

[guess]

The coordinates of the bead is

\displaystyle{\left( \frac{d}{2} \cos \theta, \frac{d}{2} \sin \theta, \frac{\theta}{2 \pi} h \right)},

where the \displaystyle{x}-direction is horizontal, the \displaystyle{y}-direction points upwards, and the \displaystyle{z}-direction is along the axis of the helix.

.

Lagrangian \displaystyle{L = T - V}, where the kinetic energy, \displaystyle{T = \frac{1}{2} m \left( \dot x^2 + \dot y^2 + \dot z^2 \right)} and the potential energy, \displaystyle{V = m g y}.

\displaystyle{ (\dot x, \dot y, \dot z) = \left( \frac{-d}{2} (\sin \theta) \dot \theta, \frac{d}{2} (\cos \theta) \dot \theta, \frac{h}{2 \pi} \dot \theta \right)}

\displaystyle{L = \frac{1}{2} m \left( \frac{d^2}{4} + \frac{h^2}{4 \pi^2} \right) \dot \theta^2 - \frac{d}{2} m g \sin \theta}

\displaystyle{\frac{\partial L}{\partial \dot \theta} = m \left( \frac{d^2}{4} + \frac{h^2}{4 \pi^2} \right) \dot \theta}
\displaystyle{\frac{\partial L}{\partial \theta} = - \frac{d}{2} m g \cos \theta}

.

The Lagrange equation:

\displaystyle{ \begin{aligned} D ( \partial_2 L \circ \Gamma[q]) - (\partial_1 L \circ \Gamma[q]) &= 0 \\ \frac{d}{dt} \frac{\partial L}{\partial \dot \theta} - \frac{\partial L}{\partial \theta} &= 0 \\ m \left( \frac{d^2}{4} + \frac{h^2}{4 \pi^2} \right) \ddot{\theta} + \frac{d}{2} m g \cos \theta &= 0 \\ \end{aligned}}

.


(define ((T-hl d h m g) local)
  (let ((t (time local))
        (thetadot (velocity local)))    
    (* 1/8 m (square thetadot) 'H)))

;; (+ (square d) 
;; (/ (square h) (square 'pi))))))

(show-expression
  ((T-hl 'd 'h 'm 'g)
    (->local 't
             'theta
             'thetadot)))

(define ((V-hl d h m g) local)
  (let ((t (time local))
        (theta (coordinate local)))
    (let ((y (* 1/2 d (sin theta))))
      (* m g y))))


(show-expression
  ((V-hl 'd 'h 'm 'g)
    (->local 't
             'theta
             'thetadot)))

(define L-hl (- T-hl V-hl))

(show-expression
  ((L-hl 'd 'h 'm 'g)
    (->local 't
             'theta
             'thetadot)))

(show-expression
 (((Lagrange-equations
    (L-hl 'd 'h 'm 'g))
   (literal-function 'theta))
  't))

[guess]

— Me@2021-02-05 04:23:02 PM

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2021.02.06 Saturday (c) All rights reserved by ACHK

Ex 1.15 Central force motion

Posted on by

Structure and Interpretation of Classical Mechanics

.

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.

~~~

The Lagrangians in rectangular coordinates:

\displaystyle{ \begin{aligned} &L (t; x, y, x; v_x, v_y, v_z) \\ &= \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) \\ \end{aligned}}

The Lagrangians in spherical coordinates:

\displaystyle{ \begin{aligned} &L (t; r, \theta, \phi; \dot r, \dot \theta, \dot \phi) \\ &= \frac{1}{2} m r^2 \dot \phi^2 (\sin \theta)^2 + \frac{1}{2} m \dot \theta^2 r^2 + \frac{1}{2} m \dot r^2 - U (r) \\ &= \frac{1}{2} m \left( \dot r^2 + r^2 \dot \theta^2 + r^2 (\sin^2 \theta) \dot \phi^2 \right) - U (r) \\ \end{aligned}}

.


(define ((F->C F) local)
  (->local (time local)
           (F local)
           (+ (((partial 0) F) local)
              (* (((partial 1) F) local) 
                 (velocity local)))))

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

(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-polar m U)
  (compose (L-central-rectangular m U) (F->C p->r)))

(show-expression
  ((L-central-polar 'm (literal-function 'U))
   (->local 't 
             (up 'r 'theta 'phi) 
             (up 'rdot 'thetadot 'phidot))))

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— Me@2021-01-22 02:59:11 PM

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2021.01.22 Friday (c) All rights reserved by ACHK

Ex 1.14 Lagrange equations for L’, 2

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Structure and Interpretation of Classical Mechanics

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Show by direct calculation that the Lagrange equations for \displaystyle{L'} are satisfied if the Lagrange equations for \displaystyle{L} are satisfied.

~~~

\displaystyle{x = F(t, x')}

\displaystyle{x' = G(t, x)}

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\displaystyle{ \begin{aligned} &\partial_2 L \circ \Gamma[q] \\ &= \frac{\partial}{\partial v} L(t, x, v) \\ &= \frac{\partial}{\partial v} L'(t, x', v') \\ &= \frac{\partial}{\partial x'} L'(t, x', v') \frac{\partial x'}{\partial v} + \frac{\partial}{\partial v'} L'(t, x', v') \frac{\partial v'}{\partial v} \\ &= \frac{\partial}{\partial v'} L'(t, x', v') \frac{\partial v'}{\partial v} \\ \end{aligned}}

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\displaystyle{v' = \partial_0 G(t, x) + \partial_1 G(t,x) v}

\displaystyle{ \begin{aligned} \frac{\partial v'}{\partial v} &= \partial_1 G(t,x) = \frac{\partial x'}{\partial x} \\ \end{aligned}}

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\displaystyle{ \begin{aligned} &\partial_2 L \circ \Gamma[q] \\ &= \frac{\partial L'}{\partial v'} \frac{\partial x'}{\partial x} \\ \end{aligned}}

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The Lagrange equation:

\displaystyle{ \begin{aligned} D ( \partial_2 L \circ \Gamma[q]) - (\partial_1 L \circ \Gamma[q]) &= 0 \\ \end{aligned}}

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\displaystyle{ \begin{aligned} &\partial_1 L \circ \Gamma[q] \\ &= \partial_1 L' \circ \Gamma[q'] \\ &= \frac{\partial}{\partial x} L' (t, x', v') \\ &= \frac{\partial L'}{\partial x'} \frac{\partial x'}{\partial x} + \frac{\partial L'}{\partial v'} \frac{\partial v'}{\partial x} \\ \end{aligned}}

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\displaystyle{ \begin{aligned} D ( \partial_2 L \circ \Gamma[q]) - (\partial_1 L \circ \Gamma[q]) &= 0 \\ D \left( \frac{\partial L'}{\partial v'} \frac{\partial x'}{\partial x} \right) - \left( \frac{\partial L'}{\partial x'} \frac{\partial x'}{\partial x} + \frac{\partial L'}{\partial v'} \frac{\partial v'}{\partial x} \right) &= 0 \\ D \left( \frac{\partial L'}{\partial v'} \right) \frac{\partial x'}{\partial x} + \frac{\partial L'}{\partial v'} D \left( \frac{\partial x'}{\partial x} \right) - \left( \frac{\partial L'}{\partial x'} \frac{\partial x'}{\partial x} + \frac{\partial L'}{\partial v'} \frac{\partial v'}{\partial x} \right) &= 0 \\ \end{aligned}}

\displaystyle{ \begin{aligned} \left[ D \left( \frac{\partial L'}{\partial v'} \right) - \frac{\partial L'}{\partial x'} \right] \frac{\partial x'}{\partial x} + \frac{\partial L'}{\partial v'} D \left( \frac{\partial x'}{\partial x} \right) - \frac{\partial L'}{\partial v'} \frac{\partial v'}{\partial x} &= 0 \\ \end{aligned}}

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\displaystyle{ \begin{aligned} &D \left( \frac{\partial x'}{\partial x} \right) \\ &= \frac{d}{dt} \left( \frac{\partial x'}{\partial x} \right) \\ &= \frac{\partial}{\partial t} \left( \frac{\partial x'}{\partial x} \right) \frac{\partial t}{\partial t} + \frac{\partial}{\partial x'} \left( \frac{\partial x'}{\partial x} \right) \frac{\partial x'}{\partial t} + \frac{\partial}{\partial v'} \left( \frac{\partial x'}{\partial x} \right) \frac{\partial v'}{\partial t} \\ &= \frac{\partial}{\partial t} \left( \frac{\partial x'}{\partial x} \right) + \frac{\partial}{\partial x} \left( \frac{\partial x'}{\partial x'} \right) \frac{\partial x'}{\partial t} + \frac{\partial}{\partial x} \left( \frac{\partial x'}{\partial v'} \right) \frac{\partial v'}{\partial t} \\ &= \frac{\partial}{\partial t} \left( \frac{\partial x'}{\partial x} \right) + \frac{\partial}{\partial x} \left( 1 \right) \frac{\partial x'}{\partial t} + \frac{\partial}{\partial x} \left( 0 \right) \frac{\partial v'}{\partial t} \\ &= \frac{\partial}{\partial x} \left( \frac{\partial x'}{\partial t} \right) + 0 + 0 \\ &= \frac{\partial v'}{\partial x} \\ \end{aligned}}

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\displaystyle{ \begin{aligned} \left[ D \left( \frac{\partial L'}{\partial v'} \right) - \frac{\partial L'}{\partial x'} \right] \frac{\partial x'}{\partial x} + \frac{\partial L'}{\partial v'} \frac{\partial v'}{\partial x} - \frac{\partial L'}{\partial v'} \frac{\partial v'}{\partial x} &= 0 \\ \left[ D \left( \frac{\partial L'}{\partial v'} \right) - \frac{\partial L'}{\partial x'} \right] \frac{\partial x'}{\partial x} &= 0 \\ D \left( \frac{\partial L'}{\partial v'} \right) - \frac{\partial L'}{\partial x'} &= 0 \\ \end{aligned}}

— Me@2021-01-06 08:10:46 PM

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2021.01.09 Saturday (c) All rights reserved by ACHK

Ex 1.14 Lagrange equations for L’

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Structure and Interpretation of Classical Mechanics

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Show by direct calculation that the Lagrange equations for \displaystyle{L'} are satisfied if the Lagrange equations for \displaystyle{L} are satisfied.

~~~

Equation (1.69):

\displaystyle{C \circ \Gamma[q'] = \Gamma[q]}

Equation (1.70):

\displaystyle{L' = L \circ C}

Equation (1.71):

\displaystyle{L' \circ \Gamma[q'] = L \circ C \circ \Gamma[q'] = L \circ \Gamma[q]}

The Lagrange equation:

\displaystyle{ \begin{aligned} D ( \partial_2 L \circ \Gamma[q]) - (\partial_1 L \circ \Gamma[q]) &= 0 \\ \end{aligned}}

.

\displaystyle{ \begin{aligned} &\partial_2 L \circ \Gamma[q] \\ &= \frac{\partial}{\partial v} L(t, x, v) \\ &= \frac{\partial}{\partial v} L'(t, x', v') \\ &= \frac{\partial}{\partial x'} L'(t, x', v') \frac{\partial x'}{\partial v} + \frac{\partial}{\partial v'} L'(t, x', v') \frac{\partial v'}{\partial v} \\ \end{aligned}}

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Since it is just a coordinate transformation \displaystyle{x = F(t, x')}, \displaystyle{x} has no explicitly dependent on \displaystyle{v'}. Similarly, if we consider the coordinate transformation \displaystyle{x' = G(t, x)}, \displaystyle{x'} has no explicitly dependent on \displaystyle{v}. So

\displaystyle{ \begin{aligned} \frac{\partial x'}{\partial v} &= 0 \\ \end{aligned}}

.

\displaystyle{ \begin{aligned} &\partial_2 L \circ \Gamma[q] \\ &= \frac{\partial}{\partial v'} L'(t, x', v') \frac{\partial v'}{\partial v} \\ \end{aligned}}

…

— Me@2020-12-28 04:03:24 PM

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2020.12.28 Monday (c) All rights reserved by ACHK

How to Find Lagrangians

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Lagrange’s equations are a system of second-order differential equations. In order to use them to compute the evolution of a mechanical system, we must find a suitable Lagrangian for the system. There is no general way to construct a Lagrangian for every system, but there is an important class of systems for which we can identify Lagrangians in a straightforward way in terms of kinetic and potential energy. The key idea is to construct a Lagrangian L such that Lagrange’s equations are Newton’s equations \displaystyle{\vec F = m \vec a}.

— 1.6 How to Find Lagrangians

— Structure and Interpretation of Classical Mechanics

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2020.12.06 Sunday ACHK

Possion’s Lagrange Equation

Posted on by

Structure and Interpretation of Classical Mechanics

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Ex 1.10 Higher-derivative Lagrangians

Derive Lagrange’s equations for Lagrangians that depend on accelerations. In particular, show that the Lagrange equations for Lagrangians of the form \displaystyle{L(t, q, \dot q, \ddot q)} with \displaystyle{\ddot{q}} terms are

\displaystyle{D^2(\partial_3L \circ \Gamma[q]) - D(\partial_2 L \circ \Gamma[q]) + \partial_1 L \circ \Gamma[q] = 0}

In general, these equations, first derived by Poisson, will involve the fourth derivative of \displaystyle{q}. Note that the derivation is completely analogous to the derivation of the Lagrange equations without accelerations; it is just longer. What restrictions must we place on the variations so that the critical path satisfies a differential equation?

Varying the action

\displaystyle{ \begin{aligned} S[q] (t_1, t_2) &= \int_{t_1}^{t_2} L \circ \Gamma [q] \\ \eta(t_1) &= \eta(t_2) = 0 \\ \end{aligned}}

\displaystyle{ \begin{aligned} \delta_\eta S[q] (t_1, t_2) &= 0 \\ \end{aligned}}

\displaystyle{ \begin{aligned} \delta_\eta S[q] (t_1, t_2) &= \int_{t_1}^{t_2} \delta_\eta \left( L \circ \Gamma [q] \right) \\ \end{aligned}}

\displaystyle{ \begin{aligned} \delta_\eta I [q] &= \eta \\ \delta_\eta g[q] &= D \eta~~~\text{with}~~~g[q] = Dq \\ \end{aligned}}

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Let \displaystyle{h[q] = D^2 q}.

\displaystyle{ \begin{aligned} \delta_\eta h[q] &= \lim_{\epsilon \to 0} \frac{h[q+\epsilon \eta] - h[q]}{\epsilon} \\ &= \lim_{\epsilon \to 0} \frac{D^2 (q+\epsilon \eta) - D^2 q}{\epsilon} \\ &= \lim_{\epsilon \to 0} \frac{D^2 q + D^2 \epsilon \eta - D^2 q}{\epsilon} \\ &= \lim_{\epsilon \to 0} \frac{D^2 \epsilon \eta}{\epsilon} \\ &= D^2 \eta \\ \end{aligned}}

\displaystyle{ \begin{aligned} \Gamma [q] (t) &= (t, q(t), D q(t), D^2 q(t)) \\ \delta_\eta \Gamma [q] (t) &= (0, \eta (t), D \eta (t), D^2 \eta (t)) \\ \end{aligned}}

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Chain rule of functional variation

\displaystyle{ \begin{aligned} &\delta_\eta F[g[q]] \\ &= \delta_\eta (F \circ g)[q] \\ &= \delta_{ \left( \delta_\eta g[q] \right)} F[g] \\ \end{aligned}}

Since variation commutes with integration,

\displaystyle{ \begin{aligned} \delta_\eta S[q] (t_1, t_2) &= \delta_\eta \int_{t_1}^{t_2} L \circ \Gamma [q] \\ &= \int_{t_1}^{t_2} \delta_\eta \left( L \circ \Gamma [q] \right) \\ \end{aligned}}

By the chain rule of functional variation:

\displaystyle{ \begin{aligned} \delta_\eta L \circ \Gamma [q] = \delta_{ \left( \delta_\eta \Gamma[q] \right)} L[\Gamma[q]] \\ \end{aligned}}

If \displaystyle{L} is path-independent,

\displaystyle{ \begin{aligned} \delta_\eta \left( L \circ \Gamma [q] \right) = \left( DL \circ \Gamma[q] \right) \delta_\eta \Gamma[q] \\ \end{aligned}}

But is \displaystyle{L} path-independent?

The \displaystyle{L \circ \Gamma [.]} is path-dependent. Its input is a path \displaystyle{q}, not just \displaystyle{q(t)}, the value of \displaystyle{q} at the time \displaystyle{t}. However, \displaystyle{L(.)} itself is a path-independent function, because its input is not a path \displaystyle{q}, but a quadruple of values \displaystyle{(t, q(t), Dq(t), D^2 q(t))}.

\displaystyle{ \begin{aligned} L \circ \Gamma [q] = L(t, q(t), Dq(t), D^2 q(t)) \\ \end{aligned}}

Since \displaystyle{L} is path-independent,

\displaystyle{ \begin{aligned} \delta_\eta \left( L \circ \Gamma [q] \right) = \left( DL \circ \Gamma[q] \right) \delta_\eta \Gamma[q] \\ \end{aligned}}

\displaystyle{ \begin{aligned} &\delta_\eta S[q] (t_1, t_2) \\ &= \int_{t_1}^{t_2} \delta_\eta L \circ \Gamma [q] \\ &= \int_{t_1}^{t_2} \left( D \left( L \circ \Gamma[q] \right) \right) \delta_\eta \Gamma[q] \\ &= \int_{t_1}^{t_2} \left( D \left( L(t, q, D q, D^2 q) \right) \right) (0, \eta (t), D \eta (t), D^2 \eta (t)) \\ &= \int_{t_1}^{t_2} \left[ \partial_0 L \circ \Gamma[q], \partial_1 L \circ \Gamma[q], \partial_2 L \circ \Gamma[q], \partial_3 L \circ \Gamma[q] \right] (0, \eta (t), D \eta (t), D^2 \eta (t)) \\ &= \int_{t_1}^{t_2} (\partial_1 L \circ \Gamma[q]) \eta + (\partial_2 L \circ \Gamma[q]) D \eta + (\partial_3 L \circ \Gamma[q]) D^2 \eta \\ &= \int_{t_1}^{t_2} (\partial_1 L \circ \Gamma[q]) \eta + \left[ \left. (\partial_2 L \circ \Gamma[q]) \eta \right|_{t_1}^{t_2} - \int_{t_1}^{t_2} D(\partial_2 L \circ \Gamma[q]) \eta \right] + \int_{t_1}^{t_2} (\partial_3 L \circ \Gamma[q]) D^2 \eta \\ \end{aligned}}

Since \displaystyle{\eta(t_1) = 0} and \displaystyle{\eta(t_2) = 0},

\displaystyle{ \begin{aligned} \delta_\eta S[q] (t_1, t_2) &= \int_{t_1}^{t_2} (\partial_1 L \circ \Gamma[q]) \eta - \int_{t_1}^{t_2} D(\partial_2 L \circ \Gamma[q]) \eta + \int_{t_1}^{t_2} (\partial_3 L \circ \Gamma[q]) D^2 \eta \\ \end{aligned}}

Here is a trick for integration by parts:

As long as the boundary term \displaystyle{\left. u(t)v(t) \right|_{t_1}^{t_2} = 0},

\displaystyle{\int_{t_1}^{t_2} u(t) dv(t) = - \int_{t_1}^{t_2} v(t) du(t)}

So if \displaystyle{D \eta(t_1) = 0} and \displaystyle{D \eta(t_2) = 0},

\displaystyle{ \begin{aligned} \delta_\eta S[q] (t_1, t_2) &= \int_{t_1}^{t_2} (\partial_1 L \circ \Gamma[q]) \eta - \int_{t_1}^{t_2} D(\partial_2 L \circ \Gamma[q]) \eta - \int_{t_1}^{t_2} D(\partial_3 L \circ \Gamma[q]) D \eta \\ \end{aligned}}

Since \displaystyle{\eta(t_1) = 0} and \displaystyle{\eta(t_2) = 0},

\displaystyle{ \begin{aligned} \delta_\eta S[q] (t_1, t_2) &= \int_{t_1}^{t_2} (\partial_1 L \circ \Gamma[q]) \eta - \int_{t_1}^{t_2} D(\partial_2 L \circ \Gamma[q]) \eta + \int_{t_1}^{t_2} D^2 (\partial_3 L \circ \Gamma[q]) \eta \\ \end{aligned}}

\displaystyle{ \begin{aligned} \delta_\eta S[q] (t_1, t_2) &= \int_{t_1}^{t_2} \left[ (\partial_1 L \circ \Gamma[q]) - D(\partial_2 L \circ \Gamma[q]) + D^2 (\partial_3 L \circ \Gamma[q]) \right] \eta \\ \end{aligned}}

By the principle of stationary action, \displaystyle{ \delta_\eta S[q] (t_1, t_2) = 0}. So

\displaystyle{ \begin{aligned} 0 &= \int_{t_1}^{t_2} \left[ (\partial_1 L \circ \Gamma[q]) - D(\partial_2 L \circ \Gamma[q]) + D^2 (\partial_3 L \circ \Gamma[q]) \right] \eta \\ \end{aligned}}

Since this is true for any function \displaystyle{\eta(t)} that satisfies \displaystyle{\eta(t_1) = \eta(t_2) = 0} and \displaystyle{D\eta(t_1) = D\eta(t_2) = 0},

\displaystyle{ \begin{aligned} (\partial_1 L \circ \Gamma[q]) - D(\partial_2 L \circ \Gamma[q]) + D^2 (\partial_3 L \circ \Gamma[q]) &= 0 \\ D^2 (\partial_3 L \circ \Gamma[q]) - D(\partial_2 L \circ \Gamma[q]) + \partial_1 L \circ \Gamma[q] &= 0 \\ \end{aligned}}

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Note:

The notation of the path function \displaystyle{\Gamma} is \displaystyle{\Gamma[q](t)}, not \displaystyle{\Gamma[q(t)]}.

The notation \displaystyle{\Gamma[q](t)} means that \displaystyle{\Gamma} takes a path \displaystyle{q} as input. And then returns a path-independent function \displaystyle{\Gamma[q]}, which takes time \displaystyle{t} as input, returns a value \displaystyle{\Gamma[q](t)}.

The other notation \displaystyle{\Gamma[q(t)]} makes no sense, because \displaystyle{\Gamma[.]} takes a path \displaystyle{q}, not a value \displaystyle{q(t)}, as input.

— Me@2020-11-11 05:37:13 PM

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2020.11.11 Wednesday (c) All rights reserved by ACHK

Ex 1.9 Lagrange’s equations

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Structure and Interpretation of Classical Mechanics

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Derive the Lagrange equations for the following systems, showing all of the intermediate steps as in the harmonic oscillator and orbital motion examples.

b. An ideal planar pendulum consists of a bob of mass \displaystyle{m} connected to a pivot by a massless rod of length \displaystyle{l} subject to uniform gravitational acceleration \displaystyle{g}. A Lagrangian is \displaystyle{L(t, \theta, \dot \theta) = \frac{1}{2} m l^2 \dot \theta^2 + mgl \cos \theta}. The formal parameters of \displaystyle{L} are \displaystyle{t}, \displaystyle{\theta}, and \displaystyle{\dot \theta}; \displaystyle{\theta} measures the angle of the pendulum rod to a plumb line and \displaystyle{\dot \theta} is the angular velocity of the rod.

~~~

\displaystyle{ \begin{aligned} L (t, \xi, \eta) &= \frac{1}{2} m l^2 \eta^2 + m g l \cos \xi \\ \end{aligned}}

\displaystyle{ \begin{aligned} \partial_1 L (t, \xi, \eta) &= - m g l \sin \xi \\ \partial_2 L (t, \xi, \eta) &= m l^2 \eta \\ \end{aligned}}

Put \displaystyle{q = \theta},

\displaystyle{ \begin{aligned} \Gamma[q](t) &= (t; \theta(t); D\theta(t)) \\ \end{aligned}}

\displaystyle{ \begin{aligned} \partial_1 L \circ \Gamma[q] (t) &= - m g l \sin \theta \\ \partial_2 L \circ \Gamma[q] (t) &= m l^2 D \theta \\ \end{aligned}}

The Lagrange equation:

\displaystyle{ \begin{aligned} D ( \partial_2 L \circ \Gamma[q]) - (\partial_1 L \circ \Gamma[q]) &= 0 \\ D ( m l^2 D \theta ) - ( - m g l \sin \theta ) &= 0 \\ D^2 \theta + \frac{g}{l} \sin \theta &= 0 \\ \end{aligned}}

— Me@2020-09-28 05:40:42 PM

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2020.09.30 Wednesday (c) All rights reserved by ACHK

Ex 1.8.2.5 Implementation of $delta$

Posted on by

Structure and Interpretation of Classical Mechanics

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Verify that the operators \displaystyle{D} (differentiation) and \displaystyle{\delta} (variation) commute (Equation 1.27) using the scmutils software library:

\displaystyle{D \delta_{\eta} f [q] = \delta_\eta g[q]} with \displaystyle{g [q] = D ( f[q] )}

~~~

(define (((delta eta) f) q)
  (define (g epsilon)
    (f (+ q (* epsilon eta))))
  ((D g) 0))

(define q (literal-function 'q (-> Real (UP Real))))

(define eta (literal-function 'eta (-> Real (UP Real))))

.

(define (f q)
  (compose (literal-function 'f
             (-> (UP Real (UP* Real) (UP* Real)) Real))
           (Gamma q)))

(define (g q)
  (compose (literal-function 'g
             (-> (UP Real (UP* Real) (UP* Real)) Real))
           (Gamma q)))

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(define (g q) (D (f q)))

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(define LHS ( (D (((delta eta) f) q)) 't))

(define RHS ((((delta eta) g) q) 't))

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(print-expression LHS)

(show-expression LHS)

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\displaystyle{\partial_1 \partial_1 f \left( \begin{bmatrix} t \\ q(t) \\ Dq(t) \end{bmatrix} \right) D q(t) \eta(t) + D^2 q \partial_2 \partial_2 f D \eta + D^2 q \partial_1 \partial_2 f \eta + ... }

\displaystyle{... \partial_1 \partial_2 f D \eta D q + D^2 \eta \partial_2 f + \partial_0 \partial_2 f D \eta + \partial_0 \partial_1 f \eta + D \eta \partial_1 f}

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(print-expression RHS)

(show-expression RHS)

(- LHS RHS)

— Me@2020-08-24 03:18:21 PM

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2020.08.24 Monday (c) All rights reserved by ACHK

Ex 1.8.2.4 Implementation of $\delta$

Posted on by

Structure and Interpretation of Classical Mechanics

.

Verify the product rule of variation (Equation 1.23) using the
scmutils software library:

\displaystyle{\delta_\eta \left(f [q] g [q] \right) = \left( \delta_\eta f[q] \right) g[q] + f[q] \delta_\eta g[q]}

~~~

(define (((delta eta) f) q)
  (define (g epsilon)
    (f (+ q (* epsilon eta))))
  ((D g) 0))

(define q (literal-function 'q (-> Real (UP Real))))

(define eta (literal-function 'eta (-> Real (UP Real))))

.

(define (f q)
   (compose (literal-function 'f
              (-> (UP Real (UP* Real) (UP* Real)) Real))
            (Gamma q)))

(define (g q)
   (compose (literal-function 'g
              (-> (UP Real (UP* Real) (UP* Real)) Real))
            (Gamma q)))

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(define (f_times_g q) (* (f q) (g q)))

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(define LHS ((((delta eta) f_times_g) q) 't))

(define RHS (+ (* ((((delta eta) f) q) 't) ((g q) 't))
               (* ((f q) 't) ((((delta eta) g) q) 't))))

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(print-expression LHS)

(show-expression LHS)

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\displaystyle{g D \eta \partial_2 f + g \eta \partial_1 f + f D \eta \partial_2 g + f \eta \partial_1 g}

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(print-expression RHS)

(show-expression RHS)

(- LHS RHS)

— Me@2020-08-06 07:23:27 PM

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2020.08.06 Thursday (c) All rights reserved by ACHK