### Exercise 1.14: Coordinate-independence of Lagrange equations

Check that the Lagrange equations for central force motion in polar coordinates and in rectangular coordinates are equivalent.

In order to do this, compute the expressions for accelerations $a_x(t)$ and $a_y(t)$ in terms of the polar coordinates and substitute into the lagrange’s equations in cartesian coordinates.

(defn L-central-rectangular [m U]
(fn [[_  [x y] [v_x v_y]]]
(let [r (sqrt (+ (square x) (square y)))]
(- (* 1/2 m (+ (square v_x) (square v_y))) (U r)))))

(def eom-central-rectangular (let [L (L-central-rectangular 'm (literal-function 'U))]
(Lagrange-equations L)))

(def eom-central-r2p (let [r (literal-function 'r)
phi (literal-function 'varphi)
rt (literal-function 'r)
x (* r (cos phi))
y (* r (sin phi))
v_x (D x)
v_y (D y)
a_x (D v_x)
a_y (D v_y)
state (up x y)
eom ((eom-central-rectangular state) 't)
]
(apply up eom)
))
(rendertex eom-central-r2p)
\begin{pmatrix}\displaystyle{- m\,\cos\left(\varphi\left(t\right)\right)\,{\left(D\varphi\left(t\right)\right)}^{2}\,r\left(t\right) -2\,m\,D\varphi\left(t\right)\,\sin\left(\varphi\left(t\right)\right)\,Dr\left(t\right) - m\,r\left(t\right)\,\sin\left(\varphi\left(t\right)\right)\,{D}^{2}\varphi\left(t\right) + m\,\cos\left(\varphi\left(t\right)\right)\,{D}^{2}r\left(t\right) + \cos\left(\varphi\left(t\right)\right)\,DU\left(r\left(t\right)\right)} \cr \cr \displaystyle{- m\,{\left(D\varphi\left(t\right)\right)}^{2}\,r\left(t\right)\,\sin\left(\varphi\left(t\right)\right) + 2\,m\,\cos\left(\varphi\left(t\right)\right)\,D\varphi\left(t\right)\,Dr\left(t\right) + m\,\cos\left(\varphi\left(t\right)\right)\,r\left(t\right)\,{D}^{2}\varphi\left(t\right) + m\,\sin\left(\varphi\left(t\right)\right)\,{D}^{2}r\left(t\right) + \sin\left(\varphi\left(t\right)\right)\,DU\left(r\left(t\right)\right)}\end{pmatrix}
(let [eom eom-central-r2p
eom1 (ref eom 0)
eom2 (ref eom 1)]
(rendertexvec (up (simplify (+ eom1 eom2)))))
\begin{pmatrix}\displaystyle{- m\,\cos\left(\varphi\left(t\right)\right)\,{\left(D\varphi\left(t\right)\right)}^{2}\,r\left(t\right) - m\,{\left(D\varphi\left(t\right)\right)}^{2}\,r\left(t\right)\,\sin\left(\varphi\left(t\right)\right) + 2\,m\,\cos\left(\varphi\left(t\right)\right)\,D\varphi\left(t\right)\,Dr\left(t\right) + m\,\cos\left(\varphi\left(t\right)\right)\,r\left(t\right)\,{D}^{2}\varphi\left(t\right) -2\,m\,D\varphi\left(t\right)\,\sin\left(\varphi\left(t\right)\right)\,Dr\left(t\right) - m\,r\left(t\right)\,\sin\left(\varphi\left(t\right)\right)\,{D}^{2}\varphi\left(t\right) + m\,\cos\left(\varphi\left(t\right)\right)\,{D}^{2}r\left(t\right) + m\,\sin\left(\varphi\left(t\right)\right)\,{D}^{2}r\left(t\right) + \cos\left(\varphi\left(t\right)\right)\,DU\left(r\left(t\right)\right) + \sin\left(\varphi\left(t\right)\right)\,DU\left(r\left(t\right)\right)}\end{pmatrix}

Rewriting the above equations, we get

\begin{align*} -mr\dot{\varphi}^2 \cos{\varphi} + (mr\ddot{\varphi} + 2m\dot{r}\dot{\varphi})(-\sin{\varphi}) + m\ddot{r}\cos{\varphi} + DU(r)\cos{\varphi} &= 0 \\ -mr\dot{\varphi}^2 \sin{\varphi} + (mr\ddot{\varphi} + 2m\dot{r}\dot{\varphi})\cos{\varphi} + m\ddot{r}\sin{\varphi} + DU(r)\sin{\varphi} &= 0 \end{align*}

Multiply first eqn by $\cos{\varphi}$ and the second by $\sin{\varphi}$ and add

$$-mr\dot{\varphi}^2 + m\ddot{r} + DU(r) = 0\\$$

Multiply first eqn by $\sin{\varphi}$ and the second by $\cos{\varphi}$ and subtract to get

$$mr \ddot{\varphi} + 2m\dot{r}\dot{\varphi} = 0\\$$

These are the same equations of motion that we obtained by directly building the Lagrangian from polar coordinates.

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