SICM Exercise 1.14
Solution to exercise 1.14 of Structure and Interpretation of Classical Mechanics by Gerald Jay Sussman and Jack Wisdom.
Coordinate-independence of Lagrange equations
🛈 Note
Check that the Lagrange equations for central force motion in polar coordinates and in rectangular coordinates are equivalent. Determine the relationship among the second derivatives by substituting paths into the transformation equations and computing derivatives, then substitute these relations into the equations of motion.
The path transformations are as follows:
\[ x(t) = r(t) \cos \varphi(t) \]\[ y(t) = r(t) \sin \varphi(t) \]\[ v_x(t) = Dx(t) = \dot r(t) \cos \varphi(t) - r(t) \dot \varphi(t) \sin \varphi(t) \]
\[ v_y(t) = Dy(t) = \dot r(t) \sin \varphi(t) - r(t) \dot \varphi(t) \cos \varphi(t) \]\[ a_x(t) = D^2x(t) = \ddot r(t) \cos \varphi(t) - 2 \dot r(t) \dot \varphi(t) \sin \varphi(t) - r(t) \left( \ddot \varphi(t) \sin \varphi(t) + {\dot \varphi}^2(t) \cos \varphi(t) \right) \]
\[ a_y(t) = D^2y(t) = \ddot r(t) \sin \varphi(t) + 2 \dot r(t) \dot \varphi(t) \cos \varphi(t) + r(t) \left( \ddot \varphi(t) \cos \varphi(t) - {\dot \varphi}^2(t) \sin \varphi(t) \right) \]
The Lagrange equations in the carthesian coordinate system are (eqs. 1.62 and 1.63 from the book):
\[ m D^2 x(t) = - \frac{x(t)}{r(t)} DU(r(t)) \]\[ m D^2 y(t) = - \frac{y(t)}{r(t)} DU(r(t)) \]
We substitute the path transforms into the Lagrange equations.
\begin{equation} m \left(\ddot r(t) \cos \varphi(t) - 2 \dot r(t) \dot \varphi(t) \sin \varphi(t) - r(t) \left( \ddot \varphi(t) \sin \varphi(t) + {\dot \varphi}^2(t) \cos \varphi(t) \right) \right) = - \cos \varphi(t) DU(r(t)) \label{eq:1.14_lag_1} \end{equation}
\begin{equation} m \left(\ddot r(t) \sin \varphi(t) + 2 \dot r(t) \dot \varphi(t) \cos \varphi(t) + r(t) \left(\ddot \varphi(t) \cos \varphi(t) - {\dot \varphi}^2(t) \sin \varphi(t) \right) \right) = - \sin \varphi(t) DU(r(t)) \label{eq:1.14_lag_2} \end{equation}
To get the first Lagrange equation we have to multiply Equation \ref{eq:1.14_lag_1} by \( \cos \varphi(t) \) and Equation \ref{eq:1.14_lag_2} by \( \sin \varphi(t) \). We can also swap the signs of the last term of the left hand side of the second equation to make things simpler.
\begin{equation*} m \bigl(\ddot r(t) \cos^2 \varphi(t) - 2 \dot r(t) \dot \varphi(t) \sin \varphi(t) \cos \varphi(t) - r(t) \left( \ddot \varphi(t) \sin \varphi(t) \cos \varphi(t) + {\dot \varphi}^2(t) \cos^2 \varphi(t) \right) \bigr) = - \cos^2 \varphi(t) DU(r(t)) \end{equation*}
\begin{equation*} m \bigl(\ddot r(t) \sin^2 \varphi(t) + 2 \dot r(t) \dot \varphi(t) \sin \varphi(t) \cos \varphi(t) - r(t) \left(- \ddot \varphi(t) \sin \varphi(t) \cos \varphi(t) + {\dot \varphi}^2(t) \sin^2 \varphi(t) \right) \bigr) = - \sin^2 \varphi(t) DU(r(t)) \end{equation*}
Now we add up the two equations. Using the trignometric identity \( \sin^2 a + \cos^2 b = 1 \) we can simplify most terms. The other terms that contain \( \sin \varphi (t) \cos \varphi (t) \) have opposing signs and drop out.
\[ m (\ddot r - r {\dot \varphi}^2) = - DU(r(t)) \]To get the second Lagrange equation we have to multiply Equation \ref{eq:1.14_lag_1} by \( \sin \varphi(t) \) and Equation \ref{eq:1.14_lag_2} by \( \cos \varphi(t) \).
\begin{equation*} m \bigl(\ddot r(t) \sin \varphi(t) \cos \varphi(t) - 2 \dot r(t) \dot \varphi(t) \sin^2 \varphi(t) - r(t) \left( \ddot \varphi(t) \sin^2 \varphi(t) + {\dot \varphi}^2(t) \sin \varphi(t) \cos \varphi(t) \right) \bigr) = - \sin \varphi(t) \cos \varphi(t) DU(r(t)) \end{equation*}
\begin{equation*} m \bigl(\ddot r(t) \sin \varphi(t) \cos \varphi(t) + 2 \dot r(t) \dot \varphi(t) \cos^2 \varphi(t) - r(t) \left(- \ddot \varphi(t) \cos^2 \varphi(t) + {\dot \varphi}^2(t) \sin \varphi(t) \cos \varphi(t) \right) \bigr) = - \sin \varphi(t) \cos \varphi(t) DU(r(t)) \end{equation*}
Now we subtact the first equation from the second equation. Doing the same simplifications as in the previous equation we get the second Lagrange equation simplified by \( r(t) \).
\[ 2 m \dot r \dot \varphi + r \ddot \varphi = 0 \]