SICM Exercise 1.15
Solution to exercise 1.15 of Structure and Interpretation of Classical Mechanics by Gerald Jay Sussman and Jack Wisdom.
Equivalence
🛈 Note
Show by direct calculation that the Lagrange equations for \(L′\) are satisfied if the Lagrange equations for \(L\) are satisfied.
We have to prove that if
\[ D \left( \partial_2 L \circ \Gamma [q] \right) - \partial_1 L \circ \Gamma [q] = 0 \]then
\[ D \left( \partial_2 L' \circ \Gamma [q'] \right) - \partial_1 L' \circ \Gamma [q'] = 0 \]Start by substituting \( L' = L \circ C \) into the second equation.
\[D \left( \partial_2 \left( L \circ C \right) \circ \Gamma [q'] \right) - \partial_1 \left( L \circ C \right) \circ \Gamma [q'] = 0\]Now we can apply the chain rule for partial derivatives
\[D \left( \left(DL \circ C \cdot \partial_2 C \right) \circ \Gamma [q'] \right) - \left( DL \circ C \cdot \partial_1 C \right) \circ \Gamma [q'] = 0\]Expand the composition operator to the factors
\[D \left(DL \circ C \circ \Gamma [q'] \cdot \partial_2 C \circ \Gamma [q'] \right) - DL \circ C \circ \Gamma [q'] \cdot \partial_1 C \circ \Gamma [q'] = 0\]Now we can use the \(\Gamma [q] = C \circ \Gamma [q']\) identitiy.
\[D \left(DL \circ \Gamma [q] \cdot \partial_2 C \circ \Gamma [q'] \right) - DL \circ \Gamma [q] \cdot \partial_1 C \circ \Gamma [q'] = 0\]Now let us look at the derivatives inside the equation. The derivative \(DL\) is a down tuple of the partial derivatives of \(L\).
\[DL = [\partial_0 L,\, \partial_1 L,\, \partial_2 L,\, ...] \]\[DL \circ \Gamma[q] = [\partial_0 L \circ \Gamma[q],\, \partial_1 L \circ \Gamma[q],\, \partial_2 L \circ \Gamma[q],\, ...] \]
The partial derivatives of \(C\) are the partial derivatives of the up tuples. For example \( \partial_2 C \) is the derivative of the up tuple with respect to the third argument of \(C\).
\[\partial_2 C = \left(0,\, 0,\, \partial_1F(t,x'),\, ...\right)\]\[\partial_1 C = \left(0,\, \partial_1 F(t, x'),\, \partial_1 \partial_0 F(t, x') + \partial_1^2 F(t, x')v',\, ...\right)\]\[\partial_2 C \circ \Gamma[q'] = \left(0,\, 0,\, \partial_1F(t,q'),\, ...\right)\]
\[\partial_1 C \circ \Gamma[q'] = \left(0,\, \partial_1 F(t, q'),\, \partial_1 \partial_0 F(t, q') + \partial_1^2 F(t, q')Dq',\, ...\right)\]
Substituting these into the lagrange equation and using the rule for the multiplicatioin of a down tuple and an up tuple we get:
\begin{equation*} D \left(\partial_2 L \circ \Gamma [q] \cdot \partial_1 F(t, q’) \right) - \left( \partial_1 L \circ \Gamma [q] \cdot \partial_1 F(t, q’) + \partial_2 L \circ \Gamma[q] \cdot (\partial_1 \partial_0 F(t, q’) + \partial_1^2 F(t, q’)Dq’ \right) = 0 \end{equation*}
Now we can take the derivative of the first expression and use the product rule.
\begin{equation*} D \left(\partial_2 L \circ \Gamma [q] \right) \cdot \partial_1 F(t, q’) + \partial_2 L \circ \Gamma [q] \cdot \left(\partial_0 \partial_1 F(t, q’) + \partial_1^2 F(t, q’) Dq’ \right) - \left( \partial_1 L \circ \Gamma [q] \cdot \partial_1 F(t, q’) + \partial_2 L \circ \Gamma[q] \cdot (\partial_1 \partial_0 F(t, q’) + \partial_1^2 F(t, q’)Dq’ \right) = 0 \end{equation*}
\[ D \left(\partial_2 L \circ \Gamma [q] \right) \cdot \partial_1 F(t, q') - \partial_1 L \circ \Gamma [q] \cdot \partial_1 F(t, q') = 0 \]Now we can divide by \( \partial_1 F(t, q') \) to get the lagrange equation for the unprimed system.
\[ D \left(\partial_2 L \circ \Gamma [q] \right) - \partial_1 L \circ \Gamma [q] = 0 \]