Ex 1.14 Lagrange equations for L’

Structure and Interpretation of Classical Mechanics

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

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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:

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

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

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

Physics Town

continues

Ex 1.14 Lagrange equations for L’

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