Possion’s Lagrange Equation

Structure and Interpretation of Classical Mechanics

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

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

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

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

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

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

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

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

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 $\displaystyle{D\eta(t_1) = D\eta(t_2) = 0}$ ,

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

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

Physics Town

continues

Possion’s Lagrange Equation

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