# Varying the action, 2.1

Equation (1.28): $\displaystyle{S[q](t_1, t_2) = \int_{t_1}^{t_2} L \circ \Gamma[q]}$

Equation (1.30): $\displaystyle{h[q] = L \circ \Gamma[q]}$ $\displaystyle{\delta_\eta S[q](t_1, t_2) = \int_{t_1}^{t_2} \delta_\eta h[q]}$

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Let $\displaystyle{F}$ be a path-independent function and $\displaystyle{g}$ be a path-dependent function; then $\displaystyle{\delta_\eta h[q] = \left( DF \circ g[q] \right) \delta_\eta g[q]~~~~~\text{with}~~~~~h[q] = F \circ g[q].~~~~~(1.26)}$ $\displaystyle{\delta_\eta F \circ g[q] = \left( DF \circ g[q] \right) \delta_\eta g[q]}$

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— 1.5.1 Varying a path

— Structure and Interpretation of Classical Mechanics

. \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}}

Assume that $\displaystyle{L}$ is a path-independent function, so that we can use Eq. 1.26: \displaystyle{ \begin{aligned} &= \int_{t_1}^{t_2} (D L \circ \Gamma[q]) \delta_\eta \Gamma[q] \\ \end{aligned}} \displaystyle{ \begin{aligned} &= \int_{t_1}^{t_2} (D L \circ \Gamma[q]) (0, \eta(t), D\eta(t)) \\ &= \int_{t_1}^{t_2} (D L \left[ \Gamma[q] \right]) (0, \eta(t), D\eta(t)) \\ \end{aligned}}

Assume that $\displaystyle{L}$ is a path-independent function, so that any value of $\displaystyle{L}$ depends on the value of $\displaystyle{\Gamma}$ at that moment only, instead of depending on the whole path $\displaystyle{\Gamma}$: \displaystyle{ \begin{aligned} &= \int_{t_1}^{t_2} (D L (\Gamma[q])) (0, \eta(t), D\eta(t)) \\ &= \int_{t_1}^{t_2} (D L (t, q, v)) (0, \eta(t), D\eta(t)) \\ &= \int_{t_1}^{t_2} [\partial_0 L (t, q, v), \partial_1 L (t, q, v), \partial_2 L (t, q, v)] (0, \eta(t), D\eta(t)) \\ \end{aligned}}

What kind of product is it here? Is it just a dot product? Probably not. \displaystyle{ \begin{aligned} &= \int_{t_1}^{t_2} [\partial_1 L (t, q, v) \eta(t) + \partial_2 L (t, q, v) D\eta(t)] \\ \end{aligned}}

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— Me@2019-10-12 03:42:01 PM

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# Introduction to Differential Equations

llamaz 1 hour ago [-]

I think the calculus of variations might be a better approach to introducing ODEs in first year.

You can show that by generalizing calculus so the values are functions rather than real numbers, then trying to find a max/min using the functional version of $\displaystyle{\frac{dy}{dx} = 0}$, you end up with an ODE (viz. the Euler-Lagrange equation).

This also motivates Lagrange multipliers which are usually taught around the same time as ODEs. They are similar to the Hamiltonian, which is a synonym for energy and is derived from the Euler-Lagrange equations of a system.

Of course you would brush over most of this mechanics stuff in a single lecture (60 min). But now you’ve motivated ODEs and given the students a reason to solve ODEs with constant coefficients.

— Hacker News

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2019.10.02 Wednesday ACHK

# 少補習

— Me@2017-11-09 01:26:59 PM

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