# The Principle of Least Action

$$\frac{\d x_t}{\d t} = f_t(x_t)$$

# Principle of Least Action

$$\htmlClass{fragment}{ S(q) = \int_{t_0}^{t_1} L_t(q_t,\dot{q}_t) \d t } \qquad \htmlClass{fragment}{ \delta S(q) = 0 }$$

# What does this tell us?

Letting $q^\eps = q + \eps v$, write $$\htmlClass{fragment}{ \delta S(q) } \htmlClass{fragment}{ = \left.\frac{\d}{\d\eps} \int_{t_0}^{t_1} L(q_t^\eps, \dot{q}_t^\eps) \d t\right|_{\eps=0} } \htmlClass{fragment}{ = \left.\int_{t_0}^{t_1} \frac{\d}{\d\eps}L(q_t^\eps, \dot{q}_t^\eps) \right|_{\eps=0} \d t }$$ and, letting $L^\eps = L(q_t^\eps, \dot{q}_t^\eps)$, the total derivative is $$\htmlClass{fragment}{ \frac{\d L^\eps}{\d\eps} } \htmlClass{fragment}{ = \frac{\d t}{\d\eps} \frac{\pd L^\eps}{\pd t} + \frac{\d q^\eps}{\d\eps} \frac{\pd L^\eps}{\pd q^\eps} + \frac{\d \dot{q}^\eps}{\d\eps} \frac{\pd L^\eps}{\pd \dot{q}^\eps} } \htmlClass{fragment}{ = v \frac{\pd L^\eps}{\pd q^\eps} + \dot{v} \frac{\pd L^\eps}{\pd \dot{q}^\eps} . }$$

# Euler-Lagrange Equations

Hence, choosing variations which vanish at the boundary, $$\htmlClass{fragment}{ \delta S(q) } \htmlClass{fragment}{ = \int_{t_0}^{t_1} v \frac{\pd L}{\pd q} + \dot{v} \frac{\pd L}{\pd \dot{q}} \d t } \htmlClass{fragment}{ = \int_{t_0}^{t_1} v \del{\frac{\pd L}{\pd q} - \frac{\d}{\d t} \frac{\pd L}{\pd \dot{q}} }\d t } \htmlClass{fragment}{ = 0 }$$ using integration by parts, we obtain the differential equation $$\htmlClass{fragment}{ \frac{\d}{\d t} \frac{\pd L}{\pd \dot{q}} - \frac{\pd L}{\pd q} = 0 }$$ by applying the fundamental lemma of variational calculus.

# Newton's Second Law

Consider the Lagrangian $$\htmlClass{fragment}{ L(q,\dot{q}) } \htmlClass{fragment}{ = \frac{1}{2}m\dot{q}^2 - V(q) }$$ and write $\htmlClass{fragment}{\frac{\pd L}{\pd q}}\htmlClass{fragment}{ = - \frac{\pd V}{\pd q}}\htmlClass{fragment}{ = F}$ along with $\htmlClass{fragment}{\frac{\d}{\d t} \frac{\pd L}{\pd \dot{q}}}\htmlClass{fragment}{ = \frac{\d}{\d t}m\dot{q}}\htmlClass{fragment}{ = ma}$. Then $$\htmlClass{fragment}{ \frac{\d}{\d t} \frac{\pd L}{\pd \dot{q}} - \frac{\pd L}{\pd q} = 0 } \qquad \htmlClass{fragment}{ \implies } \qquad \htmlClass{fragment}{ F = ma . }$$

# Properties of Euler-Lagrange Equations

$$\frac{\d}{\d t} \frac{\pd L}{\pd \dot{q}} - \frac{\pd L}{\pd q} = 0$$

# Contact Dynamics

$$\htmlClass{fragment}{ \delta S(q,\dot{q}) = 0 } \qquad \htmlClass{fragment}{ S(q,\dot{q}) = \int_{t_0}^{t_1} L(q_t,\dot{q}_t) \d t }$$