Euler-Lagrange and Hamilton's Equations

Newton's second law of motion is  

$${\bf F} = \frac{d{\bf p}}{dt}$$ (139)

or in component form (for each component F_i)  

$$F_{i} = \frac{d{p_{i}}}{dt}$$ (140)

where $p_{i} = m \dot{q}_{i}$ (with q_i being the generalized position coordinate) so that $\frac{d{p_{i}}}{dt} = \dot{m} \dot{q}_{i} + m \ddot{q}_{i}$. If $\dot{m} = 0$ then $F_{i} = m \ddot{q}_{i} = m a_{i}$. For conservative forces ${\bf F} = - \bigtriangledown V$ where V is the scalar potential. Rewriting Newton's law we have  

$$-\frac{dV}{d q_{i}} = \frac{d}{dt}(m \dot{q}_{i})$$ (141)

Let us define the Lagrangian $L(q_{i} , \dot{q}_{i}) \equiv T - V$ where T is the kinetic energy. In freshman physics $T = T(\dot{q}_{i}) = \frac{1}{2} m \dot{q}_{i}^{2}$ and V=V(q_i) such as the harmonic oscillator $V(q_{i}) = \frac{1}{2} k q_{i}^{2}$. That is in freshman physics T is a function only of velocity $\dot{q}_{i}$ and V is a function only of position q_i. Thus $L(q_{i} , \dot{q}_{i}) = T(\dot{q}_{i}) - V(q_{i})$. It follows that $\frac{\partial L}{\partial q_{i}} =-\frac{dV}{d q_{i}}$ and $\frac{\partial L}{\partial \dot{q}_{i}} =\frac{dT}{d \dot{q}_{i}} = m \dot{q}_{i} = p_{i}$. Thus Newton's law is

with the canonical momentum [1] defined as  

$$p_{i} \equiv \frac{\partial L}{\partial \dot{q}_{i}}$$ (142)

The second equation of (4.4) is known as the Euler-Lagrange equations of motion and serves as an alternative formulation of mechanics [1]. It is usually written  

$$\frac{d}{dt}(\frac{\partial L}{\partial \dot{q}_{i}}) - \frac{\partial L}{\partial q_{i}} = 0$$ (143)

or just  

$$\dot{p}_{i} = \frac{\partial L}{\partial q_{i}}$$ (144)

We have obtained the Euler-Lagrange equations using simple arguments. A more rigorous derivation is based on the calculus of variations [1] as discussed in Section 7.3.

We now introduce the Hamiltonian H defined as a function of p and q as  

$$H( p_{i} , q_{i}) \equiv p_{i} \dot{q}_{i} - L( q_{i} , \dot{q}_{i} )$$ (145)

For the simple case $T= \frac{1}{2} m \dot{q}_{i}^{2}$ and $V \neq V(\dot{q}_{i})$ we have $p_{i}\frac{\partial L}{\partial \dot{q}_{i}} = m \dot{q}_{i}$ so that $T = \frac{p_{i}^{2}}{2m}$and $p_{i} \dot{q}_{i} = \frac{p_{i}^{2}}{m}$ so that $H( p_{i} , q_{i}) = \frac{p_{i}^{2}}{2m} + V(q_{i}) = T + V$ which is the total energy. Hamilton's equations of motion immediately follow from (4.8) as  

$$\frac{\partial H}{\partial p_{i}} = \dot{q}_{i}$$ (146)

because $L \neq L(p_{i})$ and $\frac{\partial H}{\partial q_{i}} = - \frac{\partial L}{\partial q_{i}}$ so that from (4.4)  

$$- \frac{\partial H}{\partial q_{i}} = \dot{p}_{i} .$$ (147)