Hamiltonian mechanics ed

More extreme then Lagrangian mechanics!

Canonical momentum ed

Having the Langrangian \(L\) on \(TQ\), we can define\[ p_i(q, \dot q, t) = \frac{\partial L}{\partial \dot q^i}(q, \dot q, t) \]to be the canonical momentum variables. This is actually a map from the tangential bundle \(TQ\) into the dual bundle \(T^\star Q\).

We also need the inverse map \(\dot q(q, p, t)\), so ...assuming invertability!

Having coordinates \(q^i\) on \(Q\), we have the basis vectors \(\partial_i = \frac{\partial}{\partial q^i}\) on \(TQ\). We can use the dual basis of \(dq^i\) on \(T^\star Q\) with \(dq^i(\partial_j) = \delta^i_j\). So the basis expansion of a dual vector is written as\[ p = p_i \, dq^i \]and we can find vector components by\[ dq^i(\dot q) = \dot q^i \]

Hamiltonian ed

Define the Hamiltonian to be the Legendre transform:\[ H(q,p,t) = p_i \, \dot q^i(q, p, t) - L(q, \dot q(q, p, t), t) = p(\dot q) - L(\dots) \]

Let's assume, \((q(t), p(t))\) is a solution of the Euler-Lagrange-Equations. Then\[ \frac{\partial H}{\partial q^i}(\dots) = p \frac{\partial \dot q}{\partial qi} - \frac{\partial L}{\partial q^i} - \underbrace{\frac{\partial L}{\partial \dot q}}_{=p} \frac{\partial \dot q}{\partial q^i} = - \frac{\partial L}{\partial q^i} = - \frac{d}{dt} \frac{\partial L}{\partial \dot q^i} = - \dot p_i \]and\[ \frac{\partial H}{\partial p_i} = \dot q^i + p \frac{\partial \dot q}{\partial p_i} - \underbrace{\frac{\partial L}{\partial \dot q}}_{=p} \frac{\partial \dot q}{\partial p_i} = \dot q^i \]

In short, we have the canonical equations\[ \dot q = \frac{\partial H}{\partial p}, \quad \dot p = - \frac{\partial H}{\partial q} \]