Brief Review of Undergraduate Classical Mechanics (8) - Basic Hamiltonian Mechanics

이 글은 2020.10.14에 작성됨.

This post was written on Oct 14, 2020.

 

If you want to use dynamical degrees of freedom $(p,q)$ instead of $(q,\dot{q})$, we now should conduct a Legendre transformation.

$$\begin{align} dL(q^i, \dot{q}^i) &=\frac{\partial L}{\partial q^i}dq^i + \frac{\partial L}{\partial \dot{q}^i} d\dot{q}^i + \frac{\partial L}{\partial t} dt \\ &= \dot{p}_i dq^i + p_i d\dot{q}^i + \frac{\partial L}{\partial t} dt \\ &=d(p_i \dot{q}^i)-\dot{q}^i dp_i + \dot{p}_i dq^i + \frac{\partial L}{\partial t} dt  \\ d(p_i \dot{q}^i -L) & =\dot{q}^i dp_i - \dot{p}_i dq^i - \frac{\partial L}{\partial t} dt\end{align}$$

We can define another quantity $H=pq-L$ parameterized with $p$ and $q$, which is called 'Hamiltonian'. It is different from Energy function $h(q,\dot{q}$. Fundamentally its parameters cover $q$ and $\dot{q}$. Think about thermodynamic potentials.

$$dU(S,V,N)=TdS-PdV+\mu dN$$

$$dF(T,V,N)=-SdT-PdV+\mu dN$$

It is also an example of Legendre transformation. Anyway, If we define Hamiltonian like this, we can get Hamilton's equation of motion.

$$\dot{q}^i = \frac{\partial H}{\partial p_i},\;\;\;\;\;\; \dot{p}_i=-\frac{\partial H}{\partial q^i}$$

This is the 2 sets of 1st order differential equation while Lagrangian mechanics yields a set of 2nd order differential equation. Let's consider total time-derivative of the Hamiltonian.

$$\frac{dH}{dt}=\frac{\partial H}{\partial p_i} dp_i + \frac{\partial H}{\partial q^i} dq^i + \frac{\partial H}{\partial t}=\dot{q}^i \dot{p}_i - \dot{p}_i \dot{q}^i + \frac{\partial H}{\partial t}=\frac{\partial H}{\partial t}$$

This is a key concept of Hamiltonian mechanics. If Hamiltonian doesn't depend on time explicitly, where we call it has a time-translational symmetry, $H$ is conserved quantity. We saw above that Noether current(charge) about time-translation symmetry was Energy function.

 

We can also construct Hamiltonian mechanics from action extremization.

$$S=\int L dt =\int dt \left(p_i\dot{q}^i-H \right)$$

Varying the action, where our boundary variation is always 0, we get

$$\delta S = \int dt \left( \dot{q}^i \delta p_i + p_i \delta \dot{q}^i - \frac{\partial H}{\partial p_i} \delta p_i - \frac{\partial H}{\partial q^i} \delta q^i \right) = 0$$

and after integration by parts,

$$\int dt \left( \left( \dot{q}^i - \frac{\partial H}{\partial p_i} \right) \delta p_i - \left( \dot{p}_i + \frac{\partial H}{\partial q_i} \right) \delta q^i \right) =0$$

from which we can derive Hamilton's equations of motion also.