# Wiki Wiki Web

## Lagrangian

The classical lagrangian is: $$L := T-V$$

where $T=\frac{1}{2}mv^2$ is the kinetic energy and $V$ is the potential field $F = - grad V$.

We can have $C$ holonomic constraints $f_i(r,t)=0$ $i \in {1, \ldots C}$.

### Lagrange equations of the first kind

$$\frac{\partial L}{\partial r} - \frac{d}{dt} \frac{\partial L}{\partial \dot{r}} + \sum_{i=1}^C \frac{\partial f_i}{\partial r} = 0$$

### Euler-Lagrange equations of the second kind

$r_k = r(q,t)$ where $q$ is a point in configuration space which takes into account constraints.

$\dot{q}_j = \frac{d q_j}{dt}$, $v_k = \sum_j \frac{\partial r_k}{\partial q_j} \dot{q}_j + \frac{\partial r_k}{\partial t}$

The euler-Lagrange equation is: $$\frac{d}{dt} \frac{\partial L}{\partial \dot{q}_j} = \frac{\partial L}{\partial q_j}$$

### Principle of least action or Hamilton Principle

This states that the action $S$ of the lagrangian is minimal.

$S = \int_t L dt$ is minimal, maximal or at a saddle point.

We check that the action does not change for a small perturbation of the lagrangian $\delta L$:

$$\delta L = \sum_j \frac{\partial L}{\partial q_j} \delta q_j+ \frac{\partial L}{\partial \dot{q}_j} \delta \dot{q}_j$$

$\delta \dot{q}_j = \frac{d}{dt} \delta q_j$

$\frac{\partial L}{\partial \dot{q}_j} \frac{d}{dt} \delta q_j = \frac{d}{dt} (\frac{\partial L}{\partial \dot{q}_j} \delta q_j) - \frac{d}{dt} \frac{\partial L}{\partial \dot{q}_j} \delta q_j$

$$\int \delta L dt = \left[\frac{\partial L}{\partial \dot{q}_j} \delta q_j\right]_0^T + \int \left(\sum_j \frac{\partial L}{\partial q_j} - \frac{d}{dt} \frac{\partial L}{\partial \dot{q}_j} \right) \delta q_j dt$$

The first term cancels as the perturbation has to be 0 at initial and final value, and the integral cancels on all path if and only if the Eurler-Lagrange equation holds.

### Hamiltonian Equation

We define $$H = \sum p_i \dot{q}_i - L$$

where $p_i$ is the canonical momentum $p_i := \frac{\partial L}{\partial \dot{q}_i}$

We can derive Hamilton's equations from Eurler-Lagrange equations.

\begin{eqnarray} dL &=& \sum_i{\frac{\partial L}{\partial \dot{q}_j} d \dot{q}_j + \frac{\partial L}{\partial q_j} d q_j } + \frac{\partial L}{\partial t} dt \\ &=& \sum_i{{p_i d \dot{q}_j} + \frac{\partial L}{\partial q_j} d q_j } + \frac{\partial L}{\partial t} dt \\ \end{eqnarray} integrating by parts: \begin{eqnarray} dL &=& \sum_i{d({p_i \dot{q}_j}) - \dot{q}_j dp_i + \frac{\partial L}{\partial q_j} d q_j } + \frac{\partial L}{\partial t} dt \\ d(\sum_i p_i \dot{q}_i -L) &=& \sum_i{ \dot{q}_j dp_i - \frac{\partial L}{\partial q_j} d q_j } - \frac{\partial L}{\partial t} dt \\ dH &=& \sum_i{\frac{\partial H}{\partial p_j} d \dot{q}_j + \frac{\partial H}{\partial q_j} d q_j } + \frac{\partial H}{\partial t} dt \\ \end{eqnarray} Equalising the $dp_i$, $dq_i$ and $dt$ terms, we get following the $2n+1$ equations: $\frac{\partial H}{\partial p_j} = \dot{q}_j$, $\frac{\partial H}{\partial q_j} = - \frac{\partial L}{\partial q_j}$ $\frac{\partial H}{\partial t} = -\frac{\partial L}{\partial t}$

To get rid of the Lagrangian in the Hamiltonian evolution, we can then use the Euler-Lagrange equation to reinject the value of $\frac{\partial L}{\partial q_j} = \frac{d}{dt} \frac{\partial L}{\partial \dot{q}_j}= \frac{dp_i}{dt} = \dot{p}_i$:

We then get the Hamiltonian evolution: \begin{eqnarray} \frac{\partial H}{\partial p_j} &=& \dot{q}_j \\ \frac{\partial H}{\partial q_j} &=& - \dot{p}_i \\ \frac{\partial H}{\partial t} &=& -\frac{\partial L}{\partial t} \\ \end{eqnarray} The hamiltonian is equivalent to the lagrange formulation, one has $n$ equations of the second order, the other one, a system of $2n$ equations of the first order.

## Legendre Transform

We had $L=T-V$ whereas $E=T+V$. Energy is related to the Hamiltonian of the system.

Another way to see it is as a Legendre transform. Let: \begin{eqnarray} df &=& u dx + v dy \\ u &=& \frac{\partial f}{\partial x} \\ v &=& \frac{\partial f}{\partial y} \end{eqnarray} Using $u,v$ as free variables, we define the legendre transform $g$: $$g = f - ux$$ We have $$dg = -x du + v dy$$

Similarly, in thermodynamics, the free energy $F=U-TS$ is a function of $T,V$ whereas energy is a function of extensive variables $S,V$.

## Poisson Brackets

The poisson bracket is defined as: $$[f,g]_P = \sum_k \frac{\partial f}{\partial q_k} \frac{\partial g}{\partial p_k} - \frac{\partial f}{\partial p_k} \frac{\partial g}{\partial q_k}$$ The bracket is antisymetric, like the commutator: $[f,g] = -[g,f]$

Properties for the coordinate functions are: \begin{eqnarray} [p_i,q_j] &=& \delta_{ij} \\ [p_i,p_j] &=& 0 \\ [q_i,q_j] &=& 0 \end{eqnarray}

Hamiltonian equations expressed as Poisson brackets are: $$\dot{q}_i = [q_i, H ]_P, \dot{p}_i = [p_i, H ]_P$$