Recalling Lagrangian Mechanics

The Lagrangian is a function of coordinates \bgroup\color{black}$ q_i$\egroup and their time derivatives \bgroup\color{black}$ \dot{q}_i$\egroup.

$\displaystyle L(q_i,\dot{q}_i) = T-U$    
$\displaystyle {\partial L\over\partial q_i} = {d\ \over dt}{\partial L\over\partial \dot{q}_i}$    

This gives one second order Lagrange equation for each coordinate.

We may define the conjugate momentum to the coordinate \bgroup\color{black}$ q_i$\egroup as

\bgroup\color{black}$\displaystyle p_i\equiv {\partial L\over\partial \dot{q}_i}. $\egroup

Then Lagrange's equation simply states

\bgroup\color{black}$\displaystyle \dot{p}_i={\partial L\over\partial q_i} .$\egroup

This is one of two ``Hamilton's equations'' which we will discus in the next section where we will use the momentum rather than the velocity \bgroup\color{black}$ \dot{q}$\egroup to analyze a problem.

Recall that symmetries or invariances give us some conserved quantities in physics. An important example of this is the time independence of the Lagrangian.

\bgroup\color{black}$\displaystyle {\partial L\over \partial t}=0 $\egroup

Of course we think the laws of physics are time independent so this is true in the big picture. We may use it calculate the conserved quantity.

$\displaystyle {d L\over d t}$ $\displaystyle ={\partial L\over \partial t} + \sum\limits_j {\partial L\over \partial q_j}\dot{q}_j+ \sum\limits_j {\partial L\over \partial \dot{q}_j}\ddot{q}_j$ chain rule    
$\displaystyle {\partial L\over\partial q_i}$ $\displaystyle = {d\ \over dt}{\partial L\over\partial \dot{q}_i}$ Lagrange eq.    
$\displaystyle {d L\over d t}$ $\displaystyle =\sum\limits_j\dot{q}_j{d\ \over dt}{\partial L\over \partial \dot{q}_j} + \sum\limits_j {\partial L\over \partial \dot{q}_j}\ddot{q}_j$ plug and $\displaystyle {\partial L\over \partial t}=0$    
$\displaystyle {d L\over d t}$ $\displaystyle =\sum\limits_j {d\ \over dt}\left(\dot{q}_j{\partial L\over \partial \dot{q}_j}\right)$    
$\displaystyle {d L\over d t}$ $\displaystyle ={d\ \over dt}\sum\limits_j \dot{q}_j{\partial L\over \partial \dot{q}_j}$ sum is total derivative    
$\displaystyle {0}$ $\displaystyle ={d\ \over dt}\left(\sum\limits_j \dot{q}_j{\partial L\over \partial \dot{q}_j} -L\right)$    

We may define the quantity in parentheses to be the Hamiltonian.

\bgroup\color{black}$\displaystyle H\equiv \sum\limits_j \dot{q}_j{\partial L\over \partial \dot{q}_j} -L=const. $\egroup

With this definition, the Hamiltonian is a conserved quantity. (Recall that in relativity, the momentum conjugate to the time coordinate is the total energy.)

Recall that by Euler's theorem

\bgroup\color{black}$\displaystyle \sum\limits_j \dot{q}_j{\partial L\over \partial \dot{q}_j} =2T $\egroup

if \bgroup\color{black}$ U$\egroup does not depend on the velocities \bgroup\color{black}$ \dot{q}_j$\egroup. (This is rather simple.)

So, if the Potential is velocity independent (remember this) then

\bgroup\color{black}$\displaystyle H=2T-L=T+U .$\egroup

This simplifies the calculation of \bgroup\color{black}$ H$\egroup for most problems. If the Potential is velocity independent, The Hamiltonian is the total energy and the total energy is conserved if the Lagrangian is time independent.

An important exception to this is Electromagnetism where the magnetic force is velocity dependent and hence the Hamiltonian is not simply \bgroup\color{black}$ T+U$\egroup, however, it does represent the total energy including the energy in the EM field. The Hamiltonian for an electron in an electromagnetic field is.

\bgroup\color{black}$\displaystyle H_{EM}= {1\over 2m}\left(\vec{p}+{e\over c}\vec{A}\right)^2 -e\phi $\egroup

Jim Branson 2012-10-21