$\newcommand{\pp}[2][]{\frac{\partial#1}{\partial #2}}$
Lagrangian, canonical momenta and Hamiltonian are given for a variety of systems. It is suggested that the reader may verify all the results given below.
- Free particle \begin{eqnarray} L &=& \frac{1}{2}m \dot x^2 - V(x)\\ p &=& \pp[L]{\dot x}=m\dot x\\ H &=& \frac{p^2}{2m} + V(x). \end{eqnarray} In this case the canonical momentum is ordinary momentum \(m\dot x\) and \(H\) has the form (K.E + P.E.). Similar statements are not always true.
- From Hamiltonian to Lagrangian \begin{eqnarray} H(q,p) = \dfrac{p^\beta}{\beta} + V(q) \end{eqnarray} The canonical momentum is \begin{equation} \dot{q} = \beta \Big(\frac{p^{\beta-1}}{\beta}\Big) \Longrightarrow p\dot{q} = p^{\beta-1} \end{equation} and the Lagrangian works out to be \[L = p\dot q-H=\frac{\dot{q}^\alpha}{\alpha} -V(q)\] where \(\alpha\) is given by \[ \frac{1}{\alpha} + \frac{1}{\beta}=1\]
- For a hypothetical system \begin{eqnarray} L &=&\frac{1}{2}M \dot x^2 \exp(x/\alpha) -V(x)\\ p &=& M \dot x \exp(x/\alpha)\\ H &=& \frac{p^2}{2M}\, \exp(-x/\alpha) + V(x) \end{eqnarray}
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