1
$\begingroup$

Consider a Lagrangian $L(x,\dot x,t)$ and a corresponding Hamiltonian $H=\dot xp-L$ where $p=\partial L/\partial \dot x$ which satisfies Hamilton's equations $$\frac{\partial H}{\partial x}=-\dot p$$ $$\frac{\partial H}{\partial p}=\dot x.$$ I'm trying to show that Hamilton's equations are unchanged by a gauge transformation of the Lagrangian $L'=L+ \frac{dF}{dt}$ where $F(x,t)$ is a function of the position and time only. I first expand the derivative of $F$ $$\frac{dF}{dt}=\frac{\partial F}{\partial t}+ \frac{\partial F}{\partial x}\dot x$$ the new conjugate momentum is $$p'=\frac{\partial L'}{\partial \dot x}=\frac{\partial L}{\partial \dot x}+\frac{\partial}{\partial \dot x} \frac{dF}{dt}=p+ \frac{\partial F}{\partial x}$$ and so $$\frac{\partial}{\partial p'}= \frac{\partial p}{\partial p'} \frac{\partial}{\partial p}=\frac{\partial}{\partial p}$$ The new Hamiltonian is $$H'=p'\dot x-L' = p \dot x-L+\frac{\partial F}{\partial x} \dot x - \frac{dF}{dt}=H- \frac{\partial F}{\partial t}$$ Hamilton's equations are then $$\frac{\partial H'}{\partial p'}=\frac{\partial H}{\partial p}- \frac{\partial}{\partial p} \frac{\partial F}{\partial t}=\dot x-0=\dot x$$ and $$\frac{\partial H'}{\partial x}=\frac{\partial H}{\partial x}- \frac{\partial}{\partial x} \frac{\partial F}{\partial t}= -\dot p-\frac{\partial }{\partial t} \frac{\partial F}{\partial x}$$ It is this last equation where I'm having trouble. To satisfy Hamilton's equations, the right side should be equal to $-\dot p'= -\frac{d}{dt}(p+\frac{\partial F}{\partial x})$ however I end up with a partial derivative on the last term rather than a total derivative as it should be. How can one justify this as satisfying Hamilton's equations?

$\endgroup$
1
$\begingroup$

Assuming $x,\dot x$ and $p'$ are independent we have $$\frac{\partial H'}{\partial x}= \frac{\partial}{\partial x } \left( p' \dot x-L'\right)= -\frac{\partial L'}{\partial x}=-\frac{\partial L}{\partial x}-\frac{\partial}{\partial x}\frac{dF}{dt}$$ By Euler-Lagrange: $$\frac{\partial L}{\partial x}= \frac{d}{dt} \left( \frac{\partial L}{\partial \dot x}\right) = \frac{d p}{dt}$$ Commuting the partial in $x$ and the total time derivative we get $$\frac{\partial H'}{\partial x}= -\frac{d}{dt} \left( p+ \frac{\partial F}{\partial x}\right)=-\dot p$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.