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?


It seems that the error is arising when taking the partial of $H'$. When applying $\partial/\partial x$ to $H'(x,p')$ we are implicitly holding $p'$ constant, where as $\partial H/\partial x=- \dot p$ is true only when we are holding $p$ constant, not $p'$. A workaround is to use the form $H'=p' \dot x-L'$ as the functional dependence is clear. Then $$\frac{\partial H'}{\partial x}= -\frac{\partial L'}{\partial x} = -\frac{\partial L}{\partial x}- \frac{\partial}{\partial x} \frac{d F}{dt}= -\frac{d}{dt} \left( \frac{\partial L}{\partial \dot x} + \frac{\partial F}{\partial x}\right) = -\dot p'$$ where Euler-Lagrange was used on the second last equality. I'm still unsure of how to apply the chain rule to my original expression to get the correct result.


This is a situation where it helps to be very careful about what everything means. First, (being very persnickety) the right hand side should be equal to $-\dot{p}'$ (since $F=-\nabla_x V$.)

Secondly, (hint) the operator $\partial_x$ has a slightly different meaning when acting on functions of $p,x$ and functions of $p',x$.


Applying the chain rule explicitly, \begin{align*} \partial_x|_{p'} H' &= \frac{\partial H}{\partial x}\Big|_p+\frac{\partial H}{\partial p}\Big|_x\frac{\partial p}{\partial x}\Big|_{p'}-\frac{\partial}{\partial x}\frac{\partial F}{\partial t}\quad \text{(since $F$ depends only on $x,t$)}\\ &=-\dot{p}+\dot{x}\Big(-\frac{\partial}{\partial x}\Big|_{p'}\frac{\partial}{\partial x}F(x,t)\Big)-\frac{\partial}{\partial x}\frac{\partial F}{\partial t}\\ &=-\dot p-\dot x\partial_x^2|_{p'}F(x,t)-\partial_x\partial_tF \\ &=-\dot p' \end{align*}

  • $\begingroup$ Thanks, I've found a workaround to get the result. I'm still unsure of how one would apply $\partial_x$ to $H'$ though. Can you elaborate? $\endgroup$ – gene Oct 30 '17 at 15:47
  • $\begingroup$ Elaboration complete. $\endgroup$ – TotallyRhombus Oct 31 '17 at 2:02

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.