# Hamiltonian formalism (with symplectic form) for time-dependent Lagrangian

I have been working on some results that work for time-independent Lagrangians $$L\Big(q,\dot{q}\Big)$$ and return a Hamiltonian function

$$H(q,\dot{q})=\dot{q}^i \frac{\partial L}{\partial \dot{q}^i}-L$$

and a symplectic form

$$\Omega=\delta \Big(\frac{\partial L}{\partial \dot{q}^i }\Big)\wedge \delta q^i$$

where the idea is that we can read off the canonical momentum $$p_i=\frac{\partial L}{\partial \dot{q}^i}$$ from the symplectic form.

When the Lagrangian has an explicit time dependence $$L\Big(t,q,\dot{q}\Big)$$, I'm not sure how this works. What I tried to do was to treat $$t$$ as a new coordinate and define $$q^\mu=(t,q^i)$$ and parametrize everything with some new parameter $$s$$ such that

\begin{align} S[q]&=\int dt L\Big(t,q(t),\dot{q}(t)\Big)\\ &=\int ds\ \frac{dt}{ds} L\Big(t(s),q(s),\big(\frac{dt}{ds}\big)^{-1} q'(s)\Big) &= \int ds \tilde{L}(q^\mu,q'^\mu) \end{align} where we defined $$q'=\frac{dq}{ds}$$ as opposed to $$\dot{q}=\frac{dq}{dt}$$. The problem with this new Lagrangian is that the Legendre transformation vanishes identically. Indeed

\begin{align} \tilde{H}&=q'^\mu \frac{\partial \tilde{L}}{\partial q'^\mu}-\tilde{L}\\ &=0 \end{align} Instead, the Hamiltonian is simply the canonical momentum conjugate to $$t$$

\begin{align} H(t,q^i,\dot{q}^i)&=-\frac{\partial \tilde{L}}{\partial t'}\\ &=\dot{q}^i \frac{\partial L}{\partial \dot{q}^i}-L(t,q^i,\dot{q}^i) \end{align}

My question is: In this framework where we treat $$t$$ as a new coordinate and parametrize it with a new time $$s$$, how do we calculate the symplectic form for $$H$$?

• OP's question seems to be partly inspired by the covariant Hamiltonian/phase space formalism, cf. e.g. OP's previous question here.

• OP's method of generating a worldline (WL) reparametrization invariant formulation is also considered in my Phys.SE answer here, which is summarized below.

1. Let us first consider the Lagrangian formalism. Given a possibly time-dependent Lagrangian $$L(t,q,v)$$ with an $$n$$ generalized positions $$(q^1,\ldots,q^n)$$ and $$n$$ generalized velocities $$(v^1,\ldots,v^n)$$ and time $$t$$. The corresponding action $$S[q]~=~\int\! dt~L(t,q,\frac{dq}{dt}) \tag{A}$$ has $$n$$ dynamical variables. There are hence $$n$$ EOMs.

Define a WL reparametrization invariant action with $$n+1$$ dynamical variables \begin{align}\widetilde{S}[t,q]~=~&\int\! ds~\widetilde{L}(t,q,\dot{q}), \cr \widetilde{L}(t,q,\dot{t},\dot{q})~:=~&\dot{t}L(t,q,\frac{\dot{q}}{\dot{t}}),\end{align} \tag{B} where dot denotes differentiation wrt. $$s$$. There are hence $$n+1$$ EOMs but only $$n$$ independent.

The corresponding symplectic form is defined as $$\widetilde{\Omega}~:=~\delta \Big(\frac{\partial \widetilde{L}}{\partial \dot{t} }\Big)\wedge \delta t +\delta \Big(\frac{\partial \widetilde{L}}{\partial \dot{q}^j }\Big)\wedge \delta q^j,\tag{C}$$ cf. OP's question.

2. Next let us consider the Hamiltonian formalism. Given a possibly time-dependent Hamiltonian Lagrangian $$L_H(t,q,p,v)~=~p_jv^j-H(t,q,p). \tag{D}$$ The corresponding Hamiltonian action $$S[q,p]~=~\int\! dt~L_H(t,q,p,\frac{dq}{dt}) \tag{E}$$ has $$2n$$ dynamical variables. There are hence $$2n$$ EOMs.

Define a worldline (WL) reparametrization invariant Hamiltonian action with $$2n+3$$ dynamical variables \begin{align}\widetilde{S}_H[t,q,p_t,p,e]~=~&\int\! ds~\widetilde{L}_H(t,q,\dot{t},\dot{q},p_t,p,e), \cr \widetilde{L}_H(t,q,\dot{t},\dot{q},p_t,p,e)~:=~&p_j\dot{q}^j+p_t\dot{t}-\widetilde{H}(t,q,p_t,p,e),\cr\widetilde{H}(t,q,p_t,p,e)~:=~&e(p_t+H(t,q,p)), \end{align} \tag{E} where the einbein field $$e$$ is a Lagrange multiplier. There are hence $$2n+3$$ EOMs but only $$2n+1$$ independent.

The corresponding symplectic form is defined as $$\widetilde{\Omega}_H~:=~\delta p_t\wedge \delta t +\delta p_j\wedge \delta q^j,\tag{F}$$ cf. OP's question.

• Thanks for your answer! There's one thing I don't understand. What Hamiltonian function are you using to get Hamilton's equations? If we define a variable $Q^A=(t,q^j,p_t,p_j)$ are they still $\frac{dQ^A}{ds}=\Omega^{AB}H_{,B}$? Which $H$ goes there? The one that vanishes identically? Or just the one that is defined as the canonical momentum conjugated to t? Commented Mar 23 at 15:40
• I updated the answer. Commented Mar 24 at 7:15
• I'm sorry but I still don't understand how to calculate Hamilton's equations from this formalism. The Hamiltonian in the new system vanishe identically so it doesn't matter how the symplectic form looks, the equations of motion won't work. Am I supposed to use Dirac's theory of Hamiltonian systems with constraints? Commented Apr 10 at 3:40
• FWIW the EOMs are worked out in the linked answer. Commented Apr 10 at 6:25