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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 $H$?

Qmechanic
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P. C. Spaniel
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1 Answers1

2

  • 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.

Qmechanic
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