Monge-gauge Hamiltonian of a three-dimensional string
= Monge-gauge Hamiltonian of a three-dimensional string
{c}
{title2=$\mathcal H=\sqrt{(1+\Phi'^2)(1+\Pi^2)}$}
With unit tension and $X^0=t$, $X^1=\sigma$, the <Virasoro constraints> give $P_1=-\Pi\Phi'$ and $P_0=-\sqrt{(1+\Phi'^2)(1+\Pi^2)}$. The reduced energy density is $\mathcal H=-P_0$. On the chiral branch $\Pi=\Phi'$, the <Hamiltonian field equation> becomes $\dot\Phi=\Phi'$.