For the wave equation Hamiltonian density is $2H=\phi_t^2+\phi_x^2$ while the Lagrangian density is $2L=\phi_t^2-\phi_x^2$. I can easily compute the pde from the Lagrangian density but how does one do it from Hamiltonian density?

  • $\begingroup$ By using Hamilton's equations. $\endgroup$
    – Qmechanic
    Commented Jul 17, 2019 at 5:20

1 Answer 1


You need to write the Hamiltonian as a function of $\phi,\,\phi_x,\,\pi:=\frac{\partial L}{\partial\phi_t}=\phi_t$ but not $\phi_t$ before you can obtain Hamilton's equations. Since $H=\frac12\left(\pi^2+\phi_x^2\right)$, $\phi_t=\frac{\partial H}{\partial\pi}=\pi$ while $\pi_t=-\frac{\partial H}{\partial\phi}+\partial_x\frac{\partial H}{\partial\phi_x}=\phi_{xx}$. We can combine these to give $\phi_{tt}=\phi_{xx}$, the same equation obtainable from the Lagrangian.

  • $\begingroup$ But this required knowledge of the Lagrangian beforehand is there some other way that involves converting the halmiltonian into Lagrangian and then applying EL e.g in this case I recognize phi_x^2 as potential density so for the Lagrangian density I simply had to put a negative sign and then apply EL but I am not sure how the method generalizes $\endgroup$ Commented Jul 17, 2019 at 5:41
  • $\begingroup$ @bentenyson You can only use a Hamiltonian once you've written it in terms of $\pi$ instead of $\phi_t$ (although $\phi_x$ is allowed), so you do need to know enough about the Lagrangian to write your Hamiltonian in terms of $\pi$. You do not, however, have to obtain the Euler-Lagrange equation from the Lagrangian; you can use the Hamiltonian instead, but they will agree. $\endgroup$
    – J.G.
    Commented Jul 17, 2019 at 8:03

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