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Given a Lagrangian density $\mathcal L(\phi_r,\partial_\mu\phi_r,\phi_n,\partial_\mu\phi_n)$, for which we find out that for some $\phi_n$ its conjugate momentum vanishes:

$$\pi_n=\frac{\partial\mathcal L}{\partial(\partial_0\phi_n)}=0.$$

  1. What does this imply?

  2. And if you will say that this implies that $\phi_n$ is a field depending on other fields, what does this mean?

  3. Does this mean that we can construct another Lagrangian density $\hat{\mathcal L}(\phi_r,\partial_\mu\phi_r)$, which has no $\phi_n$ as its argument, so that all the fields involved are independent, but still gives a Euler-Lagrange equation having the same solution as that of ${\mathcal L}(\phi_r,\partial_\mu\phi_r,\phi_n,\partial_\mu\phi_n)$?

  4. Another question, is it true that the Hamiltonian density $\mathcal H(\phi_r,\nabla\phi_r,\pi_r,\nabla\pi_r,\phi_n,\nabla\phi_n,\pi_n,\nabla\pi_n)$ constructed using $\mathcal L(\phi_r,\partial_\mu\phi_r,\phi_n,\partial_\mu\phi_n)$ doesn't contain variables $\phi_n$ and $\pi_n$?

  5. Yet another question, suppose $\mathcal H(\phi_r,\nabla\phi_r,\pi_r,\nabla\pi_r,\phi_n,\nabla\phi_n,\pi_n,\nabla\pi_n)$ does contain $\phi_n$ and $\pi_n$, but under the solution of equation of motion, $\mathcal H(\phi_r,\nabla\phi_r,\pi_r,\nabla\pi_r,\phi_n,\nabla\phi_n,\pi_n,\nabla\pi_n)=\hat{\mathcal H}(\phi_r,\nabla\phi_r,\pi_r,\nabla\pi_r)$, which $\hat{\mathcal H}$ being seen as another Hamiltonian density depents only on independent variables $\phi_r$ and $\pi_r$. Now if we derive the equation of motion of $\phi_r$ and $\pi_r$ from $\hat{\mathcal H}$, will it be consistent with the original one derived from $\mathcal L$, looking only at the variables $\phi_r$?

The motivation for me to consider these questions is that only algebraic manipulations that preserve the equations of motion should make sense, so that the symmetry examinations of $\mathcal L$ will indeed profit us finally.

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Related: – Qmechanic Nov 22 '13 at 18:06

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