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Take an inviscid, incompressible fluid, ignore external forces for the sake of simplicity.

The Lagrangian density is

$$ \mathcal{L} = \frac{\rho}{2} {\vec v}\cdot \vec v $$

I'm trying to solve Euler-Lagrange like so: $$ \frac{\partial \mathcal{L}}{\partial \vec v} = \partial_\mu \left( \frac{\partial \mathcal{L}}{\partial(\partial_\mu \vec v)} \right) $$

The RHS is $0$ because $\mathcal{L}$ does not depend on any derivatives of $\vec v$.

The LHS is $\rho \vec v$.

This is nonsense, I was expecting to get the momentum equations for an inviscid incompressible fluid.

Sancol.
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user2958456
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    If you use position as your coordinate instead of velocity, you recover the continuity equation – DanDan0101 Mar 23 '24 at 22:39
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    Closely related: https://physics.stackexchange.com/q/14652/226902 https://physics.stackexchange.com/q/138149/226902. This paper contains a lot of references for the variational principle for the perfect fluid (and perfect multifluids) in special and general relativity https://arxiv.org/abs/1906.03140 – Quillo Mar 23 '24 at 23:54
  • That's not at all what a Lagrangian for hydrodynamics would look like, though. See https://arxiv.org/abs/2204.06006 and https://arxiv.org/abs/1805.09331 for examples of EFTs for hydrodynamics. They involve the density $\rho$ and momentum $\pi$ (whose thermodynamic conjugate is the velocity $v$). – just a phase Mar 25 '24 at 17:58

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