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I am struggling to understand why the principle of least action which is derived in classical mechanics from d'Alembert's principle continues to be valid in a regime that treats a relativistic field. What is it that tells us that the equations of motion can still be obtained from the principle of least action (applied on a lagrangian that is relativistically invariant)?

Qmechanic
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Ratul Thakur
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    I'd kind of turn this around, and say "The Lagrangian formulation of a Newtonian theory is shown to be equivalent to Newtonian mechanics, and being equivalent, can itself be taken to be the first principle. Knowing this, it is the first principle that generalizes to relativistic theories when the Newtonian formalism fails." – Zo the Relativist Aug 15 '21 at 11:38
  • Related: https://physics.stackexchange.com/q/15899/2451 and links therein. – Qmechanic Aug 15 '21 at 14:52

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Here is one possible line of reasoning:

  1. In the case of relativistic point mechanics, one can still use d'Alembert's principle and the relativistic version of Newton's laws to derive Lagrange equations. (The main difference compared to the non-relativistic case is the form of the kinetic term in the Lagrangian.)

  2. In the case of e.g. a relativistic scalar field with EOM $$ \mp\Box\phi~=~{\cal V}^{\prime}(\phi)$$ with Minkowski signature $(\pm,\mp,\mp,\mp)$, it is not hard to see that a Lagrangian density is given as $$ {\cal L}~=~\pm\frac{1}{2}\partial_{\mu}\phi\partial^{\mu}\phi - {\cal V}(\phi). $$ In fact, in relativistic field theory, the Lagrangian formulation is often taken as a starting point/first principle, cf. above comment by Jerry Schirmer.

Qmechanic
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