When transforming a field in classical field theory the transformation of the four-gradient of this field follows automatically. At least this is what i have learned in my lectures.

This circumstance kind of contradicts my understanding of the Lagrange formalism in classical mechanics. In classical mechanics the generalized coordinates and their derivatives are understood as coordinates in configuration space. Therefore i can perform a transformation in configuration space only concerning the generalized coordinates and not their derivatives.

So there is probably a strong difference between the Lagrange formalism of fields and the Lagrange formalism of classical mechanics which I do not understand yet.

Question: Why is it possible to transform $q(t)$ in classical mechanics without transforming $\dot{q}(t)$, but it is not possible to transform $\phi(x)$ without transforming $\partial_\mu \phi(x)$?


1 Answer 1


That seems to be a misunderstanding. The field-theoretic case behaves in essentially the same way as the point mechanical situation.

  1. On one hand, the arguments of the Lagrangian density $${\cal L}(\phi,\partial_t\phi,\partial_x\phi,\partial_y\phi,\partial_z\phi,t,x,y,z)$$ are independent variables in the same way as the arguments of the Lagrangian $$L(q,\dot{q},t)$$ are independent variables.

  2. On the other hand, inside the actions $S[\phi]$ (and $S[q]$), the derivatives $\partial_{\mu}\phi$ (and $\dot{q}$) depend on $\phi$ (and $q$), respectively.

The proof is a straightforward generalization of my Phys.SE answer here.

  • $\begingroup$ So actually one would have to give also an explicit transformation for the derivatives? I am just confused since it always seems to be implicitly understood to transform the field derivatives according to transformation of the field configuration. $\endgroup$ Nov 22, 2020 at 7:34
  • 1
    $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    Nov 22, 2020 at 7:51

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