Invariance of the action under a transformation

[div curl F= 0]

If the action of a theory is invariant under a transformation (i.e. a lorentz transformation or a spacetime translation), does this imply that the Lagrangian is also invariant under the transformation?

$$L \to L + \delta L \;\;;\;\; \delta L = 0$$?

 

[Bob_for_short]

In some cases - yes. In some other cases the Lagrangian acquires a term expressed as a total derivative and thus it is inessantial for the variational principle.

 

[RedX]

If the action of a theory is invariant under a transformation (i.e. a lorentz transformation or a spacetime translation), does this imply that the Lagrangian is also invariant under the transformation?

$$L \to L + \delta L \;\;;\;\; \delta L = 0$$?

You are asking two different questions. One is about transformations in general, and the other is Lorentz transformations.

When you vary the action, you vary the integrand, the limits of the integral, and the measure. All 3 of those things vary. To 1st order, you can vary just one, and leave the rest constant, and add over all such variations (3 of them).

Take the variation of the measure, $$\delta (d^4x)$$. This is 0 if your transformation is Lorentz. If it's not, who knows?

Take the variation of the limits of the integral, say under a time translation: $$\int d^3x \int_{t_1+\delta t}^{t_2+\delta t} dt \mathcal L -\int d^3x \int_{t_1}^{t_2} dt \mathcal L=L(t_2)\delta t-L(t_1)\delta t$$ where L=$$\int d^3x \mathcal L$$. This term in general is not zero, and is what gives the term L in the conserved quantity $$H=px'-L$$. So I don't think $$\int d^4x \mathcal \delta L$$ can be zero, since it must cancel this term so that the total action is zero.