Noetherian currents and the Surface term

[cpsinkule]

When deriving the conserved currents from continuous symmetries, my book states that we can also add a 4-divergence to the lagrangian density which does not change the action under variation. The four divergence can be transformed into a boundary integral by stokes theorem. However, my book fails to mention any assumptions we are making about this extra term so that it does not contribute to the variation of the action. The final result is the current
J^μ=Π^μδΦ-W^μ where the W^μ is from the four divergence mentioned above ∂_μW^μ
My question is: What assumptions, if any, are imposed on the W^μ so that it doesn't affect the action under variation so that we can include it in the current?

 

[Orodruin]

The usual assumption is that the fields go to zero sufficiently fast for the boundary terms to vanish.

 

[samalkhaiat]

When deriving the conserved currents from continuous symmetries, my book states that we can also add a 4-divergence to the lagrangian density which does not change the action under variation. The four divergence can be transformed into a boundary integral by stokes theorem. However, my book fails to mention any assumptions we are making about this extra term so that it does not contribute to the variation of the action. The final result is the current
J^μ=Π^μδΦ-W^μ where the W^μ is from the four divergence mentioned above ∂_μW^μ
My question is: What assumptions, if any, are imposed on the W^μ so that it doesn't affect the action under variation so that we can include it in the current?

Yes, it is always possible to add a total divergence to the Lagrangian without affecting the dynamics. Indeed, you can show that $$S = \int_{\Omega} d^4 x \ \mathcal{L},$$ and $$S = \int_{\Omega} d^4 x \ (\mathcal{L} + \partial_{\mu} V^{\mu}) ,$$ lead to the same Euler-Lagrange equation provided that the “arbitrary” vector $V^{\mu}$ vanishes on the boundary of the region of integration: $$\int_{\Omega} d^{4}x \ \partial_{\mu}V^{\mu} = \int_{\partial \Omega} d\Sigma_{\mu} \ V^{\mu} = 0 .$$ However, your book is very wrong in saying that this arbitrary vector $V^{\mu}$ is the same vector $W^{\mu}$ which appears as part of the Noether current. The vector $W^{\mu}$ is not arbitrary, rather it depends on the form of the Lagrangian. The $W^{\mu}$ simply tells you how the Lagrangian transforms under the action of the symmetry group and, therefore, it cannot be arbitrary.