# If the lagrangian density changes by a total derivative of the lagrangian density

When we derive energy momentum tensor current by actively transforming field. We see that lagrangian ( density) changes by a total derivative of the lagrangian. If a total derivative of the function is added to a lagrangian, i know the action will remain invariant becausr its variation will be zero because at the boundaries variation of field is zero. But in the energy momentum tensor case we have a total derivative of the lagrangian itself which is a function of fields AND THEIR DERIVATIVES, why should action still remain invariant? I mean the variation of surface term is not zero in this case because we have derivatives of fields whose variations on the boundary is not necessarily zero?

• It's not clear what you're asking. If the Lagrangian density $\mathcal L$ exits, then it's not unique since $\mathcal L + \partial_{\mu}\Lambda^{\mu}$ where $\Lambda^{\mu}$ is an arbitrary function of $x$ and the fields - but not of the derivatives of the fields - leads to a surface term - and produces the same Euler-Lagrange equations as $\mathcal L$ since the dynamics are determined by the boundary conditions. – Cinaed Simson Oct 12 at 8:42
• I am saying when we derive energy momentum tensor current ( by making an active transformation ), we see that lagrangian density changes by a total derivative of the lagrangian density, which does not satisfy the conditions satisfied by Lamba in your comment since lambda as u say is a function of fields only but the lagrangian density is a function of derivatives of the field also – Farman Ullah Oct 12 at 8:46
• I don't know where the $\int d^4x\;\partial_{\mu}\mathcal{L}$ term is coming from. But if the divergence terms looked something like $\int d^{4}x\; \partial_{\mu}(\frac{\delta \mathcal{L}}{\delta\partial_{\mu }\phi^{\alpha}}\delta \phi^{\alpha})$ then it may a conserved current. – Cinaed Simson Oct 13 at 4:25
• Refer to schwartz qft,, page 35 – Farman Ullah Oct 13 at 16:05