 Solution

Take a constant infinitesimal translation with fixed-coordinate field variation $\delta \varphi ={ϵ}^{\nu }{\partial }_{\nu }\varphi$. The associated change in the Lagrangian density is $\delta \mathcal{L}={ϵ}^{\nu }{\partial }_{\nu }\mathcal{L}={\partial }_{\mu }({ϵ}^{\nu }{\delta }^{\mu }{}_{\nu }\mathcal{L})$. Substituting this total derivative into the Noether current gives $J^{\mu }={ϵ}^{\nu }{T}^{\mu }{}_{\nu }$, whereThis is the canonical stress-energy tensor. The chosen sign of the translation parameter produces precisely this sign of the Noether current; reversing the parameter reverses the current without changing its conservation law.

To check current conservation directly, differentiate the tensor. The terms containing ${\partial }_{\nu }{\partial }_{\rho }\varphi$ cancel against the derivative of the kinetic term, leavingThus ${\partial }_{\mu }{T}^{\mu }{}_{\nu }=0$ on shell. Raising the second Lorentz index gives $T^{\mu \nu }=T^{\nu \mu }$: this scalar field already has a symmetric canonical stress-energy tensor.