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As I checked, the energy-momentum tensor defined as ${T^\mu}_\nu=\frac{\partial {\cal L}}{\partial(\partial_\mu \phi)}\partial_\nu \phi-{\cal L}{\delta^\mu}_\nu$ at the solution $\phi$ of equation of motion(Euler-Lagrange equation) satisfies automatically the conservation law: $\partial_\mu{T^\mu}_\nu=0$, without any reference to the translation symmetry under $x^\mu\rightarrow x^\mu-a^\mu$.
So, what is the need of this symmetry?
Or, could there be something wrong with my calculation or conceptual issues?

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1 Answer 1

up vote 3 down vote accepted

Oh, I made a mistake: In deriving $\partial_\mu{T^\mu}_\nu=0$, I had assumed $L$ doesn't depend on $x$ explicitly but solely on $\phi$ and $\partial_\mu \phi$, and this is just the condition of translation symmetry!

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