Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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!

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.