# Why do we need to symmetrize the stress-energy tensor?

An elementary definition of the stress-energy tensor $$T_{\mu\nu}$$ in terms of the Lagrangian for flat spacetime is $$T^{\mu\nu}=\frac{\partial \mathcal{L}}{\partial(\partial_\mu\phi)}(\partial^\nu\phi)-\eta^{\mu\nu}\mathcal{L}.$$ For translational symmetries of spacetime both $$T^{\mu\nu}$$ is conserved: $$\partial_\mu T^{\mu\nu}=0.$$

But $$T_{\mu\nu}$$ is not always automatically symmetric in the indices $$\mu,\nu$$. Therefore, one considers a modified stress-energy tensor $$\Theta_{\mu\nu}$$ defined as $$\Theta^{\mu\nu}:=T^{\mu\nu}+\partial_\kappa A^{\kappa\mu\nu}$$ where $$A^{\kappa\mu\nu}$$ is an arbitrary tensor antisymmetric in the indices $$\kappa,\mu$$. This too satifies $$\partial_\mu\Theta^{\mu\nu}=0.$$

• What is the need of defining a symmetric stress-energy tensor? What happens if I continue to work with $$T^{\mu\nu}$$ instead of the modified tensor $$\Theta^{\mu\nu}$$?

• It is also unclear how one would choose the tensor $$A^{\kappa\mu\nu}$$ uniquely to symmetrize $$T^{\mu\nu}$$ and find a unique $$\Theta^{\mu\nu}$$.

• First, we don't need the symmetrimized tensor be unique, we can choose arbitrary second order tensor whose divergence is zero to symmetrimize energy-stress tensor. Second, the reason we want to symmetrimize energy-stress tensor is due to the consideration of conservation law of angular momentum. Please check the last chapter of Classical Mechanics by Goldstein. Apr 10, 2020 at 5:53
• @y255yan Answers should not belong to the comments. Consider expanding that into a proper answer. Apr 10, 2020 at 7:16
• Related: physics.stackexchange.com/q/68564/2451 and links therein. Apr 10, 2020 at 9:08
• See Belinfante–Rosenfeld stress–energy tensor in Wikipedia. A motivation is to agree with the stress energy tensor used in general relativity. Jun 26 at 14:27