# How to find Belinfante-Rosenfeld SEM tensor?

Using definition of SE tensor as a response to the infinitesimal coordinate change $$\delta_{\epsilon} S=\int\partial_{\mu}\epsilon_{\nu}T_{\mu\nu}d^Dx;\quad \partial_{\mu}T_{\mu\nu}=0;\quad (7)$$ We can prove that for $$SO(D)$$ invariant theories $$T_{\mu \nu}$$ can always be made symmetric. It can be seen as follows. Consider $$\epsilon_\nu=\omega_{\nu \lambda} x_\lambda$$, where $$\omega_{\mu \nu}=-\omega_{\nu \mu}$$, then from (7) we have $$\delta_\epsilon S=\frac{1}{2} \int_{\mathbb{R}^D}\left[\partial_\mu \omega_{\nu \lambda}\left(x_\lambda T_{\mu \nu}-x_\nu T_{\mu \lambda}\right)-\omega_{\mu \nu}\left(T_{\mu \nu}-T_{\nu \mu}\right)\right] d^D \boldsymbol{x} .$$ For constant $$\omega_{\mu \nu}$$ this variation should vanish for rotationally invariant theories, which implies $$T_{\mu \nu}-T_{\nu \mu}=\partial_\lambda f_{\lambda \mu \nu}, \quad f_{\lambda \mu \nu}=-f_{\lambda \nu \mu} , \tag{i}$$ cf. my related Phys.SE question here.

We can build Belinfante-Rosenfeld SEM tensor by: $$\tilde{T}_{\mu \nu} \stackrel{\text { def }}{=} T_{\mu \nu}-\partial_\lambda B_{\lambda \mu \nu} \tag{ii}$$ where $$B_{\lambda \mu \nu}=\frac{1}{2}\left(f_{\lambda \mu \nu}-f_{\mu \lambda \nu}-f_{\nu \lambda \mu}\right) \text {. } \tag{iii}$$ But nobody actually says how to find this tensor, do I have to just find noether current under rotations or what?

• So normally the stress energy tensor is just defined as the conserved current due to translation not rotation (In flat theories) Mar 9, 2023 at 12:25
• but rotation can be described as local translations that depend of x so my equations don't change Mar 9, 2023 at 13:02

1. Comparing eqs. (i) and (ii) suggests that $$B_{\lambda \mu \nu} -(\mu\leftrightarrow \nu)~=~ -f_{\lambda \mu \nu}.\tag{iv}$$
• I am asking how to find particular $B_{\lambda\mu\nu}$ or particular $f_{\lambda\mu\nu}$ for known lagrandian or SE tensor Mar 9, 2023 at 13:14