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As I know Stress-Energy tensor is defined as Noether current under arbitrary coordinate transformations $\boldsymbol{x} \rightarrow \boldsymbol{x}+\epsilon(\boldsymbol{x})$. $$ \delta_\epsilon S=\int_{\mathbb{R}^D}\left[\epsilon_\nu\left(\frac{\partial \mathcal{L}}{\partial \Phi} \partial_\nu \Phi+\frac{\partial \mathcal{L}}{\partial\left(\partial_\mu \Phi\right)} \partial_\mu \partial_\nu \Phi\right)+\partial_\mu \epsilon_\nu \frac{\partial \mathcal{L}}{\partial\left(\partial_\mu \Phi\right)} \partial_\nu \Phi\right] d^D \boldsymbol{x}. $$ First term here equals to $\epsilon_\nu \partial_\mu\left(\delta_{\mu \nu} \mathcal{L}\right)$ as a reflection of the fact that the action does not depend on $\boldsymbol{x}$ explicitly and hence for constant $\epsilon_\mu$ the variation should vanish. After integrating by parts we get $$ \delta_\epsilon S=\int_{\mathbb{R}^D} \partial_\mu \epsilon_\nu\left(\frac{\partial \mathcal{L}}{\partial\left(\partial_\mu \Phi\right)} \partial_\nu \Phi-\delta_{\mu \nu} \mathcal{L}\right) d^D \boldsymbol{x} \stackrel{\text { def }}{=} \int_{\mathbb{R}^D} \partial_\mu \epsilon_\nu T_{\mu \nu} d^D \boldsymbol{x}=-\int_{\mathbb{R}^D} \epsilon_\nu \partial_\mu T_{\mu \nu} d^D \boldsymbol{x} . $$ That's where we are getting the definition of SE tensor as a response to the infinitesimal coordinate change $$ \delta_\epsilon S=\int_{\mathbb{R}^D} \partial_\mu \epsilon_\nu T_{\mu \nu} d^D \boldsymbol{x};\quad \partial_\mu T_{\mu \nu}=0;\tag{7} $$ So for integration by parts we used the fact that on boundaries $\epsilon(\boldsymbol{x})$ equals 0. But for special case with rotational invarience my teacher writes:

In $SO(D)$ 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 $$ T_{\mu \nu}-T_{\nu \mu}=\partial_\lambda f_{\lambda \mu \nu}, \quad f_{\lambda \mu \nu}=-f_{\lambda \nu \mu} . $$

So we demand $\omega$ to be constant which, for ex. in $D=(1,3)$ spacetime definitely means that $\epsilon(x)$ will not vanish on edges of integration and all mentioned theory before will not work because we no more can integrate by parts without getting additional term: $$\int\epsilon_\nu \partial_\mu\left(\delta_{\mu \nu} \mathcal{L}\right)d^Dx=$$ $$=\int\partial_\mu\left[\epsilon_\nu\left(\delta_{\mu \nu} \mathcal{L}\right)\right]d^Dx-\int\left(\delta_{\mu \nu} \mathcal{L}\right)\partial_\mu\epsilon_\nu d^Dx\neq_{just} -\int\left(\delta_{\mu \nu} \mathcal{L}\right)\partial_\mu\epsilon_\nu d^Dx.$$

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  1. OP and their teacher are using the trick of $x$-dependent parameters in Noether's first theorem.

  2. The variation $\delta x^{\mu}~=~\epsilon^{\mu}(x)=\omega^{\mu}{}_{\nu}(x)~x^{\nu}$ is never independent of $x$ for infinitesimal Lorentz transformations; it is instead $\omega^{\mu}{}_{\nu}(x)$ that becomes $x$-independent.

  3. OP's calculation is still valid modulo boundary terms. For $x$-independent $\omega_{\mu\nu}$, the teacher's formula becomes $$ \delta S~\sim~-\frac{1}{2}\omega_{\mu\nu} \int_V\! d^Dx(T^{\mu\nu}-T^{\nu\mu}), \tag{A}$$ where $\sim$ means equality modulo boundary terms.

  4. In fact it is enough to assume that the Lorentz transformation is a quasi-symmetry of the action under Lorentz transformations.

  5. From eq. (A) one can infer the existence of the Belinfante-Rosenfeld SEM tensor part $f_{\lambda\mu\nu}$ using arguments laid out between eqs. (9) & (10) in e.g. this Phys.SE post (i.e. either by assuming an arbitrary integration region $V$, or by using an algebraic Poincare Lemma).

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  • $\begingroup$ Sorry but I don't understand, how $\omega$ can vanish on boundaries if it's constant $\endgroup$
    – islam
    Mar 8, 2023 at 19:59
  • $\begingroup$ so $\omega$ becomes zero on the boundaries because of small buffering? $\endgroup$
    – islam
    Mar 9, 2023 at 11:58
  • $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    Mar 9, 2023 at 12:45

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