Let us consider a QFT on a curved Riemannian manifold and use the following definition for the stress tensor: \begin{eqnarray} \delta\langle\cdots\rangle_e=-\int e\delta e^a_\mu\langle T^\mu{}_a\cdots\rangle_e, \end{eqnarray} where $e^a_\mu$ is the tetrad solved from \begin{eqnarray} e^a_\mu e^b_\nu\delta_{ab}=g_{\mu\nu}, \end{eqnarray} and $e=$det$\{e^a_\mu\}$ and $e=\sqrt{g}$ with det$\{g_{\mu\nu}\}=g$.

I want to derive the Ward-Takahashi identity due to the diffeomorphism invariance. For simplicity, we take the insertion $\cdots$ as some world scalar $\psi(x)$ which may carry nontrivial local Lorentz representation, e.g., Dirac spinor. The invariance reads as \begin{eqnarray} \langle\psi(x+\xi)\rangle_{e^a_\mu+\delta e^a_\mu}=\langle\psi(x)\rangle_{e^a_\mu}, \end{eqnarray} with \begin{eqnarray} \delta e^a_\mu&=&-\xi^\nu\partial_\nu e^a_\mu-e^a_\nu\partial_\mu\xi^\nu\\ &=&-\xi^\nu\triangledown_\nu e^a_\mu-e^a_\nu\triangledown_\mu\xi^\nu, \end{eqnarray} where $\triangledown$ is the conventional covariant derivative only taking care of the Einstein indices $\mu$. Putting into the definition of the stress tensor, I obtain for the infinitesimal and compactly supported $\xi$, \begin{eqnarray} \xi^\nu\partial_\nu\langle \psi(x)\rangle_e&=&\int dy e(y)\delta e^a_\mu(y)\langle T^\mu{}_a(y)\psi(x)\rangle_e \\ &=&-\int dy \sqrt{g}[\xi^\nu\triangledown_\nu e^a_\mu+e^a_\nu\triangledown_\mu\xi^\nu]\langle T^\mu{}_a(y)\psi(x)\rangle_e. \end{eqnarray} Defining \begin{eqnarray} T^{\mu}{}_\nu\equiv T^\mu{}_ae^a_\nu, \end{eqnarray} and using the arbitrariness of $\xi$ and integration by parts, we have \begin{eqnarray} \sqrt{g}\triangledown_\mu T^\mu{}_\nu(y)\psi(x)=\delta(y-x)\partial_\nu\psi(x)+\sqrt{g}[\triangledown_\nu e^a_\mu] T^\mu{}_a(y)\psi(x). \end{eqnarray}

I think the second term on the right-hand side should be absent in the correct answer. I have tried to prove the second term vanishes, e.g., using $T^{\mu\nu}=T^{\nu\mu}$, but this classical equation does not hold in the QFT due to possible nontrivial contact term.

My question is whether my derivation above is correct or not? If correct, how to interpret the second additional term which sounds like an additional local Lorentz rotation?


Now I explain why $[T^{\mu\nu}-T^{\nu\mu}]$ has a nonzero contact term with $\psi$ when, for instance, $\psi$ transforms nontrivially under local Lorentz (actually "orthogonal" in the current Riemannian case) transformation.

Let us assume that the generator matrix for $\psi$ from the local Lorentz transformation is $S_{ab}$ and $S_{ab}=-S_{ba}$. Then \begin{eqnarray} \langle(1-\frac{i}{2}\lambda^{ab}(x)S_{ab})\psi(x)\rangle_{e^a_\mu+\lambda^{ab}e_{b\mu}}=\langle\psi(x)\rangle_{e^a_\mu}, \end{eqnarray} where $\lambda^{ab}(x)$ is an infinitesimal transformation $\lambda^{ab}=-\lambda^{ba}$.

Then, by the definition of the stress tensor, we obtain \begin{eqnarray} \sqrt{g}e^{\mu}_a e^\nu_b[T_{\mu\nu}-T_{\nu\mu}](y)\psi(x)=iS_{ab}\psi(x)\delta(y-x). \end{eqnarray} It also implies that when our theory contains spinors, we must define the stress tensor by variation of tetrad $e^a_\mu$ rather than the metric $\delta g_{\mu\nu}$, the later of which is automatically symmetric.

  • $\begingroup$ The canonical stress tensor has a non-trivial contact term in $T^{[\mu\nu]}$. The energy momentum tensor obtained from the metric variation does not have any such contact term. $\endgroup$
    – Prahar
    Feb 22, 2023 at 12:47
  • $\begingroup$ @Prahar Actually, when the theory contains spinors, we cannot define stress tensor by variation of $g_{\mu\nu}$, e.g., checking the Dirac spinor Lagrangian on a curved manifold. I have added an appendix to calculate such a contact term. $\endgroup$
    – Yuan Yao
    Feb 22, 2023 at 13:15

1 Answer 1


I realize that my calculation is correct and the above second term is necessary to have a consistent equation.

Let us first define the spin connection: \begin{eqnarray} \omega^{ab}_\nu\equiv e^{\mu a}\triangledown_\nu e^b_\mu, \end{eqnarray} which is antisymmetric about $a\leftrightarrow b$.

Using the result in the Appendix, we combine the two terms on the right-hand side and obtain \begin{eqnarray} \sqrt{g}\triangledown_\mu T^\mu{}_\nu(y)\psi(x)=\delta(y-x)D_\nu\psi(x), \end{eqnarray} where \begin{eqnarray} D_\nu=\partial_\nu+\frac{i}{2}\omega^{ab}_\nu S_{ab}, \end{eqnarray} when $\psi(x)$ is a spinor.

$D_\nu$ is the covariant derivative due to the local Lorentz symmetry. On the other hand, $\partial_\nu$ is not a well-defined derivative just as the situation when we have a U(1)-gauge symmetry. By the above grouping, both sides transform linearly about the local Lorentz transformation.


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