# Why does a certain covariant derivative of the stress-energy tensor vanish due to diffeomorphism invariance?

In Polchinski's string theory volume 1, at chap 1.2, after equation 1.2.22, he says $$\nabla_{a}T^{ab}=0$$ as a consequence of diffeomorphism invariance.

But I cannot derive it. My derivation is as follows:

Under the reparametrization:${\sigma^{a}}'=\sigma^a+\epsilon^a$ $$\delta X^\mu=X'^\mu(\tau',\sigma')-X^{\mu}(\tau',\sigma')=-\epsilon^a \partial_{a}X^\mu$$

$$\delta \gamma_{ab}=\gamma'_{a,b}(\tau',\sigma')-\gamma_{ab}(\tau ', \sigma')=-\nabla_{a}\epsilon_b-\nabla_b \epsilon_a$$ Since this is a symmetry, it leave the action invariant up to a total derivative. But in this case, if we choose $\epsilon=0$ at the boundary of the worldsheet, the action is invariant: $$\delta S=0$$

Thus we have: $$\delta S=\frac{\delta S}{\delta{\gamma_{ab}}}\delta \gamma_{ab}+\frac{\delta S}{\delta X^{\mu}}\delta X^{\mu}=0$$

By definition:

$$\frac{\delta S}{\delta \gamma_{ab}}=-\frac{1}{4\pi}(-\gamma)^{1/2}T^{ab}$$

Thus we have

$$\int d^2 \sigma (\frac{1}{4\pi}(-\gamma)^{1/2}T^{ab}(\nabla_a \epsilon_b+\nabla_b \epsilon_a)+\frac{\delta S}{\delta X^\mu}\delta X^\mu)=0$$

In order to have $\nabla_a T^{ab}=0$, we need to have $\frac{\delta S}{\delta X^\mu}=0$, which can be obtained by imposing the the equation of motion of $X^{\mu}$.There is no wrong in imposing the equation of motion to get conserved currents.

But my confusion is that if we have to impose the equation of motion of $X^\mu$ to get $\nabla_a T^{ab}=0$, why we can't use equation of $\gamma_{ab}$ ,namely $T_{ab}=0$ to get the conclusion. Why bother using eom of $X^{\mu}$?

I will appreciate that if someone can help check my argument to point out my mistake or explain the reason.

• There is a Noether current associated with the world-sheet translation $\delta\sigma^a=\epsilon v^a$, given by $j_a=iv^bT_{ab}$. Since the translation is a symmetry for any $v^a$, this current is conserved, and so is the energy-momentum tensor. Have a look at the end of section 2.3, and at exercise 2.5. – Haz Jul 7 '15 at 6:22
• I probably misunderstood something, but, yes, if you have $T_{ab} = 0$ everywhere this implies $\nabla_a T^{ab} = 0$. The point is that diffeomorphism invariance guarantees that the divergence of $T_{ab}$ is $0$ even when $T_{ab}$ itself is not. – Gabriel Cozzella Jul 7 '15 at 15:17
• I think these are two different problem. In exercise2.5, and at the end of sec 2.3,the symmetry is a global symmetry, we removed the metric using the diffeomorphism and weyl rescaling. But my problem concerns with diffeomorphism invariance,which is a local symmetry. – xjtan Jul 7 '15 at 15:19
• Related: physics.stackexchange.com/q/175186/2451 and links therein. – Qmechanic Jul 11 '15 at 23:04

Details have to be filled in, but I think the general idea went along these lines: the variation of the action with respect to the metric $g_{\mu\nu}$ is given by $$\delta_gS \sim \int T^{\mu\nu}\delta g_{\mu\nu}\,.$$ Now specialize to particular variations of $g_{\mu\nu}$, the diffeomorphisms. For an infinitesimal diffeomorphism of the form $x^\mu\to x^\mu+\xi^\mu(x)$ the variation of the metric is given by the Lie-derivative: $$\delta_\xi g_{\mu\nu} = \mathcal L_\xi g_{\mu\nu} = \nabla_\mu\xi_\nu + \nabla_\nu\xi_\mu\,.$$ Stick this into the first expression and use $T_{\mu\nu}=T_{\nu\mu}$ to get $$\delta S \sim \int T^{\mu\nu}\nabla_\mu\xi_\nu$$ Integrate by parts and discard the surface term to get $\nabla_\mu T^{\mu\nu}=0$.