Non-trivial components of the stress-energy tensor of the bosonic string ghost action

Question

The stress-energy tensor derived from the ghost action of a bosonic string is:

$$ T_{\alpha \beta} = \frac{i}{4 \pi} \left ( b_{\alpha \gamma} \nabla_{\beta} c^{\gamma} + b_{\beta \gamma} \nabla_{\alpha} c^{\gamma} - c^\gamma \nabla_\gamma b_{\alpha \beta} - g_{\alpha \beta} b_{\gamma \delta} \nabla_\gamma c^{\delta} \right) $$

My book says, but doesn't give the proof, that in light-cone coordinates, the only two non-trivial components are the diagonal ones, given by:

$$ T_{++} = \frac{i}{4 \pi} \left ( 2b_{++} \partial_+ c^+ - c^+ \partial b_{++} \right) $$

and the same thing for $T_{--}$ but replacing all plus signs in the indices with minus signs.

But I can't derive this. I know that in light-cone coordinates, $g_{++} = g_{--} = 0$ and $g_{+-} = g_{-+} = -\tfrac{1}{2}$ and the covariant derivative is just the partial derivative.

Now for example, consider $T_{+-}$ and focus on the $c^\gamma \nabla_\gamma b_{\alpha \beta}$ term. In light-cone coordinates, it's:

$$ c^+ \partial_+ b_{+-} + c^- \partial_- b_{+-} $$

Since these are the only terms involving partial derivatives of $b$, they must kill each other for $T_{+-}$ to be zero. I know that $b$ is symmetric and traceless, and that $b$ and $c$ anti-commute, but that's not enough to produce

$$ c^+ \partial_+ b_{+-} + c^- \partial_- b_{+-} = 0 $$

I don't see how it's possible.

This post imported from StackExchange Physics at 2014-05-21 08:30 (UCT), posted by SE-user user46242

Answer 1

The point is that a traceless symmetric tensor has its (+-) and (-+) components equal to zero. This follows directly from the expression of the light-cone components of a tensor in terms of its $(x^0,x^1)$-components. This implies that $T_{+-}=T_{-+}=0$ and $b_{+-}=b_{-+}=0$. Using that, it is easy to compute $T_{++}$ and $T_{--}$.