Where this definition $T_{\alpha\beta}=-\frac{2}{T}\frac{1}{\sqrt{-h}}\frac{\delta S}{\delta h^{\alpha \beta}}$ come from?

Question

In the book "String theory and M-theory" by Becker, Becker and Schwarz, the author says that the Nambu-Goto action $$S_{NG}=-T\int d\sigma\, \tau \sqrt{(\dot{X}\cdot X')^2 -\dot{X}^2X'^2}$$ is equivalent to the sigma model action $$S_{\sigma}=-\frac{1}{2}T\int d^2 \sigma\, \sqrt{-h}\,h^{\alpha \beta}\partial_{\alpha}X\cdot \partial_{\beta}X$$ where $h_{\alpha\beta}$ is an auxiliary world-sheet metric and energy-momentum tensor is zero: $$T_{\alpha\beta}=-\frac{2}{T}\frac{1}{\sqrt{-h}}\frac{\delta S}{\delta h^{\alpha \beta}}=0.$$

Where this definition $T_{\alpha\beta}=-\frac{2}{T}\frac{1}{\sqrt{-h}}\frac{\delta S}{\delta h^{\alpha \beta}}$ come from? The definition of energy-momentum tensor was $$T^{\mu}_{\nu}=\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi)}\partial_{\nu}\phi-\mathcal{L}\delta^{\mu}_{\nu}.$$

Answer 1

Because by construction, it satisfies 3 key properties one expects from the energy-momentum tensor:

1. It is a symmetric rank-2 tensor,
2. By an infinitesimal coordinate transformation $x^\mu\to x^\mu+\xi^\mu$, you can show that it is conserved in the sense that $\nabla_\mu T^{\mu\nu}=0$,
3. In a gravitational setup, in which the metric is a dynamical variable, it correctly identifies the energy-momentum tensor as a source of gravity. In other words, it is consistent with Einstein equations.

These are nicely explained in section 6.2.4 of Padmanabhan's book on Gravitation.

Note that this definition could in principle differ from a canonical definition based on the Noether's procedure, in which the energy-momentum tensor is identified with the generator of Poincaré translations. This happens for example for Maxwell's action. However, the difference is a total derivative term which can be taken into account by the Belinfante–Rosenfeld procedure.