Since
$$
\delta\gamma_{\alpha\beta}
=2\partial_{(\alpha}X_\mu\,\partial_{\beta)}\delta X^\mu,
$$
the determinant variation gives
$$
\delta S_{\rm NG}
=-\frac T2\int d^2\sigma\,\sqrt{-\gamma}\,
\gamma^{\alpha\beta}\delta\gamma_{\alpha\beta}
=-T\int d^2\sigma\,\sqrt{-\gamma}\,
\gamma^{\alpha\beta}\partial_\alpha X_\mu\partial_\beta\delta X^\mu.
$$
Integrating by parts and taking $\delta X^\mu$ to vanish on the boundary yields
$$
\delta S_{\rm NG}
=T\int d^2\sigma\,\delta X_\mu\,
\partial_\alpha\left(\sqrt{-\gamma}\,
\gamma^{\alpha\beta}\partial_\beta X^\mu\right).
$$
The <principle of stationary action> therefore gives
$$
\partial_\alpha\left(\sqrt{-\det\gamma}\,
\gamma^{\alpha\beta}\partial_\beta X^\mu\right)=0.
$$

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 306 / 1 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-306/1/b/solution