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  Symmetry of the Polyakov action?

+ 6 like - 0 dislike

Let us look at the Polyakov action for a string moving in a spacetime with metric $g_{\mu \nu}(X)$:$$S_P = -{1\over{4\pi \alpha'}} \int d^2 \sigma \sqrt{-\gamma} \gamma^{ab} \partial_a X^\mu \partial_b X^\nu g_{\mu\nu}(X) \tag{1}$$ and suppose there exists a Killing vector $k_\mu$ in spacetime satisfying Killing's equation $$\nabla_\mu k_\nu + \nabla_\nu k_\mu = 0.\tag{2}$$ Does this lead to a symmetry of the Polyakov action?

This post imported from StackExchange Physics at 2016-01-17 15:57 (UTC), posted by SE-user Ham
asked Jan 7, 2016 in Theoretical Physics by Ham (30 points) [ no revision ]

1 Answer

+ 1 like - 0 dislike

The answer is Yes. Hints:

  1. Perform an infinitesimal variation $$ \delta X^{\mu}~=~\varepsilon K^{\mu}(X) \tag{A} $$ of the Polyakov action, where $\varepsilon$ is an infinitesimal parameter. Then $$ \partial_a \delta X^{\lambda} ~=~ \varepsilon\partial_a X^{\mu}~\partial_{\mu}K^{\lambda}, \qquad \delta G_{\mu\nu}~=~\varepsilon K^{\lambda}~\partial_{\lambda}G_{\mu\nu}.\tag{B}$$

  2. It is easier to use the definition of a Killing vector field $$0~=~({\cal L}_K G)_{\mu\nu}~=~ K^{\lambda}~\partial_{\lambda}G_{\mu\nu} + \partial_{\mu}K^{\lambda}~G_{\lambda\nu}+G_{\mu\lambda}~\partial_{\nu}K^{\lambda} \tag{C} $$ rather than the equivalent eq. (2).

This post imported from StackExchange Physics at 2016-01-17 15:57 (UTC), posted by SE-user Qmechanic
answered Jan 7, 2016 by Qmechanic (2,860 points) [ no revision ]

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