Diffeomorphism invariance of the Polyakov action

[maverick280857]

[SOLVED] Diffeomorphism invariance of the Polyakov action

Hi,

I'm struggling with something that is quite elementary. I know that the Polyakov action is diffeomorphism invariant and Weyl invariant. Denoting the world-sheet coordinates \sigma^0 = \sigma and \sigma^1 = t and the independent world-sheet metric by h_{\alpha\beta}(\sigma), the Polyakov action can be written as

S_{P} = -\frac{T}{2}\int d^{2}\sigma \, \sqrt{-h}h^{\alpha\beta}\partial_\alpha X^\mu \partial_\beta X_\mu

where T is the string tension and \alpha, \beta run over 0 and 1 (labeling \sigma and t).

Now, if I define

\delta_{E}h_{\alpha\beta}(\sigma) = h_{\alpha\beta}'(\sigma) - h_{\alpha\beta}(\sigma)

as the Einstein variation of the independent world-sheet metric, where the primed coordinates denote the transformed world-sheet coordinates under a diffeomorphism defined by

{\sigma'}^{\gamma} = \sigma^\gamma - \xi^\gamma(\sigma)

then I would like to show that

\delta_E h_{\alpha\beta} = \partial_\alpha \xi^\gamma h_{\gamma\beta} + \partial_\beta \xi^\gamma h_{\alpha\gamma} + \xi^\gamma \partial_\gamma h_{\alpha\beta}

This is supposed to follow from chain rule, but I do not quite get all the terms.

My working

h'_{\gamma\delta}(\sigma') = h_{\alpha\beta}(\sigma)\frac{d\sigma^\alpha}{d{\sigma'}^\gamma}\frac{d{\sigma}^\beta}{d{\sigma'}^\delta}

Now by definition,

\sigma^\alpha = {\sigma'}^\alpha + \xi^\alpha(\sigma)

So,

\frac{d\sigma^\alpha}{d{\sigma'}^\beta} = \delta^\alpha_\beta + \frac{d\sigma^\delta}{d{\sigma'}^\beta}\frac{d{\xi}^\alpha(\sigma)}{d {\sigma}^{\delta}}

But this seems to give only the first two terms of \delta_E h and not the third one, which is the usual derivative (change along gradient) term that I'd expect to get.

Any suggestions will be greatly appreciated.

Thanks!

 

[maverick280857]

Solved. Thanks anyway!

 

[samalkhaiat]

Hi,
h'_{\gamma\delta}(\sigma') = h_{\alpha\beta}(\sigma)\frac{d\sigma^\alpha}{d{\sigma'}^\gamma}\frac{d{\sigma}^\beta}{d{\sigma'}^\delta}

Thanks!

Expand the left-hand-side as well
h^{'}_{ \mu \nu } ( \sigma^{'} ) = h^{'}_{ \mu \nu } ( \sigma - \xi ) \approx h^{'}_{ \mu \nu } ( \sigma ) - \xi^{ \rho } \partial_{ \rho } h^{'}_{ \mu \nu } ( \sigma ) .
But infinitesimally
\xi^{ \rho } \partial_{ \rho } h^{'}_{ \mu \nu } ( \sigma ) \approx \xi^{ \rho } \partial_{ \rho } h_{ \mu \nu } ( \sigma ) .

 

[maverick280857]

Thanks samalkhaiat, yes, I got that.

I would also like to show that

(a) \delta_E(h^{\alpha\beta}\partial_\alpha X^\mu \partial_\beta X_\mu) = \xi^\rho \partial_\rho(h^{\alpha\beta}\partial_\alpha X^\mu \partial_\beta X_\mu) because "h^{\alpha\beta}\partial_\alpha X^\mu \partial_\beta X_\mu is an Einstein scalar". What is an Einstein scalar?

My guess is that h^{\alpha\beta}\partial_\alpha X^\mu \partial_\beta X_\mu transforms as a scalar function.

(b) \delta_E(\sqrt{-h}) = \partial_\rho(\xi^\rho \sqrt{-h})

This one worries me a little because of the placement of parenthesis...any ideas?

How does one prove these relations?

 

[samalkhaiat]

Thanks samalkhaiat, yes, I got that.

I would also like to show that

(a) \delta_E(h^{\alpha\beta}\partial_\alpha X^\mu \partial_\beta X_\mu) = \xi^\rho \partial_\rho(h^{\alpha\beta}\partial_\alpha X^\mu \partial_\beta X_\mu) because "h^{\alpha\beta}\partial_\alpha X^\mu \partial_\beta X_\mu is an Einstein scalar". What is an Einstein scalar?

My guess is that h^{\alpha\beta}\partial_\alpha X^\mu \partial_\beta X_\mu transforms as a scalar function.

That is correct.

(b) \delta_E(\sqrt{-h}) = \partial_\rho(\xi^\rho \sqrt{-h})

This one worries me a little because of the placement of parenthesis...any ideas?

How does one prove these relations?

Use the following fact. For any operator \hat{ \delta }, satisfying the Leibnz rule such as ( \partial , \delta ), the following can be proven
\frac{ \hat{ \delta }( \sqrt{ - g } ) }{ \sqrt{ - g } } = \frac{1}{2} g^{ \mu \nu } \hat{ \delta } g_{ \mu \nu } .

Sam