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How to show that the polyakov action of a massless particle is conformally invariant?

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I am trying to show that the Polyakov action of a massless particle in a $D$ dimensional Minkowski spacetime $$S=\int d\tau(e^{-1}\dot{x}^{\mu}\dot{x}^{\nu}\eta_{\mu\nu})$$
is invariant under the conformal transformation generated by a vector field $\xi$ in the Minkowski spacetime, i.e. $\mathcal{L}_{\xi}\eta=\lambda\eta$, where $\mathcal{L}_{\xi}$ is the Lie derivative.

It turns out that in order to have conformal invariance, the auxiliary field $e$ has two transform in the following way:
$$\mathcal{L}_{\xi}e=\frac{2}{D}e(\partial_{\mu}\xi^{\mu})$$
But the above transformation is counter-intuitive to me. I cannot understand the physical or geometrical meaning of that transformation. Why should the field $e$ transform in that way under conformal transformation?

asked Oct 3, 2016 in Theoretical Physics by XIaoyiJing (50 points) [ revision history ]

1 Answer

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Using the explicit for of the conformal Killing vector you can see that the divergence vanishes for translations and Lorentz rotations,  but it is not zero for dilations. 

So, the transformation accounts for the contraction of the volume under those conformal transformations that change it. 

answered Oct 3, 2016 by anonymous [ no revision ]

Could you elaborate? What do you mean by "contraction of the volume under those conformal transformations that change it"?

Some of the conformal transformations rescale the volume. The divergence measures this effect, which is is then taken into account as a compensating transformation for the auxiliary field

Why does auxiliary field transform as a compensation of the rescaling effect?

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