Wednesday, March 8, 2017

A calculation on moduli stabilization

In section 21.6 "Moduli stabilization and the landscape" in Zwiebach's string theory book, I read the sentence "Deriving the potential $V(R)$ associated with $R$ is a straightforward but technical calculation in general relativity". At this point I did not understand what the calculation was. I vaguely remembered a paper by Witten about instabilities in Kaluza-Klein spacetimes related to instantons. A calculation with instantons is indeed technical, but perhaps straightforward for experts. I found more information in a paper by Denef [1]. The calculation has nothing to do with instantons, but is indeed a straightforward calculation in differential geometry. In the rest of this blog post I set out the calculation in the form of a new exercise for Zwiebach's book.

Problem 21.7 Calculation of the potential $V(R)$
This exercise steps through the calculation of equation (21.68) in Zwiebach. It is based on the paper Physics of String Flux Compactifications, Denef et al, 2007
(a) The Einstein-Hilbert action in 6 dimensions is \begin{equation*} S = \int\!\! d^6 x\ M_6^4 \sqrt{-G}\ \mathcal{R} \end{equation*} with $\mathcal{R}$ the Ricci scalar of the 6-dimensional metric. Suppose the 6-dimensional metric is of the form \begin{equation}\label{eq:20170225b} G_{MN}dx^M dx^N = g_{\mu\nu}dx^{\mu} dx^{\nu} + R^2(x) g_{ab} dx^a dx^b \end{equation} with $M,N$: $0,\ldots, 5$ and $\mu,\nu$: $0,\ldots, 3$ and $a,b$: $4,5$. Mathematicians call a metric of this form a warped product. A physical interpretation is a compactified space where the compactification radius $R(x)$ depends on the 4-dimensional spacetime. Calculate that the 6-dimensional Ricci scalar of the metric \eqref{eq:20170225b} has the form \begin{equation*} \mathcal{R} =\mathcal{R}_4- 4 \frac{\Delta R}{R} - \frac{2}{R^2} g^{\mu\nu} \partial_{\mu}R\ \partial_{\nu} R + \frac{1}{R^2} \mathcal{R}_2 \end{equation*} with $\mathcal{R}_4$ the Ricci-scalar of the 4-dimensional metric $g_{\mu\nu}$, $\mathcal{R}_2$ the Ricci-scalar of the 2-dimensional metric $g_{ab}$ and $\Delta$ the Laplacian with respect to the metric $g_{\mu\nu}$.
(b) If the volume of the compactified space is \begin{equation*} \int\!\! d^2 y\ \sqrt{g_{(2)}} = V_2 \end{equation*} use the Gauss-Bonnet theorem \begin{equation*} \int\!\! d^2 y\ \sqrt{g_{(2)}}\ \mathcal{R}_2 = 4 \pi ( 2 - 2 g) \end{equation*} to show that \begin{equation}\label{eq:20170225f} S = \int\!\! d^4 x\ M_6^4 V_2 \sqrt{-g_{(4)}} \ \left(R^2 \mathcal{R}_4 - 4 R \Delta R - 2 g^{\mu\nu}\partial_{\mu}R\ \partial_{\nu} R\right) + \int\!\! d^4 x\ M_6^4 \sqrt{-g_{(4)}} 4 \pi ( 2 - 2 g) \end{equation} I have used the abbreviation $\det(-g_{\mu\nu}) = g_{(4)}$ and $\det(g_{ab}) = g_{(2)}$
(c) In the action \eqref{eq:20170225f}, the graviton and scalar field are mixed because of the term $R^2 \mathcal{R}_4$, we will separate them by writing $g_{\mu\nu} = R^{-2} h_{\mu\nu}$. Calculate that under this Weyl transformation the action becomes \begin{equation}\label{eq:20170225g} S = \int\!\! d^4 x\ M_6^4 V_2 \sqrt{-h} \ \left(\mathcal{R}_h +2 \frac{\Delta R}{R} - \frac{6}{R^2} h^{\mu\nu}\partial_{\mu}R\ \partial_{\nu} R - V(R) \right) \end{equation} with \begin{equation}\label{eq:20170225h} V(R) = - ( 2 - 2 g ) \frac{4 \pi}{V_2 R^4} \end{equation} In equations \eqref{eq:20170225g}and \eqref{eq:20170225h}, the Laplacian and the Ricci scalar $\mathcal{R}_h$ are with respect to the metric $h_{\mu\nu}$. This concludes the calculation of equation (21.68) in Zwiebach. We have also calculated that $a_g = \dfrac{4 \pi}{V_2}$.

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