M-Theory: Bosonic Fields - Need Help With Part III

[ergospherical]

Need help with part iii)

i) Under $C \rightarrow C + d\Lambda$, and since $dG = d^2C = 0 \implies d(\Lambda \wedge G \wedge G) = d\Lambda \wedge G \wedge G$, then neglecting the surface terms\begin{align*}
\int_D d\Lambda \wedge G \wedge G = \int_D d(\Lambda \wedge G \wedge G) &= \int_{\partial D} \Lambda \wedge G \wedge G = 0
\end{align*}ii) Varying with respect to the metric\begin{align*}
\delta S = \dfrac{1}{2}M^9 \int d^{11}x \left[\delta\sqrt{-g} \left(R - \dfrac{1}{48}G_{\mu \nu \rho \sigma}G^{\mu \nu \rho \sigma} \right) + \sqrt{-g} \delta R \right]
\end{align*}Using the classic trick for diagonalisable matrices $\delta \sqrt{-g} = \dfrac{1}{2\sqrt{-g}} (-g) \mathrm{tr}(g^{-1} \delta g) = \dfrac{-1}{2} \sqrt{-g} g_{\mu \nu} \delta g^{\mu \nu}$. Meanwhile for the Ricci scalar\begin{align*}

\delta R = \delta (g^{\mu \nu} R_{\mu \nu}) &= g^{\mu \nu} \delta R_{\mu \nu} + \delta g^{\mu \nu}R_{\mu \nu} \\

&= \nabla_{\mu} [g^{\rho \nu} \delta \Gamma^{\mu}_{\rho \nu} - g^{\mu \nu} \delta \Gamma^{\rho}_{\nu \rho}] + \delta g^{\mu \nu} R_{\mu \nu}

\end{align*}Therefore the Einstein equation should be\begin{align*}
\dfrac{-1}{2} g_{\alpha \beta} \left(R - \dfrac{1}{48}G_{\mu \nu \rho \sigma}G^{\mu \nu \rho \sigma} \right) + R_{\alpha \beta} &= 0 \\

\end{align*}Is this correct?

iii) Little progress made, hints appreciated.

 

[MathematicalPhysicist]

From which paper/book/lecture notes is this taken from?

 

[ergospherical]

It was one of Prof Tong's example sheets, I can't remember which one though. It might be one of these: https://www.damtp.cam.ac.uk/user/examples/indexP3.html

 

[samalkhaiat]

Need help with part iii)
Therefore the Einstein equation should be $$\begin{align*}<br /> \dfrac{-1}{2} g_{\alpha \beta} \left(R - \dfrac{1}{48}G_{\mu \nu \rho \sigma}G^{\mu \nu \rho \sigma} \right) + R_{\alpha \beta} &= 0 \\<br /> \end{align*}$$ Is this correct?

No. The field equation should be of the form $$R_{\mu\nu} - \frac{1}{2}g_{\mu\nu}R \propto T_{\mu\nu}(G),$$ where, $T_{\mu\nu}(G)$ is the energy-momentum tensor of the 4-form field $G$. Try to practise with the action $$S = \int d^{4}x \sqrt{-g} \left( R - \frac{1}{4}F^{2}\right),$$ where $$F^{2} = F_{\mu\nu}F^{\mu\nu} = g^{\mu\rho}g^{\nu\sigma}F_{\mu\nu}F_{\rho\sigma} .$$

iii) Little progress made, hints appreciated.

The relevant part of the action is $$- \frac{1}{2(4!)}\int d^{11}x \ \sqrt{-g} \ G^{2} - \frac{1}{6} \int C \wedge dC \wedge dC .$$ For the first integral, the variation gives you $$- \frac{1}{4!} \int d^{11} \sqrt{-g} \ G^{\mu_{1} \cdots \mu_{4}} (d \delta C)_{\mu_{1} \cdots \mu_{4}},$$ which is same as $$- \int \ \star G \wedge d(\delta C) = - \int \ d (\delta C) \wedge \star G .$$ Now, since $$d (\delta C \wedge \star G) = d(\delta C) \wedge \star G - \delta C \wedge d (\star G) ,$$ then the variation of the first integral (ignoring boundary integral) is $$- \int \ \delta C \wedge d \star G .$$
Similarly, for the second integral, you get $$-\frac{1}{6}\left( \int \delta C \wedge G \wedge G + 2 \int d(\delta C) \wedge C \wedge G \right) .$$ Integrating the second term by parts, the result of the variation becomes $$- \frac{3}{6} \int \delta C \wedge G \wedge G .$$ Thus, the vanishing variation of the action with respect to the 3-form $C$ gives $$\int \delta C \wedge \left( d \star G + \frac{1}{2} G \wedge G \right) = 0.$$ So, for arbitrary 3-form $\delta C$, you get $$d \star G + \frac{1}{2} G \wedge G = 0.$$

 

[ergospherical]

Try to practise with the action $$S = \int d^{4}x \sqrt{-g} \left( R - \frac{1}{4}F^{2}\right),$$ where $F^{2} = F_{\mu\nu}F^{\mu\nu} = g^{\mu\rho}g^{\nu\sigma}F_{\mu\nu}F_{\rho\sigma}$

I will re-write\begin{align*}
S = \int d^4 x \sqrt{-g} \left( g^{\mu \nu} R_{\mu \nu} - \dfrac{1}{4} g^{\mu\rho}g^{\nu\sigma}F_{\mu\nu}F_{\rho\sigma} \right)

\end{align*}then vary $S$ with respect to the metric,\begin{align*}
\delta S &= \int d^4 x \left( R- \dfrac{1}{4}F^2 \right)\delta \sqrt{-g} + \int d^4 x \sqrt{-g} \left( \delta R - \dfrac{1}{4} F_{\mu\nu}F_{\rho\sigma} \delta(g^{\mu\rho}g^{\nu\sigma})\right)
\end{align*}We have the results:\begin{align*}
\delta \sqrt{-g} &= -\dfrac{1}{2} \sqrt{-g} g_{\mu \nu} \delta g^{\mu \nu} \\ \\

\delta R &= \delta (g^{\mu \nu} R_{\mu \nu}) = \nabla_{\mu} [g^{\rho \nu} \delta \Gamma^{\mu}_{\rho \nu} - g^{\mu \nu} \delta \Gamma^{\rho}_{\nu \rho}] + \delta g^{\mu \nu} R_{\mu \nu} \\ \\

\delta(g^{\mu\rho}g^{\nu\sigma}) &= g^{\mu \rho} \delta g^{\nu \sigma} + g^{\nu \sigma} \delta g^{\mu \rho}

\end{align*}Ignoring the total derivative using the divergence theorem, the second integral becomes\begin{align*}
I_2 &= \int d^4 x \sqrt{-g} \left( \delta g^{\mu \nu} R_{\mu \nu} -\dfrac{1}{4} F_{\mu\nu}F_{\rho\sigma}(g^{\mu \rho} \delta g^{\nu \sigma} + g^{\nu \sigma} \delta g^{\mu \rho}) \right) \\

&= \int d^4 x \sqrt{-g} \left( R_{\mu \nu} -\dfrac{1}{4} {F^{\rho}}_{\mu}F_{\rho\nu} -\dfrac{1}{4} {F_{\mu}}^{\sigma}F_{\nu\sigma} \right)\delta g^{\mu \nu}
\end{align*}due to the antisymmetry of $F$ it follows that ${F^{\rho}}_{\mu}F_{\rho\nu} = {F_{\mu}}^{\rho} F_{\nu \rho}$, hence putting $\delta S = 0$ gives\begin{align*}
-\dfrac{1}{2} g_{\mu \nu} \left( R - \frac{1}{4}F^2 \right) + R_{\mu \nu} -\frac{1}{2} {F_{\mu}}^{\rho}F_{\nu\rho} = 0 \\ \\
\end{align*}which may be rearranged to \begin{align*}
R_{\mu \nu} - \frac{1}{2} Rg_{\mu \nu} &= \dfrac{1}{2} \left( {F_{\mu}}^{\rho}F_{\nu\rho} - \frac{1}{4} g_{\mu \nu} F^2 \right)
\end{align*}which does appear to be the stress energy tensor $T_{\mu \nu}(F)$ up to perhaps an erroneous proportionality constant?

$$- \frac{1}{4!} \int d^{11} \sqrt{-g} \ G^{\mu_{1} \cdots \mu_{4}} (d \delta C)_{\mu_{1} \cdots \mu_{4}},$$ which is same as $$- \int \ \star G \wedge d(\delta C) = - \int \ d (\delta C) \wedge \star G .$$

I'm having some trouble with this part. Defining the tensor $\epsilon_{\mu_1 \dots \mu_{11}} = \sqrt{-g} [\mu_1 \dots \mu_{11}]$ such that the volume element is $\boldsymbol{\epsilon} = \epsilon_{1\dots 11} d^{11}x = \sqrt{-g} d^{11} x$, \begin{align*}
\int {\star G} \wedge d(\delta C) &= \int d^{11}x \dfrac{11!}{4! 7!} (\star G)_{[1\dots 7} d(\delta C)_{8\dots 11]} \\

&= \int d^{11}x \dfrac{11!}{4! 7!} \dfrac{1}{7!} \epsilon_{[1\dots 7| \alpha \beta \gamma \delta} G^{\alpha \beta \gamma \delta} (d\delta C)_{|8\dots 11]}
\end{align*}I'm not sure if there's an identity I could use to tidy up the antisymmetrisation?

 

[samalkhaiat]

I'm having some trouble with this part. Defining the tensor $\epsilon_{\mu_1 \dots \mu_{11}} = \sqrt{-g} [\mu_1 \dots \mu_{11}]$ such that the volume element is $\boldsymbol{\epsilon} = \epsilon_{1\dots 11} d^{11}x = \sqrt{-g} d^{11} x$, \begin{align*}
\int {\star G} \wedge d(\delta C) &= \int d^{11}x \dfrac{11!}{4! 7!} (\star G)_{[1\dots 7} d(\delta C)_{8\dots 11]} \\

&= \int d^{11}x \dfrac{11!}{4! 7!} \dfrac{1}{7!} \epsilon_{[1\dots 7| \alpha \beta \gamma \delta} G^{\alpha \beta \gamma \delta} (d\delta C)_{|8\dots 11]}
\end{align*}I'm not sure if there's an identity I could use to tidy up the antisymmetrisation?

That mess does not take you anywhere. The proof of following identity can be found in many textbooks $$\alpha \wedge \star \beta = \beta \wedge \star \alpha = (\alpha , \beta) \ \epsilon ,$$ where $$(\alpha , \beta) (x) = \frac{1}{p!} \alpha_{\mu_{1} \cdots \mu_{p}}(x) \beta^{\mu_{1} \cdots \mu_{p}}(x) ,$$$$\epsilon (x) = \sqrt{-g(x)} dx^{0} \wedge \cdots \wedge dx^{n-1} \equiv \sqrt{-g(x)} \ d^{n}x .$$