Wednesday, July 19, 2017

Emergent Gravity Explained

In order to exhibit the hidden gravity in noncommutative $U(1)$ gauge theory we apply the Seiberg-Witten map \cite{Seiberg:1999vs,Douglas:2001ba}.
We start with noncommutative Moyal-Weyl $U(1)$ gauge theory coupled to a complex scalar field in the fundamental representation given by the action
\begin{eqnarray}
S=-\frac{1}{4g^2}\int d^Dx \hat{F}_{\mu\nu}*\hat{F}^{\mu\nu}+\frac{1}{2}\int d^Dx(\hat{D}_{\mu}\hat{\phi})^{\dagger}*(\hat{D}^{\mu}\hat{\phi}).
\end{eqnarray}
The star $U(1)$ gauge transformations are given explicitly by
\begin{eqnarray}
\hat{A}_{\mu}\longrightarrow \hat{A}_{\mu}^{\prime}=\hat{A}_{\mu}+\partial_{\mu}\hat{\lambda}-i[\hat{A}_{\mu},\hat{\lambda}]_*~,~\hat{\phi}\longrightarrow \hat{\phi}^{'}=\hat{\phi}+i\hat{\lambda}*\hat{\phi}.
\end{eqnarray}
Following Seiberg and Witten we will now construct an explicit map between the noncommutative vector potential ${\hat{A}}_{\mu}$ and a commutative vector potential $A_{\mu}$ which will implement explicitly the perturbative equivalence of the above noncommutative gauge theory to a conventional gauge theory. Clearly, this map must depend both on the gauge parameter as well as on the vector potential in order to be able to achieve equivalence between the physical orbits in the two theories. Also, since the gauge field is coupled to a scalar field, the noncommutative scalar field will also be mapped to a conventional scalar field. We write then
\begin{eqnarray}
&&\hat{A}_{\mu}(A+\delta_{\lambda}A)=\hat{A}_{\mu}(A)+\delta_{\hat{\lambda}}\hat{A}_{\mu}(A)\nonumber\\
&&\hat{\phi}(A+\delta_{\lambda}A,\phi+\delta_{\lambda}\phi)=\hat{\phi}(A,\phi)+\delta_{\hat{\lambda}}\hat{\phi}(A,\phi).
\end{eqnarray}
\begin{eqnarray}
\delta_{\hat{\lambda}}\hat{A}_{\mu}=\partial_{\mu}\hat{\lambda}-i[\hat{A}_{\mu},\hat{\lambda}]_*~,~\delta_{\hat{\lambda}}\hat{\phi}=i\hat{\lambda}*\hat{\phi}.
\end{eqnarray}
\begin{eqnarray}
\delta_{{\lambda}}{A}_{\mu}=\partial_{\mu}{\lambda}~,~\delta_{{\lambda}}{\phi}=i{\lambda}{\phi}.
\end{eqnarray}
To solve the above equation we write
\begin{eqnarray}
\hat{A}^{\mu}=A^{\mu}+A^{\mu\prime}(A)~,~\hat{\lambda}=\lambda+\lambda^{\prime}(\lambda,A).
\end{eqnarray}
It reduces then to
\begin{eqnarray}
A_{\mu}^{\prime}(A+\delta A)-A_{\mu}^{\prime}(A)-\partial_{\mu}\lambda^{\prime}&=&-i[A_{\mu},\lambda]_*\nonumber\\
&=&\theta^{\alpha\beta}\partial_{\alpha}A_{\mu}\partial_{\beta}\lambda.
\end{eqnarray}
We have
\begin{eqnarray}
A^{\mu\prime}(A+\delta A)=A^{\mu\prime}(A)+\frac{\delta}{\delta A_{\beta}}A^{\mu\prime}.\partial_{\beta}\lambda+\frac{\delta}{\delta (\partial_{\nu}A_{\beta})}A^{\mu\prime}.\partial_{\nu}\partial_{\beta}\lambda.
\end{eqnarray}
$A^{\mu\prime}$ and $\lambda^{\prime}$ must be both of order $\theta$. Furthermore, by thinking along the lines of a derivative expansion, we know that $A^{\mu\prime}$ is quadratic in $A^{\mu}$ of the form $A\partial A$ whereas $\lambda^{\prime}$ is linear in $A^{\mu}$. The last equation above also suggests that $\lambda^{\prime}$ is proportional to $\partial_{\alpha}\lambda$. For constant $A$ the above condition reduces then to
\begin{eqnarray}
A_{\mu}(A+\delta A)-A_{\mu}(A)-\partial_{\mu}\lambda^{\prime}&=&0\Rightarrow
\frac{\delta}{\delta (\partial_{\nu}A_{\beta})}A_{\mu}^{\prime}.\partial_{\nu}\partial_{\beta}\lambda+\frac{1}{2}\theta^{\alpha\beta}\partial_{\mu}\partial_{\beta}\lambda.A_{\alpha}=0,\nonumber\\
\end{eqnarray}
\begin{eqnarray}
\lambda^{\prime}=\frac{1}{2}\theta^{\alpha\beta}\partial_{\alpha}\lambda A_{\beta}.
\end{eqnarray}
The coefficient $1/2$ is fixed by the requirement that the second derivative in $\lambda$ cancels. The condition becomes
\begin{eqnarray}
\frac{1}{2}\frac{\delta}{\delta
(\partial_{\nu}A_{\beta})}A_{\mu}^{\prime}+\frac{1}{2}\frac{\delta}{\delta
(\partial_{\beta}A_{\nu})}A_{\mu}^{\prime}+\frac{1}{4}\theta^{\alpha\beta}\eta^{\nu}_{\mu}A_{\alpha}+\frac{1}{4}\theta^{\alpha\nu}\eta^{\beta}_{\mu}A_{\alpha}=0.
\end{eqnarray}
A solution is given by

\begin{eqnarray}
A_{\mu}^{\prime}=-\frac{1}{2}\theta^{\alpha\beta}A_{\alpha}(2\partial_{\beta}A_{\mu}-\partial_{\mu}A_{\beta}).
\end{eqnarray}
It is now very easy to verify that this solves the Seiberg-Witten condition also for non-constant $A$.
We do now the same for the scalar field. We write
\begin{eqnarray}
\hat{\phi}=\phi+\phi^{\prime}(A,\phi).
\end{eqnarray}
We get immediately the condition
\begin{eqnarray}
\phi^{\prime}(A+\delta A,\phi+\delta \phi)-\phi^{\prime}(A,\phi)&=&i\hat{\lambda}*\hat{\phi}-\delta_{\lambda}\phi\nonumber\\
&=&i{\lambda}{\phi}^{\prime}+\frac{i}{2}\theta^{\mu\nu}\partial_{\mu}\lambda A_{\nu}\phi-\frac{1}{2}\theta^{\mu\nu}\partial_{\mu}\lambda\partial_{\nu}\phi.
\end{eqnarray}
Again, we note that $\phi^{\prime}$ is of order $\theta$ and it must be proportional to $A\partial\phi$. It is not difficult to convince ourselves that the solution is given by

\begin{eqnarray}
\phi^{\prime}=-\frac{1}{2}\theta^{\mu\nu}A_{\mu}\partial_{\nu}\phi.
\end{eqnarray}
We compute the expression of the action in the new variables. We compute first
\begin{eqnarray}
\hat{F}_{\mu\nu}&=&\partial_{\mu}\hat{A}_{\nu}-\partial_{\nu}\hat{A}_{\mu}-i[\hat{A}_{\mu},\hat{A}_{\nu}]_*\nonumber\\
&=&F_{\mu\nu}+\partial_{\mu}A_{\nu}^{\prime}-\partial_{\nu}A_{\mu}^{\prime}+\theta^{\alpha\beta}\partial_{\alpha}A_{\mu}\partial_{\beta}A_{\nu}\nonumber\\
&=&F_{\mu\nu}+\theta^{\alpha\beta}F_{\mu\alpha}F_{\nu\beta}-\theta^{\alpha\beta}A_{\alpha}\partial_{\beta}F_{\mu\nu}.
\end{eqnarray}
We get immediately the action
\begin{eqnarray}
S_F&=&-\frac{1}{4g^2}\int d^Dx \hat{F}_{\mu\nu}\hat{F}^{\mu\nu}\nonumber\\
&=&-\frac{1}{4g^2}\int d^Dx \bigg[{F}_{\mu\nu}{F}^{\mu\nu}+2\theta^{\alpha\beta}F_{\beta}~^{\nu}\bigg(F_{\alpha}~^{\sigma}F_{\sigma\nu}+\frac{1}{4}\eta_{\nu\alpha}F^{\rho\sigma}F_{\rho\sigma}\bigg)\bigg].
\end{eqnarray}
Also we compute the covariant derivative
\begin{eqnarray}
\hat{D}_{\mu}\hat{\phi}&=&\partial_{\mu}\hat{\phi}-i\hat{A}_{\mu}*\hat{\phi}\nonumber\\
&=&D_{\mu}\phi-\frac{1}{2}\theta^{\alpha\beta}A_{\alpha}\partial_{\beta}D_{\mu}\phi-\frac{\theta^{\alpha\beta}}{2}D_{\alpha}\phi F_{\beta\mu}.
\end{eqnarray}
\begin{eqnarray}
(\hat{D}_{\mu}\hat{\phi})^{\dagger}&=&\partial_{\mu}\hat{\phi}^{\dagger}+i\hat{\phi}^{\dagger}*\hat{A}_{\mu}\nonumber\\
&=&(D_{\mu}\phi)^{\dagger}-\frac{1}{2}\theta^{\alpha\beta}A_{\alpha}\partial_{\beta}(D_{\mu}\phi)^{\dagger}-\frac{\theta^{\alpha\beta}}{2}(D_{\alpha}\phi)^{\dagger} F_{\beta\mu}.
\end{eqnarray}
The charged scalar action becomes
\begin{eqnarray}
S_{\phi}&=&\frac{1}{2}\int d^Dx(\hat{D}_{\mu}\hat{\phi})^{\dagger}*(\hat{D}^{\mu}\hat{\phi})\nonumber\\
&=&\frac{1}{2}\int d^Dx\bigg[({D}_{\mu}{\phi})^{\dagger}({D}^{\mu}{\phi})-\frac{1}{2}\bigg(\theta^{\mu\alpha}F_{\alpha}~^{\nu}+\theta^{\nu\alpha}F_{\alpha}~^{\mu}+\frac{1}{2}\eta^{\mu\nu}\theta^{\alpha\beta}F_{\alpha\beta}\bigg)({D}_{\mu}{\phi})^{\dagger}({D}_{\nu}{\phi})\bigg].\nonumber\\
\end{eqnarray}
The main observation of Rivelles \cite{Rivelles:2002ez} is that we can rewrite the above $\theta$-expanded actions of the noncommutative $U(1)$ gauge field  $\hat{A}_{\mu}$ and the noncommutative charged scalar field $\hat{\phi}$ as a coupling of a commutative $U(1)$ gauge field $A_{\mu}$ and a commutative charge scalar field $\phi$ to a metric $g_{\mu\nu}=\eta_{\mu\nu}+h_{\mu\nu}+\eta_{\mu\nu} h$ with $h_{\mu\nu}\eta^{\mu\nu}=0$. This metric itself is determined by the commutative $U(1)$ gauge field $A_{\mu}$ and the noncommutativity structure $\theta_{\mu\nu}$. Indeed, the dynamics of a commutative Maxwell field $A_{\mu}$ and the charged scalar field $\phi$ in a linearized gravitational field is given by the actions

\begin{eqnarray}
S_F&=&-\frac{1}{4g^2}\int d^Dx \sqrt{-g}{F}_{\mu\nu}{F}^{\mu\nu}\nonumber\\
&=&-\frac{1}{4g^2}\int d^Dx \bigg[{F}_{\mu\nu}{F}^{\mu\nu}+2h^{\mu\nu}F_{\mu}~^{\rho}F_{\rho\nu}\bigg].
\end{eqnarray}
\begin{eqnarray}
S_{\phi}&=&\frac{1}{2}\int \sqrt{-g}d^Dx({D}_{\mu}{\phi})^{\dagger}*({D}^{\mu}{\phi})\nonumber\\
&=&\frac{1}{2}\int d^Dx\bigg[({D}_{\mu}{\phi})^{\dagger}({D}^{\mu}{\phi})-h^{\mu\nu}({D}_{\mu}{\phi})^{\dagger}({D}_{\nu}{\phi})+2h({D}_{\mu}{\phi})^{\dagger}({D}^{\mu}{\phi})\bigg].
\end{eqnarray}
We get immediately the traceless metric
\begin{eqnarray}
h^{\mu\nu}=\frac{1}{2}(\theta^{\mu\beta}F_{\beta}~^{\nu}+\theta^{\nu\beta}F_{\beta}~^{\mu})+\frac{1}{4}\theta^{\alpha\beta}F_{\alpha\beta}\eta^{\mu\nu}.
\end{eqnarray}
The gauge field has in this setting a dual role. It couples minimally to the charged scalar field as usual but also it sources the gravitational field. In other words, the gravitational field is not just a background field since it is determined by the dynamical gauge field. We write the interval
\begin{eqnarray}
ds^2=(1+\frac{1}{4}\theta^{\alpha\beta}F_{\alpha\beta})dx_{\mu}dx^{\mu}+\theta^{\nu\beta}F_{\beta\mu}dx_{\mu}dx^{\nu}.
\end{eqnarray}
We can compute the Riemann tensor, the Ricci tensor and the Ricci scalar and find them of order one, two and two in $\theta$  respectively. Thus, to the linear order in $\theta$, the Riemann tensor is non zero while the Ricci tensor and the Ricci scalar are zero. In other words, the above metric can not describe a flat spacetime. In fact it describes a gravitational plane wave since at zero order in $\theta$ we must have ordinary electromagnetism and thus $F_{\mu\nu}$ (and hence the metric $h_{\mu\nu}$) depends on the plane wave $\exp(ikx)$ with $k^2=0$. At first order a plane wave solution $A_{\mu}=F_{\mu}\exp(ikx)$ with $k^2=0$ and $k^{\mu}F_{\mu}=0$ can also be constructed explicitly \cite{Rivelles:2002ez}.

On the other hand, we can see that this metric describes a spacetime with a covariantly constant symplectic form $\theta$ which is also null, since to the linear order in $\theta$ we must have the equations $D_{\mu}\theta^{\alpha\beta}=0$ and $\theta_{\mu\nu}\theta^{\mu\nu}=0$, and as a consequence this spacetime is indeed a pp-wave spacetime \cite{EK}, which is precisely a gravitational plane wave as we have checked explicitly above to the first order in $\theta$.