We find static spherically symmetric solutions of scale invariant
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0$) or anti-de Sitter ($\lambda<0$) spacetimes and satisfies eq.~(\ref{dd}). For $\lambda=0$, the spacetime is asymptotically flat and they describe an ordinary black hole. \item The vanishing of the first factor in~(\ref{factor}) gives rise to the solutions of class (i) characterized by $R=0$ and $R_{\mu\nu} \neq0$, \begin{eqnarray} ds^2=-\left(1-\frac{M}{r}+\frac{K}{r^2}\right)dt^2+\frac{dr^2}{\displaystyle 1-\frac{M}{r}+\frac{K}{r^2}}+r^2 d\Omega_2^2 \label{BH}\, , \end{eqnarray} describing more general asymptotically flat solutions which are reduced to the ordinary Scharzschild black holes for $K=0$. In the next subsection we will investigate the scalar-flat solutions in more details. \end{itemize} ]]>
0$, pure $R^2$ gravity has a positive action since \begin{eqnarray} S_{\rm R^2}=\frac{1}{16\mu^2}\int d^4 x\sqrt{-g}R^2\geq 0 \, .\label{bb} \end{eqnarray} Therefore, backgrounds with $R=0$ saturate the above bound and are global minima of~(\ref{bb}). For $\mu^2<0$, of course such curvature scalar-flat spacetimes are global maxima. In both cases, scalar-flat spacetimes are extrema of~(\ref{bb}) and it can trivially be checked that indeed they satisfy the vacuum field equations~(\ref{jmn}). For a general static spherically symmetric spacetime with metric of the form~(\ref{dsg}), the curvature scalar is \begin{eqnarray} R= -\left(\frac{\lambda '(r)}{2 \lambda (r)}+\frac{2}{r}\right)\nu'(r) +\left(\frac{\lambda '(r)^2}{2 \lambda (r)^2}-\frac{2}{r^2}-\frac{\lambda ''(r)}{\lambda (r)}-\frac{2 \lambda '(r)}{r \lambda (r)}\right)\nu(r)+\frac{2}{r^2}\, . \end{eqnarray} Hence, the equation $R=0$ is written as \begin{eqnarray} \nu'(r)+p(r)\nu(r)+q(r)=0\, , \label{M} \end{eqnarray} where \begin{eqnarray} p(r)&=&\frac{2 r^2 \lambda \lambda ''-r^2 \lambda '^2+4 r \lambda \lambda '+4 \lambda ^2}{r^2 \lambda \lambda '+4 r \lambda ^2}, \nonumber \\ ~~~q(r)&=&-\frac{4 \lambda }{r^2 \lambda '+4 r \lambda }, \end{eqnarray} and $\lambda(r),\nu(r)$ are subject to the conditions \begin{eqnarray} \lambda(r)\to 1, ~~~\nu(r)\to 1, ~~~\mbox{for $r\to \infty$}\, . \end{eqnarray} The general solution to~(\ref{M}) is then \begin{eqnarray} \nu(r)=\frac{1}{u(r)}\int u(r)q(r) dr, ~~~~u(r)=e^{\int p(r) dr}\, . \end{eqnarray} In other words, there is no corresponding Birkhoff theorem in pure $R^2$ theory: given any particular profile function $\lambda(r)$ (or $\nu(r)$), the other is uniquely determined by the asymptotic conditions and the singularity structure. For example, by choosing \begin{eqnarray} \lambda(r)=\left(1-\frac{M}{r}\right)^2 \end{eqnarray} we find that \begin{eqnarray} \nu(r)=1+\frac{\left(M-2 c_1\right) (M-2 r)}{2 r^2}\, . \end{eqnarray} However, it is easy to see that \begin{eqnarray} C^{\mu\nu\rho\sigma}C_{\mu\nu\rho\sigma}=\frac{3 \left(-4 M^2 \left(c_1+r\right)+M r \left(8 c_1+3 r\right)-2 c_1 r^2+2 M^3\right){}^2}{r^8 (M-r)^2} \label{c2} \end{eqnarray} and therefore there are singularities at $r=0$ and $r=M$. Nevertheless, the second singularity at $r=M$ is removed for $c_1=-M/2$. For this particular value of $c_1$ we find \begin{eqnarray} \nu(r)=\left(1-\frac{M}{r}\right)^2 \end{eqnarray} so that \begin{eqnarray} ds^2=-\left(1-\frac{M}{r}\right)^2dt^2+\frac{dr^2}{\left(1-\frac{M}{r}\right)^2}+r^2\d\Omega^2_2 \label{s1}\, , \end{eqnarray} which is just the extremal RN black hole. However, there are also other asymptotically flat solutions with the correct Newtonian limit. For example, we find that for \begin{eqnarray} \lambda(r)=\left(1-\frac{M}{r}\right)^4 \end{eqnarray} the solution of~(\ref{M}) is \begin{eqnarray} \nu(r)=\left(1-\frac{M}{r}\right)^{-2} \left(1+\frac{M^2}{3 r^2}-\frac{10 M}{9 r}+\frac{c_1 e^{-\frac{3 M}{r}}}{r}\right)\, . \end{eqnarray} Note that there is a singularity at $r=0$ and a second one at $r=M$ as can be seen by a direct calculation of a the scalar invariant~(\ref{c2}) for example. This second singularity can be removed for $c_1=-2Me^3/9$ and hence the metric is written as \begin{eqnarray} ds^2=-\left(1-\frac{M}{r}\right)^4dt^2+\frac{\left(1-\frac{M}{r}\right)^2}{1+\frac{M^2}{3r^2}-\frac{10M}{9r}\left(1+\frac{1}{5}e^{3(1-\frac{M}{r})}\right)}dr^2+r^2\d\Omega^2_2 \label{s2} \end{eqnarray} is also a vacuum static spherically symmetric solution of the $R^2$ theory. Note that the metric~(\ref{s2}) has no singularity at $r=M$ as the geometry there is, like in~(\ref{s1}), $AdS_2\times S^2$. Another class of solutions is provided for example for \begin{eqnarray} \lambda=e^{-M/r}\, . \end{eqnarray} In this case, we find that \begin{eqnarray} \nu(r)= \frac{r^2 \left[r^2 \left(c_1 e^{M/r}+1024 r\right)+4 M^3+72 M^2 r+456 M r^2\right]}{(M+4 r)^5}\, , \end{eqnarray} which is asymptotically flat and singular only at $r=0$. Therefore, the vacuum solutions of the $R^2$ theory are scalar flat $R=0$ backgrounds. In particular, these scalar flat backgrounds are global minima of the action and clearly there is no corresponding Birkhoff theorem for the $R^2$ theory. ]]>
0$, the corresponding solution ($\bar g_{\mu\nu},\phi$) of eqs.~(\ref{E1}) and~(\ref{E2}) is given by \begin{eqnarray} && e^{a\phi}=\frac{1}{8\mu^2} R\, , \label{tr1}\\ && \bar g_{\mu\nu}=\frac{ R}{8\mu^2}\, g_{\mu\nu}\, .\label{tr2} \end{eqnarray} Clearly, this transformation works as long as $ R\neq 0$, otherwise the transformation is singular and the solution of the $R^2$ action~(\ref{S1}) is not a solution of~(\ref{E}). For the spherical symmetric vacuum solution \begin{eqnarray}\label{nonricci} ds^2=-\left(1-\frac{M}{r}+\frac{K}{r^2}\right)dt^2+\frac{dr^2}{\displaystyle 1-\frac{M}{r}+\frac{K}{r^2}}+r^2 d\Omega_2^2 \label{BH3} \, , \end{eqnarray} we find that, although it is not Ricci flat, it has vanishing Ricci scalar \begin{eqnarray} R=0\, . \end{eqnarray} As a result, this asymptotically flat, spherical symmetric vacuum solution of the pure $R^2$ theory does not exist in the dual Einstein gravity with cosmological constant. This can be easily seen by varying the quadratic action of $R^2$ \begin{eqnarray} \delta S=0~\longrightarrow~~\left(R\, \delta\sqrt{-g}+ 2 \sqrt{-g} \, \delta R\right)R=0\, , \end{eqnarray} which is solved either by $R=0$ or by $ \left(\delta\sqrt{-g}+ 2 \sqrt{-g} \delta R\right)=0$. In the Einstein frame only the second class of solutions are captured. The first class with $R=0$ appears singular since it gives rise to a singular conformal transformation with $\Phi=0$. Differently stated, it corresponds to the tensionless limit (i.e.\ strong coupling limit) of Einstein gravity with $M_P\rightarrow 0$. Therefore the space of general solutions of the $R^2$ theory is that of the dual Einstein theory but including also its tensionless limits. Hence we conclude that in the tensionless limit of Einstein gravity with $M_P \rightarrow0$, there are vacuum solutions of the type eq.~(\ref{nonricci}) with non-zero Ricci tensor but with vanishing scalar curvature. These are however not solutions of weakly coupled gravity with finite $M_P$. This reminds us to a similar situation in the framework of tensionless string in six dimensions where one has an equation of motion which is not possible to be obtained from an action. It also reminds us to the strong coupling limit in string theory written in the sigma-model frame, with $e^{-2\phi}R$ in the limit $g^2_s=e^{2\phi}\rightarrow \infty$. From eq.~(\ref{S44}), the additional class of solutions with $R=0$ exist in all dimensions with $n>2$. Therefore the extension to the tensionless solutions does not only appear in four-dimensions but it is a common property valid in all dimensions with $n>2$. The asymptotically de Sitter solutions like eq.~(\ref{BHds}) with $\lambda>0$ having $R\ne 0$ corresponds to the second class of solutions \begin{eqnarray} R_{\mu\nu}=\lambda g_{\mu\nu}\, , ~~~~ R=12 \lambda \, , \end{eqnarray} with \begin{eqnarray} d s^2=-\left(1-\frac{M}{r}-\lambda r^2\right)dt^2+\frac{dr^2}{\displaystyle 1-\frac{M}{r}-\lambda r^2}+r^2 d\Omega_2^2 \label{BH4}\, , \end{eqnarray} and therefore, the corresponding solution of~(\ref{E}) is \begin{eqnarray} &&e^{a\phi}=\frac{3}{2\mu^2} \lambda \, ,\\ &&d\bar s^2=\frac{3\lambda}{2\mu^2} \left\{-\left(1-\frac{M}{r}-\lambda r^2\right)dt^2+\frac{dr^2}{\displaystyle 1-\frac{M}{r}-\lambda r^2}+r^2 d\Omega_2^2\right\} \label{BHds1}\, . \end{eqnarray} It is easy to see that~(\ref{BHds1}) satisfies \begin{eqnarray} \bar R_{\mu\nu}=2\mu^2 \bar g_{\mu\nu}\, , \end{eqnarray} as expected. Note that for Ricci flat solutions with $R_{\mu\nu}=R=0$, the transformation from the $R^2$ action eq.~(\ref{S1}) to the Einstein frame action eq.~(\ref{E}) is still possible. In this case one has also to take the limit $\mu=0$, while keeping $e^{a\phi}=\frac{R}{8\mu^2} $ finite. Finally, one would like to know how to find the sub-class of solutions of~(\ref{S1}) once a solution of Einstein gravity eq.~(\ref{E}) is known. For this, let us suppose that we have a solution $(\bar g_{\mu\nu},\phi)$ to eqs.~(\ref{E1}) and~(\ref{E2}) of the Einstein theory eq.~(\ref{E}). Then, it turns out that \begin{eqnarray} R=e^{-2b \phi}\left(\bar R-6b\bar \nabla^2\phi-6b^2(\bar \nabla\phi)^2\right)\, , \end{eqnarray} and the corresponding particular solution to the pure $R^2$ theory will be given by \begin{eqnarray} g_{\mu\nu}=\frac{8\mu^2}{ R} \bar g_{\mu\nu}\, . \end{eqnarray} ]]>
0$. It can straightforwardly be verified that, due to the conformal invariance of the Maxwell Lagrangian, the dual theory is written as \begin{eqnarray} S=\int d^4 x\sqrt{-g}\left(\frac{1}{2}R-\partial_\mu\phi\partial^\mu\phi-\mu^2-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}\right)\, . \label{Sed} \end{eqnarray} Then since \begin{eqnarray} \bar R=\frac{-2\mu^2Q^2}{Z}\, , \end{eqnarray} by using the transformations~(\ref{tr1}),~(\ref{tr2}), we find that the corresponding solution in the Einstein frame is \begin{eqnarray} d s^2&=&\frac{-Q^2}{4 m^2Z}\left[-\left(1-\frac{M}{r}-\frac{Z}{r^2}+\frac{Q^2\mu^2}{6Z}r^2\right)dt^2+\frac{dr^2}{1-\frac{M}{r}-\frac{Z}{r^2}+\frac{Q^2\mu^2}{6Z}r^2}+r^2 d\Omega_2^2\ \right]\qquad\\ A_t&=&\frac{Q}{r}\, , \end{eqnarray} and it is acceptable for only for $Z<0$ (for $\mu^2>0$). In other words, $R^2$ theory coupled to electromagnetism admits solutions of dS-RN and AdS-RN type~(\ref{sol1}). However, only (dS-RN) are solutions of the dual Einstein gravity. Again, the dual Einstein theory can only capture part of the possible solutions of the $R^2$ theory. ]]>
0$) backgrounds with $R=0$ saturate the bound~(\ref{bb}) and minimize the action, they resemble the BPS condition and the associated zero entropy of the corresponding BPS~states. We also note that backgrounds with $R<0$ have negative entropy.\footnote{The fact that higher curvature gravity may have negative entropy has been noticed also in~\cite{Cvetic}.} The interpretation of such backgrounds is not very clear. Note in particular that such backgrounds cannot be conformally mapped to backgrounds with Lorentzian signature in the Einstein frame as long as $\mu^2>0$. The conformal mapping gives spacetime with wrong signature indicating a strong instability. The question is if we can still map backgrounds with $R<0$ to metrics with Lorentzian signature in the Einstein frame. This indeed can be done for negative $\Phi<0$ in eq.~(\ref{t1}), which can be parametrized as \begin{eqnarray} \Phi=-\frac{1}{2}e^{a\phi}\, , ~~~~e^{a\phi}=-\frac{1}{8\mu^2}R\, . \end{eqnarray} Then Einstein frame action turns out to be \begin{eqnarray} S'=\int d^4 x \sqrt{-g}\left(-\frac{1}{2}R-\frac{1}{2}\partial_\mu \phi \partial_\nu \phi -\mu^2\right) \label{Eprime}\, , \end{eqnarray} and the conformal transformation to the Einstein frame is only possible for $e^{a\phi}>0$, i.e.\ $R<0$. However, although the metric in the Einstein frame is of Lorentzian signature now, gravity has been turned repulsive. Thus, the negative entropy in the $R^2$ theory associated to backgrounds with $R<0$, corresponds to repulsive gravity in the Einstein frame. In other words, gravity is repulsive on backgrounds with negative entropy in the $R^2$ theory. This is a further indication that $R^2$ theory, although it can be conformally transformed to Einstein frame, it is quite different form it. ]]>