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Our starting point is the general action of spatial covariant gravity coupled with a non-dynamical scalar field [85]:
$ S=\int{\rm{d}}t{\rm{d}}^{3}xN\sqrt{h}{\cal{L}}\left(h_{ij},K_{ij},R_{ij},N,\phi,{\rm{D}}_{i}\right), $
(1) where
$ h_{ij} $ is the spatial metric, N is the lapse function,$ R_{ij} $ is the spatial Ricci tensor, and$ K_{ij} $ is the extrinsic curvature defined by$ K_{ij}=\frac{1}{2N}\left(\partial_{t}h_{ij}-{\rm{D}}_{i}N_{j}-{\rm{D}}_{j}N_{i}\right), $
(2) where
$ N_{i} $ is the shift vector, and$ {\rm{D}}_{i} $ is the spatial derivative adapted to$ h_{ij} $ . Scalar field ϕ has no kinetic term and thus acts as an auxiliary variable. We emphasize that although this auxiliary scalar field was originally motivated from a scalar field with a spacelike gradient in the so-called spatial gauge, here (and also in the analysis in [85]), we do not make such an assumption and merely treat it as an auxiliary field, i.e., without any time derivative terms. Ref. [85] showed that the theory in (1) generally propagates three degrees of freedom, i.e., two tensorial and one scalar degrees of freedom. The aim of this study is to find the condition on the Lagrangian to eliminate the scalar mode.The most strict approach to analyzing the number of degrees of freedom is the constraint analysis, either in the Hamiltonian or Lagrangian formalism. In the absence of auxiliary scalar field ϕ, the exact conditions for the TTDOFs were found in [78]. These conditions are variation-differential equations for the Lagrangian, which are difficult (if not impossible) to solve in order to produce the concrete Lagrangian systematically. To address this problem, a perturbation method was employed in [79] (and previously in [58] with a dynamical lapse function). The advantage of the perturbation method is that one can determine the conditions on the Lagrangian order by order in perturbations, which by themselves are much simpler than the fully nonlinear conditions.
In this study, we use this perturbation method to determine the conditions on action (1) such that no scalar mode propagates. To be precise, our task is to study the perturbations of action (1) around a cosmological background and eliminate the scalar modes at the linear order in perturbations. The difference between this study and previous reports [58, 79] is that we first perform the perturbation analysis on the general form of the Lagrangian instead of a specific Lagrangian (e.g., in the polynomial form). Therefore, the resulting conditions are generic and can be applied to more general Lagrangians.
To eliminate the unwanted scalar degree of freedom, we focus solely on the scalar perturbations. We consider perturbations around a Friedmann-Robertson-Walker background. After the residual gauge freedom of spatial diffeomorphism is fixed, the Arnowitt-Deser-Misner (ADM) variables are parameterized as follows:
$ N = \bar{N} {\rm e}^{A}, $
(3) $ N_{i} = \bar{N}a\partial_{i}B, $
(4) $ h_{ij} = a^{2}e^{2\zeta}\delta_{ij}, $
(5) where a is the scale factor. Here, A, B, and ζ are the scalar perturbations of the metric. The auxiliary scalar field is perturbed as usual:
$ \phi = \bar{\phi} + \delta\phi. $
(6) All perturbation variables are set to zero at the background level, i.e.,
$ A=B=\zeta=\delta\phi=0 $ .$ \bar{N} $ and$ \bar{\phi} $ are the background values of the lapse function and scalar field, respectively, which are functions of time. Note that because of the loss of time reparametrization invariance, we cannot fix the background value of lapse function$ \bar{N} $ to unity. In this paper, we use the same shorthand as that in [72]:$ \dot{X} = \frac{1}{\bar{N}}\frac{\partial X}{\partial t}, $
(7) and
$ f'=\bar{N}\left.\frac{\partial f}{\partial N}\right|_{N=\bar{N}},\quad f''=\bar{N}^{2}\left.\frac{\partial^{2}f}{\partial N^{2}}\right|_{N=\bar{N}}, $
(8) etc.
To obtain the action up to the second order in perturbations, we first expand all the quantities in the following form:
$ Q=Q^{(0)}+\delta Q^{(1)}+\delta Q^{(2)}, $
(9) where
$ Q^{(0)} $ represents the background value of any quantity Q, and$ \delta Q^{(n)} $ represents the n-th order perturbation of Q. Precisely, we obtain$ \phi^{(0)}=\bar{\phi},\quad\delta\phi^{(1)}=\delta\phi, $
(10) $ N^{(0)}=\bar{N},\quad\delta N^{(1)}=\bar{N}A,\quad\delta N^{(2)}=\frac{1}{2}\bar{N}A^{2}, $
(11) $ N^{i(0)}=0,\quad\delta N^{i(1)}=\bar{N}a^{-1}\partial^{i}B, $
(12) $ h_{ij}^{(0)}=a^{2}\delta_{ij},\quad\delta h_{ij}^{(1)}=2\zeta a^{2}\delta_{ij},\quad\delta h_{ij}^{(2)}=2\zeta^{2}a^{2}\delta_{ij}, $
(13) $ K_{ij}^{(0)}=a\dot{a}\delta_{ij},\quad\delta K_{ij}^{(1)}=a\dot{a}\delta_{ij}\left(-A+2\zeta\right)+a^{2}\delta_{ij}\dot{\zeta}-a\partial_{i}\partial_{j}B, $
(14) $ \begin{aligned}[b] \delta K_{ij}^{(2)} = \; & a\dot{a}\delta_{ij}\left(-2\zeta A+\frac{1}{2}A^{2}+2\zeta^{2}\right)+a^{2}\delta_{ij}\dot{\zeta}\left(2\zeta-A\right) \\ & +aA\partial_{i}\partial_{j}B+a\left(\partial_{i}\zeta\partial_{j}B+\partial_{j}\zeta\partial_{i}B-\delta_{ij}\partial^{l}B\partial_{l}\zeta\right), \end{aligned} $
(15) $ R_{ij}^{(0)}=0,\quad\delta R_{ij}^{(1)}=-\partial_{i}\partial_{j}\zeta-\delta_{ij}\partial^{2}\zeta, $
(16) and
$ \delta R_{ij}^{(2)}=\partial_{i}\zeta\partial_{j}\zeta-\delta_{ij}\partial_{l}\zeta\partial^{l}\zeta. $
(17) In the above and henceforth, spatial indices are raised and lowered by
$ \delta^{ij} $ and$ \delta_{ij} $ , respectively. Note that only linear-order terms are considered for the perturbations of auxiliary scalar field ϕ and shift vector$ N^{i} $ . With the above results, we can proceed to calculate the perturbations of the action in the subsequent sections. -
We first derive the background equations of motion, which can be used to simplify calculations at higher orders. Hence, we must obtain linear-order action
$ S_{1} $ :$ S_{1}=\int{\rm{d}}t{\rm{d}}^{3}x\left.\frac{\delta S}{\delta\Phi_{I}}\right|_{{\rm{bg}}}\delta\Phi_{I}^{(1)}, $
(18) where
$ \Phi_{I}\equiv\left\{ h_{ij},K_{ij},R_{ij},N,\phi\right\} $
(19) stands for the basic quantities in our model (1), and ''
$ Q|_{{\rm{bg}}} $ '' denotes the background value of any quantity Q. We obtain$ \begin{aligned}[b] S_{1} = & \int{\rm{d}}t{\rm{d}}^{3}x\left[\bar{N}a^{3}\left.\frac{\delta{\cal{L}}}{\delta K_{ij}}\right|_{{\rm{bg}}}\delta K_{ij}^{(1)}+\left(\bar{N}\frac{1}{2}a\delta^{ij}\bar{{\cal{L}}}+\bar{N}a^{3}\left.\frac{\delta{\cal{L}}}{\delta h_{ij}}\right|_{{\rm{bg}}}\right)\right. \\ &\times\delta h_{ij}^{(1)}+\bar{N}a^{3}\left.\frac{\delta{\cal{L}}}{\delta\phi}\right|_{{\rm{bg}}}\delta\phi +\left(a^{3}\bar{{\cal{L}}}+\bar{N}a^{3}\left.\frac{\delta{\cal{L}}}{\delta N}\right|_{{\rm{bg}}}\right)\delta N^{(1)}\\ &\left.+\bar{N}a^{3}\left.\frac{\delta{\cal{L}}}{\delta R_{ij}}\right|_{{\rm{bg}}}\delta R_{ij}^{(1)}\right],\end{aligned} $
(20) where
$ \bar{{\cal{L}}} $ represents the background Lagrangian density. We emphasize that in (20) (and henceforth), when making the variation,$ h_{ij} $ ,$ K_{ij} $ ,$ R_{ij} $ , N, and ϕ are treated as independent variables2 . Note that at the background level,$ N=\bar{N} $ ,$ N_{i}=0 $ ,$ h_{ij}=a^{2}\delta_{ij} $ , and$ \phi=\bar{\phi} $ . Substituting (10)− (17) into (20), after some simplifications, we obtain$ S_{1}[A,\zeta,\delta\phi]=\int{\rm{d}}t{\rm{d}}^{3}x\bar{N}a^{3}\left({\cal{E}}_{A}A+{\cal{E}}_{\zeta}\zeta+{\cal{E}}_{\delta\phi}\delta\phi\right), $
(21) where we define
$ {\cal{E}}_{A}=\bar{{\cal{L}}}+\bar{N}\left.\frac{\delta{\cal{L}}}{\delta N}\right|_{{\rm{bg}}}-a^{2}H\delta_{ij}\left.\frac{\delta{\cal{L}}}{\delta K_{ij}}\right|_{{\rm{bg}}}, $
(22) with
$ H\equiv\dot{a}/a\equiv\partial_{t}a/(a\bar{N}) $ ,$ \begin{aligned}[b] {\cal{E}}_{\zeta}=\;&3\bar{{\cal{L}}}+a^{2}\delta_{ij}\Bigg[2\left.\frac{\delta{\cal{L}}}{\delta h_{ij}}\right|_{{\rm{bg}}}-3H\left.\frac{\delta{\cal{L}}}{\delta K_{ij}}\right|_{{\rm{bg}}}\\&-\frac{1}{\bar{N}}\frac{\partial}{\partial t}\left(\left.\frac{\delta{\cal{L}}}{\delta K_{ij}}\right|_{{\rm{bg}}}\right)-\frac{\dot{\bar{N}}}{\bar{N}}\left.\frac{\delta{\cal{L}}}{\delta K_{ij}}\right|_{{\rm{bg}}}\Bigg],\end{aligned} $
(23) and
$ {\cal{E}}_{\delta\phi}=\left.\frac{\delta{\cal{L}}}{\delta\phi}\right|_{{\rm{bg}}}. $
(24) Requiring the linear variation of the background action to be vanishing, i.e.,
$ S_{1}[A,\zeta,\delta\phi]=0 $ , we obtain$ {\cal{E}}_{A}=0,\quad{\cal{E}}_{\zeta}=0,\quad{\cal{E}}_{\delta\phi}=0, $
(25) which are the background equations of motion. At this point, note that because SCG explicitly breaks the time diffeomorphism, all three of these equations are independent. This differs from generally covariant theories, in which only two of the three equations are independent owing to the time diffeomorphism.
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The quadratic action for the perturbations receives contributions from both first-order and second-order quantities. Its general form is given by
$\begin{aligned}[b]\\[-10pt] S_{2}[A,B,\zeta,\delta\phi]=\int{\rm{d}}t{\rm{d}}^{3}x\left(\left.\frac{\delta S}{\delta\Phi_{I}}\right|_{{\rm{bg}}}\delta\Phi_{I}^{(2)}+\frac{1}{2}\left.\frac{\delta^{2}S}{\delta\Phi_{I}\delta\Phi_{J}}\right|_{{\rm{bg}}}\delta\Phi_{I}^{(1)}\delta\Phi_{J}^{(1)}\right),\\ \end{aligned} $ (26) where (henceforth, we suppress subscript ''bg'' for simplicity)
$ \int{\rm{d}}t{\rm{d}}^{3}x\frac{\delta S}{\delta\Phi_{I}}\delta\Phi_{I}^{(2)} = \int{\rm{d}}t{\rm{d}}^{3}x\left[\frac{\delta\left(N\sqrt{h}{\cal{L}}\right)}{\delta K_{ij}}\delta K_{ij}^{(2)}+\frac{\delta\left(N\sqrt{h}{\cal{L}}\right)}{\delta h_{ij}}\delta h_{ij}^{(2)}\right. \left.+\frac{\delta\left(N\sqrt{h}{\cal{L}}\right)}{\delta N}\delta N^{(2)}+\frac{\delta\left(N\sqrt{h}{\cal{L}}\right)}{\delta R_{ij}}\delta R_{ij}^{(2)}\right], $
(27) and
$ \begin{aligned}[b] \int{\rm{d}}t{\rm{d}}^{3}x\frac{\delta^{2}S}{\delta\Phi_{I}\delta\Phi_{J}}\delta\Phi_{I}^{(1)}\delta\Phi_{J}^{(1)} =\; & \int{\rm{d}}t{\rm{d}}^{3}x\Bigg[2\frac{\delta^{2}\left(N\sqrt{h}{\cal{L}}\right)}{\delta h_{mn}\delta K_{ij}}\delta h_{mn}^{(1)}\delta K_{ij}^{(1)}+\frac{\delta^{2}\left(N\sqrt{h}{\cal{L}}\right)}{\delta h_{mn}\delta h_{ij}}\delta h_{mn}^{(1)}\delta h_{ij}^{(1)}+\frac{\delta^{2}\left(N\sqrt{h}{\cal{L}}\right)}{\delta K_{mn}\delta K_{ij}}\delta K_{mn}^{(1)}\delta K_{ij}^{(1)} \\ & +2\frac{\delta^{2}\left(N\sqrt{h}{\cal{L}}\right)}{\delta\phi\delta K_{ij}}\delta\phi\delta K_{ij}^{(1)}+2\frac{\delta^{2}\left(N\sqrt{h}{\cal{L}}\right)}{\delta N\delta K_{ij}}\delta N^{(1)}\delta K_{ij}^{(1)}+2\frac{\delta^{2}\left(N\sqrt{h}{\cal{L}}\right)}{\delta\phi\delta h_{ij}}\delta\phi\delta h_{ij}^{(1)} \end{aligned} $
$ \begin{align} \quad\quad & +2\frac{\delta^{2}\left(N\sqrt{h}{\cal{L}}\right)}{\delta N\delta h_{ij}}\delta N^{(1)}\delta h_{ij}^{(1)}+2\frac{\delta^{2}\left(N\sqrt{h}{\cal{L}}\right)}{\delta\phi\delta N}\delta\phi\delta N^{(1)}+\frac{\delta^{2}\left(N\sqrt{h}{\cal{L}}\right)}{\delta\phi^{2}}\delta\phi\delta\phi +\frac{\delta^{2}\left(N\sqrt{h}{\cal{L}}\right)}{\delta N^{2}}\delta N^{(1)}\delta N^{(1)}+2\frac{\delta^{2}\left(N\sqrt{h}{\cal{L}}\right)}{\delta h_{mn}\delta R_{ij}}\delta h_{mn}^{(1)}\delta R_{ij}^{(1)}\\ & +2\frac{\delta^{2}\left(N\sqrt{h}{\cal{L}}\right)}{\delta K_{mn}\delta R_{ij}}\delta K_{mn}^{(1)}\delta R_{ij}^{(1)} +2\frac{\delta^{2}\left(N\sqrt{h}{\cal{L}}\right)}{\delta N\delta R_{ij}}\delta N^{(1)}\delta R_{ij}^{(1)}+\frac{\delta^{2}\left(N\sqrt{h}{\cal{L}}\right)}{\delta R_{mn}\delta R_{ij}}\delta R_{mn}^{(1)}\delta R_{ij}^{(1)}+2\frac{\delta^{2}\left(N\sqrt{h}{\cal{L}}\right)}{\delta\phi\delta R_{ij}}\delta\phi\delta R_{ij}^{(1)}\Bigg]. \end{align} $
(28) The quadratic action for the scalar perturbations takes the general form
$ \begin{aligned}[b] S_{2}\left[A,B,\zeta,\delta\phi\right] =\; & \int{\rm{d}}t{\rm{d}}^{3}x\bar{N}a^{3}\left(A\hat{{\cal{O}}}_{AA}A+B\hat{{\cal{O}}}_{BB}B+\zeta\hat{{\cal{O}}}_{\zeta\zeta}\zeta+\delta\phi\hat{{\cal{O}}}_{\phi\phi}\delta\phi+A\hat{{\cal{O}}}_{AB}B+A\hat{{\cal{O}}}_{A\zeta}\zeta+A\hat{{\cal{O}}}_{A\phi}\delta\phi\right. \\ & \left.+\delta\phi\hat{{\cal{O}}}_{\phi\zeta}\zeta+\delta\phi\hat{{\cal{O}}}_{\phi B}B+\zeta\hat{{\cal{O}}}_{\zeta B}B+\dot{\zeta}\hat{{\cal{O}}}_{\dot{\zeta}A}A+\dot{\zeta}\hat{{\cal{O}}}_{\dot{\zeta}B}B+\dot{\zeta}\hat{{\cal{O}}}_{\dot{\zeta}\zeta}\zeta+\dot{\zeta}\hat{{\cal{O}}}_{\dot{\zeta}\phi}\delta\phi+\dot{\zeta}\hat{{\cal{O}}}_{\dot{\zeta}\dot{\zeta}}\dot{\zeta}\right), \end{aligned} $
(29) where
$ \hat{{\cal{O}}}_{XY} $ represents the operator involving perturbation variables X and Y (e.g.,$ \hat{{\cal{O}}}_{A\phi} $ represents the operator involving A and$ \delta\phi $ ). Generally,$ \hat{{\cal{O}}}_{XY} $ is time dependent and may contain spatial derivatives. By substituting (10)−(17) into (27) and (28) and after some tedious manipulations, we obtain$ \hat{{\cal{O}}}_{AA}=\frac{1}{2}\bar{{\cal{L}}}-\frac{1}{2}a\dot{a}\delta_{ij}\frac{\delta{\cal{L}}}{\delta K_{ij}}+\frac{1}{2}a^{2}\dot{a}^{2}\delta_{mn}\delta_{ij}\frac{\delta^{2}{\cal{L}}}{\delta K_{mn}\delta K_{ij}}+\frac{3}{2}\bar{N}\frac{\delta{\cal{L}}}{\delta N}-\bar{N}a\dot{a}\delta_{ij}\frac{\delta^{2}{\cal{L}}}{\delta N\delta K_{ij}}+\frac{1}{2}\bar{N}^{2}\frac{\delta^{2}{\cal{L}}}{\delta N^{2}}, $
(30) $ \hat{{\cal{O}}}_{BB}=\frac{1}{2}a^{2}\frac{\delta^{2}{\cal{L}}}{\delta K_{mn}\delta K_{ij}}\partial_{m}\partial_{n}\partial_{i}\partial_{j}, $
(31) $ \begin{aligned}[b] \hat{{\cal{O}}}_{\zeta\zeta} =\; & \frac{9}{2}\bar{{\cal{L}}}+8a^{2}\delta_{ij}\frac{\delta{\cal{L}}}{\delta h_{ij}}+2a^{4}\delta_{mn}\delta_{ij}\frac{\delta^{2}{\cal{L}}}{\delta h_{mn}\delta h_{ij}}+8a\dot{a}\delta_{ij}\frac{\delta{\cal{L}}}{\delta K_{ij}} +4a^{3}\dot{a}\delta_{ij}\delta_{mn}\frac{\delta^{2}{\cal{L}}}{\delta h_{mn}\delta K_{ij}}+2a^{2}\dot{a}^{2}\delta_{mn}\delta_{ij}\frac{\delta^{2}{\cal{L}}}{\delta K_{mn}\delta K_{ij}} \\ & +\frac{1}{2}\frac{\delta^{2}{\cal{L}}}{\delta R_{mn}\delta R_{ij}}\left(\partial_{i}\partial_{j}\partial_{m}\partial_{n}+2\delta_{ij}\partial_{m}\partial_{n}\partial^{2}+\delta_{ij}\delta_{mn}\partial^{4}\right)-2a\dot{a}\frac{\delta^{2}{\cal{L}}}{\delta K_{mn}\delta R_{ij}}\delta_{mn}\partial_{i}\partial_{j} -2a^{2}\delta_{mn}\frac{\delta^{2}{\cal{L}}}{\delta h_{mn}\delta R_{ij}}\left(\partial_{i}\partial_{j}+\delta_{ij}\partial^{2}\right) \\ & -2\frac{\delta{\cal{L}}}{\delta R_{ij}}\left(2\partial_{i}\partial_{j}+\delta_{ij}\partial^{2}\right), \end{aligned} $
(32) $ \hat{{\cal{O}}}_{\phi\phi}=\frac{1}{2}\frac{\delta^{2}{\cal{L}}}{\delta\phi^{2}}, $
(33) $ \hat{{\cal{O}}}_{AB}=\left(-\bar{N}a\frac{\delta^{2}{\cal{L}}}{\delta N\delta K_{ij}}+a^{2}\dot{a}\delta_{mn}\frac{\delta^{2}{\cal{L}}}{\delta K_{mn}\delta K_{ij}}\right)\partial_{i}\partial_{j}, $
(34) $ \begin{aligned}[b] \hat{{\cal{O}}}_{A\zeta} =\; & 3\bar{{\cal{L}}}+2a^{2}\delta_{ij}\frac{\delta{\cal{L}}}{\delta h_{ij}}+3\bar{N}\frac{\delta{\cal{L}}}{\delta N}+2\bar{N}a^{2}\delta_{ij}\frac{\delta^{2}{\cal{L}}}{\delta N\delta h_{ij}}-3a\dot{a}\delta_{ij}\frac{\delta{\cal{L}}}{\delta K_{ij}} -2a^{3}\dot{a}\delta_{ij}\delta_{mn}\frac{\delta^{2}{\cal{L}}}{\delta h_{mn}\delta K_{ij}}+2\bar{N}a\dot{a}\delta_{ij}\frac{\delta^{2}{\cal{L}}}{\delta N\delta K_{ij}}\\ &-2a^{2}\dot{a}^{2}\delta_{mn}\delta_{ij}\frac{\delta^{2}{\cal{L}}}{\delta K_{mn}\delta K_{ij}} -\left(\frac{\delta{\cal{L}}}{\delta R_{ij}}+\bar{N}\frac{\delta^{2}{\cal{L}}}{\delta N\delta R_{ij}}\right)\left(\partial_{i}\partial_{j}+\delta_{ij}\partial^{2}\right)+a\dot{a}\delta_{mn}\frac{\delta^{2}{\cal{L}}}{\delta K_{mn}\delta R_{ij}}\partial_{i}\partial_{j}, \end{aligned} $
(35) $ \hat{{\cal{O}}}_{A\phi}=\frac{\delta{\cal{L}}}{\delta\phi}+\bar{N}\frac{\delta^{2}{\cal{L}}}{\delta\phi\delta N}-a\dot{a}\delta_{ij}\frac{\delta^{2}{\cal{L}}}{\delta\phi\delta K_{ij}}, $
(36) $ \hat{{\cal{O}}}_{\phi\zeta}=3\frac{\delta{\cal{L}}}{\delta\phi}+2a^{2}\delta_{ij}\frac{\delta^{2}{\cal{L}}}{\delta\phi\delta h_{ij}}+2a\dot{a}\delta_{ij}\frac{\delta^{2}{\cal{L}}}{\delta\phi\delta K_{ij}}-\frac{\delta^{2}{\cal{L}}}{\delta\phi\delta R_{ij}}\left(\partial_{i}\partial_{j}+\delta_{ij}\partial^{2}\right), $
(37) $ \hat{{\cal{O}}}_{\phi B}=-a\frac{\delta^{2}{\cal{L}}}{\delta\phi\delta K_{ij}}\partial_{i}\partial_{j}, $
(38) $ \begin{align} \hat{{\cal{O}}}_{\zeta B} = -5a\frac{\delta{\cal{L}}}{\delta K_{ij}}\partial_{i}\partial_{j}+a\frac{\delta{\cal{L}}}{\delta K_{ij}}\delta_{ij}\partial^{2}-2a^{3}\delta_{mn}\frac{\delta^{2}{\cal{L}}}{\delta h_{mn}\delta K_{ij}}\partial_{i}\partial_{j}-2a^{2}\dot{a}\delta_{mn}\frac{\delta^{2}{\cal{L}}}{\delta K_{mn}\delta K_{ij}}\partial_{i}\partial_{j} +a\frac{\delta^{2}{\cal{L}}}{\delta K_{mn}\delta R_{ij}}\left(\partial_{i}\partial_{j}\partial_{m}\partial_{n}+\delta_{ij}\partial^{2}\partial_{m}\partial_{n}\right), \end{align} $
(39) $ \hat{{\cal{O}}}_{\dot{\zeta}A}=-a^{3}\dot{a}\delta_{mn}\delta_{ij}\frac{\delta^{2}{\cal{L}}}{\delta K_{mn}\delta K_{ij}}+\bar{N}a^{2}\delta_{ij}\frac{\delta^{2}{\cal{L}}}{\delta N\delta K_{ij}}, $
(40) $ \hat{{\cal{O}}}_{\dot{\zeta}B}=-a^{3}\frac{\delta^{2}{\cal{L}}}{\delta K_{mn}\delta K_{ij}}\delta_{mn}\partial_{i}\partial_{j}, $
(41) $ \hat{{\cal{O}}}_{\dot{\zeta}\zeta}=5a^{2}\delta_{ij}\frac{\delta{\cal{L}}}{\delta K_{ij}}+2a^{4}\delta_{ij}\delta_{mn}\frac{\delta^{2}{\cal{L}}}{\delta h_{mn}\delta K_{ij}}+2a^{3}\dot{a}\delta_{mn}\delta_{ij}\frac{\delta^{2}{\cal{L}}}{\delta K_{mn}\delta K_{ij}}-a^{2}\delta_{mn}\frac{\delta^{2}{\cal{L}}}{\delta K_{mn}\delta R_{ij}}\left(\partial_{i}\partial_{j}+\delta_{ij}\partial^{2}\right), $
(42) $ \hat{{\cal{O}}}_{\dot{\zeta}\phi}=a^{2}\delta_{ij}\frac{\delta^{2}{\cal{L}}}{\delta\phi\delta K_{ij}}, $
(43) and
$ \hat{{\cal{O}}}_{\dot{\zeta}\dot{\zeta}}=\frac{1}{2}a^{4}\frac{\delta^{2}{\cal{L}}}{\delta K_{mn}\delta K_{ij}}\delta_{mn}\delta_{ij}. $
(44) By deriving the above results, we utilize integrations by parts to simplify the expressions.
By varying the quadratic action (29), one can easily obtain the equations of motion for perturbation variables
$ \left\{ A,B,\delta\phi,\zeta\right\} $ as$ 2\hat{{\cal{O}}}_{AA}A+\hat{{\cal{O}}}_{AB}B+\hat{{\cal{O}}}_{A\phi}\delta\phi+\hat{{\cal{O}}}_{A\zeta}\zeta+\hat{{\cal{O}}}_{\dot{\zeta}A}\dot{\zeta} = 0, $
(45) $ \hat{{\cal{O}}}_{AB}A+2\hat{{\cal{O}}}_{BB}B+\hat{{\cal{O}}}_{\phi B}\delta\phi+\hat{{\cal{O}}}_{\zeta B}\zeta+\hat{{\cal{O}}}_{\dot{\zeta}B}\dot{\zeta} = 0, $
(46) $ \hat{{\cal{O}}}_{A\phi}A+\hat{{\cal{O}}}_{\phi B}B+2\hat{{\cal{O}}}_{\phi\phi}\delta\phi+\hat{{\cal{O}}}_{\phi\zeta}\zeta+\hat{{\cal{O}}}_{\dot{\zeta}\phi}\dot{\zeta} = 0, $
(47) $ \begin{aligned}[b] &\frac{1}{\bar{N}}\Bigg[2\partial_{t}\left(a^{3}\hat{{\cal{O}}}_{\dot{\zeta}\dot{\zeta}}\dot{\zeta}\right)+ \partial_{t}\left(a^{3}\hat{{\cal{O}}}_{\dot{\zeta}\phi}\delta\phi\right)\\&+\partial_{t}\left(a^{3}\hat{{\cal{O}}}_{\dot{\zeta}B}B\right)+\partial_{t}\left(a^{3}\hat{{\cal{O}}}_{\dot{\zeta}A}A\right)+\partial_{t}\left(a^{3}\hat{{\cal{O}}}_{\dot{\zeta}\zeta}\zeta\right)\Bigg] \\ & -a^{3}\left(2\hat{{\cal{O}}}_{\zeta\zeta}\zeta+\hat{{\cal{O}}}_{A\zeta}A+\hat{{\cal{O}}}_{\phi\zeta}\delta\phi+\hat{{\cal{O}}}_{\zeta B}B\right) = 0. \end{aligned} $
(48) Equations (45)−(48) show that A, B, and
$ \delta\phi $ are all auxiliary variables, as the equations of motion do not contain their first-order time derivative terms. In the case of a non-degenerate coefficient matrix for perturbation variables A, B, and$ \delta\phi $ , i.e.,$ \varOmega\equiv\det\left(\begin{array}{*{20}{c}}{ 2\hat{{\cal{O}}}_{AA} }&{ \hat{{\cal{O}}}_{AB} }&{ \hat{{\cal{O}}}_{A\phi}}\\ {\hat{{\cal{O}}}_{AB} }&{ 2\hat{{\cal{O}}}_{BB} }&{ \hat{{\cal{O}}}_{\phi B}}\\ {\hat{{\cal{O}}}_{A\phi} }&{ \hat{{\cal{O}}}_{\phi B} }&{ 2\hat{{\cal{O}}}_{\phi\phi}} \end{array}\right)\neq0, $
(49) they can be formally solved from their equations of motion (45)−(47) as
$ \begin{aligned}[b] A =\; & -\frac{1}{2\varOmega}\Big[\Big(\hat{{\cal{O}}}_{A\zeta}\hat{{\cal{O}}}_{\phi B}^{2}-\hat{{\cal{O}}}_{AB}\hat{{\cal{O}}}_{\phi B}\hat{{\cal{O}}}_{\phi\zeta}-\hat{{\cal{O}}}_{A\phi}\hat{{\cal{O}}}_{\phi B}\hat{{\cal{O}}}_{\zeta B}+2\hat{{\cal{O}}}_{A\phi}\hat{{\cal{O}}}_{BB}\hat{{\cal{O}}}_{\phi\zeta}+2\hat{{\cal{O}}}_{AB}\hat{{\cal{O}}}_{\phi\phi}\hat{{\cal{O}}}_{\zeta B} -4\hat{{\cal{O}}}_{A\zeta}\hat{{\cal{O}}}_{BB}\hat{{\cal{O}}}_{\phi\phi}\Big)\zeta\\ & +\Big(\hat{{\cal{O}}}_{\dot{\zeta}A}\hat{{\cal{O}}}_{\phi B}^{2}-\hat{{\cal{O}}}_{AB}\hat{{\cal{O}}}_{\phi B}\hat{{\cal{O}}}_{\dot{\zeta}\phi}-\hat{{\cal{O}}}_{A\phi}\hat{{\cal{O}}}_{\phi B}\hat{{\cal{O}}}_{\dot{\zeta}B} +2\hat{{\cal{O}}}_{A\phi}\hat{{\cal{O}}}_{BB}\hat{{\cal{O}}}_{\dot{\zeta}\phi}+2\hat{{\cal{O}}}_{AB}\hat{{\cal{O}}}_{\phi\phi}\hat{{\cal{O}}}_{\dot{\zeta}B}-4\hat{{\cal{O}}}_{\dot{\zeta}A}\hat{{\cal{O}}}_{BB}\hat{{\cal{O}}}_{\phi\phi}\Big)\dot{\zeta}\Big], \end{aligned} $ (50) $ \begin{aligned}[b] B =\; & -\frac{1}{2\varOmega}\Big[\Big(\hat{{\cal{O}}}_{\zeta B}\hat{{\cal{O}}}_{A\phi}^{2}-\hat{{\cal{O}}}_{AB}\hat{{\cal{O}}}_{A\phi}\hat{{\cal{O}}}_{\phi\zeta}-\hat{{\cal{O}}}_{A\phi}\hat{{\cal{O}}}_{\phi B}\hat{{\cal{O}}}_{A\zeta}+2\hat{{\cal{O}}}_{\phi B}\hat{{\cal{O}}}_{AA}\hat{{\cal{O}}}_{\phi\zeta}+2\hat{{\cal{O}}}_{AB}\hat{{\cal{O}}}_{\phi\phi}\hat{{\cal{O}}}_{A\zeta} -4\hat{{\cal{O}}}_{\zeta B}\hat{{\cal{O}}}_{AA}\hat{{\cal{O}}}_{\phi\phi}\Big)\zeta\\ & +\Big(\hat{{\cal{O}}}_{\dot{\zeta}B}\hat{{\cal{O}}}_{A\phi}^{2}-\hat{{\cal{O}}}_{AB}\hat{{\cal{O}}}_{A\phi}\hat{{\cal{O}}}_{\dot{\zeta}\phi}-\hat{{\cal{O}}}_{A\phi}\hat{{\cal{O}}}_{\phi B}\hat{{\cal{O}}}_{\dot{\zeta}A} +2\hat{{\cal{O}}}_{\phi B}\hat{{\cal{O}}}_{AA}\hat{{\cal{O}}}_{\dot{\zeta}\phi}+2\hat{{\cal{O}}}_{AB}\hat{{\cal{O}}}_{\phi\phi}\hat{{\cal{O}}}_{\dot{\zeta}A}-4\hat{{\cal{O}}}_{\dot{\zeta}B}\hat{{\cal{O}}}_{AA}\hat{{\cal{O}}}_{\phi\phi}\Big)\dot{\zeta}\Big], \end{aligned} $
(51) $ \begin{aligned}[b] \delta\phi =\; & -\frac{1}{2\varOmega}\Big[\Big(\hat{{\cal{O}}}_{\phi\zeta}\hat{{\cal{O}}}_{AB}^{2}-\hat{{\cal{O}}}_{AB}\hat{{\cal{O}}}_{A\zeta}\hat{{\cal{O}}}_{\phi B}-\hat{{\cal{O}}}_{AB}\hat{{\cal{O}}}_{A\phi}\hat{{\cal{O}}}_{\zeta B}+2\hat{{\cal{O}}}_{\phi B}\hat{{\cal{O}}}_{AA}\hat{{\cal{O}}}_{\zeta B}+2\hat{{\cal{O}}}_{BB}\hat{{\cal{O}}}_{A\phi}\hat{{\cal{O}}}_{A\zeta} -4\hat{{\cal{O}}}_{BB}\hat{{\cal{O}}}_{AA}\hat{{\cal{O}}}_{\phi\zeta}\Big)\zeta\\ & +\Big(\hat{{\cal{O}}}_{\dot{\zeta}\phi}\hat{{\cal{O}}}_{AB}^{2}-\hat{{\cal{O}}}_{AB}\hat{{\cal{O}}}_{\dot{\zeta}A}\hat{{\cal{O}}}_{\phi B}-\hat{{\cal{O}}}_{AB}\hat{{\cal{O}}}_{A\phi}\hat{{\cal{O}}}_{\dot{\zeta}B} +2\hat{{\cal{O}}}_{\phi B}\hat{{\cal{O}}}_{AA}\hat{{\cal{O}}}_{\dot{\zeta}B}+2\hat{{\cal{O}}}_{BB}\hat{{\cal{O}}}_{A\phi}\hat{{\cal{O}}}_{\dot{\zeta}A}-4\hat{{\cal{O}}}_{BB}\hat{{\cal{O}}}_{AA}\hat{{\cal{O}}}_{\dot{\zeta}\phi}\Big)\dot{\zeta}\Big],\end{aligned} $
(52) with
$ \begin{aligned}[b] \varOmega=\;&\hat{{\cal{O}}}_{A\phi}^{2}\hat{{\cal{O}}}_{BB}+\hat{{\cal{O}}}_{AA}\hat{{\cal{O}}}_{\phi B}^{2}+\hat{{\cal{O}}}_{AB}^{2}\hat{{\cal{O}}}_{\phi\phi}-\hat{{\cal{O}}}_{AB}\hat{{\cal{O}}}_{A\phi}\hat{{\cal{O}}}_{\phi B}\\&-4\hat{{\cal{O}}}_{AA}\hat{{\cal{O}}}_{BB}\hat{{\cal{O}}}_{\phi\phi}. \end{aligned} $
(53) Although we assume that
$ \varOmega\neq0 $ to ensure that auxiliary variables A, B, and$ \delta\phi $ are solvable (at least formally), note that even if$ \varOmega\neq0 $ is not satisfied, we can still obtain conditions to eliminate the unwanted scalar degree of freedom. We will show this explicitly through a detailed analysis of a specific example in the next section.Finally, by substituting solutions (50)−(52) into the equation of motion for ζ (48), we obtain an equation of motion for single variable ζ. If no further conditions are assumed, the effective equation of motion for ζ will contain its second-order time derivative term,
$ \ddot{\zeta} $ , indicating a propagating degree of freedom carried by ζ. For our purpose, to eliminate the scalar degree of freedom, we must ensure that ζ is not propagating, at least at linear order in a cosmological background. To this end, we must make the coefficient of$ \ddot{\zeta} $ vanish. After some manipulations, the coefficient of$ \ddot{\zeta} $ is found to be$ \begin{aligned}[b] \varDelta =\; & 4\hat{{\cal{O}}}_{\dot{\zeta}\dot{\zeta}}\left(\hat{{\cal{O}}}_{A\phi}^{2}\hat{{\cal{O}}}_{BB}+\hat{{\cal{O}}}_{AA}\hat{{\cal{O}}}_{\phi B}^{2}+\hat{{\cal{O}}}_{AB}^{2}\hat{{\cal{O}}}_{\phi\phi}-\hat{{\cal{O}}}_{AB}\hat{{\cal{O}}}_{A\phi}\hat{{\cal{O}}}_{\phi B}-4\hat{{\cal{O}}}_{AA}\hat{{\cal{O}}}_{BB}\hat{{\cal{O}}}_{\phi\phi}\right) -\hat{{\cal{O}}}_{\phi B}^{2}\hat{{\cal{O}}}_{\dot{\zeta}A}^{2}-\hat{{\cal{O}}}_{A\phi}^{2}\hat{{\cal{O}}}_{\dot{\zeta}B}^{2}-\hat{{\cal{O}}}_{AB}^{2}\hat{{\cal{O}}}_{\dot{\zeta}\phi}^{2}+2\hat{{\cal{O}}}_{A\phi}\hat{{\cal{O}}}_{\phi B}\hat{{\cal{O}}}_{\dot{\zeta}A}\hat{{\cal{O}}}_{\dot{\zeta}B}\\ &+2\hat{{\cal{O}}}_{AB}\hat{{\cal{O}}}_{\phi B}\hat{{\cal{O}}}_{\dot{\zeta}A}\hat{{\cal{O}}}_{\dot{\zeta}\phi}+2\hat{{\cal{O}}}_{AB}\hat{{\cal{O}}}_{A\phi}\hat{{\cal{O}}}_{\dot{\zeta}B}\hat{{\cal{O}}}_{\dot{\zeta}\phi} +4\hat{{\cal{O}}}_{BB}\hat{{\cal{O}}}_{\phi\phi}\hat{{\cal{O}}}_{\dot{\zeta}A}^{2}+4\hat{{\cal{O}}}_{AA}\hat{{\cal{O}}}_{BB}\hat{{\cal{O}}}_{\dot{\zeta}\phi}^{2}+4\hat{{\cal{O}}}_{AA}\hat{{\cal{O}}}_{\phi\phi}\hat{{\cal{O}}}_{\dot{\zeta}B}^{2} \\ & -4\hat{{\cal{O}}}_{AA}\hat{{\cal{O}}}_{\phi B}\hat{{\cal{O}}}_{\dot{\zeta}B}\hat{{\cal{O}}}_{\dot{\zeta}\phi}-4\hat{{\cal{O}}}_{AB}\hat{{\cal{O}}}_{\phi\phi}\hat{{\cal{O}}}_{\dot{\zeta}A}\hat{{\cal{O}}}_{\dot{\zeta}B}-4\hat{{\cal{O}}}_{A\phi}\hat{{\cal{O}}}_{BB}\hat{{\cal{O}}}_{\dot{\zeta}A}\hat{{\cal{O}}}_{\dot{\zeta}\phi}.\end{aligned} $ (54) Therefore, we require that
$ \varDelta=0, $
(55) which is a necessary condition to eliminate the scalar degree of freedom (i.e., the TTDOF conditions), at least at the linear order in perturbations around a cosmological background.
-
The analysis and conditions obtained in the above section are general but quite formal. In this section, we apply condition (54) to a concrete model as an example. According to the classification of SCG monomials [60, 61], we consider a polynomial-type Lagrangian built of monomials of
$ d=2 $ , where d is the total number of derivatives. Precisely, the action is given by$ S_{2}=\int {\rm d} t {\rm d}^{3}xN\sqrt{h}\left({\cal{L}}-\Lambda\right), $
(56) with
$ {\cal{L}}=c_{1}K_{ij}K^{ij}+c_{2}K^{2}+c_{3}R+c_{4}a_{i}a^{i}+d_{1}{\rm{D}}_{i}\phi{\rm{D}}^{i}\phi+d_{2}a_{i}{\rm{D}}^{i}\phi, $
(57) where
$ c_{i} $ and$ d_{i} $ are the general functions of N and ϕ, and Λ is the cosmological constant. Additionally, acceleration$ a_{i} $ is defined as$ a_{i}=\partial_{i}\ln N $ . We make the assumption that both ϕ and N are homogeneous and isotropic at the background, i.e.,$ \bar{\phi}=\bar{\phi}(t) $ ,$ \bar{N}=\bar{N}(t) $ . To simplify our calculation, we denote$ {\cal{L}}_{c}={\cal{L}}-\Lambda. $
(58) We will observe that non-vanishing cosmological constant Λ, which may be time dependent, is necessary to have a cosmological background solution.
-
As both ϕ and N depend only on time at the background level, the background values of each quantity in the Lagrangian are given by
$ \bar{K}_{ij}=a\dot{a}\delta_{ij},\quad\bar{K}^{ij}=Ha^{-2}\delta^{ij},\quad\bar{K}=3H, $
(59) and
$ \bar{R}=0,\quad\bar{a}_{i}=0,\quad\partial_{i}\bar{\phi}=0. $
(60) The background value of the Lagrangian is
$ \bar{{\cal{L}}_{c}}=3H^{2}b-\Lambda, $
(61) where we define
$ b=c_{1}+3c_{2} $
(62) for short. Thus, the non-vanishing derivatives of Lagrangian
$ {\cal{L}}_{c} $ with respect to various quantities are$ \frac{\delta{\cal{L}}_{c}}{\delta N}=\frac{3}{\bar{N}}H^{2}b', $
(63) $ \frac{\delta{\cal{L}}_{c}}{\delta K_{ij}}=2Ha^{-2}\delta^{ij}b, $
(64) $ \frac{\delta{\cal{L}}_{c}}{\delta\phi}=3H^{2}\frac{\partial b}{\partial\phi}, $
(65) $ \frac{\delta^{2}{\cal{L}}_{c}}{\delta N^{2}}=\frac{3}{\bar{N}^{2}}H^{2}b''-\frac{2}{\bar{N}^{2}}c_{4}\partial^{2}, $
(66) $ \frac{\delta^{2}{\cal{L}}_{c}}{\delta K_{mn}\delta K_{ij}}=2a^{-4}\left(\frac{1}{2}c_{1}\left(\delta^{im}\delta^{jn}+\delta^{in}\delta^{jm}\right)+c_{2}\delta^{ij}\delta^{mn}\right), $
(67) $ \frac{\delta^{2}{\cal{L}}_{c}}{\delta\phi^{2}}=3H^{2}\frac{\partial^{2}b}{\partial\phi^{2}}-2d_{1}\partial^{2}, $
(68) $ \frac{\delta^{2}{\cal{L}}_{c}}{\delta N\delta K_{ij}}=\frac{2}{\bar{N}}Ha^{-2}\delta^{ij}b', $
(69) $ \frac{\delta^{2}{\cal{L}}_{c}}{\delta\phi\delta N}=3H^{2}\frac{\partial^{2}b}{\partial\phi\partial N}-\frac{d_{2}}{\bar{N}}\partial^{2}, $
(70) and
$ \frac{\delta^{2}{\cal{L}}_{c}}{\delta\phi\delta K_{ij}}=2Ha^{-2}\delta^{ij}\frac{\partial b}{\partial\phi}. $
(71) We do not present derivatives with respect to
$ R_{ij} $ , as they are irrelevant to the degeneracy analysis. According to (61)−(71), the background equations of motion (25) are explicitly given by$ {\cal{E}}_{A}=3H^{2}\left(-b+b'\right)-\Lambda=0, $
(72) $ {\cal{E}}_{\zeta}=-9H^{2}b-6\dot{H}b-\frac{6H}{\bar{N}}\dot{\bar{N}}b-3\Lambda=0, $
(73) $ {\cal{E}}_{\delta\phi}=3H^{2}\frac{\partial b}{\partial\phi}=0, $
(74) From (74), to have a homogeneous and isotropic background, b defined in (62) must be a function of lapse function N only. Background equations of motion (72)−(74) can significantly simplify the calculation of the quadratic action for the perturbations in the following.
By substituting (61)−(71) into (30)−(44), we can evaluate operators
$ \hat{{\cal{O}}}_{XY} $ that are relevant to eliminating the scalar mode, which are given by$ \hat{{\cal{O}}}_{AA}=\frac{3}{2}H^{2}\left(2b-2b'+b''\right)-c_{4}\partial^{2}, $
(75) $ \hat{{\cal{O}}}_{BB}=a^{-2}\left(c_{1}+c_{2}\right)\partial^{4}, $
(76) $ \hat{{\cal{O}}}_{\phi\phi}=\frac{3}{2}H^{2}\frac{\partial^{2}b}{\partial\phi^{2}}-d_{1}\partial^{2}, $
(77) $ \hat{{\cal{O}}}_{AB}=2Ha^{-1}\left(-b'+b\right)\partial^{2}, $
(78) $ \hat{{\cal{O}}}_{A\phi}=3H^{2}\frac{\partial\left(b'-b\right)}{\partial\phi}-d_{2}\partial^{2}, $
(79) $ \hat{{\cal{O}}}_{\phi B}=-2Ha^{-1}\frac{\partial b}{\partial\phi}\partial^{2}, $
(80) $ \hat{{\cal{O}}}_{\dot{\zeta}A}=6H\left(-b+b'\right), $
(81) $ \hat{{\cal{O}}}_{\dot{\zeta}B}=-2a^{-1}b\partial^{2}, $
(82) $ \hat{{\cal{O}}}_{\dot{\zeta}\phi}=6H\frac{\partial b}{\partial\phi}, $
(83) and
$ \hat{{\cal{O}}}_{\dot{\zeta}\dot{\zeta}}=3b. $
(84) Subsequently, by substituting (75)−(84) into the determinant of coefficient matrix (53), and recalling that
$ \dfrac{\partial b}{\partial\phi}=0 $ from background equation of motion (74), we find$ \begin{aligned}[b] \varOmega =\;& a^{-2}\left(c_{1}+c_{2}\right)\left(d_{2}^{2}-4d_{1}c_{4}\right)\partial^{8} \\ & +2H^{2}a^{-2}\left[3\left(2b-2b'+b''\right)\left(c_{1}+c_{2}\right)-2\left(-b'+b\right)^{2}\right]d_{1}\partial^{6}. \end{aligned} $
(85) Degeneracy condition (55) is expressed as
$ \begin{aligned}[b] 0\equiv\varDelta = \;& 8a^{-2}bc_{1}\left(d_{2}^{2}-4c_{4}d_{1}\right)\partial^{8} \\ & +48H^{2}a^{-2}d_{1}c_{1}\left(2bb'-2b'^{2}+bb''\right)\partial^{6}. \end{aligned} $
(86) In deriving (85) and (86), we use background equations of motion (72)−(74) to simplify the expressions.
In the following, we discuss two cases based on whether
$ \varOmega $ is vanishing or not. -
When
$ \varOmega\neq0 $ , all auxiliary variables A, B,$ \delta\phi $ are solvable; thus, the degeneracy condition$ \varDelta=0 $ holds. According to (86), to eliminate the unwanted scalar degree of freedom, we require the coefficients of operators$ \partial^{6} $ and$ \partial^{8} $ to be vanishing simultaneously, i.e.,$ \left(2bb'-2b'^{2}+bb''\right)c_{1}d_{1}=0, $
(87) and
$ bc_{1}\left(d_{2}^{2}-4c_{4}d_{1}\right)=0, $
(88) which are two constraints among
$ c_{1} $ ,$ c_{4} $ ,$ d_{1} $ ,$ d_{2} $ , and b. Before solving these two equations, let us conduct a brief analysis.First, (87) and (88) are trivially satisfied if
$b\equiv c_{1}+ 3c_{2}=0$ . However, this conflicts with background equation of motion (72), which leads to$ \Lambda=0 $ . As mentioned above, a non-vanishing (and actually positive) cosmological constant is required to obtain a cosmological background solution. Therefore, we must require$ b\neq0 $ . Second, (87) and (88) are also satisfied if$ c_{1}=0 $ . However, although we focus only on the scalar perturbations in this study, further calculation reveals that only$ c_{1}K_{ij}K^{ij} $ will contribute to the kinetic term for the tensor perturbations (i.e., the gravitational waves). As a result, vanishing$ c_{1} $ would also eliminate the gravitational waves, which is unacceptable based on the observation of gravitational waves. Thus, we must also require$ c_{1}\neq0 $ . After considering that$ b,c_{1}\neq0 $ , (88) holds only if$d_{2}^{2}- 4c_{4}d_{1}=0.$ Finally, if$ d_{1}=0 $ , it follows that$ d_{2}=0 $ . From (85), this will lead to$ \varOmega=0 $ , which conflicts with our requirement$ \varOmega\neq0 $ . Thus, we must have$ d_{1}\neq0 $ .Based on the above arguments, equations (87)−(88) hold only if
$ \begin{align} d_{2}^{2}-4c_{4}d_{1} & =0, \end{align} $
(89) $ \begin{align} 2bb'-2b'^{2}+bb'' & =0, \end{align} $
(90) are satisfied. The general solutions for
$ c_{4} $ and b to (89)−(90) are$ c_{4}=\frac{d_{2}^{2}}{4d_{1}}, $
(91) $ b=\frac{C_{2}N}{1+C_{1}N}, $
(92) where
$ C_{1} $ and$ C_{2} $ are constants. Finally, the Lagrangian containing no dynamical scalar degree of freedom at linear order in a cosmological background is$ \begin{aligned}[b] {\cal{L}}=\;&c_{1}\hat{K}_{ij}\hat{K}^{ij}+\frac{1}{3}\frac{C_{2}N}{1+C_{1}N}K^{2}+c_{3}R+\frac{d_{2}^{2}}{4d_{1}}a_{i}a^{i}\\&+d_{1}{\rm{D}}_{i}\phi{\rm{D}}^{i}\phi+d_{2}a_{i}{\rm{D}}^{i}\phi, \end{aligned}$
(93) where
$\hat{K}_{ij}=K_{ij}-\dfrac{1}{3}Kh_{ij}$ is the traceless part of$ K_{ij} $ , and coefficients$ c_{1} $ ,$ c_{3} $ ,$ d_{1} $ , and$ d_{2} $ are the general functions of N and ϕ.It is interesting to compare (93) with the result obtained in [79]. In [79] (see Eq. (61)), only the first three terms in (93) are present. In particular, acceleration term
$ a_{i}a^{i} $ is not allowed. Here, owing to the presence of auxiliary field ϕ, the acceleration term can be introduced, together with two other terms involving$ {\rm{D}}_{i}\phi $ . By applying appropriate limit$ d_{1},d_{2}\rightarrow0 $ , we can return to the result in [79]. -
If
$ \varOmega=0 $ , not all of auxiliary variables A, B,$ \delta\phi $ are solvable. Nevertheless, we can still determine the conditions on the coefficients to eliminate the scalar degree of freedom. First, according to (85),$ \varOmega=0 $ implies two equations:$ \left(c_{1}+c_{2}\right)\left(d_{2}^{2}-4d_{1}c_{4}\right)=0, $
(94) and
$ \left[3\left(2b-2b'+b''\right)\left(c_{1}+c_{2}\right)-2\left(-b'+b\right)^{2}\right]d_{1}=0. $
(95) Several branches of solutions to (94) and (95) exist, which we discuss below.
-
The simplest case is to assume
$ d_{1}=d_{2}=0 $ . Here, Lagrangian (57) reduces to$ {\cal{L}}=c_{1}K_{ij}K^{ij}+c_{2}K^{2}+c_{3}R+c_{4}a_{i}a^{i}, $
(96) which is simply the case that has already been discussed in [79]. The Lagrangian that propagates only two tensor degrees of freedom (up to the linear order in perturbations) is given by
$ {\cal{L}}=c_{1}\hat{K}_{ij}\hat{K}^{ij}+\frac{1}{3}\frac{C_{2}N}{1+C_{1}N}K^{2}+c_{3}R, $
(97) where
$ c_{1} $ and$ c_{3} $ are general functions of N and ϕ. -
If
$ d_{1}\neq0 $ , from (95) we must have$ 3\left(2b-2b'+b''\right)\left(b-2c_{2}\right)-2\left(-b'+b\right)^{2}=0. $
(98) For (94), we assume
$ c_{1}+c_{2}\neq0 $ ; thus, it follows that$ d_{2}^{2}-4d_{1}c_{4}=0. $
(99) Because background equation of motion (72) must hold, we require
$ -b+b'\neq0 $ to guarantee non-vanishing cosmological constant Λ and thus a cosmological background solution. Thus, (98) indicates$ b-2c_{2}\neq0,\quad2b-2b'+b'' \neq 0. $
(100) Subsequently, (98) implies that
$ c_{2}=\frac{1}{2}b-\frac{\left(b-b'\right)^{2}}{3\left(2b-2b'+b''\right)}, $
(101) and
$ c_{1}=-\frac{1}{2}b+\frac{\left(b-b'\right)^{2}}{2b-2b'+b''}, $
(102) where b is a general function of N only. The Lagrangian now becomes
$ \begin{aligned}[b] {\cal{L}}=\;&\left[-\frac{1}{2}b+\frac{\left(b-b'\right)^{2}}{2b-2b'+b''}\right]K_{ij}K^{ij}\\&+\left[\frac{1}{2}b-\frac{\left(b-b'\right)^{2}}{3\left(2b-2b'+b''\right)}\right]K^{2}\\&+c_{3}R+c_{4}a_{i}a^{i}+d_{1}{\rm{D}}_{i}\phi{\rm{D}}^{i}\phi+d_{2}a_{i}{\rm{D}}^{i}\phi, \end{aligned} $
(103) where
$ c_{3} $ ,$ c_{4} $ ,$ d_{1} $ , and$ d_{2} $ are general functions of N and ϕ.We can easily check that operators
$ \hat{{\cal{O}}}_{XY} $ is the same as those in (75)−(84). The equations of motion for the perturbation variables are$ 2\hat{{\cal{O}}}_{AA}A+\hat{{\cal{O}}}_{AB}B+\hat{{\cal{O}}}_{A\phi}\delta\phi+\hat{{\cal{O}}}_{A\zeta}\zeta+\hat{{\cal{O}}}_{\dot{\zeta}A}\dot{\zeta} = 0, $
(104) $ \hat{{\cal{O}}}_{AB}A+2\hat{{\cal{O}}}_{BB}B+\hat{{\cal{O}}}_{\zeta B}\zeta+\hat{{\cal{O}}}_{\dot{\zeta}B}\dot{\zeta} = 0, $
(105) $ 2\hat{{\cal{O}}}_{\phi\phi}\delta\phi+\hat{{\cal{O}}}_{A\phi}A = 0, $
(106) $\begin{aligned}[b] & \frac{1}{\bar{N}}\Bigg[2\partial_{t}\left(a^{3}\hat{{\cal{O}}}_{\dot{\zeta}\dot{\zeta}}\dot{\zeta}\right)\\&+\partial_{t}\left(a^{3}\hat{{\cal{O}}}_{\dot{\zeta}B}B\right)+ \partial_{t}\left(a^{3}\hat{{\cal{O}}}_{\dot{\zeta}A}A\right)+\partial_{t}\left(a^{3}\hat{{\cal{O}}}_{\dot{\zeta}\zeta}\zeta\right)\Bigg] \\ & -a^{3}\left(2\hat{{\cal{O}}}_{\zeta\zeta}\zeta+\hat{{\cal{O}}}_{A\zeta}A+\hat{{\cal{O}}}_{\zeta B}B\right) = 0. \end{aligned} $
(107) We rewrite the first three equations in the matrix form:
$ {\boldsymbol{D}}{\boldsymbol{\alpha}}={\boldsymbol{\beta}}, $
(108) where
$ {\boldsymbol{D}}=\left[\begin{array}{*{20}{c}} {2\hat{{\cal{O}}}_{AA} }&{ \hat{{\cal{O}}}_{AB} }&{ \hat{{\cal{O}}}_{A\phi}}\\ {\hat{{\cal{O}}}_{AB} }&{ 2\hat{{\cal{O}}}_{BB} }&{ 0}\\ {\hat{{\cal{O}}}_{A\phi} }&{ 0 }&{ 2\hat{{\cal{O}}}_{\phi\phi} }\end{array}\right], $
(109) $ {\boldsymbol{\alpha}}=\left[\begin{array}{c} A\\ B\\ \delta\phi \end{array}\right],\quad{\boldsymbol{\beta}}=\left[\begin{array}{c} -\hat{{\cal{O}}}_{A\zeta}\zeta-\hat{{\cal{O}}}_{\dot{\zeta}A}\dot{\zeta}\\ -\hat{{\cal{O}}}_{\zeta B}\zeta-\hat{{\cal{O}}}_{\dot{\zeta}B}\dot{\zeta}\\ 0 \end{array}\right]. $
(110) Determinant of coefficient matrix
$ {\boldsymbol{D}} $ of auxiliary perturbation variables A, B, and$ \delta\phi $ is$ \begin{aligned}[b] \varOmega & =\det{\boldsymbol{D}} \\ & =\hat{{\cal{O}}}_{A\phi}^{2}\hat{{\cal{O}}}_{BB}+\hat{{\cal{O}}}_{AB}^{2}\hat{{\cal{O}}}_{\phi\phi}-4\hat{{\cal{O}}}_{AA}\hat{{\cal{O}}}_{BB}\hat{{\cal{O}}}_{\phi\phi}=0. \end{aligned} $
(111) By multiplying the third row of
$ {\boldsymbol{D}} $ by$ \hat{{\cal{O}}}_{A\phi}\hat{{\cal{O}}}_{BB} $ , the second row by$ \hat{{\cal{O}}}_{AB}\hat{{\cal{O}}}_{\phi\phi} $ , and the first row by$ -2\hat{{\cal{O}}}_{BB}\hat{{\cal{O}}}_{\phi\phi} $ and by adding the third and second rows to the first row, we derive the following matrix:$ \left(\begin{array}{*{20}{c}} {0 }&{ 0 }&{ 0}\\{ \hat{{\cal{O}}}_{AB} }&{ 2\hat{{\cal{O}}}_{BB} }&{ 0}\\ {\hat{{\cal{O}}}_{A\phi} }&{ 0 }&{ 2\hat{{\cal{O}}}_{\phi\phi}} \end{array}\right), $
(112) which shows that
$ {\rm{rank}}{\boldsymbol{D}}=2<3 $ . The augmented matrix of the set of equations is defined as$ \begin{aligned}[b]\\[-8pt] {\boldsymbol{E}}=\left(\begin{array}{*{20}{c}} {2\hat{{\cal{O}}}_{AA} }&{ \hat{{\cal{O}}}_{AB} }&{ \hat{{\cal{O}}}_{A\phi} }&{ -\hat{{\cal{O}}}_{A\zeta}\zeta-\hat{{\cal{O}}}_{\dot{\zeta}A}\dot{\zeta}}\\ {\hat{{\cal{O}}}_{AB} }&{ 2\hat{{\cal{O}}}_{BB} }&{ 0 }&{ -\hat{{\cal{O}}}_{\zeta B}\zeta-\hat{{\cal{O}}}_{\dot{\zeta}B}\dot{\zeta}}\\ {\hat{{\cal{O}}}_{A\phi} }&{ 0 }&{ 2\hat{{\cal{O}}}_{\phi\phi} }&{ 0}\end{array}\right).\\ \end{aligned} $
(113) By performing the same operations to
$ {\boldsymbol{D}} $ , the augmented matrix becomes$ \left(\begin{array}{*{20}{c}} {0 }&{ 0 }&{ 0 }&{ \left(2\hat{{\cal{O}}}_{BB}\hat{{\cal{O}}}_{\phi\phi}\hat{{\cal{O}}}_{A\zeta}-\hat{{\cal{O}}}_{\zeta B}\hat{{\cal{O}}}_{AB}\hat{{\cal{O}}}_{\phi\phi}\right)\zeta+\left(2\hat{{\cal{O}}}_{BB}\hat{{\cal{O}}}_{\phi\phi}\hat{{\cal{O}}}_{\dot{\zeta}A}-\hat{{\cal{O}}}_{\dot{\zeta}B}\hat{{\cal{O}}}_{AB}^{2}\hat{{\cal{O}}}_{\phi\phi}\right)\dot{\zeta}}\\ {\hat{{\cal{O}}}_{AB} }&{ 2\hat{{\cal{O}}}_{BB} }&{ 0 }&{ -\hat{{\cal{O}}}_{\zeta B}\zeta-\hat{{\cal{O}}}_{\dot{\zeta}B}\dot{\zeta}}\\ {\hat{{\cal{O}}}_{A\phi} }&{ 0 }&{ 2\hat{{\cal{O}}}_{\phi\phi} }&{ 0 }\end{array}\right), $ (114) which shows that the crucial term relevant to our analysis is
$ \begin{aligned}[b] \Xi\equiv\;&\left(2\hat{{\cal{O}}}_{BB}\hat{{\cal{O}}}_{\phi\phi}\hat{{\cal{O}}}_{A\zeta}-\hat{{\cal{O}}}_{\zeta B}\hat{{\cal{O}}}_{AB}\hat{{\cal{O}}}_{\phi\phi}\right)\zeta\\&+\left(2\hat{{\cal{O}}}_{BB}\hat{{\cal{O}}}_{\phi\phi}\hat{{\cal{O}}}_{\dot{\zeta}A}-\hat{{\cal{O}}}_{\dot{\zeta}B}\hat{{\cal{O}}}_{AB}^{2}\hat{{\cal{O}}}_{\phi\phi}\right)\dot{\zeta}.\end{aligned} $
(115) Fortunately, the explicit expression for Ξ is not required in the following analysis.
Based on whether Ξ is identically vanishing or not, we have two cases. If
$ \Xi\neq0, $
(116) we obtain
$ {\rm{rank}}{\boldsymbol{E}}=3>{\rm{rank}}{\boldsymbol{D}}=2 $ , which implies that the set of equations for the auxiliary variables has no solution.In contrast, if
$ \Xi=0, $
(117) we obtain
$ {\rm{rank}}{\boldsymbol{E}}={\rm{rank}}{\boldsymbol{D}}=2<3 $ , which means that the set of equations is solvable. In this case, equations (104)−(106) become$ 2\hat{{\cal{O}}}_{BB}B+\hat{{\cal{O}}}_{AB}A+\hat{{\cal{O}}}_{\zeta B}\zeta+\hat{{\cal{O}}}_{\dot{\zeta}B}\dot{\zeta}=0, $
(118) $ 2\hat{{\cal{O}}}_{\phi\phi}\delta\phi+\hat{{\cal{O}}}_{A\phi}A=0, $
(119) from which we can solve B as
$ B=-\frac{\hat{{\cal{O}}}_{AB}A+\hat{{\cal{O}}}_{\zeta B}\zeta+\hat{{\cal{O}}}_{\dot{\zeta}B}\dot{\zeta}}{2\hat{{\cal{O}}}_{BB}}. $
(120) Substituting solution (120) into the last equation of motion (107) yields the second order time derivative term of ζ,
$ \Delta_{1}\ddot{\zeta}, $
(121) where
$ \Delta_{1}=\frac{4\hat{{\cal{O}}}_{\dot{\zeta}\dot{\zeta}}\hat{{\cal{O}}}_{BB}-\hat{{\cal{O}}}_{\dot{\zeta}B}^{2}}{2\hat{{\cal{O}}}_{BB}}. $
(122) To eliminate the scalar mode, we must require
$ 4\hat{{\cal{O}}}_{\dot{\zeta}\dot{\zeta}}\hat{{\cal{O}}}_{BB}-\hat{{\cal{O}}}_{\dot{\zeta}B}^{2}=0, $
(123) which corresponds to
$ 4a^{-2}b\left[b-3\left(c_{1}+c_{2}\right)\right]\partial^{4}=0. $
(124) Recall that we have assumed
$ c_{1}+c_{2}\neq0 $ from the beginning; thus, the only feasible solution to (124) is$ c_{1}=0 $ . Again, this will lead to the disappearance of$ \hat{K}_{ij}\hat{K}^{ij} $ ; thus, the kinetic term for the gravitational wave, which is unacceptable.In conclusion, no viable Lagrangian is obtained in Case 2.
-
In this case, we assume
$ c_{1}+c_{2}=0,\quad-b'+b=0. $
(125) As mentioned above, this case conflicts with the background equation of motion (72), and thus is unphysical.
-
In this case, we assume
$ c_{1}+c_{2}=0,\quad d_{1}=0. $
(126) The Lagrangian reduces to be
$ {\cal{L}}=-\frac{1}{2}bK_{ij}K^{ij}+\frac{1}{2}bK^{2}+c_{3}R+c_{4}a_{i}a^{i}+d_{2}a_{i}{\rm{D}}^{i}\phi. $
(127) Operators
$ \hat{{\cal{O}}}_{XY} $ are given by$ \hat{{\cal{O}}}_{AA}=\frac{3}{2}H^{2}\left(2b-2b'+b''\right)-c_{4}\partial^{2}, $
(128) $ \hat{{\cal{O}}}_{AB}=2Ha^{-1}\left(-b'+b\right)\partial^{2}, $
(129) $ \hat{{\cal{O}}}_{A\phi}=-d_{2}\partial^{2}, $
(130) $ \hat{{\cal{O}}}_{\dot{\zeta}A}=6H\left(-b+b'\right), $
(131) $ \hat{{\cal{O}}}_{\dot{\zeta}B}=-2a^{-1}b\partial^{2}, $
(132) $ \hat{{\cal{O}}}_{\dot{\zeta}\dot{\zeta}}=3b, $
(133) and
$ \hat{{\cal{O}}}_{BB}=\hat{{\cal{O}}}_{\phi\phi}=\hat{{\cal{O}}}_{\phi B}=\hat{{\cal{O}}}_{\dot{\zeta}\phi}=\hat{{\cal{O}}}_{\phi\zeta}=0 $ .Consequently, the quadratic action for the perturbation variables is
$ \begin{aligned}[b] S_{2}[A,B,\zeta,\delta\phi] =\;& \int{\rm{d}}t{\rm{d}}^{3}x\bar{N}a^{3}\Big(A\hat{{\cal{O}}}_{AA}A+\zeta\hat{{\cal{O}}}_{\zeta\zeta}\zeta+A\hat{{\cal{O}}}_{AB}B\\&+A\hat{{\cal{O}}}_{A\zeta}\zeta+A\hat{{\cal{O}}}_{A\phi}\delta\phi +\zeta\hat{{\cal{O}}}_{\zeta B}B+\dot{\zeta}\hat{{\cal{O}}}_{\dot{\zeta}A}A\\&+\dot{\zeta}\hat{{\cal{O}}}_{\dot{\zeta}B}B+\dot{\zeta}\hat{{\cal{O}}}_{\dot{\zeta}\zeta}\zeta+\dot{\zeta}\hat{{\cal{O}}}_{\dot{\zeta}\dot{\zeta}}\dot{\zeta}\Big). \end{aligned} $
(134) The equations of motion for the perturbation variables are
$ 2\hat{{\cal{O}}}_{AA}A+\hat{{\cal{O}}}_{AB}B+\hat{{\cal{O}}}_{A\zeta}\zeta+\hat{{\cal{O}}}_{A\phi}\delta\phi+\hat{{\cal{O}}}_{\dot{\zeta}A}\dot{\zeta} =0, $
(135) $ \hat{{\cal{O}}}_{AB}A+\hat{{\cal{O}}}_{\zeta B}\zeta+\hat{{\cal{O}}}_{\dot{\zeta}B}\dot{\zeta} =0, $
(136) $ \hat{{\cal{O}}}_{A\phi}A =0, $
(137) $ \begin{aligned}[b] &\frac{1}{\bar{N}}\left[2\partial_{t}\Big(a^{3}\hat{{\cal{O}}}_{\dot{\zeta}\dot{\zeta}}\dot{\zeta}\right)+ \partial_{t}\left(a^{3}\hat{{\cal{O}}}_{\dot{\zeta}B}B\right)\\&+\partial_{t}\left(a^{3}\hat{{\cal{O}}}_{\dot{\zeta}A}A\right)+\partial_{t}\left(a^{3}\hat{{\cal{O}}}_{\dot{\zeta}\zeta}\zeta\right)\Big] \\ &-a^{3}\left(2\hat{{\cal{O}}}_{\zeta\zeta}\zeta+\hat{{\cal{O}}}_{A\zeta}A+\hat{{\cal{O}}}_{\phi\zeta}\delta\phi+\hat{{\cal{O}}}_{\zeta B}B\right) =0. \end{aligned} $
(138) To obtain a nontrivial solution for A, from (137), we require that
$ \hat{{\cal{O}}}_{A\phi}=0, $
(139) which implies
$ d_{2}=0. $
(140) Thus, we can solve A and B from (135)−(136) to obtain
$ A=-\frac{\hat{{\cal{O}}}_{\zeta B}\zeta+\hat{{\cal{O}}}_{\dot{\zeta}B}\dot{\zeta}}{\hat{{\cal{O}}}_{AB}}, $
(141) and
$ B=\frac{\left(2\hat{{\cal{O}}}_{AA}\hat{{\cal{O}}}_{\zeta B}-\hat{{\cal{O}}}_{AB}\hat{{\cal{O}}}_{A\zeta}\right)\zeta+\left(2\hat{{\cal{O}}}_{AA}\hat{{\cal{O}}}_{\dot{\zeta}B}-\hat{{\cal{O}}}_{AB}\hat{{\cal{O}}}_{\dot{\zeta}A}\right)\dot{\zeta}}{\hat{{\cal{O}}}_{AB}^{2}}, $
(142) where
$ \hat{{\cal{O}}}_{AB} $ given in (129) is not vanishing.Finally, by substituting solutions (141) and (142) into (138), we obtain an effective equation of motion for ζ (with undetermined
$ \delta\phi $ ), in which the second-order time derivative term can be expressed as$ \Delta_{2}\ddot{\zeta}, $
(143) with
$ \Delta_{2}=2\frac{\hat{{\cal{O}}}_{\dot{\zeta}\dot{\zeta}}\hat{{\cal{O}}}_{AB}^{2}+\hat{{\cal{O}}}_{AA}\hat{{\cal{O}}}_{\dot{\zeta}B}^{2}-\hat{{\cal{O}}}_{AB}\hat{{\cal{O}}}_{\dot{\zeta}A}\hat{{\cal{O}}}_{\dot{\zeta}B}}{\hat{{\cal{O}}}_{AB}^{2}}. $
(144) Note that although
$ \delta\phi $ is left undetermined, it will not contribute to the$ \ddot{\zeta} $ term in its equation of motion. To avoid the unwanted scalar mode, we must impose condition$ \Delta_{2}=0 $ , which is equivalent to$ \hat{{\cal{O}}}_{\dot{\zeta}\dot{\zeta}}\hat{{\cal{O}}}_{AB}^{2}+\hat{{\cal{O}}}_{AA}\hat{{\cal{O}}}_{\dot{\zeta}B}^{2}-\hat{{\cal{O}}}_{AB}\hat{{\cal{O}}}_{\dot{\zeta}A}\hat{{\cal{O}}}_{\dot{\zeta}B}=0. $
(145) After substituting (128)−(133) into (145), we obtain
$ 6H^{2}a^{-2}b\left(bb''+2bb'-2b'^{2}\right)\partial^{4}-4a^{-2}b^{2}c_{4}\partial^{6}=0, $
(146) which implies two equations:
$ b\left(bb'' +2bb'-2b'^{2}\right) =0, $
(147) $ b^{2}c_{4} =0. $
(148) Because the background equations of motion must be satisfied,
$ b\neq0 $ . Thus, the only solutions to$ \Delta_{2}=0 $ are$ c_{4}=0,\quad b=\frac{C_{2}N}{1+C_{1}N}. $
(149) As a result, in this case, the Lagrangian that does not propagate any scalar degree of freedom at the linear order is
$ {\cal{L}}=-\frac{1}{2}\frac{C_{2}N}{1+C_{1}N}\hat{K}_{ij}\hat{K}^{ij}+\frac{1}{3}\frac{C_{2}N}{1+C_{1}N}K^{2}+c_{3}R, $
(150) where
$ c_{3} $ is a general function of N and ϕ.Here, if we select
$ c_{3}=1 $ , let$ \left|C_{2}\right|\rightarrow \infty $ ,$ \left|C_{1}\right|\rightarrow \infty $ , and maintain$ \dfrac{C_{2}}{C_{1}}=-2 $ , Lagrangian (150) reduces to$ {\cal{L}}=\hat{K}_{ij}\hat{K}^{ij}-\frac{2}{3}K^{2}+R, $
(151) which is nothing but GR.
Spatial covariant gravity with two degrees of freedom in the presence of an auxiliary scalar field: Perturbation analysis
- Received Date: 2024-03-23
- Available Online: 2024-08-15
Abstract: We investigate a class of gravity theories respecting only spatial covariance, termed spatially covariant gravity, in the presence of an auxiliary scalar field. We examine the conditions on the Lagrangian required to eliminate scalar degrees of freedom, allowing only two tensorial degrees of freedom to propagate. Instead of strict constraint analysis, in this paper, we employ the perturbation method and focus on the necessary conditions to evade the scalar mode at the linear order in perturbations around a cosmological background. Beginning with a general action and solving the auxiliary perturbation variables in terms of a would-be dynamical scalar mode, we derive the condition to remove its kinetic term, thus ensuring that no scalar mode propagates. As an application of the general condition, we study a polynomial-type Lagrangian as a concrete example, in which all monomials are spatially covariant scalars containing two derivatives. We find that the auxiliary scalar field is essential, and new terms in the Lagrangian are allowed. Our analysis provides insights into constructing gravity theories with two degrees of freedom in the extended framework of spatially covariant gravity.