Primordial Gravitational Waves in Bimetric Gravity

We study primordial tensor power-spectra generated during inflation in bimetric gravity. More precisely, we examine a homogeneous expanding spacetime in a minimal bimetric model with an inflaton and calculate tensor perturbations on the homogeneous background under slow-roll approximation. In terms of the mass eigenstates, only the power-spectrum of the massless state remains constant and both the power-spectrum of the massive state and the cross power-spectrum rapidly decay during inflation. The amplitude of the physical power-spectrum is suppressed due to the flavor mixing. All power-spectra in the flavor eigenstates coincide with each other up to the first order of the slow-roll parameter.


I. INTRODUCTION
Although no one doubt the existence of gravitons, no one really observe them.Indeed, we do not know if gravitons have mass or how many species there are.What we can confidently say is at least one of them must be sufficiently light in order to realize the Newtonian potential.Thus, we have room to suppose two or more graviton species exist.It is well known that two interacting massless graviton can not exist.However, it was recently found that a massless graviton and a massive graviton can exist at the same time [1][2][3][4][5].The form of interaction terms is highly constrained in order to avoid ghosts.Such theory necessarily includes another metric in addition to the physical metric and is called bimetric gravity.In general, the interaction terms include five theoretical parameters.
We are interested in if the bimetric theory is theoretically consistent and reconciles with known experiments.One of important check points is if the predictions coming from inflation in bimetric gravity are consistent with cosmological observations which are now getting more precise.We examined the homogeneous expanding solutions and their stability in the slowroll limit in our previous paper [6].We studied minimal bimetric models and concluded that the only branch is stable.Now, we construct homogeneous inflationary solutions under slow-roll approximation which correspond to the stable branch.Then we consider tensor perturbations on the homogeneous solutions and we calculate primordial tensor spectra generated during inflation to the first order of the slow-roll parameter.This paper is organized as follows.In section II, we introduce an inflaton to a minimal bimetric model and we construct inflationary background solutions.We explain the properties of the functions which are specific to bimetric gravity.We briefly mention slow-roll approximation and introduce a slow-roll parameter we use in the following calculation.In section III, we derive the second order Lagrangian for tensor perturbations.There are two views of this system: the flavor eigenstates and the mass eigenstates, and we use the mass eigenstates for simplifying calculations.In section IV, we calculate the primordial tensor spectra up to the first order of the slow-roll parameter by making use of the interaction picture.In the final section, we discuss the features of the tensor power-spectra in bimetric theory.

II. INFLATIONARY BACKGROUND SOLUTIONS IN BIMETRIC GRAVITY
In this section, we construct inflationary background solutions with a scalar field coupled to the physical metric in bimetric gravity.We write a minimal bimetric action with a canonical scalar field.We substitute the homogeneous isotropic metric ansatz into the action and derive the equations of motion by using the variational principle.We mention the features of the solutions based on our analysis in our previous paper [6].

A. Action and Ansatz
We consider a minimal bimetric action including a scalar field coupled to the physical metric and we substitute a homogeneous ansatz for metrics and the scalar field into the action.
We use g µν as the physical metric, f µν as the other metric and ϕ as the scalar field.
When we consider a bimetric theory [4,5], the form of the interaction terms are restricted in order to avoid the Boulware-Deser ghost [7], which include five theoretical parameters 4).For simplicity, we set α 2 = 1 and other four theoretical parameters are equal to zero.Then we have where M g and M f are the Planck scales of the physical metric and the other metric respectively and we defined the reduced Planck scale as Indices written in Greek letter run over 0, • • • , 3. The last part in the action includes the interaction terms of the physical metric and the other metric where we defined We impose the homogeneous isotropic ansatz for these metrics as where γ ij is three-dimensional flat metric and indices written in Roman letter run over spatial coordinates, i.e. i = 1, • • • , 3. N and M are lapse functions and e α and e β are scale factors which depends only on time.We also require the scalar field is also homogeneous When we substitute these ansatzes into the Lagrangian, we find where dots denote the time derivative and ǫ is the ratio of the scale factors of the physical metric and the other metric, i.e.
We can see that α, β and ϕ are dynamical variables and N and M are non-dynamical variables included in the Lagrangian linearly.

B. Equations of motion
We derive the equations of motion of the dynamical variables α, β and ϕ and two constraints from the variational principle.We have a relation between the lapse functions in order for the two constraints to hold during time evolution [8].We also describe the behavior of the solutions of the equations.
From the variations of the Lagrangian with respect to the dynamical variables α, β and ϕ, we obtain respectively, where we defined a new parameter tan a := M g /M f .Since N and M are nondynamical and included only linearly in the Lagrangian, we obtain two constraints from the variations of the action with respect to them.One of them is which comes from the variation of the action with respect to N. The variation of the action with respect to M yields the other one β We set the time derivative of eq.( 12) is equal to zero so that the constraint is satisfied during time evolution.We combine it with eq.( 9) and obtain the following equation We can obtain the same equation also by using eq.( 13) and eq.(10).When the first factor is equal to zero, the solutions are known to be pathological [9][10][11][12][13].Therefore, we assume the second factor is equal to zero, where we defined Since N is a gauge variable, we can set an arbitrary value for N. Substituting this relation into eq.(13) leads to This is another expression for the Hubble expansion of the physical metric.We can see that the Hubble expansion is a function of ǫ.By equating eq.( 12) and eq.( 17), we obtain an equation This equation says that we find the value of ǫ if we determine the energy density on the physical spacetime.We have examined the properties of the roots of this equation in the slow-roll limit in our previous paper [6].In the slow-roll limit, we neglect φ contribution and this equation reduces to algebraic equation with constant coefficients, therefore, the roots are constant.This equation basically has three roots since it is a cubic equation.One of the roots is always negative, therefore it is not appropriate solution when we take into account the definition of ǫ.Another is always larger than 1, therefore it is also inappropriate because the Hubble expansion, which is written as eq.( 17), becomes imaginary.The other one has the value between 0 and 1, therefore it is the only adoptable root as de Sitter spacetime.
We have confirmed that the root satisfies the Higuchi bound [14,15], which is the stability condition of de Sitter spacetime with a massive graviton.
Substituting eq.( 15) into eq.(9) and eq.( 10), we obtain the equation determining the value where we defined the effective mass In the slow-roll limit, ζ is equal to 1 and therefore α and β are different only by a constant.
Using the definition of ζ, we have a relation between ζ and ǫ as We can see that δζ vanishes in the slow-roll limit.

C. Slow-roll approximation
We introduce a slow-roll parameter and explain the assumptions used in the following analysis.We use the N = 1 gauge in this subsection.
We define a slow-roll parameter as where H := α.From eq.( 19), we find The third expression is obtained by eliminating φ with eq.( 9) and the final expression is obtained by using eq.( 17) and eq.( 20).From this equation, we can read off an expression of The third expression is obtained by using the definition of s and the last expression is obtained by using eq.( 17) and eq.( 20).We can see that the expression for φ is slightly changed from the conventional case .
We assume the slow-roll parameter is small and constant.We neglect the higher order contribution O(s 2 ) in the following.Under this approximation, we can see δζ is constant but ǫ has time dependence.We can easily obtain the explicit form of ǫ by integrating the differential equation ǫ = 2sǫH(1 − ǫ) obtained from eq.( 21) and eq.( 23): ǫ 0 is the value of ǫ in the slow-roll limit.We note that the functions with the subscript 0 have the values in the slow-roll limit and all of them are constant in the following.

III. SECOND ORDER LAGRANGIAN FOR TENSOR PERTURBATIONS
We derive the second order Lagrangian for tensor perturbations on the homogeneous isotropic inflationary background.First, we obtain the Lagrangian in the flavor eigenstates, and then, we move to the massive eigenstates.

A. Flavor eigenstates
We give the tensor perturbations to the original Lagrangian and derive the second order Lagrangian in the flavor eigenstates.We will see the variables in the flavor eigenstates do not decouple with each other in the slow-roll limit.
We consider perturbations such as which satisfy the following transverse traceless conditions: We can decompose the tensor perturbations using polarization tensors as where the subscripts I correspond to the plus mode and the cross mode, i.e.I = {+, ×}.
When we substitute the tensor perturbations into the action, the second order Lagrangian is reduced to where k is the norm of a three-dimensional wave vector since we have used Fourier decomposition.We drop the subscripts I in the following and we finally sum up the two polarization states, or multiply results by two.Substituting background equations, we obtain Furthermore, if we substitute the consistency relation M = ζǫN, this becomes The interaction term is proportional to (p − q) 2 and do not disappear in the slow-roll limit where ǫ = ǫ 0 and δζ = 1.Thus, it is difficult to treat it analytically.Hence, we change the variables in the next subsection in order to have decoupled equations in the slow-roll limit, which allows us to calculate power-spectra analytically.

B. Mass eigenstates
We calculate the second order Lagrangian in the mass eigenstates by changing the variables in the flavor eigenstates to those in the mass eigenstates.Then, the calculation becomes much simpler because a massive eigenstate stemming from the interaction terms and a massless eigenstate due to the general covariance do not interact with each other in de Sitter spacetime in bimetric gravity [16].
We make a transformation of the variables (q, p) into (x, y) as where κ = ζ 1/2 tan a.We note that κ is constant since δζ is constant but ǫ is time dependent as we mentioned in the last part of the previous section.Therefore, this mixing matrix is time dependent.By substituting this relation into eq.(31), we find the second order Lagrangian in the mass eigenstates up to the first order of the slow-roll parameter as in the N = 1 gauge, where and We have used the relation ǫ = ǫHδζ in eq.( 33) and also eq.( 23) in the last line of eq.( 34).If we use the N = e α gauge, this Lagrangian is rewritten as The primes denotes a time derivative with respect to the conformal time defined as dt = e α dη and H := α ′ .Furthermore, we make a scale transformation to make kinetic terms canonical as Then, we obtain the following Lagrangian.
We have used the relations H ′ = (1−s)H and H = H/e α in the first line and H = −1/(1−s)η in the second line.By substituting eq.( 23) and neglecting O(s 2 ), we finally obtain where κ 0 = tan a.We note that the cross terms of X and Y vanish in the slow-roll limit and the Lagrangian becomes diagonal in the mass eigenstates.

IV. TENSOR POWER-SPECTRA IN BIMETRIC GRAVITY
We calculate the tensor power-spectra generated during inflation in bimetric gravity.
In the mass eigenstates, we can solve the free part.Then, we calculate the first order corrections for them by using the interaction picture and obtain tensor power-spectra in the mass eigenstates.Finally, we obtain tensor power-spectra in the flavor eigenstates by using the relation between the variables in the flavor eigenstates and those in the mass eigenstates.

A. Interaction picture
First, we transform the Lagrangian to the Hamiltonian and quantize the system.Then, we calculate the correlations of the massless state and the massive state by making use of the interaction picture.
The conjugate momenta of X and Y are calculated as where we defined By performing Legendre transformation, we obtain the following Hamiltonian.
We note that we have defined X(k, η) and Y (k, η) as We define free fields X I and Y I where time evolution is governed by the Hamiltonian We separated the interaction Hamiltonian from the free one The fields X I and Y I can be expanded by creation and annihilation operators where mode functions satisfy and The correlation functions of these variables can be deduced as and The correlation functions of the original variables are calculated by using the interaction picture, up to the first order of the slow roll parameter, as and In the lowest order, there is no effect of interactions.However, the cross correlation gets corrections as We will calculate these correlations explicitly in the long wavelength limit in the following.

B. Power-spectra in the mass eigenstates
We calculate the mode functions up to the order needed for calculation and obtain the tensor power-spectra in the mass eigenstates.We will see that the amplitude of the powerspectrum of the massless state remains at the end of inflation, while those of the cross spectrum and the power-spectrum of the massive state rapidly decay and are negligible.

Power-spectrum of the massless state
First, we calculate the power-spectrum of the massless state.We see that the powerspectrum looks like that of general relativity.
In the remote past η → −∞, the differential equations of motion for the mode functions become where Thus, we adopt the Bunch-Davies vacuum states as initial conditions Then, we obtain the solution of eq.( 48) as where The Hankel function H ν X (x) is defined by This solution reduces to for η ∼ 0. In the large scale limit, the power-spectrum of the massless state is given by Therefore, we obtain where and H 0 is the constant Hubble in the slow-roll limit.We note that we have used the relation e α = (−H 0 η) −1−s obtained from the definition of s and this amplitude is almost constant as in the conventional case, i.e. xx ∼ O(η 0 ) .

Cross power-spectrum
Next, we calculate the cross spectrum of the massless state and the massive state.We conclude the cross spectrum is negligible compared with the power-spectrum of the massless state.
For calculating the cross spectrum in the first order of the slow roll parameter, we only have to know the mode functions in the lowest order.The mode functions of X which we have calculated reduces to in the slow-roll limit and the solution of eq.( 49) is written as where Note that µ is a real number since we have proven m 2 eff (ǫ 0 ) > 3H 2 0 in our previous paper.The cross spectrum of the massless state and the massive state is and X(z)Y (z) scales as η in the leading order.Therefore, Similarly, y(z)x(z) = ( x(z)y(z) ) * = O(η 3 ).The amplitude decays as e −3α during inflation and we can neglect this amplitude compared with that of the power-spectrum of the massless state.

Power-spectrum of the massive state
Finally, we calculate the power-spectrum of the massive state.It turns out that it can also be neglected compared with the power-spectrum of the massless state.
The mode functions of Y satisfy where Here, C 0 is defined as C 0 := m 2 eff (ǫ 0 )/H 2 0 − 2 and C 1 (η) is a function of η since ǫ has time dependence ǫ = ǫ(η) under slow-roll approximation.We define y = log(−η), then, this equation becomes From the definition of y, we can see y → −∞ as η → −0 .If we decompose the solutions like u Y = e y/2 f (y), then we obtain We can neglect the term which is proportional to e 2y because we are interested in the behavior of solutions around |η| ∼ 0. We obtain WKB solutions of this equation as Finally, we obtain up to the overall coefficient.From this, we can calculate the power-spectrum of the massive state Thus, we reach the final result As mentioned before, we can neglect this amplitude compared with that of the powerspectrum of the massless state as in the case of the cross spectrum.

C. Power-spectra in the flavor eigenstates
We calculate the tensor spectra in the flavor eigenstates by making use of the results in the previous subsection.We will see all of them agree with each other and these amplitudes are conserved on super-horizon scales.We also see the spectral index and the amplitudes in the leading order.
Using the relation between the mass eigenstates and the flavor eigenstates: we can obtain the following power-spectra The power-spectra of the physical metric, that of the other metric and the cross spectrum are the same in the first order of the slow-roll parameter at the end of inflation.When we include both plus mode and cross mode, the results should be multiplied by two.Superficially, these spectra do not seem to conserve.However, we can verify their conservation as follows We can read off the spectral index of the tensor power-spectrum from eq.(84) as It means that the spectra are red-tilted since ǫ 0 has a value between 0 and 1 as we mentioned in section II.The amplitudes of them in the leading order, i.e. in the slow-roll limit, are When we take the general relativity limit (M f /M g → 0), the amplitude and the spectral index of the physical metric are smoothly connected to those in the conventional case since κ 0 becomes infinity.On the other hand, when we take the massive gravity limit (M f /M g → ∞), the amplitude vanishes since κ 0 becomes zero.
We emphasize that the physical tensor modes and the other tensor modes are maximally correlated and they identically behave at the end of inflation.