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and Two Mirror Systems

Philippe Brax, Mario Pitschmann

To cite this version:

Philippe Brax, Mario Pitschmann. Exact Solutions to Non-Linear Symmetron Theory: One and Two

Mirror Systems. 2018. �cea-01686571�

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arXiv:1712.09852v1 [gr-qc] 28 Dec 2017

One and Two Mirror Systems

Philippe Brax∗

Institut de Physique Th´eorique, Universit´e Paris-Saclay, CEA, CNRS, F-91191 Gif/Yvette Cedex, France

Mario Pitschmann†

Atominstitut, Technische Universit¨at Wien, Stadionallee 2, A-1020 Wien, Austria

(Dated: December 16, 2017)

We derive the exact analytical solutions to the symmetron field theory equations in the presence of a one or two mirror system. The one dimensional equations of motion are integrated exactly for both systems and their solutions can be expressed in terms of Jacobi elliptic functions. Surprisingly, in the case of two parallel mirrors the equations of motion generically provide not a unique solution but a discrete set of solutions with increasing number of nodes and energies. The solutions obtained herein can be applied to qBOUNCE experiments, neutron interferometry and for the calculation of the symmetron field induced ”Casimir force” in the CANNEX experiment.

PACS numbers: 98.80.-k, 04.80.Cc, 04.50.Kd, 95.36.+x

I. INTRODUCTION

The accelerated expansion of the universe may require the introduction of additional degrees of freedom (see [1] for a recent review). Such new degrees of freedom, in par-ticular light scalars, are theoretically well motivated irre-spectively of their role for the acceleration of the expan-sion of the Universe. If they exist in Nature, they must appear in some screened form in order to prevent detec-tion in all past experiments and observadetec-tions involving scalar fifth forces. A number of screening mechanisms ex-ist [1], the chameleon [2–4] and Damour-Polyakov mech-anisms [5], the K-mouflage [6–8] and Vainshtein ones [9], allowing such hypothetical new fields to remain unseen in local tests of gravity.

In the Damour-Polyakov mechanism the coupling to matter weakens in regions of high density or high New-tonian potential. A particular representative of this effect is the symmetron mechanism [10, 11] (for earlier work see [12, 13]), in which the coupling of the scalar to matter is proportional to the vacuum expectation value (VEV) of the field. The effective potential of the scalar field is such that it acquires a nonzero VEV in low-density regions, while the symmetry is restored in high density regions. Thus, in such high density regions, the field effectively disappears as it is screened from any observation or mea-surement. In regions of low density on the other hand, the field spontaneously breaks symmetry and acquires a non-vanishing VEV, couples to matter and mediates a fifth force.

When the mass of the symmetron field in the cosmo-logical vacuum is small, e.g. of order of 103H

0 where H0

is the Hubble rate now, the symmetron field could have

philippe.brax@ipht.fr

mario.pitschmann@tuwien.ac.at

cosmological implications [11] in particular on the growth of perturbations and large-scale structure [14, 15]. These effects can be captured using N -body simulations in or-der to obtain non-linear properties of the matter power spectrum and the halo mass function, see [16, 17]. Al-though the symmetron is not directly coupled to photons as its matter coupling is conformal, quantum effects can give a direct interaction to electromagnetism [18] which could lead to signatures of cosmological domain walls [19, 20]. Symmetron fields for inflation have also been investigated in [21]. In [22–25], atomic interferometry was used to constrain symmetrons with masses below the dark energy scale. It was found that, whilst symmetrons with masses m slightly larger than the present Hubble rate m ∼ 103H

0 have implications cosmologically [26],

they cannot be tested by atomic interferometry, whereas symmetrons with masses of order of the dark energy scale m ∼ 10−3 eV are within reach of the E¨otwash types of

experiments [27]. Recent bounds on symmetrons have been obtained by Jaffe et al. [28].

In [29], gravity resonance spectroscopy [30, 31] is used for the first time to put new bounds on symmetrons. The experimental analysis depends heavily on the field profile of symmetrons in both the one mirror or two mir-ror setups, where the field is either present over an in-finite plane of high density or is confined between two such parallel planes. Here, we show that an idealized 1-dimensional setup of a single mirror covering an in-finite half-space or two parallel mirrors of in-finite sepa-ration each covering an infinite half-space admit exact analytical solutions, which can be expressed in terms of Jacobi elliptic functions. Moreover, we find that, very surprisingly for highly non-linear differential equations, solutions are generically not unique but rather show a discrete spectrum of possible solutions with an increas-ing number of nodes and energies. For a similar study applied to chameleon field theory see [32].

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informa-tion on symmetrons, which will provide the relevant def-initions for the field theory analysis. In section III the solutions for the one mirror case will be derived, while in sections IV and V the symmetric and anti-symmetric two mirror solutions will be given and discussed in section VI. As an illustration, in section VII a particular case study is carried out for an arbitrary choice of parameters and the complete spectrum of solutions is derived. Section VIII provides relevant information on the qBOUNCE experi-ment, where the symmetron induced resonance frequency shift for the case of a single mirror has also been sum-marized for a large range of parameters. A conclusion in section IX will be followed by an Appendix providing additional technical details on the screening of a neutron.

II. BACKGROUND

Following [1], the symmetron potential is given by

V (φ) = −µ 2 2 φ 2+λ 4 φ 4, (1)

with a parameter µ of dimension mass and the dimen-sionless self-interaction coupling λ. Together with the coupling to matter this induces an effective potential

Veff(φ) = V (φ) + A(φ) ρ , (2)

where for the symmetron we have

A(φ) = 1 + φ 2 2M2+ O(φ 4/M4) , (3) and hence Veff(φ) = 1 2  ρ M2 − µ 2φ2+λ 4 φ 4. (4)

Here, we have neglected an additional term ρ, which does not affect the equations of motion.

In the ”symmetric phase” where ρs≥ M2µ2we use

ρeff = ρs− M2µ2≥ 0 , (5)

while in the ”broken symmetry phase” where ρb< M2µ2

we use

µ2

eff= µ2−

ρb

M2 > 0 . (6)

For the relevant experimental situations such as the qBOUNCE experiments, neutron interferometry or Casimir experiments like CANNEX, we typically have ρeff≃ ρsand µ2eff ≃ µ2, still the exact solutions hold for

more general situations.

The 1-dimensional Hamiltonian is given by

H =12 dφdz 2 + Veff(φ) =1 2  dφ dz 2 +1 2  ρ M2 − µ 2φ2+λ 4φ 4. (7)

The two vacuum values of φ, given by the equation Veff,φ(φ) φ=±φV = 0, read ±φV where φV := r µ2 eff λ . (8)

For static solutions we have d2φ

dz2 = Veff,φ(φ) . (9)

Multiplication by dφ

dz and integration with respect to z provides the important relation

1 2  dφ dz 2 −12 dφdz 2 z=z 0 = Veff(φ) − Veff(φ) z=z 0, (10) which determines the solutions provided both the value of the field and its first derivative are given at a point z0.

III. 1 MIRROR

In this section, we treat the case of a single mirror filling the infinite half-space z ≤ 0.

A. ”Broken Symmetry Phase”

First, we consider the case of low density ρb < M2µ2

corresponding to the medium above the mirror and search for a solution that asymptotically for z → ∞ goes as φ(z) → ±φV implying dφ

dz → 0. Without loss of gen-erality we consider φ(z) → +φV and find for z0= 0 and

z → ∞ −1 2  dφ dz 2 z=0 = Veff(φV) − Veff(φ) z=0. (11)

Subtracting Eq. (11) from (10) gives 1 2  dφ dz 2 = Veff(φ) − Veff(φV) , (12) leading to Z φ(z) φ0 dφ q −µ2 eff φ2− φ2V + λ/2 φ4− φ4V  = z . (13) With y := φ(z) φV and k := φ0 φV

, the latter being the ratio between the value of φ taken for z = 0 and the vacuum

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value φV, we find z = 1 µeff Z y k dy′ q 1 − y′2+ 1/2 y′4− 1 = √ 2 µeff tanh−1y − tanh−1k = √ 2 µeff tanh−1y − k 1 − ky  . (14)

Inverting the relation straightforwardly leads to

φ(z) = φV k |k| |k| + tanhµ√effz 2  1 + |k| tanhµ√effz 2  , (15) or equivalently φ(z) = φV k |k|tanh µeffz √ 2 + tanh −1|k| . (16) B. ”Symmetric Phase”

Here, we consider the case of high density ρs≥ M2µ2,

as inside the mirror. Clearly, for z → −∞ we have φ(z) → 0 and hence φ′ → 0. Therefore, we find for

z0= 0 and z → −∞ −1 2  dφ dz 2 z=0 = −Veff(φ) z=0. (17)

Subtracting the latter equation of Eq. (10) gives 1 2  dφ dz 2 = Veff(φ) . (18)

Without loss of generality we consider the positive solu-tion φ(z) ≥ 0. Then, for z ≤ 0 we obviously have

dφ dz = r ρeff M2φ2+ λ 2 φ 4, (19) or with φ0:= φ(0) Z φ(z) φ0 dφ φ s 1 +λM 2 2ρeff φ2 = √ρ eff M z . (20)

Since for positive x d dx  − ln1 + √ 1 + a x2 x  = 1 x√1 + ax2, (21) we find φ(z) = φ0 cosh √ρ eff M z  − s 1 +λM 2 2ρeff φ2 0 sinh √ρeff M z  , (22) or equivalently φ(z) = −r 2ρeff λ 1 M 1 sinh √ρ eff M z − sinh −1r 2ρeff λ 1 M φ0  . (23) This can also be expressed in terms of Jacobi elliptic func-tionssince

sinh u = sc(u, 1) = sd(u, 1) . (24)

C. Boundary Conditions

Using the boundary conditions 1 2  dφ dz 2 z=0 − = 1 2  dφ dz 2 z=0 + , (25) we find ρeff 2M2φ 2 0+ λ 4φ 4 0= − µ2 eff 2 φ 2 0− φ2V + λ 4 φ 4 0− φ4V , (26) or |φ0| = φV √ 2 1 r 1 + ρeff M2µ2 eff , (27) leading to |k| = √1 2 1 r 1 + ρeff M2µ2 eff . (28)

The second boundary condition

φ(0−) = φ(0+) , (29)

is trivially satisfied.

D. Final Solution

Finally, we obtain the solution φ(z) = Θ(+z) φV k |k|tanh µeffz √ 2 + tanh −1|k| − Θ(−z) ×r 2ρλeffM1 1 sinh √ρ eff M z − sinh −1r 2ρeff λ 1 M kφV  , (30) where k is given by k = ±√1 2 1 r 1 + ρeff M2µ2 eff , (31)

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which determines the boundary value of the scalar field. This solution has been used in [29] in order to evaluate the energy levels of a neutron in a qBOUNCE experi-ment.

IV. 2 MIRRORS: SYMMETRIC SOLUTION

In this section, we treat the case of two parallel in-finitely thick mirrors separated at distance 2d in z-direction, with z = 0 being the center between the two mirrors. First, we consider symmetric solutions only.

A. ”Broken Symmetry Phase”

Here, we consider the case of low density ρb < M2µ2,

as between the mirrors and choose z0 = 0. Due to the

symmetry of the setup the derivative of the field has to vanish there yielding

1 2  dφ dz 2 = Veff(φ) − Veff(φ0) , (32)

where φ0 := φ(0). Without loss of generality we take

φ0> 0. Then, for 0 ≤ z ≤ d we have

dφ dz = − q 2 Veff(φ) − Veff(φ0) , (33) and Z φ(z) φ0 dφ q −µ2 eff φ2− φ20 + λ/2 φ4− φ40  = −z . (34) With x := φ(z) φ0

, and taking µeff> 0, we have

−z = 1 µeffp1 − k2/2 Z x 1 dx′ s 1 − x′2  1 − k 2/2 1 − k2/2x′2  = 1 µeffp1 − k2/2 ×  F  arcsin(x), k/ √ 2 p1 − k2/2  − F π2, k/ √ 2 p1 − k2/2  , (35) where we have employed the Elliptic Integral of the first kind F (φ, ℓ) = Z sin φ 0 dt p(1 − t2)(1 − ℓ2t2). (36)

Here, k gives the ratio between the maximum value of the field between the two plates and the vacuum value φV in

the absence of the two plates, i.e. the ratio k captures

the effect of confinement of the field between the plates. Hence, we obtain φ(z) = φ(0) sin  F−1  F π 2, k/√2 p1 − k2/2  − µeffp1 − k2/2 z, k/√2 p1 − k2/2  . (37)

With the Jacobi elliptic functions sn(u, ℓ) = sin F−1(u, ℓ) , cn(u, ℓ) = cos F−1(u, ℓ) , dn(u, ℓ) =

q

1 − ℓ2sin2 F−1(u, ℓ) , (38)

we can write this as

φ(z) = φ(0) sn  F π 2, k/√2 p1 − k2/2  − µeffp1 − k2/2 z, k/√2 p1 − k2/2  . (39) Since snu + Fπ 2, ℓ  , ℓ= cn(u, ℓ) dn(u, ℓ)= cd(u, ℓ) , (40) and sn(−u, ℓ) = −sn(u, ℓ) , cn(−u, ℓ) = +cn(u, ℓ) , dn(−u, ℓ) = +dn(u, ℓ) , (41) we have, also for −d ≤ z ≤ d and positive as well as negative φ(0) φ(z) = φVk cd  µeffp1 − k2/2 z, |k|/ √ 2 p1 − k2/2  . (42) B. ”Symmetric Phase”

We can read off the solution inside the mirrors directly from the corresponding solution in the 1-mirror case

φ(z) =r 2ρeff λ 1 M × 1 sinh √ρ eff M (|z| − d) + sinh −1r 2ρeff λ 1 M φd  , (43) where φd:= φ(d) and hence can be positive or negative.

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C. Boundary Conditions

Using the boundary conditions at the mirror surface 1 2  dφ dz 2 z=d − = 1 2  dφ dz 2 z=d + , (44) we find −µ 2 eff 2 φ 2 d− φ20 + λ 4 φ 4 d− φ40 = ρeff 2M2φ 2 d+ λ 4 φ 4 d, (45) or |φd| = φV|k| r 1 −k 2 2 r 1 + ρeff M2µ2 eff . (46)

Using this in the second boundary condition

φ(d−) = φ(d+) , (47)

provides the relation r 1 − k 2 2 = r 1 + ρeff M2µ2 eff × cd  µeffp1 − k2/2 d, |k|/ √ 2 p1 − k2/2  . (48) The solution of the equation above gives possible val-ues of k, i.e. it determines the value φ0 of the field

be-tween the two plates. It will give all solutions for which φ′(d

−) = ±φ′(d+), hence one still has to extract the

sub-set of solutions satisfying φ′(d

−) = φ′(d+).

D. Final Solution

Finally, we obtain the solution

φ(z) = Θ(d − |z|) φVk cd  µeffp1 − k2/2 z, |k|/ √ 2 p1 − k2/2  + Θ(|z| − d)r 2ρeff λ 1 M × sgnS sinh √ρ eff M (|z| − d) + sinh −1r 2ρeff λ 1 M φd  , (49) where sgnS:= cd  µeffp1 − k2/2 d, |k|/ √ 2 √ 1−k2/2  cd  µeffp1 − k2/2 d, |k|/ √ 2 √ 1−k2/2  , (50) k is solution of r 1 −k 2 2 = r 1 + ρeff M2µ2 eff × cd  µeffp1 − k2/2 d, |k|/ √ 2 p1 − k2/2  , (51) and φd= ±φVk r 1 −k 2 2 r 1 + ρeff M2µ2 eff . (52)

As mentioned already, Eq. (51) will give all solutions for which φ′(d

−) = ±φ′(d+), hence one still has to extract

the subset of solutions satisfying φ′(d

−) = φ′(d+).

Clearly, the expression on the left-hand side of Eq. (51) is O(1). The first factor on the right-hand side is for most cases of interest, where typically ρeff ≫ M2µ2eff,

comparably large. This implies that for a solution the second factor must be small, viz. it will be close to a zero of the Jacobi function cd. The Taylor expansion around the first zero of the Jacobi function cd reads

cd(u, ℓ) = − u − F (π/2, ℓ) + O u − F (π/2, ℓ)3 . (53) Hence to linear order we find

r 1 − k 2 2 = r 1 + ρeff M2µ2 eff × µeffp1 − k2/2 d − F  π 2, |k|/√2 p1 − k2/2  , (54) respectively µeffd = 1 r 1 − k 2 2 F π 2, |k|/√2 p1 − k2/2  ±r 1 1 + ρeff M2µ2 eff ≃ r 1 1 − k 2 2 F π 2, |k|/√2 p1 − k2/2  . (55) Since 1 r 1 −k 2 2 F π 2, |k|/√2 p1 − k2/2  ∈ [π/2, ∞) , (56) we have µeffd & π 2 . (57)

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In the limit ρeff → ∞ the inequality above becomes

rig-orous, viz. for µeffd < π/2 no solution can exist in this

limit. This result has been anticipated in [27], where also the particular solution without any nodes has been ap-proximated by an iteration procedure. Furthermore, in [33], the zero-node solution has been obtained in the ap-proximation of vanishing field values inside the mirrors.

In this work, exact solutions are presented for the first time both inside and outside the mirrors. This is im-portant for weak couplings, viz. for large values of M as well as large values of µ. In those cases the field in-side the mirrors is less suppressed. Another novelty, not anticipated before, is the existence of a whole set of so-lutions for larger values of d due to the periodicity of the Jacobi function. These solutions will be instrumen-tal in analyzing the results of neutron interferometry ex-periments where the integral of the field across the space between two mirrors measures the phase shift that a neu-tron experiences on its path to the detectors and also for Casimir experiments and the corresponding calculation of the ”Casimir force” induced by the symmetron field. The precise calculations of these effects are left for future work.

V. 2 MIRRORS: ANTI-SYMMETRIC SOLUTION

In this section, we treat again the case of two par-allel infinitely thick mirrors separated at distance 2d in z-direction, with z = 0 being the center between the two mirrors. But here, we consider anti-symmetric solutions only.

A. ”Broken Symmetry Phase”

Here, we consider the case of low density ρb < M2µ2

as between the mirrors and take z0 = 0. Due to

anti-symmetry the field has to vanish there (φ0= 0) yielding

1 2  dφ dz 2 −12φ′20 = Veff(φ) . (58) where φ′0 := dφ dz

z=0. Without loss of generality we take φ′

0> 0, then for −d ≤ z ≤ d we have

dφ dz = ±

q φ′2

0 + 2 Veff(φ) , (59)

and since Veff(φ) ≤ 0 this gives a real solution only for

φ′2 0 ≥ −2 Veff(φ). We have Z φ(z) 0 dφ pφ′2 0 − µ2effφ2+ λ/2 φ4 = ±z . (60) Defining ˜µ(±):=qµ2

eff±pµ4eff− 2λφ′20 we find

±|φ′ 0|z = √ 2 |φ′ 0| ˜ µ(+) Z Φ(z)˜ 0 d ˜φ s 1 − ˜φ2  1 − µ˜(−)2 ˜ µ(+)2φ˜ 2  = √ 2 |φ′ 0| ˜ µ(+) F  arcsin ˜Φ(z),µ˜(−) ˜ µ(+)  , (61) and obtain φ(z) = √ 2 |φ′ 0| ˜ µ(+) sin  F−1  ±µ˜ (+) √ 2 z, ˜ µ(−) ˜ µ(+)  . (62)

With the Jacobi elliptic function

sn(u, ℓ) = sin F−1(u, ℓ) , (63) and

sn(−u, ℓ) = −sn(u, ℓ) , (64) we can write this as

φ(z) = ± √ 2 |φ′ 0| ˜ µ(+) sn  ˜µ(+) √ 2 z, ˜ µ(−) ˜ µ(+)  . (65) B. ”Symmetric Phase”

Again, we can read off the solution inside the mirrors directly from the corresponding solution in the 1-mirror case φ(z) =r 2ρeff λ 1 M z |z| × 1 sinh √ρ eff M (|z| − d) + sinh −1r 2ρeff λ 1 M φd  . (66) C. Boundary Conditions

Using the boundary conditions at the surface of the mirror 1 2  dφ dz 2 z=d− = 1 2  dφ dz 2 z=d+ , (67) we find 1 2φ ′2 0 − µ2 eff 2 φ 2 d+ λ 4 φ 4 d= ρeff 2M2φ 2 d+ λ 4 φ 4 d, (68) or |φd| = |φ ′ 0| µeff r 1 + ρeff M2µ2 eff . (69)

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Using this in the second boundary condition φ(d) = φ(d+) , (70) gives ˜ µ(+)=√2 µeff r 1 + ρeff M2µ2 eff sn ˜µ (+) √ 2 d, ˜ µ(−) ˜ µ(+)  . (71)

The solution of the equation above gives the absolute value of φ′

0. It will give all solutions for which φ′(d−) =

±φ′(d

+), hence one still has to extract the subset of

so-lutions satisfying φ′(d

−) = φ′(d+).

D. Final Solution

Finally, we obtain the solution

φ(z) = ±Θ(d − |z|) √ 2 |φ′ 0| ˜ µ(+) sn  ˜µ(+) √ 2 z, ˜ µ(−) ˜ µ(+)  ± Θ(|z| − d)r 2ρλeffM1 z |z| × sgnA sinh √ρ eff M (|z| − d) + sinh −1r 2ρeff λ 1 M φd  . (72) where sgnA:= sn ˜µ (+) √ 2 d, ˜ µ(−) ˜ µ(+)  sn ˜µ (+) √ 2 d, ˜ µ(−) ˜ µ(+)  , (73) and |φ′ 0| is solution of ˜ µ(+)=√2 µeff r 1 + ρeff M2µ2 eff sn  ˜µ(+) √ 2 d, ˜ µ(−) ˜ µ(+)  , (74) and φd = ± |φ ′ 0| µeff r 1 + ρeff M2µ2 eff . (75)

As mentioned before, Eq. (74) will give all solutions for which φ′(d

−) = ±φ′(d+), hence one still has to extract

the subset of solutions satisfying φ′(d

−) = φ′(d+).

For most situations of interest, where ρeff ≫ M2µ2eff,

the left-hand side of Eq. (74) is typically small, while the factor q1 + ρeff

M2µ2

eff is comparably large. Hence, a

solu-tion demands that the Jacobi funcsolu-tion sn should be close to its zero, which is the case for a vanishing first argu-ment. For the first zero this leads to solutions obeying φ′(d

−) = −φ′(d+) rather than φ′(d−) = φ′(d+). The

-4 × 1010 -2 × 1010 2× 1010 4× 1010

1.× 10-9

FIG. 1. The 1 mirror solution is depicted for the parameters of Tab. I. One can see that the condition φ/M ≪ 1, necessitated by Eq. (3), is indeed satisfied in this case.

Taylor expansion around the second zero of the Jacobi function sn reads

sn(u, ℓ) = − u − 2F (π/2, ℓ) + O u − 2F (π/2, ℓ)3 . (76) Hence to linear order we find

˜ µ(+)=√2 µeff r 1 + ρeff M2µ2 eff ˜ µ(+)d √ 2 − 2F  π 2, ˜ µ(−) ˜ µ(+)  , (77) respectively ˜ µ(+)d =√2 2F π 2, ˜ µ(−) ˜ µ(+)  ± µ˜ (+) µeff r 1 + ρeff M2µ2 eff ≃√2 2F π 2, ˜ µ(−) ˜ µ(+)  . (78) Since √ 2 2F π 2, ˜ µ(−) ˜ µ(+)  ∈ [√2 π, ∞), (79) and ˜ µ(+)∈ [µeff, √ 2 µeff] , (80) we have µeffd & π . (81)

In the limit ρeff → ∞ the inequality above becomes

rig-orous, viz. for µeffd < π no solution can exist in this

limit. Here, d must be twice as large as in the case of the symmetric solution, which is intuitively clear.

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VI. DISCUSSION OF THE 2 MIRROR SOLUTIONS

An important question to consider is whether the sym-metric and anti-symsym-metric solutions exhaust the set of possible solutions or whether there are still others with-out definite (anti-)symmetry? One may argue that a so-lution without any symmetry must satisfy two bound-ary conditions (continuity of φ(z) and φ′(z)) for the left

boundary and a set of two different boundary conditions for the right boundary. If we start with a symmetric so-lution, then all four boundary conditions are satisfied. Moving the right mirror to generate another solution without symmetry, we keep the parameter k fixed (in or-der to still satisfy the left boundary conditions) and vary the distance dRto the right mirror. For a given value of

k, this leads to a resonance equation for dR whose

solu-tion is either the original one dR= dL or correspond to

new solutions. One may construct new solutions which are either symmetric again or anti-symmetric but will have a different number of nodes from the solution we started with. This depends on the value of dR. A more

thorough discussion of the possible existence of other so-lutions together with the symmetric and anti-symmetric ones is left for future work. On the other hand, it is not obvious how the symmetric and anti-symmetric solutions should be expected to appear in an experimental setup. In particular whether the solutions are stable or not and which one is the stablest. One can expect that the so-lution with the fewest number of zeros between the two mirrors should be the ground state, although we have no proof of that assertion and further study is certainly necessary to tackle this important question.

VII. A PARTICULAR EXAMPLE

Let us focus on a particular example for the parameters given in Tab. I for 2 mirrors. This illustrates the differ-ent cases presdiffer-ented in the previous sections and provides an example of multiple solutions. The corresponding 1 mirror solution with k = 0.214967 is depicted in Fig. 1.

Mode E [MeV] k |φ′

0| [MeV2]

0+ −1.76369 × 10−28 0.997142

1− −8.30488 × 10−29 6.9018 × 10−20

2+ −9.36839 × 10−30 0.451367

TABLE I. Values of energy E are given for the three solutions for values ρeff = 1.082 × 10−5 MeV4, M = 107 MeV, µeff =

10−10 MeV, λ = 10−2 and mirror distance d = 9.5 × 10−3 m.

The energies

E = Z ∞

−∞dz H(z) ,

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where the Hamiltonian H(z) is given by Eq. (7), for the complete set of three solutions are also given in Tab. I and the corresponding field profiles are depicted in Fig. 2. We denote solutions, viz. modes by their number of nodes 0, 1, 2 and an upper index + to denote symmetry as well as − to denote anti-symmetry. From Tab. I one can see that the 0+mode has lowest energy, 1a higher one and

2+ the highest energy. The energy level E = 0 would

correspond to the vanishing solution φ(z) = 0. These three solutions exhaust the spectrum of possible solutions for these parameter values and exhibits clearly how anti-symmetric and multiple node solutions are a crucial part of the complete set of solutions.

Intuitively, we expect the energies to increase with the order of the nodes not only in this particular example but also in general. Clearly, depending on the parameter values and distance between the mirrors one can have an arbitrary number of nodes.

VIII. SYMMETRON INDUCED FREQUENCY SHIFT IN Q BOUNCE

In this section we derive limits obtained from the qBOUNCE experiment [30, 31, 34] using the exact so-lutions obtained herein. In this experiment, ultra-cold neutrons are dropped in earth’s gravitational potential and reflected by a neutron mirror, which has been re-ported for the first time in [35]. The energy eigenstates are discrete and allow to apply the method of resonance spectroscopy. The basic setup is described in [31]. In the most recent version of the experiment, Rabi-spectroscopy has been realized with energy resolution 3×10−15 peV

[29].

The experimental setup is such, that ultra-cold neu-trons pass three regions, while being reflected on polished glass mirrors. In [29], the resonance spectroscopy tran-sition between the energy ground state E1 = 1.40672

peV and the excited states E3 = 3.32144 peV as well

as E4 = 4.08321 peV have been demonstrated. First,

the neutrons pass region I which acts as a state selector for the ground state |1i having energy E1. A polished

mirror at the bottom and a rough absorbing scatterer on top at a height of about 20 µm serve to select the ground state. Neutrons in higher, unwanted states are scattered out of the system. This region has a length of 15 cm. Subsequently, in region II, a horizontal mirror performs harmonic oscillations with a tunable frequency ω, which drives the system into a coherent superposi-tion of ground and excited states. The length of this region is 20 cm. Finally, region III is identical to the first region and hence acts again as a ground state selec-tor. The quantum-mechanical description of a neutron above a mirror in the gravitational potential is given by the Schr¨odinger equation [36]. After separation into free

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-1 × 10 5× 10 10 × 10 1× 1011 2.× 10-10 4.× 10-10 6.× 10-10 8.× 10-10 1.× 10-9 -1 × 10 1010 5 × × 1011 -5. × 10-10 5.× 10-10 -1 × 10 1010 5 × × 1011 -4. × 10-10 -2. × 10-10 2.× 10-10 4.× 10-10

FIG. 2. The field profiles for the three solutions from Tab. I are depicted. Again, one can see that the condition φ/M ≪ 1, necessitated by Eq. (3), is indeed satisfied in this case.

transversal and bound vertical states

Ψ(0)n (x, t) = e~i(p⊥·x⊥−E⊥t) 2π~v ψ (0) n (z) e− i ~Ent, (83) it reads −~ 2 2m ∂2ψ n(z) ∂z2 + mgz ψn(z) = Enψn(z) . (84)

The characteristic length scale

z0= 3

s ~2

2m2g = 5.87 µm , (85)

and energy scale E0 = ((~2mg2)/2)1/3 are given by the

mass m of the neutron and the acceleration of the earth g. With the substitution

σ = 3 r 2m2g ~2  z −mgEn  ≡z − zz n 0 , (86) Eq. (84) becomes d2ψ˜ n(σ) dσ2 − σ ˜ψn(σ) = 0 , (87)

which is Airy’s equation.

From the effective symmetron potential

Veff(φ) = 1 2  ρ M2 − µ 2φ2+λ 4φ 4, (88)

we can deduce the semi-classical neutron-symmetron cou-pling as Veff= 1 2 m M2ψ∗ψ φ 2. (89)

There are some subtleties involved here, which will be discussed in the Appendix. The corresponding quantum mechanical perturbation potential is given by

V = 1 2

m M2φ

2, (90)

and leads to a resonance frequency shift (see e.g. [37]) δE(1) mn≡ Em(1)− En(1) = 1 2 m M2 Z ∞ −∞ dz ψm(0)(z) 2 − ψ(0)n (z) 2 φ(z)2. (91) Likewise, the first order correction to the wavefunctions reads (see e.g. [37])

ψn(1)(z, t) = 1 2 m M2 × ∞ X m=1\n Z ∞ −∞ dz′ψ(0)∗m (z′) ψn(0)(z′) φ(z′)2 En(0)− Em(0) ψ(0) m(z) e− i ~Ent. (92) Hence, the correction to the density ̺(0)n (z) =

ψn(0)∗(z)ψ(0)n (z) to first order is given by

̺(1)n (z) = 2 Re ψn(0)∗(z)ψn(1)(z)  = m M2Re ∞ X m=1\n Z ∞ −∞ dz′ψm(0)∗(z′) ψ(0)n (z′) φ(z′)2 En(0)− Em(0) × ψ(0)∗ n (z)ψm(0)(z) . (93)

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In the 1-mirror case the unperturbed normalized wave-function for z > 0 reads (see e.g. [38])

ψn(0)(z) = Cn(1)Ai z − zn z0  , (94) with normalisation Cn(1)= 1 √z 0Ai′  −zzn 0  , (95) and zn = En

mg. Outside this region the wavefunction vanishes. The first few energy levels are given in table II. For a single mirror extending to z ≤ 0 we can use Eq. (30)

State Energy [peV] |1i E1= 1.40672 |2i E2= 2.45951 |3i E3= 3.32144 |4i E4= 4.08321 |5i E5= 4.77958 |6i E6= 5.42846

TABLE II. Values of the energy of the lowest six states for the neutrons in the terrestrial gravitational field.

and obtain for the resonance frequency shift δEmn(1) = 1 2 mN M2 µ2 eff z0λ Z ∞ 0 dz tanhµ√effz 2 + tanh −1k2 × (Ai z − zm z0 2 Ai′−zm z0 2 − Aiz − zn z0 2 Ai′−zn z0 2 ) . (96)

It is straightforward to find all the corresponding ex-pressions in the 2-mirror case. These exex-pressions are very elaborate in their full detail and hence we will refrain from reproducing them herein.

In Table III we summarize the resonance frequency shifts for a large range of symmetron parameters for the case of a single mirror. Notice that larger values of λ give smaller energy shifts for a given µeff. Similarly,

in-creasing M leads to smaller deviations. The whole pa-rameter space of symmetrons can therefore be efficiently constrained using the exact results obtained herein. A sophisticated χ2analysis has been carried out in parallel

to this work [29], where the solutions obtained herein for a 1 mirror setup are used to extend the exclusion region of the symmetron parameter space.

IX. CONCLUSION

We have derived exact analytical solutions to the sym-metron field theory in the presence of a one or two mirror

δE(1)14 [eV] M [MeV] λ 4.49795 × 10−14 1 10−10 4.49795 × 10−20 1 10−4 4.49795 × 10−26 1 102 4.51053 × 10−20 103 10−10 4.51053 × 10−26 103 10−4 4.51053 × 10−32 103 102 5.75763 × 10−24 105 10−10 5.75763 × 10−30 105 10−4 5.75763 × 10−36 105 102

TABLE III. δE(1)41 for 1 mirror, viz. Eq. (96), for the values

ρ = 2.51 g/cm3 and µ

eff = 10−3 meV. Notice larger values

of λ and M lead to smaller energy shifts. The current exper-imental bound at the 10−15 eV level leads to constraints on parameters which can be easily extracted using our analysis [29]. For instance for M = 1 MeV, we find that typically one can expect that λ & 10−10. The exact excluded regions in parameter space, where the screening of the neutron is taken into account, can be found in [29] (see the Appendix for a discussion).

system. The one dimensional equations of motion have been integrated in each case. Surprisingly, in the two mirror case instead of a unique solution for a given en-vironment and boundary conditions, we have obtained a discrete set of solutions with increasing number of nodes and energies. We have derived bounds for the qBOUNCE experiment and found agreement with those obtained in the study carried out in parallel to this work [29]. We have summarized the resonance frequency shift for a large range of symmetron parameters in the case of a single mirror. The solutions that we have obtained herein, and in particular their stability, should be the subject of fur-ther studies as they will play a crucial role in the analysis of neutron interferometry experiments searching for sym-metrons and for the calculation of the symmetron field induced ”Casimir force” in the CANNEX experiment.

ACKNOWLEDGMENTS

We thank Hartmut Abele, Guillaume Pignol and Ren´e Sedmik for fruitful discussions. This work is supported in part by the EU Horizon 2020 research and innova-tion programme under the Marie-Sklodowska grant No. 690575. This article is based upon work related to the COST Action CA15117 (CANTATA) supported by COST (European Cooperation in Science and Technol-ogy).

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Appendix A: Symmetron Field of a Neutron

Since neutrons are used in the search for symmetrons in qBOUNCE experiments and neutron interferometry, it is important to understand their interaction with the sym-metron. For a certain parameter regime this interaction between neutron and symmetron becomes strong. In this case, the neutron affects the background symmetron field as generated by the mirrors of the experimental setup in a non-negligible way, which in turn weakens the effect on the neutron, viz. screening of the neutron sets in. Since we treat the symmetron as a classical field theory a con-sistent description of its coupling to a quantum mechan-ical system is beyond our reach. Therefore, we employ a semi-classical treatment in which the neutron’s probabil-ity distribution times its mass acts as the source of the symmetron as defined in Eq. (89). In the following we will use a pragmatic approach, which follows what has been done in the literature so far [22–24, 39–42].

A first approach to defining the mass density is anal-ogous to ”semi-classical gravity”, which has been intro-duced by Møller [43] and Rosenfeld [44], and where the operator-valued stress-energy tensor is replaced by an ex-pectation value. Its non-relativistic Newtonian approx-imation, the so-called ”Schr¨odinger-Newton equation”, was introduced by Di´osi [45] and Penrose [46]. This equa-tion employs the probability distribuequa-tion times the mass as the source of the gravitational potential, i.e. consid-ering that the square modulus of the wave function can be used as tracing the mass density of a particle. The ”Schr¨odinger-Newton equation” has been applied to sin-gle particles by Moroz, Penrose and Tod [47], although it can only be understood in the Hartree approximation for a large ensemble of particles, see [48] for a summary of the issues related to this equation. In another approach, which is commonly used in the literature, one may con-sider that the neutron has a well-defined size provided by the quark-gluon dynamics and that in our non-relativistic treatment of the neutron at energies well-below the QCD scale this size of the order of 1 fm provides a reasonable description for its screening. Clearly, a rigorous treat-ment of this issue and distinguishing both approaches is beyond the scope of this paper. In [29], we illustrate the two cases and stress that more work needs to be done to derive completely rigorous bonds for the coupling of the symmetron to neutrons.

For a quantitative account we treat the neutron as a classical sphere. As a conservative case we take as its diameter 2Rn= 1 fm, corresponding to 5 × 10−3MeV−1

in natural units, and nuclear density ρn = 1.53 × 1010

MeV4. This is the most commonly used manner of

treat-ing the screentreat-ing of the neutron. On the other hand, if the mass distribution were related to the size of its wave function, the neutron would be approximately described as being distributed over a sphere of diameter of order z0 = 5.87 µm, which is the typical length scale for its

vertical wavefunction in the qBOUNCE experiment and leads to a much weaker screening.

200 400 600 800 1000 1200 1400 0.2 0.4 0.6 0.8 1.0 0.002 0.004 0.006 0.008 0.010 0.012 0.014 0.001 0.002 0.003 0.004 0.005

FIG. 3. Top: Here the screening charge Q is plotted as a function of the coupling parameter M . The parameters taken are for a neutron, viz. R = 2.5×10−3MeV−1, ρn= 1.53×1010

MeV4 and µ = 5 × 10−8 MeV. Bottom: The field profile of a

sphere (neutron) is plotted as a function of the radial distance of the center of the sphere. The blue line corresponds to M = 500 MeV and the yellow line to M = 50 MeV with the other parameters having the same values as those used for the upper plot. The blue square is bounded by the radius of the sphere (neutron) and the vacuum field value φ = φV.

As a result, we have to solve the field equation for a static massive sphere with radius R

d2φ dr2 + 2 r dφ dr =  ρ M2 − µ 2φ + λ φ3. (A1)

The boundary conditions are φ′(0) = 0 ,

lim

r→∞φ(r) → ±φV . (A2)

Without loss of generality we take the asymptotic value +φV. Inside the sphere we take ρ ≫ M2µ2 and since

 ρ M2 − µ 2φ λ φ3 ≃ ρ M2λ φ2 & ρ M2λ φ2V = ρ M2µ2 ≫ 1 , (A3) we find d2φ dr2 + 2 r dφ dr ≃ ρ M2φ . (A4)

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With

φ = ϕ

r , (A5)

Eq. (A4) reads

d2ϕ

dr2 ≃

ρ

M2ϕ , (A6)

providing the general solution

φ = Ae √ρ Mr r + B e−√Mρr r , (A7) with arbitrary constants A and B. The solution conver-gent for r → 0 and satisfying the boundary condition φ(0) = 0 is given by B = −A =: −C/2 φ(r) = C sinh √ρ M r  r . (A8)

Immediately outside the sphere, the potential contribu-tion is very small compared to the kinetic parts and can be approximated by the value close to the vacuum value

ρ

M2φ(R−) + λ φ(R−) 3

≫ −µ2φ(R+) + λ φ(R+)3

≃ 2µ2(φ(R+) − φV) . (A9)

Hence, for the outside solution we find d2φ dr2 + 2 r dφ dr ≃ 2µ 2 (φ − φV) . (A10)

The general solution is given by

φ − φV = E

e√2µr

r + D e−√2µr

r , (A11) with arbitrary constants E and D. The solution conver-gent for r → ∞ and satisfying the boundary condition limr→∞φ(r) → φV is given by

φ(r) = φV + D

e−√2µr

r . (A12) The boundary conditions at the surface of the sphere are given by

φ(R) = φ(R+) ,

φ′(R−) = φ′(R+) , (A13)

yielding after some algebraical manipulations

C = φV √M ρ 1 cosh √ρ M R  1 +√2µR 1 +√2µM ρ tanh √ρ M R , D = −φVR e √ 2µR 1 −√MρR1 tanh √ρ M R  1 +√2µM ρ tanh √ρ M R . (A14)

Finally, we find the solution

φ(r) =                          φV√M ρ 1 cosh √ρ M R  1 +√2µR 1 +√2µM ρ tanh √ρ M R  × sinh √ρ M r  r , for r ≤ R , φV − ρR3 3M2 QφV 1 +√2µR e−√2µ(r−R) r , for r ≥ R , where we have introduced a screening charge, which we define as Q= 1 +√2µR 3M2 ρR2 1 − √MρR1 tanh √ρ M R  1 +√2µM ρ tanh √ρ M R . (A15) With this definition

Q

0 , for screened bodies with R ≫ M/ρ , 1 , for unscreened bodies with R ≪ M/√ρ . The acceleration of a small test body, which does not disturb the field, in the outer field of the sphere is

~a = −Mφ2∇φ .~ (A16)

Asymptotically for large r we find

~a = −Q φ2V ρR3 3M4 √ 2µ 1 +√2µR e−√2µr r ~r r , (A17) justifying the definition of Q as a screening charge, which has to be multiplied to the transition energies. Hence, in order to account for the neutron’s screening one has to replace

δEpq→ Q(µ, M) δEpq, (A18)

for the extraction of the experimental limits. This re-placement has been carried out in the sophisticated anal-ysis [29] but for simplicity has been neglected for all bounds derived in this letter.

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Figure

FIG. 1. The 1 mirror solution is depicted for the parameters of Tab. I. One can see that the condition φ/M ≪ 1, necessitated by Eq
TABLE I. Values of energy E are given for the three solutions for values ρ eff = 1.082 × 10 −5 MeV 4 , M = 10 7 MeV, µ eff = 10 −10 MeV, λ = 10 −2 and mirror distance d = 9.5 × 10 −3 m.
FIG. 2. The field profiles for the three solutions from Tab. I are depicted. Again, one can see that the condition φ/M ≪ 1, necessitated by Eq
TABLE II. Values of the energy of the lowest six states for the neutrons in the terrestrial gravitational field.
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