Preheating and locked inflation: an analytic approach towards parametric resonance
PPrepared for submission to JCAP
Preheating and locked inflation: ananalytic approach towards parametricresonance
Lingfei Wang a,b a Physics Department, Lancaster University, Lancaster LA1 4YB, UK b School of Physics, Nanjing University, 22 Hankou Road, Nanjing 210093, ChinaE-mail: [email protected]
Abstract.
We take an analytic approach towards the framework of parametric resonance andapply it on preheating and locked inflation. A two-scalar toy model is analytically solved forthe λφ χ coupling for the homogenous modes. The effects of dynamic universe backgroundand backreaction are taken into account. We show the average effect of parametric resonanceto be that χ ’s amplitude doubles for each cycle of φ .Our framework partly solves the broad resonance for preheating scenario, showing twodistinct stages of preheating and making the parameters of preheating analytically calculable.It is demonstrated for slowroll inflation models, preheating is terminated, if by backreaction,typically in the 5th e-fold. Under our framework, a possible inhomogeneity amplificationeffect is also found during preheating, which both may pose strong constraints on someinflationary models and may amplify tiny existing inhomogeneities to the desired scale. Fordemonstration, we show it rules out the backreaction end of preheating of the quadraticslowroll inflation model with mass m ∼ − . For locked inflation, parametric resonance isfound to be inhibited if φ has more than one real component. a r X i v : . [ h e p - t h ] D ec ontents Inflation was first proposed in [1] 30 years ago to solve multiple problems previously encoun-tered by the Hot Big Bang theory. It soon received public attention and subsequent worksof others[2–9] have greatly perfected our understanding of inflation. Inflation has becomeso popular that almost all high energy theories have expressed their personal views on howinflation may be achieved, e.g. [10–14]. Even now, after 30 years of development, articles arestill being entitled “inflation” on a daily basis.At the same time, reheating, the immediate subsequent scenario to inflation, was more orless in oblivion. It was not until 1994 that Kofman et al. came to realize[15] that the explosiveparticle production, which they named as preheating , should be first processed before theordinary perturbative method is applied. Three years later, they proposed the analysis ofreheating[16], covering numerous factors taken into account. (See [17] for a recent review.)Besides broad resonance, the narrow resonance also came into people’s sight in the 1990sas a different non-perturbative particle production process during reheating[18, 19]. Theanalytic study of preheating however remains stalled afterwards, although numerical studiesand applications on inflationary models have been going on anyhow. Despite that someapplications of the theory in [16] have been quite successful, it is still very mathematicallyinvolved, and therefore, some of the underlying physics are not revealed.On the other hand, locked inflation[9] was proposed as a fastroll inflationary scenario.However, it was later found to be subject to multiple constraints, which together ruled out thewhole parameter space[20]. Among them, parametric resonance is a crucial one which mayterminate locked inflation much earlier than expected. Recently, locked inflation is recalledin a cyclic universe model which incorporated D-branes and curvature[21], bringing locked– 1 –nflation and locked inflationary contraction to each cycle. The proposed model howeverfaces the same parametric resonance problem as locked inflation does.In this article, we construct an analytic framework of parametric resonance for thehomogenous mode, in a way more straightforward and less involved. We then apply it onpreheating and locked inflation, and discuss the possible inhomogeneity amplification effect.First, from section 2.1 to section 2.3, we solve a half cycle of parametric resonance andshow the cause of the exponential boost effect. In section 2.4, we add up the effect cycle bycycle and give the exact solution to the Mathieu equation. We then put universe expansioninto consideration in section 3.1 and investigate how backreaction acts as the terminator ofparametric resonance in section 3.2. Such approaches then allow the discussion of preheatingand locked inflation in section 4.1 and section 4.2 respectively. A brief discussion is carriedout in section 5 on how parametric resonance may amplify the existing inhomogeneity andconstrain inflationary models. The results are summarized in section 6.
To construct the framework, we consider the simple case first in a static background withLagrangian density L = − m φ − M χ − λφ χ − ∂ µ φ∂ µ φ − ∂ µ χ∂ µ χ. (2.1) φ and χ are real scalar fields with bare masses m and M . Their interaction strength ischaracterized by the dimensionless parameter λ . Here we consider the configuration of λ | φ | (cid:29) m (cid:29) λχ and m (cid:29) M to neglect all backreactions. | | indicates the ampli-tude of oscillation when operating on a field. We allow M to be negative, but still use M to indicate the absolute value of M for convenience when we compare it with other values.In many of our calculations however, we just neglect M because it’s too small. Throughoutthis paper, we will stick to the homogenous modes of both fields, so the spatial dependencewill also be neglected in the notation. For simplicity, we will also use the word “parametricresonance” to indicate only the parametric resonance within our assumed parameter space.Please note this Lagrangian doesn’t suffer from problems like negative infinite energy fromthe negative M because it’s just the effective part we take out from the total Lagrangian,so terms like the quartic ones don’t distract us. As long as such terms are negligible, onedoesn’t need to worry about the validity of eq. (2.1).From the Lagrangian, we can write down the potential V ( φ, χ ) = 12 m φ + 12 M χ + λφ χ , (2.2)in which m φ / V ( φ, χ ) is visualized infigure 1(a), where the vertical lines also suggests the domination of m φ /
2. More details ofthe potential are displayed in figure 1(b) and figure 1(c).It is not difficult to imagine the motion of the two fields under such a configuration.Under λχ (cid:28) m , φ ’s effective mass squared would be m φ ≡ m + 2 λχ ≈ m and φ willalways be oscillating with frequency m and constant amplitude, unaffected by the motion of χ . We can easily derive its equation of motion¨ φ + ( m + 2 λχ ) φ = 0 , (2.3)– 2 – Χ (a) Landscape of V ( φ, χ ) ΦΧ (b) Magnified with M > ΦΧ (c) Magnified with M < Figure 1 . Contour-plotting of V ( φ, χ ). The first figure gives the overall landscape. The next two arezoomed in to the marked region and redrawn for details with different signs of M . A darker colorimplies a larger potential energy. where dots mean derivatives w.r.t time. After neglecting χ , we get the Simple HarmonicOscillator(SHO) solution φ ( t ) = | φ | sin mt . χ ’s effective mass squared m χ ≡ λφ + M ≈ λφ is much larger than φ ’s most of thetime, so it will oscillate much faster than φ , as long as φ is not very close to zero. When φ crosses zero however, the motion of χ would become very different from the stage of large φ . We will model the two stages in turn, trying to find out why and how χ is boostedexponentially. Still, we write the equation of motion of χ here for future reference.¨ χ + ( M + 2 λφ ) χ = 0 . (2.4) The rolling stage is when φ is not very close to zero, therefore χ is oscillating much fasterthan φ . It also requires that χ ’s effective mass varies slowly and thus can be regarded as aconstant in each cycle of χ ’s oscillation. For most cases in which the rolling stage takes upmore than half of φ ’s oscillation, one only needs to take into account the second condition,and then the first one is automatically satisfied. So the condition of this stage is (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) d m χ m χ d t (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (cid:28) . (2.5)Although this stage can be easily solved with simple approximations, we still would liketo work on it again here, mainly to arrive at the variables we are interested in and confirmthe rolling stage has a generally vanishing total effect on them. Suppose at time t = 0, χ is at maximum with χ = | χ | and ˙ χ = 0, and scalar field φ is rolling down slowly with timederivative ˙ φ . We take the effective mass m χ constant and get the zeroth order SHO solutionof eq. (2.4) χ (0) ≡ | χ | cos m χ t. (2.6)We then derive the equation of motion up to first order χ (1) by introducing ˙ φ ,¨ χ (1) + m χ χ (1) + 4 λφ ˙ φ | χ | t cos m χ t = 0 (2.7)– 3 –nd get the solution with the initial condition χ (1) (0) = 0 and ˙ χ (1) (0) = 0, χ (1) ( t ) = − λφ ˙ φ | χ | m χ (cid:18) m χ t cos m χ t + ( m χ t −
1) sin m χ t (cid:19) . (2.8)At time t = π/m χ , when χ (0) reaches maximum again, we have χ (1) (cid:16) πm χ (cid:17) = πλφ ˙ φ | χ | m χ , ˙ χ (1) (cid:16) πm χ (cid:17) = π λφ ˙ φ | χ | m χ . (2.9)If we take when φ is rolling down as an example, we will find out χ gets an additionalamplitude correction from the decrease of φ . To be precise, we consider the time interval∆ t ≡ π/m χ , during which χ rolls from one maximum to the other. χ ’s amplitude is affectedat a rate ∆ | χ | ∆ t = − χ (1) ( πm χ )∆ t = − λφ ˙ φm χ | χ | . (2.10)Replacing ∆ with differential operator gives | χ | ∝ √ m χ , (2.11)which is the effect on | χ | due to the variation of φ .Similarly, we can calculate χ ’s phase change resulting from the variation of φ , or m χ directly. The motion of φ generates an additional ˙ χ (1) ( π/m χ ) which changes the time neededto push χ back to its maximum again. The phase change is therefore ϕ χ ≡ ˙ χ (1) ( πm χ ) m χ | χ | = π λφ ˙ φm χ = π m (cid:48) χ m χ , (2.12)in which m χ and m (cid:48) χ are the initial and final effective masses of χ .The above calculation has shown how χ is affected by φ ’s motion when φ is away fromzero. Although the effects on χ ’s amplitude (eq. (2.11)) and phase (eq. (2.12)) are strong, χ can’t get exponentially boosted in this way. This is because the effects during the upwardrolling period of φ are exactly inverse from those during the downward period, so theygenerally cancel. χ is increased when φ is rolling towards zero, but it get decreased when φ moves away from zero. The same thing happens to χ ’s phase change, so the net effectvanishes. When φ is small and ˙ φ is relatively large, the condition of rolling stage eq. (2.5) breaks, andthe system enters the zero-crossing stage. The zero-crossing stage is the period φ crosseszero. During this stage, the motion of χ is very much different from that during the rollingstage. It’s because χ ’s effective mass m χ varies very fast in this stage and can become smallerthan φ ’s effective mass m φ when φ is crossing zero.Because m (cid:29) λχ , we still adopt the approximation that φ is not affected by themotion of χ , and M (cid:28) λ | φ | is still negligible. Here we set t = 0 at φ = 0. In mostapplications of parametric resonance, | mt | (cid:28) φ = | φ | mt, (2.13)– 4 –nd thus ¨ χ + 2 λ | φ | m t χ = 0 . (2.14)Here we use τ ≡ t (2 λ | φ | m ) as the scaled time, and primes as derivatives w.r.t τ . Theequation of motion of χ then becomes χ (cid:48)(cid:48) + τ χ = 0 . (2.15)The breaking of condition eq. (2.5) corresponds to | τ | ∼ <
1. If one is only interested inthe behavior of the system in | τ | <
1, one can simply expand χ in Taylor series of τ at τ = 0and only take the first few terms. Numerical simulation tells us, however, the energy booston χ spans not only | τ | <
1, but at an even broader range. To precisely calculate the boosteffect, we need to take the analytic solution of eq. (2.15) χ ( τ ) = F (cid:16) , − τ (cid:17) χ + F (cid:16) , − τ (cid:17) χ (cid:48) τ, (2.16)where F ( a, z ) is the confluent hypergeometric (limit) function , and the subscript 0 denotesthe value of variables at τ = 0.Given the initial condition χ and χ (cid:48) , one can already calculate the effect of parametricresonance. In order to enable our further calculation, we take out the leading term of χ for | τ | (cid:29) | mt | (cid:28) F ( a, − z ) in series of 1 /z . Theleading term is F ( a, − z ) ≈ Γ( a ) √ π sin (cid:16) π a + 1) − √ z (cid:17) z − a , z (cid:29) . (2.17)Here we want to calculate the energy boost effect from parametric resonance, so wecharacterize the boost rate with η ≡
12 lim τ → + ∞ log E s ( τ ) E s ( − τ ) . (2.18)The scaled energy density of χ is defined as E s ( τ ) ≡ τ χ ( τ ) + χ (cid:48) ( τ ) , (2.19)with its relation with the (unscaled) total energy density of χE ≡ λ | φ | m t χ + 12 ˙ χ = (cid:114) λ | φ | m E s . (2.20)Substituting eq. (2.16) and eq. (2.17) into eq. (2.18), we arrive at η = 12 log (cid:18) √ )Γ( ) χ χ (cid:48) Γ ( ) χ − √ )Γ( ) χ χ (cid:48) + 4Γ ( ) χ (cid:48) (cid:19) . (2.21)In the above equation, the limit τ → + ∞ has already been taken. As a result, allnon-leading terms of τ are removed. The 2 √ z in the sin function of F approximation isremoved too.From eq. (2.21) we find the boost rate is hardly dependent on any parameter of thesystem, ( m , λ , etc,) or the energy density of any component, as long as our previous assump-tions stay valid. It is also easily verified by numerical simulations. This constant propertyis very helpful when we sum the boost rates across cycles of φ in subsequent calculations.The fraction in the log function of eq. (2.21) should be of order unity, so the boost effect issignificant and exponential. The boost rate η acquires the same sign with χ χ (cid:48) , which meansthe energy of χ can either be boosted or dampened, depending on the sign of χ χ (cid:48) .– 5 – .3 Phase Delay In the above subsection, we have represented η with χ and χ (cid:48) , the motion state of χ at φ = 0. The previous calculation however only gives half of the analytic solution — we cancalculate η with any given χ and χ (cid:48) , but we haven’t developed any method in derivingthem two, except numerical ways. Therefore we will derive χ and χ (cid:48) analytically in thissubsection and provide a completely analytic solution. But before that, it’s quite importantto first clarify what “phase” is referred to here.The system has two degrees of freedom, φ and χ , which means we need four independentparameters to fully describe the motion state of the system at any specific time. We areinterested in the two parameters characterizing χ , which can either be chosen as χ and χ (cid:48) ,or its amplitude and phase. From eq. (2.21), we have already learned the boost rate η isa function of χ and χ (cid:48) . Although χ is not strictly a SHO at the zero-crossing stage, ifwe manage to represent the motion of χ at φ = 0 with an amplitude-like and a phase-likevariable however, the amplitude-like variable would cancel and η would then only dependon the phase-like variable. It’s obvious η may not stay constant in every half cycle of φ ,so reducing η to a single-parameter function allows us to track η better when studying theboost effect across many cycles of φ .In order to find the phase representation of χ at φ = 0, we attempt to solve χ and χ (cid:48) reversely. It is reasonable to believe χ is SHO-like at | τ | (cid:29) χ ( τ ) = (cid:112) E s ( τ ) | τ | sin β, χ (cid:48) ( τ ) = (cid:112) E s ( τ ) cos β (2.22)where β is the variable we use to represent the phase of χ at τ . We are only interested in β here, so we come up with the combination | τ | χ cos β − χ (cid:48) sin β = 0 . (2.23)By putting in the solution eq. (2.16) and eq. (2.17), it simplies toΓ (cid:16) (cid:17) sin (cid:16) β + 12 τ − π (cid:17) χ = 2Γ (cid:16) (cid:17) sin (cid:16) β + 12 τ − π (cid:17) χ (cid:48) . (2.24)For a specific process with initial conditions given, χ and χ (cid:48) should be definite andthus independent of the choice of τ in eq. (2.24). For this reason, we contract the τ / β and redefine β as β = β ( τ ) ≡ β − τ + 78 π, (2.25)where β is the variable for phase but it’s now independent of time τ . For short, we willhowever still use β to indicate β . In this way, eq. (2.24) also becomes independent of τ , andsimplifies to χ Γ (cid:16) (cid:17) cos (cid:16) β − π (cid:17) = 2 χ (cid:48) Γ (cid:16) (cid:17) sin β. (2.26)This immediately gives the solution χ = 2Γ (cid:16) (cid:17) sin β X, (2.27) χ (cid:48) = Γ (cid:16) (cid:17) cos (cid:16) β − π (cid:17) X, (2.28)– 6 – Π Π Π Β(cid:45) (cid:45) x max (cid:72)(cid:160) x (cid:164)(cid:76) Figure 2 . χ and χ (cid:48) as functions of β . x (in the label of y -axis) represents χ and χ (cid:48) for the blueand purple curve respectively. The max goes over all β ∈ [0 , π ) so all functions are normalized to bemaximally at unity. The dashed curve is the expected χ (cid:48) curve derived from χ if χ were a SHO allthe time. The purple curve’s deviation from the dashed curve shows the phase delay in χ ’s kineticpart at τ = 0. Parameters are chosen as λ = 1 , m = 1 , M = 10 − , | φ | = 100, and they are chosenas such by default in the following figures. where X is a function of β and E s ( τ ) and is not important in our calculation.By comparing it with the SHO solution, or the SHO-like solution eq. (2.22), one canfind out the motion of χ at τ = 0 is indeed non-SHO. The phase of its kinetic term is delayedby π/
4. This is also confirmed numerically for all β ∈ [0 , π ) in figure 2. Such a phase delaycan be understood from the decreasing transfer efficiency from χ ’s potential energy to kineticenergy as φ is approaching zero. Given the β that makes χ = 0, i.e. χ at the bottom of itspotential at τ = 0, such β doesn’t induce the largest χ (cid:48) any more. To understand it better,we compare such β with ˜ β = β + δβ , where 0 < δβ (cid:28)
1. The difference caused by δβ istwo-fold. On one hand, the phase ˜ β makes χ reach every value slightly earlier than phase β ,and such difference in timing gives ˜ β a larger effective mass of χ when compare at the samevalue of χ , and in consequence, a larger energy transfer efficiency of order δβ . On the otherhand, for ˜ β , χ gets an extra up-climbing period just before τ = 0. This effect decreases theenergy transfer efficiency of ˜ β , but only has order δβ . As a result, the net effect is ˜ β hasa larger energy transfer efficiency and therefore a larger kinetic energy. The maximum ofkinetic energy is then shifted towards a larger β , and its phase gets delayed relatively.We continue our calculation by substituting the set of solution eq. (2.27) and eq. (2.28)into the expression of η , eq. (2.21), and it becomes a very simple function of βη ( β ) = 12 log (cid:18) √ β cos (cid:16) β − π (cid:17)(cid:19) . (2.29)We are now able to compare our results with numerical ones. In figure 3, three curves of η are drawn — the analytical result eq. (2.29), the numerical solution, and the semi-analyticalresult which is acquired by substituting the numerically derived χ and χ (cid:48) into eq. (2.21).All three curves match perfectly. – 7 – Π Π Π Β(cid:45) Η Figure 3 . The energy boost effect w.r.t β . All curves are horizontally shifted to be starting at η = 0 and d η/ d β > β = 0, so they are aligned. The curves of analytical solution, numericalsimulation and semi-analytic result overlap perfectly so only one curve is visible. So now we have successfully reduced the parametric resonance effect to a completelyconstant one — it depends only on β , the phase of χ . Ths boost rate is however independentof any of the parameters of the system, λ , m , etc, or even the energy density of χ . This isindeed the case from numerical simulations. Because the boost rate is independent of theenergy of χ , we can simply sum the boost rates when calculating the total boost rate formany cycles of φ .If we naively assume β is evenly distributed in [0 , π ), and observe many enough inde-pendent half cycles of parametric resonance, the average boost rate can then be calculated,which is (cid:104) η (cid:105) ≡ π (cid:90) π η ( β )d β = 0 . . (2.30)This actually means for every half cycle of φ , χ ’s energy density becomes e (cid:104) η (cid:105) = 2 times ofits original. For future reference, here we also define the maximum boost rate η m ≡ max β ∈ [0 , π ) η ( β ) = η (cid:16) π (cid:17) = log(1 + √
2) = 0 . . (2.31) We now continue to study the long time evolution of χ across cycles of φ . Suppose the initialphase of χ is β , by which we mean initially, when φ reaches its maximum, there is χ = (cid:115) Eλ | φ | sin β, ˙ χ = √ E cos β. (2.32)We can then use the boost rate η ( β ) acquired in the last subsection to evolve the systemfrom one maximum of φ to the next. To characterize the state of χ , at every maximum of φ we use a polar coordinate system as the phase diagram of χ . The angular coordinate ischosen to be β , and the radial coordinate is e η a , where η a ≡ (cid:80) previous η is the total boost rate.– 8 – (cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45) (cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43) (cid:45) (cid:45) (cid:45) (cid:45) Figure 4 . Phase diagram of χ at the neighboring maxima of φ . The initial condition at the earliermaximum of φ is set to a smooth unit circle with mark “–”. Sign “+” marks the later maximum, i.e.after half a cycle of φ . Plots are colored according to their initial phase β , so every “–” mark evolvesto its corresponding “+” mark with the same color. The rotation of colors indicates the phase shift— the difference between the β ’s at the neighboring maxima of φ . Therefore the x and y coordinates are proportional to ˙ χ and χ respectively. If we choose theinitial condition as a constant energy density with evenly distributed β , it then correspondsto a smooth circle with unit radius on the phase diagram. For a general overview, numericalresults are given in figure 4, figure 5 and figure 6.These figures suggest a set of linear operations be applied on each “–” mark on thephase diagram, in order to get to its corresponding “+” mark with analytical methods. Thelinearity is also determined by the linear equation of motion of χ . To clearly demonstrate theoperations are linear, we represent plots on the phase diagram with (1 ,
2) matrices ( x, y ) T of their Cartesian coordinates, and model the total evolution of field χ between neighboringmaxima of φ as a (2 ,
2) linear transformation matrix, which we call here as the transformationmatrix . The transformation matrix should then be decomposed into two parts:1. During the zero-crossing stage, χ is boosted with the boost rate η a function of phase β . So for an initial circle on the phase diagram, it is distorted to an ellipse because ofthe phase dependence of η . For convenience, we rotate the ellipse by shifting β to alignits major axis with x -axis. If the initial circle has a radius e η a , we know at once thesemi-major axis has length e η a + η m , and the length of semi-minor axis is e η a − η m . Thelargest negative boost rate − η m comes from the time-reversibility of χ ’s equation of– 9 – (cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45) (cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43) (cid:43)(cid:43) (cid:43)(cid:43) (cid:43)(cid:43) (cid:43)(cid:43) (cid:43)(cid:43) (cid:43)(cid:43) (cid:43)(cid:43) (cid:43)(cid:43) (cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43) Figure 5 . Phase redistribution over half a cycle of φ . This figure aims to show how the evolutionover half a cycle of φ affects the distribution of β , i.e. the phase of χ . It is based on figure 4, andderived by scaling every “+” plot to the same radius while keeping its phase β fixed. motion. When time-reversed, the largest (positive) boost rate η m always correspondsto the largest negative boost rate − η m and proves its existence. During this step, theplot in the phase diagram is transformed to ( x (cid:48) , y (cid:48) ) T which satisfies (cid:18) x (cid:48) y (cid:48) (cid:19) = (cid:18) e η m e − η m (cid:19) (cid:18) xy (cid:19) . (2.33)2. The ellipse is rotated with an angle which is generated from the difference of the periodsof φ and χ . This is because the period of φ is not an integer times of the period of χ .Therefore χ doesn’t return to the exactly same position when φ reaches maximum (orzero) again as where it was at φ ’s last maximum (or zero). One can easily confirm therotation angle is independent of phase β , so the elliptical shape is preserved. Here wesuppose the rotation angle that should be added to β is δ , so the plot is transformed to (cid:18) x (cid:48)(cid:48) y (cid:48)(cid:48) (cid:19) = (cid:18) cos δ − sin δ sin δ cos δ (cid:19) (cid:18) x (cid:48) y (cid:48) (cid:19) . (2.34)By definition, δ is the phase χ goes through from one maximum of φ to the next. So we cansimply calculate the rotation angle δ as δ ≡ (cid:90) π m √ λ | φ | sin mt d t = 2 √ λ | φ | m , (2.35)– 10 – (cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45) (cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43) (cid:45) (cid:45) (cid:45) (cid:45) (a) (cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45)(cid:45) (cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43) (cid:43)(cid:43) (cid:43)(cid:43) (cid:43)(cid:43) (cid:43)(cid:43) (cid:43)(cid:43) (cid:43)(cid:43) (cid:43)(cid:43) (cid:43)(cid:43) (cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43)(cid:43) (b) Figure 6 . Redraw figure 4 and figure 5 with | φ | = 99 instead of 100. in which the bare mass M has been neglected. Therefore the total transformation matrix is (cid:18) cos δ − sin δ sin δ cos δ (cid:19) (cid:18) e η m e − η m (cid:19) = (cid:18) e η m cos δ − e − η m sin δe η m sin δ e − η m cos δ (cid:19) = B − (cid:18) b − √ b − b + √ b − (cid:19) B, (2.36)where b ≡ cosh η m cos δ = √ δ and B is a (2 ,
2) matrix of eigenvectors which is notimportant. We are mostly interested in the total boost rate during many cycles of φ , which isrepresented by η a = log (cid:112) x + y = log (cid:112) ( x, y )( x, y ) T . Since ( x, y ) T is multiplied on theleft by the transformation matrix for every half cycle of φ , the total boost rate is actuallydetermined by many powers of the transformation matrix. For this purpose, we have derivedeq. (2.36) which allows us to only consider the power of the eigenvalues of the transformationmatrix.For b > δ > / cosh η m , we have b − √ b − < < b + √ b −
1. Becauseone eigenvalue is larger than the other, the overall effect is the system converges to theeigenstate with the larger eigenvalue. This means all initial phases converge to the same oneafter enough long time. Because the larger eigenvalue is greater than 1, χ ’s total energy alsoincreases gradually.For b < δ < / cosh η m , the two eigenvalues have unit magnitude andopposite arguments. Consequently there is no such phase convergence, and no overall changein χ ’s total energy should be expected. To be more precise, one may want to calculate the transformation matrix via a three-step method — tworotations with angle δ/ B , butthe eigenvalues remain the same. – 11 –uch difference is also quite clear from the figures. The boost rate has dependence onthe initial phase and transforms a circle in the phase diagram to an ellipse. Because of such atransformation, the phases converge towards x -axis. On the other hand, δ rotates the ellipseand this may ruin the convergence if it’s large enough. These two effects compete and, inconsequence, the above two distinct behaviors of the system may occur.This also explains the exact solution of eq. (2.4), the so-called Mathieu equation. Thesolution of Mathieu equation has bands of stable solutions in which the amplitude of χ generally remains at its initial scale as time passes, and bands of unstable solutions whichhave an exponentially growing | χ | w.r.t time. Those bands correspond exactly to the non-convergent and convergent cases respectively.For better understanding, we can also calculate the phase that is converged to. Thefixed points under transformation matrix eq. (2.36) satisfy y (cid:48)(cid:48) /x (cid:48)(cid:48) = y/x , i.e. yx = y (cid:48)(cid:48) x (cid:48)(cid:48) = xe η m sin δ + y cos δxe η m cos δ − y sin δ . (2.37)The solutions are yx = sinh η m ± (cid:112) sinh η m − tan δe η m tan δ . (2.38)We can see how the solution is represented in the phase diagram in figure 7. From thefigure we know at once we should take the minus sign in eq. (2.38) as the phase convergedto. Based on these results, we can further derive χ ’s long-time boost rate (which takes timefar more than sufficient for convergence to take place, if there is). For the convergent case,the long time boost rate is simply the log of the larger eigenvalue in eq. (2.36), and for thenon-convergent case, no boost occurs so the long time boost rate is zero. We thus define thelong time boost rate as η l ( δ ) ≡ , cos δ < η m = 12log (cid:16) cosh η m cos δ + (cid:113) cosh η m cos δ − (cid:17) , otherwise. (2.39)It’s also visualized in figure 8.In figure 8, we can see the convergent-non-convergent boundary is at π/
4, which is exactand can be confirmed from the expression of η m . This indicates only half of the δ values givea long-time boost effect. If we consider the case with a δ slowly scanning over [0 , π ) at aconstant speed, we would get the average long-time boost rate (cid:104) η l (cid:105) ≡ π (cid:90) π η l ( δ )d δ ≈ . . (2.40)Please note there is relation η l ( − δ ) = η l ( π + δ ) = η l ( δ ).It’s quite interesting here to notice (cid:104) η l (cid:105) “happens” to be very close to (cid:104) η (cid:105) . Althoughthe expressions of (cid:104) η (cid:105) and (cid:104) η l (cid:105) are very different, this however is not a coincidence. (cid:104) η (cid:105) and (cid:104) η l (cid:105) are actually identical and numerical integrals have confirmed that. It is not yet veryclear why they are equal, but the following might be a reasonable explanation. Every valueof δ ∈ [0 , π ) is actually hit the same number of times during the calculation of (cid:104) η (cid:105) and (cid:104) η l (cid:105) . Consequently the transformation matrices of all δ ’s operate on the initial state thesame number of times, although the order of the operations is different for (cid:104) η (cid:105) and (cid:104) η l (cid:105) . The– 12 – igure 7 . The diagram showing the four fixed phases under transformation eq. (2.36), and fourregions of other phases with different directions converge towards. Here δ ≈ π/
6. The red line isthe phase converged to and is thus fixed under the transformation. The dashed curves with arrowsshow the directions other phases converge towards in different regions. The blue line stands atthe boundaries of the neighboring regions heading towards opposite directions, and therefore is alsoinvariant under the transformation. The red and blue lines are eigenstates with the larger and smallereigenvalues respectively. transformation matrices do not commute, but their commutation is only a rotation. Thecommutation could result in an either slightly larger or smaller boost rate, depending onwhether the rotation is towards the major axis of the total transformation matrix or awayfrom it. A huge number of such commutations is needed to make the two sequences thesame. Some of the commutations contributes positively to the total boost rate while othersnegatively. When summed up, their effects cancel, and maybe one might be willing to thinkthis as the reason of the equivalence between (cid:104) η (cid:105) and (cid:104) η l (cid:105) . So far, we have established the framework to consider the boost effect of parametric resonancein a static background. To apply it practically on cosmology, there are still a few steps shouldbe taken first. One is the need to put our framework into the expanding background; theother is the consideration of backreaction. We will discuss them in this section.– 13 – ∆ Η l Figure 8 . The long time boost rate under different rotation angles δ . Smaller δ typically leads tolarger boost rate because the eigenstate is more aligned with the phase of maximum boost rate. Largeenough δ ’s don’t generate any long-time boost effect because they are the non-convergent cases. When we apply this model in cosmology, a dynamic background is required. We will stillneglect the backreaction in this section, so the effect of expansion on φ is simply a dampingfactor | φ | ∝ a − ( t ), where a ( t ) is the scale factor. The equation of motion of χ howeverbecomes ¨ χ + 3 H ˙ χ + 2 λφ χ = 0 , (3.1)where H ≡ d a/a d t is the Hubble rate of the universe, and can be regarded as a constant foreach half cycle of φ . Writing χ = a Ψ simplifies eq. (3.1) to¨Ψ + 2 λφ Ψ = 0 , (3.2)in which we have neglected the effective mass contribution from universe expansion for λ | φ | (cid:29) m > H . Therefore we can see the universe expansion also gives χ an a − term.However the universe expansion has other impacts on our framework. As | φ | is dampedby expansion, the rotation angle δ no longer remains constant across cycles of φ . Thereforechanges should be made to the calculations in section 2.4. For convenience, we use thenotation x (n) to indicate the variable x of the n th half cycle of φ , or at the time when φ reaches its maximum the n th time. When all the subscripts (n) are equal in one equation,we may neglect some or all of them, as long as no confusion might be caused. Primes hereare defined as the difference x (cid:48) (n) ≡ x (n+1) − x (n) . (3.3) Besides expansion, this model is also applicable in contracting phases without any prejudice. Because weare more concerned with the expanding phase, we will only discuss the effect of universe expansion here. Theeffect of contraction however, can be derived in the same way, and stages are simply reversely sequenced inmost cases. – 14 –ifferent from the static version eq. (2.35), in an expanding background, δ (n) should beredefined as δ (n) ≡ (cid:90) t (n+1) = t (n) + π m t (n) √ λ | φ | ( t ) cos mt d t. (3.4)To finish the integral, we manually fix | φ | ( t ) at | φ | (n) , so it gives the same result with eq.(2.35). This approximation however gives δ (n) an error∆ δ (n) < (cid:90) t (n+1) t (n) √ λ || φ | (n+1) − | φ | (n) | cos mt d t = 2 √ λ || φ | (cid:48) (n) | m (3.5)where the outer | | is to take absolute values. For short, we have∆ δ < √ λ || φ | (cid:48) | m . (3.6)Similarly we get δ (cid:48) ≡ δ (n+1) − δ (n) = 2 √ λ | φ | (cid:48) m , (3.7) δ (cid:48)(cid:48) ≡ δ (cid:48) (n+1) − δ (cid:48) (n) = 2 √ λ | φ | (cid:48)(cid:48) m , (3.8)∆ δ (cid:48) < √ λ || φ | (cid:48)(cid:48) | m = | δ (cid:48)(cid:48) | . (3.9)The universe expansion redshifts | φ | , giving | φ | (cid:48) | φ | = e − πH m − − πH m , (3.10)which consequently redshifts δ . The average effect of parametric resonance is then determinedby the scales of δ , δ (cid:48) and δ (cid:48)(cid:48) .When δ (cid:48) (cid:28) δ changes very slowly compared with the oscillation of φ . The systemtherefore has enough time to converge to the eigenstate whenever there’s any. Then thesystem can be regarded as trapped in its eigenstate which is slowly varying. The boost rate η l ( δ ) changes correspondingly as δ slowly decreases, and the average boost rate in this caseshould then be taken as the long-time boost rate, (cid:104) η l (cid:105) , defined in eq. (2.40). As δ is slowlyredshifted by universe expansion, the convergent and non-convergent cases are met in turn.This makes χ ’s evolution stair-like at this stage. Half of the time, the amplitude of χ enjoysflat platforms lasting for cycles of φ during which it doesn’t grow at all. The platforms areseparated by steep growing regions which fills the other half of the time. The two distinctstages come in turn to get the stair-like curve of χ ’s amplitude.When δ (cid:48) ∼ > δ (cid:48)(cid:48) (cid:28) δ varies fast but δ (cid:48) varies very slowly. In this case, δ changes significantly between neighboring half cycles of φ , but the amount changed staysalmost fixed. For every half cycle of φ , an almost constant δ (cid:48) ∼ δ . Therefore for time long enough, the average effect is β has an even probability ateach value in [0 , π ). This corresponds to an even distribution of β in η ( β ) and, results inthe average boost rate which should be chosen as (cid:104) η (cid:105) in eq. (2.30).When δ (cid:48)(cid:48) ∼ >
1, the relevance between neighboring δ ’s or δ (cid:48) ’s is small, so one usuallyconsiders δ (n) as random in [0 , π ), being irrelevant with the history of evolution. When long– 15 – (cid:180) (cid:180) (cid:180) (cid:180) (cid:180) t (cid:45) (cid:45) (cid:45) (cid:45) (cid:45) (cid:45) (cid:45) (cid:45) (cid:45) (cid:45) (cid:200) Χ (cid:200) Figure 9 . The numerical simulation result of many preheating processes compared with our statisticalestimation. (See section 4.1.) Universal parameters are m = 10 − M p , M = 10 − M p , and γ = 10 − .The only varying parameter for different simulations is the interaction strength λ , starting at 10 − and ending at 4 × − , with a step increment 10%. The green plots are the data from numericalsimulation, taken every time φ reaches maximum. The red line is the statistical average result andthe two blue lines enclose the area within error 1 σ , derived from ∆ (cid:104) ´ η (n) (cid:105) in eq. (3.14). On the lefthalf, we see a reasonable amount of plots lying outside the blue curves because we only expect 68% ofthe plots in 1 σ . On the right half the universe actually enters the small δ (cid:48) stage and the error doesn’tgrow so much. However the blue lines are derived under the assumption that the whole evolution isthe large δ (cid:48) stage, so they are only functions of n and keep extending. Thus we find on the right half,the number of plots outside the blue lines is significantly decreased. The numerical data stop whenbackreactions start to be important. enough time is experienced, the random distribution of δ gives the same results as the caseof δ (cid:48) ∼ > δ (cid:48)(cid:48) (cid:28) (cid:104) η (cid:105) should also be chosen as the average boost rate.In realistic cases however, it is rare to tell the last two stages apart. They share thesame average boost rate (cid:104) η (cid:105) and most of their characteristics are alike, so there is no needto distinguish them. Moreover, the second category has a δ (cid:48)(cid:48) (cid:28) β needsto be shown. This however requires a prolonged constant value of δ (cid:48) , which in most cases isvery difficult to achieve with a non-vanishing δ (cid:48)(cid:48) , even though it’s much smaller than 1. Forthese reasons, we will treat them as a whole ( δ (cid:48) ∼ >
1) in our following discussions and callit the large δ (cid:48) stage , in which δ (n) is randomly distributed and independent of the history. Inthe same way, we name δ (cid:48) (cid:28) small δ (cid:48) stage .We now fall back to check the approximation we made previously in eq. (3.6) and eq.(3.9), where | φ | and | φ (cid:48) | are regarded as constants in every half cycle. They give errors∆ δ < | δ (cid:48) | and ∆ δ (cid:48) < | δ (cid:48)(cid:48) | , which fit perfectly with the three stages above. Therefore thisapproximation is applicable here.Before we end this discussion on dynamic background, we would like to briefly calculatethe statistical results from the random δ for the large δ (cid:48) stage. In n cycles of φ , the boosteffect of parametric resonance occurs n times. The total boost rate is´ η (n) ( δ ) ≡ n (cid:88) i =1 η ( δ (i) ) . (3.11)– 16 –ssuming absolute randomness of all δ (i) ∈ [0 , π ), we get the expectation value of ´ η (n) ( δ ) as (cid:104) ´ η (n) (cid:105) ≡ π ) n (cid:90) d n δ n (cid:88) i =1 η ( δ (i) ) = n π (cid:90) π η ( δ )d δ = n (cid:104) η (cid:105) ≈ . n. (3.12)This is obvious because it’s just how (cid:104) η (cid:105) is defined. To further compute the statistical error,we calculate (cid:104) ´ η (cid:105) ≡ π ) n (cid:90) d n δ n (cid:88) i =1 η ( δ (i) )= n π (cid:90) π η ( δ )d δ + n ( n − π (cid:16)(cid:90) π η ( δ )d δ (cid:17) = n (cid:104) η (cid:105) + n ( n − (cid:104) η (cid:105) . (3.13)and thus the statistical error can be estimated as∆ (cid:104) ´ η (n) (cid:105) ≡ (cid:113) (cid:104) ´ η (cid:105) − (cid:104) ´ η (n) (cid:105) = √ n (cid:112) (cid:104) η (cid:105) − (cid:104) η (cid:105) ≈ . √ n. (3.14)To justify both the statistical calculation and the definition of the large and small δ (cid:48) stages,here we also give figure 9 through multiple numerical simulations. In this section, we discuss the backreaction of χ by taking into account the λχ term in m φ ,the effective mass squared of φ . Since we assumed homogeneity, we will not discuss otherbackreactions here. In most cases of parametric resonance, the assumption λχ (cid:28) m startsvalid. However as χ ’s amplitude gets boosted by resonance, this assumption would finallybreak at | χ | ∼ m/ √ λ , provided the boost effect hasn’t yet been halted by other causes.Because | χ | ∝ m − χ ≈ / (cid:112) λφ , for every half cycle of φ , | χ | reaches its maximum when φ →
0. From this point of view, the violation of this approximation first starts small — onlyat φ → φ gradually while | χ | is being further boosted.From the relation | χ | ∝ / (cid:112) λφ , we can find the effective potential for φ , which isactually V eff ( φ ) ≡ m φ + λφ χ = 12 m φ + 12 λ | χ | m | φ || φ | abs , (3.15)where | χ | m is χ ’s amplitude at maximum φ , and | | abs means taking the absolute value(and | | still indicates taking the amplitude). Here we can see the additional part of V eff ( φ )is proportional to the absolute of φ . Compared with m φ which is proportional to φ ,the effective potential from interaction V eff ( φ ) agrees with our previous instinct that thebackreaction first comes to important at small φ .So we consider the motion of φ for half a cycle, starting from t = 0, φ = −| φ | , ˙ φ = 0,and | χ | = | χ | m . We will use γ ≡ λ | χ | m /m for simplicity. When φ < φ ’s equation ofmotion is ¨ φ + m φ − λ | χ | m | φ | = 0 . (3.16)We can analytically solve this equation and arrive at φ = 0. The solution is t = t ≡ m arccos γ γ + 1 , (3.17)– 17 – φ ( t = t ) = ( γ + 1) m | φ | sin mt . (3.18)At φ = 0, χ is boosted by parametric resonance. This gives the interaction part of V eff ( φ ) a factor of e η . Here η is still the boost rate. For this reason, the equation of motionis slightly modified for φ >
0, ¨ φ + m φ + λ | χ | m | φ | e η = 0 . (3.19)We then solve it again to arrive at ˙ φ = 0 and finish the half cycle. At the φ > φ reaches maximum again at time t = t , we have t = t + π m − m arctan γ e η (cid:112) γ + 1 , (3.20) φ ( t = t ) = | φ | ( (cid:112) γ e η + 2 γ + 1 − γ e η ) , (3.21)and | χ | = e η | χ | m | φ | φ . (3.22)We can clearly see when η = 0, there is no boost effect and the motion of the system becomessymmetric w.r.t φ = 0.When the backreaction is weak, i.e. γ (cid:28)
1, we can expand the results to first orderand get t = πm − e η + 1 m γ , (3.23) φ ( t ) = | φ | (cid:16) − ( e η − γ (cid:17) , (3.24) | χ | = e η | χ | m (cid:16) e η − γ (cid:17) , (3.25) δ = 2 √ λ | φ | m (cid:18) − (cid:16) π e η + 1) − (cid:17) γ (cid:19) . (3.26)Therefore, in case the backreaction presents, the oscillation of φ becomes faster and that of χ is compared slower. The non-vanishing η gives a boost effect. Amplitudes φ ( t ) and | χ | ( t )are thus affected and become slightly different with their original values at t = 0, suggestingan energy transfer.Whenever one takes this backreaction into account, one should adopt these corrections.However for weak backreactions in which γ (cid:28)
1, the corrections actually can be neglected.Such backreaction only comes to important when γ ∼ > .
1. In some cases, however, theexponential boost effect is terminated by other causes before γ reaches 0 .
1, so it never comesto important.If the system does evolve to γ ∼ > .
1, one then needs to take it seriously. The mostsignificant effect is φ oscillates faster and faster, due to the effective mass contribution fromthe backreaction. This has several consequences. First and most obviously, because theboost rate is counted per half cycle of φ , a faster rate of φ ’s oscillation also leads to a fasterboost to χ . The faster boost acts back on φ ’s effective mass and causes φ to oscillate evenfaster. The positive feedback gives an explosive boost effect to χ once the backreactioncomes to important. The explosive boost effect is soon shut down when | χ | m reaches thesame scale with | φ | . This is because φ and χ would then have similar frequencies and the– 18 –pproximate replacement of χ with (cid:104) χ (cid:105) would be no longer applicable. This immediatelyends the exponential boost effect of parametric resonance.Moreover, this backreaction may also offer a major contribution to δ (cid:48) . Here we givea simple estimate based on eq. (3.26). By definition, we have ( γ ) (cid:48) = 2 λ ( | χ | m ) (cid:48) /m = γ ( e η − δ (cid:48) up to first order of γ is∆ δ (cid:48) = − √ λ | φ | γ m (cid:16) π e η +2 η (cid:48) − − e η + 1 (cid:17) . (3.27)When it has the same magnitude with the originally defined δ (cid:48) in eq. (3.7) in an expandinguniverse, we can solve the magnitude of γ , γ ∼ Hm < . . (3.28)This means the impact of backreaction on rotation angle δ is even more noticeable than thaton φ ’s period.In the same way, it’s easy to derive ∆ δ (cid:48) has unit magnitude when γ ∼ m/ √ λ | φ | (cid:28)
1. Normally in an expanding universe where the backreaction is negligible, | δ (cid:48) | is alwaysdecreasing so the resonance would gradually undergo a shift from the large δ (cid:48) stage to thesmall δ (cid:48) stage. The relation of γ however tells that the parametric resonance may be subjectto a shift back to the large δ (cid:48) stage from the small δ (cid:48) stage, or even, the small δ (cid:48) stage maynot appear at all. The back shift, whenever exists, always lies ahead of the explosive boost.So for deep discussions of the ending conditions of the exponential boost, one should alwaystake it into account. Preheating is a possible stage after inflation, during which the inflaton decays into particlesthrough parametric resonance. Preheating was first realized in [15, 16], and the parameterspace is the same with what we have discussed. Therefore, during preheating, the numberof preheated particles grows exponentially, much faster than the redshift rate from universeexpansion.Although preheating should be followed by particle thermalization, and possibly a sec-ond stage of reheating between them, they are not within our consideration in this paper.Instead of calculating the reheating temperature and how long reheating and thermalizationlast, this section is devoted to showing how this framework may be applied on specific pre-heating calculations. Details about the second stage of reheating and thermalization can befound in [16] and [22]. For the same reason, we don’t care the reheating ratio between thedark and visible sector. To solve the dark radiation problem, one might want to refer toinflation models with more particle physics foundation, e.g. SUSY inspired inflation and thesubsequent reheating and thermalization studies[23–25].During preheating, the universe is dominated by φ in our model, which is usually theinflaton. As a demonstration, we here just consider the simple case of the preheating ofa generic slowroll inflation with inflaton φ . To solve problems like horizon and flatness,inflation requires an e-folding at least N ≈
60. At the beginning of such slowroll inflations,there should be λχ (cid:28) m . Otherwise there would be a significant decrease in φ ’s effectivemass and thus no inflation although φ would still be slowrolling.– 19 –n general, the end of slowroll inflation is signaled by the destruction of the slowrollcondition of φ , after which φ begins to oscillate and preheating takes place. We would liketo neglect the first cycle or so of φ in the beginning of preheating here, so m φ (cid:29) H holdsfor the rest of the preheating period. Because λχ (cid:28) m holds in the beginning of inflation,and χ is redshifted greatly during inflation, this relation is expected to still hold for a verylong period in preheating. Because the backreaction terminates parametric resonance once itcomes to important, it acts as an ending condition of the preheating scenario. Therefore wecan first neglect the presence of backreaction, and then calculate the take-over of backreactionseparately.We label the beginning of inflation as 0, the end of inflation (i.e. the start of preheating)as 1, and the point we start to consider in preheating as 2. The evolution of the universe isthen divided by such points, whose labels we use as subscripts, into intervals such as 1 → | φ | ≡ e − | φ | . (4.1) χ is redshifted during inflation (0 → →
2. We therefore get | χ | ∼ | χ | = e − N | χ | . (4.2)The universe is dominated by φ during preheating so the Hubble rate writes H = 4 πm | φ | M (4.3)in which for point 1 and after, | φ | = | φ | (cid:16) a a (cid:17) . (4.4)After point 2, the system starts at the large δ (cid:48) stage. There may be a transition fromlarge to small δ (cid:48) stage during the evolution, but the termination of parametric resonance bybackreaction is certain to take place at the large δ (cid:48) stage, which is ensured by its influenceon δ (cid:48) shown in eq. (3.27), slightly before the positive feedback ends the exponential boost.We thus also label the point after which the large δ (cid:48) is ensured by backreaction, and thepoint backreaction becomes important and consequently terminates parametric resonance aspoints 3 and 4, also in chronological order.After defining the timeline, we now work on the evolution of χ . The large and small δ (cid:48) stages share the same average boost rate (cid:104) η (cid:105) , so after point 2, field χ is affected byparametric resonance and becomes proportional to e (cid:104) η (cid:105) π m ( t − t ) . Due to universe expansion, | χ | is redshifted and thus acquires the factor ( a /a ) . As we have demonstrated in section2.1, | χ | ∝ m − φ , this gives | χ | , from the redshift of φ field, another factor ( a/a ) . Wemultiply them all together to get the statistically average evolution of field χ | χ | = | χ | (cid:16) a a (cid:17) e (cid:104) η (cid:105) π m ( t − t ) . (4.5)In a matter-like universe, we have for any time t a and t b , (cid:18) a ( t b ) a ( t a ) (cid:19) = 1 + (cid:115) πρ ( t a ) M ( t b − t a ) , (4.6)– 20 –hich, under the condition log a ( t b ) − log a ( t a ) >
1, can be simplified to t b − t a = (cid:18) a ( t b ) a ( t a ) (cid:19) M p (cid:112) πρ ( t a ) (4.7)where ρ ( t a ) is the total energy density of the universe at time t a . Because φ dominatespreheating, the evolution of | χ | finally writes | χ | = | χ | (cid:18) a a (cid:19) e B ( aa ) , (4.8)in which B ≡ (cid:104) η (cid:105) M p (3 π ) | φ | . (4.9)From the above equation, we immediately see | χ | is actually independent of λ and m .It’s obvious | χ | is λ -independent from the constant boost rate. The dependence on m , thebare mass of φ , is however canceled by the same m -dependence of ρ . This can be also viewedas m defines a time scale which both the Hubble rate and the boost rate of χ per time intervalare proportional to.We then derive the evolution of χ at points 3 and 4 which are defined from backreaction.Here we assume φ is the inflaton of a generic slowroll inflation of N e-folds. At the beginningof it, i.e. point 0, there should be λ | χ | < m to allow inflation to happen. After beingredshifted all along inflation, | χ | then becomes extremely small and this allows a safe periodof preheating before the backreaction becomes important. After that, backreaction mayget strong and, points 3 and 4 may be met in turn at any moment. To show the relation λ | χ | < m , we deploy the parameter γ ≡ λ | χ | /m <
1. Here we want to give aconservative and natural estimation, so we assume − < log γ < δ (cid:48) stage at γ ∼ > m/ √ λ | φ | , we therefore define point 3 as λ | χ | m = γ ≡ m √ λ | φ | . (4.10)Given the evolution of both fields, eq. (4.4) and eq. (4.8), this can be reduced to the equationof N → , the e-folding the universe experienced between point 2 and 3, Be N → − N → = N + 16 log m λ | φ | γ . (4.11)Also, we can work on point 4 in the same way. After defining point 4 as λ | χ | ≡ m ,the calculation is straightforward to give Be N → − N → = N −
23 log γ . (4.12)In both of the above equations, the dominant term on the l.h.s is the exponential one.We can thus see the number of e-folds of preheating is only very weakly dependent on λ and γ , because they are inside two “log”s. This is because φ oscillates faster and faster as theuniverse expands, and in consequence χ acquires a much larger boost rate per Hubble time H − . – 21 – (cid:45) (cid:45) (cid:45) (cid:45) (cid:45) (cid:45) Λ N Figure 10 . N → and N → as functions of λ . Blue and purple correspond to N → and N → respectively. It’s easy to see on the figure that N → is only weakly dependent on λ and N → isindependent of λ . Parameters are chosen to accord with the quadratic slowroll inflationary model. To provide a brief example, we take the simple case of quadratic slowroll inflation whichends at | φ | = 27 M / π , and has the number of e-folds of inflation N = 60. We also needthe value m ∼ − M p , which comes from the strength of observed CMB anisotropy. Figure10 shows the relations N → and N → have with λ at constant γ = 0 .
1. The dependenceof γ is weak, and a smaller γ simply shifts both N → and N → upwards a bit with thesame amount. For γ = e − , there is N → = 4 . N → . For N → = 4and 5 respectively, there is N − log γ = 32 and 155. As a result, for most cases where N is not very large and γ is not too small, preheating is typically ended by backreaction atthe 5th e-fold of parametric resonance i.e. the 6th e-fold after inflation, provided no othermechanism terminates preheating before that.In section 3.2, we have suggested the correction from backreaction to δ (cid:48) always comesto important earlier than that to m φ . In figure 10, it agrees with large λ . For small λ ,the figure suggests a reverse in their sequence. The reverse however actually reflects theviolation of λ | φ | (cid:29) m . For values of λ on the left of the intersection point of the two curvesin figure 10, the condition λ | φ | (cid:29) m should be broken well ahead of point 4 being reached,so the behavior of the system afterwards in fact can’t be predicted by our theory. Thisindicates another possible exit of parametric resonance, which is caused by a small | φ | that iscomparable to m/ √ λ . We label this exit as subscript a , and from the relation λ | φ a | ≡ m ,we can easily derive the e-folding N → a = 13 log λ | φ | m . (4.13)Since both point a and 4 act as exits of parametric resonance, it is important to knowwhich point comes earlier to find out by who and when parametric resonance is terminated.By setting N → = N → a , calculations give a critical value for λλ a ≈ ( N − log γ ) m B | φ | ∼ − . (4.14)– 22 –or λ > λ a , parametric resonance ends with backreaction. The amplitudes of φ and χ reachthe same scale of magnitude in the end and equilibrium may be reached between the two fields.For λ < λ a however, parametric resonance is terminated by the violation of λ | φ | (cid:29) m . Inthis case, the evolution afterwards is not characterized by our theory, and we only knowat the time this condition is broken, equilibrium has not yet been reached. Moreover, for λ < m / | φ | , there is N → a <
0. This indicates the condition of our framework is notsatisfied and such a parametric resonance does not take place immediately after inflation.Meanwhile, we are also interested in the transition point from the large δ (cid:48) stage to thesmall δ (cid:48) stage. In the context of preheating, δ (cid:48) is defined as δ (cid:48) ≡ − π √ λ | φ | mM p . (4.15)When δ (cid:48) reaches e − ∼ .
1, we think the transition takes place and label it as point b . Wethus derive the e-folding from point 2 to bN → b = 13 log 2 e π √ λ | φ | mM p . (4.16)For N → b > N → a , parametric resonance is terminated before the transition point, sothe whole process is large δ (cid:48) . Similarly if N → b > N → , the large δ (cid:48) stage is also guaranteedso no small δ (cid:48) stage takes place. For N → b <
0, parametric resonance starts from the small δ (cid:48) stage, and whether a large δ (cid:48) stage presents (although very short) depends on how preheatingis terminated.To conclude this subsection, we perform a more precise calculation of the parametricresonance of quadratic slowroll inflation, and give figure 11 which divides the parameterspace of λ and γ into five sections. The small δ (cid:48) stage may either exist or not exist forthe terminations by both backreaction and the small | φ | . There is also the case that noparametric resonance shall take place within our framework, so in all there are five sections.The numbers of the total e-folds of parametric resonance and the e-folds of the small δ (cid:48) stageare also contoured in figure 11. For the termination by small | φ | however, the reheatingis incomplete and there should be a subsequent decay of inflaton after the termination ofparametric resonance, but our theory is unable to characterize it. From the figure, we alsoconfirm that if preheating is terminated by backreaction, it should typically happen at the5th e-fold of preheating, i.e. the 6th e-fold after inflation. Locked inflation, or new old inflation[9], is a fastroll inflation scenario during which a fastrollfield φ locks the other field χ at its false vacuum. The false vacuum has a nonzero energydensity which drives inflation. Locked inflation has the same potential shape with hybridinflation[5], but the parameter configuration is different.Not long after the proposal of locked inflation, however, opposite opinions are postedsuggesting underlying problems[20, 26]. In [20], they concluded that saddle inflation, loopcorrection and parametric resonance problems eliminates part of the parameter space each,while they three together kill all possible parameter configurations. At least one of thoseproblems therefore has to be fixed to make locked inflation work again. During lockedinflation, parametric resonance is disfavored for its exponentially growing energy transfer– 23 – (cid:45) (cid:45) (cid:45) (cid:45) (cid:45) (cid:45) (cid:45) (cid:45) (cid:45) (cid:45) (cid:45) (cid:45) (cid:45) (cid:45) (cid:45) (cid:45) (cid:45) (cid:45) (cid:45) (cid:45) (cid:45) (cid:45) (cid:45) (cid:45) (cid:45) (cid:45) (cid:45) (cid:45) Λ Γ Figure 11 . The different circumstances of preheating in the parameter space of λ and γ . Thegreen and red regions are backreaction termination with and without the small δ (cid:48) stage respectively.The blue and orange regions are terminated by the small | φ | , with and without the small δ (cid:48) stagecorrespondingly. In the gray region, the preheating stage is not entered after inflation. Among thedashed lines, the number of total e-folds of parametric resonance is contoured in black and that ofthe small δ (cid:48) stage is contoured in red. from φ field to χ , so it acts as a problem here which may end locked inflation much earlierthan expected.From our above analysis of the mechanism of parametric resonance, we infer for φ withmore than one real component, parametric resonance is inhibited. This solves the parametricresonance problem of locked inflation and makes it possible. Such an argument comes fromeq. (2.5), the condition of rolling stage. For φ singlet, m χ is dominated by λφ so thiscondition breaks when φ is approaching zero. Only when eq. (2.5) breaks, φ would enter thezero-crossing stage and perform the exponential boost. If λφ isn’t the absolute dominant of m χ , eq. (2.5) would never break down and the zero-crossing stage would never be entered,even if φ = 0. For φ multiplets, every component of φ interacts with χ independently. Sowhen one component crosses zero, other components are in general nonzero and they provide χ with a large effective mass. The zero-crossing component thus never dominates and eq.(2.5) never breaks.As an example, we take a φ doublet, whose components are φ , φ . For demonstration,we simply assume they have an identical amplitude | φ | but a phase difference π − ϕ . So– 24 –hey can be written as φ = | φ | cos( m φ t − ϕ ) , (4.17) φ = | φ | sin( m φ t + ϕ ) . (4.18)The effective mass squared of χ then becomes˜ m χ ≡ M + 2 λ ( φ + φ ) = M + 2 λ | φ | (1 + sin 2 ϕ sin 2 m φ t ) , (4.19)equivalent to the singlet case with transformed M → ˜ M = M + 2 λ | φ | , λ → ˜ λ = λ sin 2 ϕ and m φ → ˜ m φ = 2 m φ .The equivalent singlet case however has the relation ˜ λ | φ | < ˜ M , which breaks theassumption λ | φ | (cid:29) m (cid:29) M . In such cases, the condition of rolling stage eq. (2.5) wouldnever break, and the parametric resonance effect would then be inhibited.In general cases where | φ | (cid:54) = | φ | but | φ | ∼ | φ | , there are also similar results, whichtypically have ˜ M ∼ . λ ( | φ | + | φ | ). For the same reason, eq. (2.5) would never breakand locked inflation can last long enough for φ multiplets. Therefore for φ multiplets, theparametric resonance “problem” of locked inflation is not actually a problem. Of course,this doesn’t solve the problem of embedding locked inflation in some particle physics theory.However, one still needs a viable particle physics setup (or a convincing argument about thedecay to visible sector) to make locked inflation promising. This is beyond the coverage ofthis paper. The cosmic inhomogeneity, if arose from the inflaton, may get amplified by parametric res-onance during the large δ (cid:48) stage. The amplification also requires a change in the equationof state of the universe that is triggered directly or determined indirectly by the total boostrate of | χ | . This process is similar to modulated perturbations discuss in [27]. The differencehere is the decay rate is determined by the relative phase difference between the inflaton andthe preheated field, i.e. both fields instead of the inflaton only.To explicate the whole process, we start with two points at a distance 1 /k . In thebeginning of inflation, they live within a Hubble radius, so homogenous χ is assumed. Afterthe mode k leaves horizon, the two points become uncorrelated and the phases of χ at the twopoints start to differ because of the inhomogeneity δφ k which causes different m χ ’s. Also, aswe choose the end of inflation at a specific value of φ on the uniform-density hypersurface, thepoint with a slightly larger φ would also gain a slightly longer inflation. For both reasons, atthe point where φ is larger, the phase that χ experiences would also be larger. This transfersinhomogeneity to χ ’s phase from φ ’s energy density.After inflation, the universe enters the large δ (cid:48) stage. Due to statistical reasons, thetotal transformation matrix of the whole large δ (cid:48) stage would be an ellipse with the aver-age total boost rate (cid:104) ´ η (cid:105) and the difference between the major and minor axes of the scale2∆ (cid:104) ´ η (cid:105) . Different initial phases of χ then correspond to different total boost rates, i.e. dif-ferent parametric resonance efficiencies. If there exists a transition of the equation of statedepending on the efficiency of parametric resonance, like the backreaction end which changesthe universe from matter-like to radiation-like when λ | χ | ∼ m , additional inhomogeneity– 25 –ay be generated. At points where the efficiency is lower for example, more time is neededto boost χ to the amplitude required for the transition in the equation of state to take place,and subsequently the entrance into a different equation of state is delayed. Therefore thereare regions which stay in the larger equation of state longer and suffer a heavier redshiftdue to universe expansion, and also regions which enjoy a weaker redshift. The difference inredshift rates finally leads to different energy densities at different places, and thus additionalcurvature perturbation on the uniform-density hypersurface. If the effect is stronger thanthe existing inhomogeneity, we will find inhomogeneity amplified in this process.To calculate, we start from inflation in which the inhomogeneity transfers from φ tothe phase of χ . We still adopt the two points at the distance 1 /k and difference in φ at thehorizon exit of mode k as δφ k . Because we use a universal ending condition for inflation, wealign the evolution history of the two points at the end of inflation. Then the only difference δφ k causes to the phase of χ is one point enjoys an additional period of inflation as thefield φ rolls from φ + δφ k to φ , and this gives an additional phase to χ . Using the slowrollapproximation, the resulting phase difference can be calculated∆ ϕ ≡ √ λ φ (cid:12)(cid:12)(cid:12)(cid:12) δφ k ˙ φ k (cid:12)(cid:12)(cid:12)(cid:12) = (cid:114) λ π M p H k m δφ k φ k , (5.1)where we use the subscript k to represent values at the horizon exit of mode k .During the large δ (cid:48) stage, the parametric resonance process is approximated as stochas-tic. So we acquire the statistical result of the average total boost rate (cid:104) ´ η (cid:105) and its statisticalerror ∆ (cid:104) ´ η (cid:105) , which are defined in eq. (3.12) and eq. (3.14) and are functions of n , the numberof half cycles of φ during the large δ (cid:48) stage. If we consider the total transformation matrixof the whole large δ (cid:48) stage which transforms a circle of unity to an ellipse, (cid:104) ´ η (cid:105) can also beunderstood as the typical average total boost rate of the ellipse and ∆ (cid:104) ´ η (cid:105) as half of the typicaldifference of the total boost rate between the major and minor axes.Now we have the ellipse as the total boost rate function of the initial phase. The totalboost rate goes from maximum to minimum with a change of initial phase π/
2, so the phasedifference ∆ ϕ would give a difference in the total boost rate, and thus a difference in thenumber of half cycles of φ needed to reach the same amplitude | χ | . For ∆ ϕ <
1, we have thenumber of additional half cycles ∆ n = 2∆ ϕπ (cid:104) ´ η (cid:105)(cid:104) η (cid:105) . (5.2)Then ∆ n leads to delayed or advanced transition of the equation of state. Let us supposethe equation of state is changed by ∆ w . The resulting inhomogeneity is then δρ k ρ k ≡ ∆ nπm wH = (cid:114) λπ wM p HH k m ∆ (cid:104) ´ η (cid:105)(cid:104) η (cid:105) δφ k φ k , (5.3)where H here is the Hubble rate at the transition of the equation of state. During preheating, the universe is matter-like because φ dominates with its energy density proportionalto a − . For preheating ended by backreaction, the dominant potential term becomes λφ χ . Therefore φ and χ have the same redshift rate. With their redshift rate a − for constant-mass fields, and their scaling relationwith effective mass eq. (2.11), we can conclude both fields are proportional to a − . This gives the energydensity λ | φ | | χ | ∝ a − and thus a radiation-like universe. This result is also numerically verified in [28]. – 26 –f we don’t want it to overwhelm the existing inhomogeneity δφ k /φ k , we then need itto be smaller than that, which gives (cid:114) λπ wM p HH k m ∆ (cid:104) ´ η (cid:105)(cid:104) η (cid:105) < , (5.4)and consequently, λ < πm w M H H k (cid:104) η (cid:105) ∆ (cid:104) ´ η (cid:105) . (5.5)For ∆ ϕ >
1, we have instead ∆ n = 2∆ (cid:104) ´ η (cid:105)(cid:104) η (cid:105) , (5.6)which generates the inhomogeneity δρ k ρ k = 6 π ∆ wHm ∆ (cid:104) ´ η (cid:105)(cid:104) η (cid:105) . (5.7)∆ (cid:104) ´ η (cid:105) / (cid:104) η (cid:105) is at least 1 and increases as the large δ (cid:48) stage continues; ∆ w ∼ . To make δρ k /ρ k < − , we therefore need at least 8 e-folds before the transition to get a small enough H , which is hardly possible for a quadratic slowroll inflation, unless we have thousands ofe-folds of inflation to redshift χ .So we will still focus on the ∆ ϕ < δφ k /φ k ∼ − and H k ∼ m , we can see from eq. (5.1) the constraint from ∆ ϕ < λ < − .However the constraint from eq. (5.5) is even stronger. At the backreaction end, theequation of state switches from 0 to , so ∆ w = . For the preheating fully consisted of thelarge δ (cid:48) stage, we can simplify part of eq. (5.5) to a function of N → only m H (cid:104) η (cid:105) ∆ (cid:104) ´ η (cid:105) = 3 √ π M p (cid:104) η (cid:105) | φ | ( (cid:104) η (cid:105) − (cid:104) η (cid:105) ) e N → . (5.8)So eq. (5.5) becomes λ < √ πm M p | φ | H k (cid:104) η (cid:105) (cid:104) η (cid:105) − (cid:104) η (cid:105) e N → . (5.9)Here we simply choose N → = 4 . H k = 10 m to get the numerical value λ < − .If we want the result λ < − , from which we can reserve a tiny living space for thebackreaction end as shown in figure 11, we would need N → >
18, which means more than10 e-folds of inflation. For the backreaction end with both the large and small δ (cid:48) stagesduring preheating but the same e-folding of the large δ (cid:48) stage, the constraint is even stronger.Such a strong constraint basically rules out the backreaction end of preheating for thequadratic slowroll inflation, unless additional techniques are taken. The attempt of using adynamic λ to solve this problem is also difficult, because that would require λ < − duringinflation, which makes χ also slowrolling. For natural initial conditions, one then needs todeal with a very large χ field after inflation which doesn’t even need parametric resonanceto become large. This however disagrees with our initial purpose of choosing χ as the firstlevel decay product of φ . Therefore, we infer that the first level of decay of the inflaton of– 27 –uadratic slowroll inflation is insufficient for reaching a chemical equilibrium between the twotypes of particles, if we still adopt δφ/φ ∼ − at the cosmic scale.One may, however, assume inflation only generated a much weaker inhomogeneity andpreheating then amplified it to the proper amount. To get a weaker inhomogeneity, we needto change the mass m . In order to benefit from the scaling property, we will still use m to indicate the old m which alone can get the proper inhomogeneity during inflation. Onthe r.h.s of eq. (5.3), H and H k are both proportional to m , and δφ k /φ k ∝ m , so the r.h.sas a whole is proportional to m , while the l.h.s is the desired inhomogeneity and should bekept unchanged. We can absorb the dependence on m into λ for the rest of the calculation,and drag it out in eq. (5.5). If nothing else is modified, we would get a similar result λm /m ∼ − . Still if we want λ ∼ − , this would require m ∼ − M p . Also, becausethe amplification rate is the term on the l.h.s of eq. (5.3) before δφ k /φ k which has H k in it,the amplification is scale dependent. Different length scales leave horizon at different Hubblerates which consequently give different inhomogeneity transfer rates. This comes from thedifferent amounts of phase differences of χ , which are only generated at their correspondinghorizon exits. Therefore, the scale dependence further adds − (cid:15) to the spectral index for thiscase, providing (cid:15) ≡ dd t H is the slowroll parameter.For generic cases, we just need to redefine ∆ ϕ . Still aligning every point at the end ofinflation and assuming the only difference the inhomogeneity of inflaton gives on χ ’s phaseis a slightly longer or shorter inflation, we can then define it as ∆ ϕ ≡ (cid:104) m χ (cid:105) ∆ t . Here (cid:104) m χ (cid:105) is the average m χ during the additional inflationary period ∆ t caused by the inhomogeneityof the inflaton. For short, we use δ k ≡ δρ k /ρ as the energy density inhomogeneity at horizonexit. The time difference of inflation can then be written as∆ t = − δρ k d t d ρ = − δ k d t d log ρ = δ k H k (cid:15) k , (5.10)where again the subscript k indicates the value at horizon exit of mode k. So the inhomo-geneity arose from parametric resonance, represented as ˜ δ k , should be˜ δ k = 6∆ wHm φ ∆ (cid:104) ´ η (cid:105)(cid:104) η (cid:105) (cid:104) m χ (cid:105) k H k (cid:15) k δ k . (5.11)The amplification rate then should be the terms on the r.h.s before δ k (for ∆ ϕ < m ∼ − . This constraint isalso quite strong in general cases, as long as all of the following criteria are met so that theconstraint is applicable.1. The inflaton has some inhomogeneity after the horizon exit.2. There is a period of large δ (cid:48) stage during the first level decay of the inflaton.3. A transition of the equation of state of the universe is a direct or indirect consequenceof the first level decay product’s reaching a critical density.This constraint however doesn’t apply to the small δ (cid:48) stage, because all phases convergeto the same one in the small δ (cid:48) stage. Nor does it apply to perturbative reheating, etc. Theconstraint doesn’t affect models without any transition in the equation of state, such as theinflaton behaves like radiation after inflation which has the same equation of state with theinteraction term. – 28 – Summary
In this article, we have established an analytic framework for the homogeneous mode ofparametric resonance in a particular parameter space. In a static background, the Mathieuequation is analytically solved to give the stable and unstable bands. In an expandinguniverse, distinct stages of large and small δ (cid:48) are separately considered. At the large δ (cid:48) stage,the phase χ experiences for every half cycle of φ is regarded as random and independentof the history, like the Markov chain. During the small δ (cid:48) stage, all states with differentphases converge to the same eigenstate which is slowly varying. In this eigenstate, | χ | enjoysa stair-like growth. Both stages share the same average boost rate (cid:104) η (cid:105) = 0 . m φ . We also find the backreaction has an impact on δ (cid:48) whichcomes to important earlier than the m φ take-over and guarantees a short large δ (cid:48) stage beforethe backreaction exit of parametric resonance.Applying it on preheating, we have constructed a timeline starting from the beginningof inflation till the backreaction exit. Taking the quadratic slowroll model as an example,the large δ (cid:48) guarantee point and the transition point from large to small δ (cid:48) are included. Theviolation of λ | φ | (cid:29) m has also been considered as another exit of parametric resonance,namely the small | φ | exit. Regarding to how parametric resonance is terminated and whethera small δ (cid:48) stage presents, we divide the parameter space of λ and γ into four parts. Thebackreaction exit typically takes place during the 5th e-fold of preheating for slowroll modelswith e-folding 60.However in a subsequent discussion, we have found parametric resonance during pre-heating can be responsible in amplifying inhomogeneity. To avoid conflict with current CMBspectrum, it acts as both a constraint on parameter space for some inflationary models, andan amplifier to the weak inhomogeneity of some other models which are otherwise disfavored.We have also set up a demonstration by ruling out the backreaction end of preheating of thequadratic slowroll inflation with mass 10 − . Amplification rate for general cases is calculated,and criteria of applicability of this effect is explained.On the other hand, the parametric resonance problem of locked inflation isn’t a problemfor φ multiplets. This is because when one component of φ crosses zero, the contributionfrom other components on m χ would be strong and consequently prevent the exponentialboost effect.This paper aims to show an alternative framework for the broad resonance. For thisreason, we don’t specify the particle physics model and only use singlets. We also show thepreheating stage for quadratic slowroll inflation only as an demonstration, without calculatingthe reheating temperature and thermalization, or discussing difficulties of inflationary models,such as the visible v.s. dark sector or how to embed locked inflation in a particle physicsfoundation. Acknowledgments
I would like to thank Yeuk-kwan Edna Cheung and Yun Zhang for extensive discussionsand Anupam Mazumdar for helpful suggestions on the draft. Exchange of views withMingzhe Li and Youhua Xu is also acknowledged. This work is supported in part by AProject Funded by the Priority Academic Program Development of Jiangsu HigherEducation Institutions (PAPD), NSFC grant No. 10775067, Research Links Programme of– 29 –wedish Research Council under contract No. 348-2008-6049, and Chinese CentralGovernment’s 985 grant for Nanjing University.
References [1] A. H. Guth,
The Inflationary Universe: A Possible Solution to the Horizon and FlatnessProblems , Phys.Rev.
D23 (1981) 347–356.[2] A. D. Linde,
A New Inflationary Universe Scenario: A Possible Solution of the Horizon,Flatness, Homogeneity, Isotropy and Primordial Monopole Problems , Phys.Lett.
B108 (1982)389–393.[3] A. Linde,
Chaotic inflation , Physics Letters B (1983), no. 3-4 177–181.[4] V. Mukhanov and H. Feldman,
Theory of cosmological perturbations , Physics Reports (Jan,1992).[5] A. D. Linde,
Hybrid inflation , Phys.Rev.
D49 (1994) 748–754, [ astro-ph/9307002 ].[6] C. Armendariz-Picon, T. Damour, and V. F. Mukhanov, k - inflation , Phys.Lett.
B458 (1999)209–218, [ hep-th/9904075 ].[7] D. H. Lyth and D. Wands,
Generating the curvature perturbation without an inflaton , Phys.Lett.
B524 (2002) 5–14, [ hep-ph/0110002 ].[8] A. D. Linde,
Fast roll inflation , JHEP (2001) 052, [ hep-th/0110195 ].[9] G. Dvali and S. Kachru,
New old inflation , hep-th/0309095 .[10] E. J. Copeland, A. R. Liddle, D. H. Lyth, E. D. Stewart, and D. Wands, False vacuuminflation with Einstein gravity , Phys.Rev.
D49 (1994) 6410–6433, [ astro-ph/9401011 ].[11] E. D. Stewart,
Inflation, supergravity and superstrings , Phys.Rev.
D51 (1995) 6847–6853,[ hep-ph/9405389 ].[12] G. Dvali and S. Tye,
Brane inflation , Phys.Lett.
B450 (1999) 72–82, [ hep-ph/9812483 ].[13] S. Kachru, R. Kallosh, A. D. Linde, J. M. Maldacena, L. P. McAllister, et. al. , Towardsinflation in string theory , JCAP (2003) 013, [ hep-th/0308055 ].[14] S. Dimopoulos, S. Kachru, J. McGreevy, and J. G. Wacker,
N-flation , JCAP (2008) 003,[ hep-th/0507205 ].[15] L. Kofman, A. D. Linde, and A. A. Starobinsky,
Reheating after inflation , Phys.Rev.Lett. (1994) 3195–3198, [ hep-th/9405187 ].[16] L. Kofman, A. D. Linde, and A. A. Starobinsky, Towards the theory of reheating after inflation , Phys.Rev.
D56 (1997) 3258–3295, [ hep-ph/9704452 ].[17] R. Allahverdi, R. Brandenberger, F.-Y. Cyr-Racine, and A. Mazumdar,
Reheating inInflationary Cosmology: Theory and Applications , Ann.Rev.Nucl.Part.Sci. (2010) 27–51,[ arXiv:1001.2600 ].[18] J. H. Traschen and R. H. Brandenberger, PARTICLE PRODUCTION DURINGOUT-OF-EQUILIBRIUM PHASE TRANSITIONS , Phys.Rev.
D42 (1990) 2491–2504.[19] Y. Shtanov, J. H. Traschen, and R. H. Brandenberger,
Universe reheating after inflation , Phys.Rev.
D51 (1995) 5438–5455, [ hep-ph/9407247 ].[20] E. Copeland and A. Rajantie,
The End of locked inflation , JCAP (2005) 008,[ astro-ph/0501668 ].[21] C. Li, L. Wang, and E. Cheung,
Bound to bounce: a coupled scalar-tachyon model for a smoothcyclic universe , arXiv:1101.0202 . * Temporary entry *. – 30 –
22] S. Davidson and S. Sarkar,
Thermalization after inflation , JHEP (2000) 012,[ hep-ph/0009078 ].[23] R. Allahverdi, A. Ferrantelli, J. Garcia-Bellido, and A. Mazumdar,
Non-perturbative productionof matter and rapid thermalization after MSSM inflation , Phys.Rev.
D83 (2011) 123507,[ arXiv:1103.2123 ].[24] R. Allahverdi and A. Mazumdar,
Reheating in supersymmetric high scale inflation , Phys.Rev.
D76 (2007) 103526, [ hep-ph/0603244 ].[25] A. Mazumdar and J. Rocher,
Particle physics models of inflation and curvaton scenarios , Phys.Rept. (2011) 85–215, [ arXiv:1001.0993 ].[26] R. Easther, J. Khoury, and K. Schalm,
Tuning locked inflation: Supergravity versusphenomenology , JCAP (2004) 006, [ hep-th/0402218 ].[27] R. Allahverdi,
Scenarios of modulated perturbations , Phys.Rev.
D70 (2004) 043507,[ astro-ph/0403351 ].[28] D. I. Podolsky, G. N. Felder, L. Kofman, and M. Peloso,
Equation of state and beginning ofthermalization after preheating , Phys.Rev.
D73 (2006) 023501, [ hep-ph/0507096 ].].