Astrophysical constraints on dark-matter Q -balls in the presence of baryon-violating operators
AAstrophysical constraints on dark-matter Q -balls in the presence of baryon-violatingoperators Eric Cotner and Alexander Kusenko
1, 2 Department of Physics and Astronomy, University of California, Los Angeles, CA, 90095-1547, USA Kavli Institute for the Physics and Mathematics of the Universe (WPI),University of Tokyo, Kashiwa, Chiba 277-8568, Japan
Supersymmetric extensions of the standard model predict the existence of non-topological solitons, Q -balls. Assuming the standard cosmological history preceded by inflation, Q -balls can form in theearly universe and can make up the dark matter. The relatively large masses of such dark-matterparticles imply a low number density, making direct detection very challenging. The strongest limitscome from the existence of neutron stars because, if a baryonic Q -ball is captured by a neutron star,the Q -ball can absorb the baryon number releasing energy and eventually destroying a neutron star.However, in the presence of baryon number violating higher-dimension operators, the growth of a Q -ball inside a neutron star is hampered once the Q -ball reaches a certain size. We re-examine thelimits and identify some classes of higher-dimensional operators for which supersymmetric Q -ballscan account for dark matter. The present limits leave a wide range of parameters available for darkmatter in the form of supersymmetric Q -balls. I. INTRODUCTION
Supersymmetric (SUSY) extensions of the standardmodel predict a scalar potential with a large number offlat directions [1]. Such potentials admit stable configu-rations, SUSY Q -balls [2–4]. Even if the scale of super-symmetry breaking is well above the reach of the presentcollider experiments, the flat directions can exist at ahigh scale and can play an important role in cosmol-ogy. If inflation took place in the early universe, a scalarcondensate can form along the flat directions, leadingto matter–antimatter asymmetry[5–7]. In general, thisscalar condensate is unstable with respect to fragmenta-tion into Q -balls [4, 8–10], which can be entirely stableand can play the role of dark matter [4, 7, 11, 12]. Thisscenario offers a common origin to ordinary matter anddark matter.Dark-matter Q -balls have relatively large masses, and,therefore, very small number densities. A direct detec-tion of such dark matter is extremely challenging [8, 13].These flat directions are only flat at tree level, and ingeneral they are lifted by non-renormalizable terms inthe potential coming from loop corrections and GUT orPlanck-scale physics, taking the form of polynomials inthe squark fields and their conjugates V lifting = g Λ n + m − φ n ( φ ∗ ) m + c.c. (1)suppressed by some energy scale Λ ∼ GeV. If n (cid:54) = m , then baryon number is no longer conserved, ful-filling one of the Sakharov conditions for baryogenesis[14]. The same operators will destabilize the Q -ball [15]and allow it to decay via processes that do not conservethe baryon number. If supersymmetric Q -balls make upthe main component of dark matter, limits on their life-times (namely τ (cid:38) H − ) restrict the set of operators inthe lifting potential in order to prevent their decay ontoo short of a timescale.However, one can set additional constraints on the types of operators in the lifting potential by examiningthe effects of a star infected with a Q -ball. A Q -ballcomposed of squarks in the presence of baryonic matterabsorbs the net baryon number and radiates pions on itssurface [16]. For a main sequence star, a Q -ball shouldpass through with a negligible change in velocity, due tothe relatively low density of the star, and high inertia ofthe Q -ball. A neutron star, however, has a high enoughdensity of baryons that a collision with a Q -ball shouldslow it to a crawl, at which point it would sink to thecenter of the star and begin to consume it from the in-side out [17, 18]. If the Q -ball is absolutely stable, itgrows without bound as it absorbs more neutrons untileither the neutron star is completely consumed, or the Q -ball collapses into a black hole, causing the neutronstar to collapse. Either way, we find the star dies rela-tively quickly on cosmological timescales, on the order of10 years.However, the baryon number violation at a high scale isboth plausible and necessary for the Affleck-Dine baryo-genesis to work. In the presence of baryon-number vio-lating operators, the growth of a Q -ball inside a neutronstar may be stymied by the baryon number destructionin the Q -ball interior, which becomes important once the Q -ball VEV reaches a certain magnitude [18, 19]. In thispaper, we will re-examine the astrophysical bounds tak-ing into account the baryon number violating operators.The paper is organized as follows: section II provides abrief review of allowed Q -ball states, section III explainsthe machinery of calculating the decay rate of the Q -ball, section IV details the interaction of the Q -ball witha neutron star, and section V explains the evolution ofthe baryon number within the Q -ball and star. SectionVI takes this analysis and applies limits to the class ofbaryon-violating operators. a r X i v : . [ h e p - ph ] D ec II. STABLE Q -BALL STATES We begin with a review of the stable ground statesof Q -balls. The minimum necessary ingredients are acomplex scalar field φ with a U(1) symmetry unbrokenat the origin φ = 0. Given a theory of multiple scalarfields with the action S = (cid:90) d x (cid:20) ∂ µ φ † i ∂ µ φ i + 12 ∂ µ χ j ∂ µ χ j − V ( φ i , χ j ) (cid:21) (2)We can perform a Legendre transformation to get theHamiltonian density of the theory, which gives us a func-tional for the energy. E = (cid:90) d x H (3) H = | ˙ φ i | + |∇ φ i | + 12 ˙ χ j + 12 ( ∇ χ j ) + V ( φ i , χ j ) (4)Explicitly adding a Lagrange multiplier ω i (cid:16) Q i − i (cid:82) d x (cid:16) ˙ φ † i φ i − φ † i ˙ φ i (cid:17)(cid:17) to enforce chargeconservation, we get a modified energy functional E = (cid:90) d x ˜ H + ω i Q i (5)˜ H = |∇ φ i | + 12 ( ∇ χ j ) + ˜ V ( φ i , χ j ) (6)˜ V ( φ i , χ j ) = V ( φ i , χ j ) − ω i | φ i | (7)where we have assumed time dependence φ i = φ i ( x ) e iω i t and χ j = χ j ( x ). If for any value of φ, χ (cid:54) = 0 and 0 <ω i < m there exists a point where ˜ V <
0, then stable Q -ball states exist. Furthermore, we can postulate thatthe stable states will be spherically symmetric, so thatthey depend only on the radial coordinate r . A. Flat direction Q -balls Assuming V ( φ ) ≈ M ∼ (1 TeV) far from the origin,the vev in the interior of the Q -ball is not well-localizedin φ -space and the thin wall approximation does nothold. Instead, one can consider a thick-wall variational ansatz φ = φ exp (cid:0) − ( r/R ) (cid:1) . While the r → ωr ) /ωr , the analysisof Ref. [20] shows that the exponential ansatz is in goodoverall agreement with a numerical solution. EvaluatingEq. (5) with the assumption that (cid:82) d x V ≈ πR M / R and using Hamilton’s equa-tion of motion ω = ˙ θ = ∂ E /∂Q , we arrive at ω = ± M (cid:104) π · / /Q (cid:105) / φ = M (cid:20) Q / π (cid:21) / R = 1 M (cid:20) / Q π (cid:21) / E = M (cid:104) π · / Q (cid:105) / (8) Λ | ϕ | M V FIG. 1. Schematic scalar potential with a flat direction whichis lifted by higher-dimension terms near | φ | ∼ Λ. Potentials ofthis form admit flat direction Q -balls which eventually growinto the curved direction type once the critical charge is sur-passed. We can see that these types of Q -balls are stable in thelarge Q limit since ω < m for large charge (the criticalcharge is Q c = 12 √ π ( M/m ) with m the mass of thescalar at φ = 0), and E ∝ Q / . Q -balls of this type are common in supersymmetrictheories where a flat direction develops in the scalar po-tential for the superpartners of the quarks and leptons[3, 17]. The conserved U(1) charge in these cases arethen lepton and/or baryon number and are referred to inthe literature as L-balls and B-balls. In addition to beingable to form stable solitons, the interior of these Q -ballscan sometimes support lepton- or baryon-violating vac-uua [3], which may be exploited in theories of baryo- orleptogenesis. Theories with charged inflatons may alsobe able to support these types of Q -balls since inflatonpotentials must be relatively “flat” to satisfy the slow-rollconditions. B. Curved direction Q -balls As the charge of a flat direction Q -ball grows, and thevalue of the scalar field vev φ slides to higher values,the corrections introduced by the lifting potential V lifting begin affecting the Q -ball (see figure 1). This happenswhen φ ∼ Λ. If the lifting potential is of a form thatrespects the baryon number conservation, it can continuegrowing, albeit in a different manner. The vev hits a wallwhen it reaches its maximum at φ = Λ and cannot climbany higher, so we can approximate the scalar potentialnear this point as V ( φ ) ≈ M + V /B =0lifting = M + 2 Re( g )Λ n − | φ | n (9)Since the vev is constrained to be near Λ, we can usethe thin wall approximation. Substituting into equation5 and fixing φ = Λ, we vary with respect to R to get ω = ± Λ (cid:112) g ) + ( M/ Λ) φ = Λ R = 12Λ (cid:34) Qπ (cid:112) g ) + ( M/ Λ) (cid:35) / (10) E = Λ (cid:112) g ) + ( M/ Λ) Q Since we expect M/ Λ (cid:28)
1, we can neglect those termsunder the square roots for simplicity. The critical chargeat which point a flat direction Q -ball will become acurved direction Q -ball is Q c ≈ . /M ) ∼ .If the lifting potential is not baryon-conserving, theU(1) symmetry is no longer respected and the Q -balldestabilizes, rapidly decaying until the lifting term isnegligible and the Q -ball has reverted back to the flatdirection type. Since curved direction Q -balls are nec-essarily more massive than the flat direction type (andtheir baryon consumption rate even faster), any limitsobtained for flat direction Q -balls will also apply to thecurved direction type, so we need only consider those be-longing to the flat direction classification from now on. III. THE DECAY RATE
We would now like to calculate the decay rate of thequanta of the Q -ball to other particles. The decay of Q -balls to neutrinos was first treated as an evaporation phe-nomenon due to the Pauli exclusion principle preventingdecays in the interior of the Q -ball [21]. Bosons presentno such obstacles, and therefore decays to scalar and vec-tor particles can occur throughout the volume of the Q -ball, provided their mass is less than ω . This may be diffi-cult to achieve in general since most coupled scalar/gaugefields will get a mass term due to the nonzero expecta-tion value in the Q -ball interior. However, the Nambu-Goldstone modes of the Q -ball field itself do not sufferthis mass term, and decays to these modes can occur ifthe U (1) symmetry is very slightly broken by the liftingpotential.Much work has been done calculating the decay andevaporation rates and energy spectra of Q -ball decays tofermions (both massless and massive) [22, 23]. However,these previous studies did not treat decay of the conden-sate to bosons, and are related, but not relevant to theproblem at hand. In this situation, we can utilize a sim-ple method of calculating the decay rate that uses regularperturbation theory (with some extra steps). A. Mathematical background
The probability for an initial state |{ φ i }(cid:105) to evolve intothe final state |{ φ f }(cid:105) is given by P = | (cid:104){ φ f }|{ φ i }(cid:105) | . Inthe case of decays from a Q -ball, we are interested inthe situation where the initial state is simply the scalar condensate describing the Q -ball: | φ c (cid:105) . Since the conden-sate is a persistent feature of the vacuum, the expectationvalue of the fields operator is simply the wave function: (cid:104) φ ( x ) (cid:105) = φ c ( x ), a c-number function. φ c ( x ) is the so-lution to the classical equations of motion in vacuum,which admit Q -ball solutions. Therefore, we can decom-pose the field operator into a classical and quantum part: φ = φ c + ˆ φ (we later employ a different decomposition inorder to properly separate the field into its mass eigen-states, but it is conceptually similar to this one).However, we are interested in how the Q -ball decays,so we must consider the state in which the scalar con-densate is in the background of an interacting vacuum: | Φ c (cid:105) . The transition probability to any set of final stateparticles { φ f } is then P = | (cid:104) Φ c | Φ c { φ f }(cid:105) | . Using thesingle-particle expansion of the final particle states |{ φ f }(cid:105) = (cid:90) (cid:89) f (cid:32) d p f (2 π ) φ f ( p f ) (cid:112) E f (cid:33) |{ p f }(cid:105) (11)the transition probability is then P = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:90) (cid:89) f (cid:32) d p f (2 π ) φ f ( p f ) (cid:112) E f (cid:33) (cid:104) Φ c | Φ c { p f }(cid:105) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (12)Instead of using arbitrary wave functions as the finalstate, we can simply use the states of definite momen-tum, as is typically done, so that φ f ( p f ) = (2 π ) δ ( p f − p ) / √ V . In this case, the differential transition probabil-ity is then dP = (cid:89) f (cid:18) d p f (2 π ) E f (cid:19) | (cid:104) Φ c | Φ c { p f }(cid:105) | (13)The matrix element M = (cid:104) Φ c | Φ c { p f }(cid:105) can be computedperturbatively just as is normally done in QFT, exceptthat we have to keep in mind the expansion of the scalarfield operator φ = φ c + ˆ φ . This leads to a bit of a com-plication, since working in the momentum space involvesa Fourier transform of φ c , introducing an additional in-tegral which consumes some of the delta functions thatnormally can be separated from the scattering amplitude M . In addition, there is also no integral over the impactparameter since there are no collisions involved in thisdecay process. Depending on the number of interactionvertices in the process, we find the matrix element canbe written schematically as M = (cid:104) Φ c | Φ c { p f }(cid:105) = A n ( { p f } )(2 π ) δ ( nω − Σ f E f ) (14)where ω is the Q -ball energy per particle (chemical po-tential), n is the number of Q -ball quanta consumed bythe decay (determined by the number of external legs at-tached to the condensate), and A is a “reduced” matrixelement. The delta function enforces global energy con-servation, and although momentum is conserved at eachvertex internal to the diagram, global momentum is not.This can be understood by the fact that the existence ofthe condensate breaks the spatial translation invarianceof the vacuum, and therefore momentum is no longer aconserved quantity, the condensate instead absorbing thedifference, similar to how a crystal lattice will absorb therecoil from a nuclear decay in the M¨ossbauer effect.Now, one will find that equation 13 implicitly containsthe square of a delta function, which is a little troubling.However, integration over the final state momenta willeat up one of the delta functions, leaving a δ (0), which isproportional to an infinite period of time T = 2 πδ (0), inthe sense that the limit of T is this quantity, so that strip-ping this from the RHS gives us a probability per unittime per unit phase space; in other words, the differentialdecay rate d Γ = (cid:89) f (cid:18) d p f (2 π ) E f (cid:19) |A n ( { p f } ) | (2 π ) δ ( nω − Σ f E f )(15)This method has wide applicability in calculating the de-cay of condensates and background fields, as the finalstate particles can be of either bosonic or fermionic type(the initial states can only be bosonic since fermions can’tform condensates). The authors have also verified in thelimit that the condensate wave function is that of a singlezero-momentum particle φ c ∼ / √ V , the Fourier trans-form of which is a zero-momentum delta function, thedecay rate reduces to that of a familiar single particledecay, as one would expect. The only drawback of thismethod is that it cannot handle decays that significantlyalter the condensate wave function since φ c would thenbe different in the initial and final states and it wouldnot be appropriate to expand around. Thankfully, wewill only be interested in decays involving ∆ Q (cid:46)
10 from Q -balls with Q ∼ , so the change in charge per decayis entirely negligible. B. Mass eigenstates and phonons
As briefly mentioned earlier, we would like to use a de-composition of the field operator that respects the masseigenstates of the theory. For a theory with an unbrokenU(1), a polar decomposition φ = ρe iθ / √ ρ .Therefore, this field is massive with the same mass asthe original complex field: m | φ | = m ρ . The po-tential is completely devoid of any terms containing θ however, due to the U(1) symmetry. This angular de-gree of freedom is therefore a massless Goldstone bosonof the theory (inside the Q -ball it picks up a small massdue to the fact that has a minimum wavelength λ ∼ R ). Therefore we need a representation of the phonon oper-ator that captures perturbations around the condensatewhile keeping the mass eigenstates separate. This leadsus to consider the decomposition of the phonon field intoa radial and angular part: φ = 1 √ ρe iθ = 1 √ ρ c + ˆ ρ ) e i ( θ c +ˆ θ ) ≈ φ c + 1 √ ρe iωt + i √ ψe iωt + · · · (16)where ˆ ψ ≡ ρ c ˆ θ , θ c ≡ ωt , and the · · · refers to the higher-order terms in the Taylor expansion of the exponential.Although there is no way to invert the full relationshipfor ˆ ρ and ˆ ψ in terms of ˆ φ and ˆ φ ∗ , the expansion to linearorder can be inverted, and this gives us an approximatedictionary between the different phonon operators:ˆ φ = 1 √ ρ + i ˆ ψ ) e iωt ˆ φ † = 1 √ ρ − i ˆ ψ ) e − iωt (17)ˆ ρ = 1 √ (cid:16) ˆ φ † e iωt + ˆ φe − iωt (cid:17) ˆ ψ = i √ (cid:16) ˆ φ † e iωt − ˆ φe − iωt (cid:17) Unfortunately, we cannot simply substitute the above re-lationships into the Lagrangian because these are onlycorrect to first order; we must expand around ρ c and θ c in each term, then do a Taylor expansion in the exponen-tial.The ˆ φ operator is complex, yet is not charged under the U (1) of the theory inside the Q -ball because ˆ φ → ˆ φe iα is not a symmetry of the Lagrangian unless φ c = 0 (inwhich case we are outside the Q -ball). Neither of the ˆ ρ orˆ ψ is charged either, so a charged current cannot exist inthe interior unless it is via bulk motion of, or interactionwith, the condensate field φ c . C. Calculation of the matrix element
We will now use the method of sections III A and III Bin order to derive the matrix element for decay of several Q -ball quanta to phonons within the Q -ball (the Feyn-man diagram representation of which is given by figure2). We consider the lifting potential discussed earlier andexpand it in polar form: L lifting = − g Λ n + m − φ n ( φ † ) m + c.c.= − g nm (cid:18) ρ √ (cid:19) n + m e i ( n − m ) θ + c.c. (18)where g nm ≡ g/ Λ n + m − . We now expand around the Q -ball condensate in the way prescribed above, giving us n + m − N ρ − N ψ N ρ N ψ FIG. 2. Feynman diagram representation of the matrix ele-ment responsible for decay of the Q -ball into phonons. Exter-nal lines on the left marked by a cross are interactions of theoperator with the condensate φ c , whereas external lines onthe right are the phonons produced from the decay. Arrowsdenote flow of momentum, not charge. L lifting = − g nm ( n + m ) / n + m (cid:88) j =0 ∞ (cid:88) k =0 (cid:18) n + mj (cid:19) i k ( n − m ) k k ! ( ρ n + m − j − kc e i ( n − m ) θ c )ˆ ρ j ˆ ψ k + c.c. (19)Now we calculate the matrix element for the decay of the condensate to N ρ ρ ’s and N ψ ψ ’s: M = i ( n + m ) / (cid:88) j,k (cid:18) n + mj (cid:19) ( n − m ) k k ! (cid:90) d x ρ n + m − j − kc (cid:104) i k g nm e i ( n − m ) ωt + c.c. (cid:105) × (cid:68) (cid:12)(cid:12)(cid:12) ˆ ρ j ˆ ψ k (cid:12)(cid:12)(cid:12) p , · · · , p N ρ , k , · · · , k N ψ (cid:69) (20)= 2 πi ( n + m ) / C nmN ρ N ψ (cid:90) (cid:32) q D (cid:89) q = q d q (2 π ) ρ c ( q ) (cid:33) (2 π ) δ ( Q − ( P + K )) × (cid:2) i N ψ g nm δ (( n − m ) ω − ( P + K )) + ( − i ) N ψ g ∗ nm δ (( m − n ) ω − ( P + K )) (cid:3) (21)where C nmjk ≡ j ! (cid:0) n + mj (cid:1) ( n − m ) k , D ≡ n + m − N ρ − N ψ and Q = (cid:80) q , P = (cid:80) p , K = (cid:80) k are the sums ofthe various 4-momenta. We substitute in the Q -ball wave function to the Fourier transform ρ c ( q ) = √ φ c ( q ) = √ π / R φ e − q R / : M = i (2 π ) − D D/ ( n + m ) / C nmN ρ N ψ (cid:16) √ π / R φ (cid:17) D (cid:90) (cid:32) q D (cid:89) q = q d q e − R q (cid:33) δ ( Q − ( P + K )) × (cid:2) i N ψ g nm δ (( n − m ) ω − ( P + K )) + ( − i ) N ψ g ∗ nm δ (( m − n ) ω − ( P + K )) (cid:3) (22)Now, we use an interesting geometric argument to solve this integral. Since d q = dq dq dq and q = q + q + q ,we observe that (besides the delta functions), the integral is hyper-spherically symmetric in the 3 D -dimensional q -space. The three delta functions each define a hyperplane in this space, the union of which is a 3( D − D -dimensional space. This hypersurface is displaced from the originby the vector v = ( P + K ) / √ D (notice the hyperplane defined by Q i − ( P i + K i ) = 0 has a unit normal vectorof ˆ n = (1 , , · · · , / √ D and is displaced from the origin by a distance of | P i + K i | / √ D ). This integral thereforerepresents a spherically symmetric Gaussian integral over a 3( D − v .We can therefore rotate our coordinate system so that v points in the new “ˆ z ” direction and transform to a type of“hypercylindrical coordinates”: ( s, φ, θ , · · · , θ D − − , x, y, z ) where the coordinates x , y , and z are Euclidean anda specification of ( x, y, z ) = (0 , , v ) constrains one to the hypersurface. Then, we simply perform the sphericallysymmetric integral over this surface: (cid:90) (cid:32) q D (cid:89) q = q d q e − R q (cid:33) δ ( Q − ( P + K )) = (cid:90) d D − se − R ( s + v ) = Ω D − − e − R v (cid:90) ∞ ds s D − − e − R s (23)where Ω n − = 2 π n/ / Γ( n/
2) is the solid angle of the ( n − D − − ). After the dust has settled, thematrix element is found to be M = 2 iπ / ( n + m ) / − N ρ − N ψ C nmN ρ N ψ R φ D e − R D ( P + K ) × (cid:2) i N ψ g nm δ (( n − m ) ω − ( P + K )) + ( − i ) N ψ g ∗ nm δ (( m − n ) ω − ( P + K )) (cid:3) (24)The number of Q -ball quanta that decay in each event (and thereby amount of charge violation) can be read off fromthe delta function, and is ∆ Q = | n − m | , as expected. One important note is that since m ρ (cid:29) ω , the condensatecannot decay to ρ ’s unless | n − m | ω > m ρ , which requires the amount of charge violation to be very high. Decaysto ψ ’s might appear to proceed unimpeded, however, because they are massless. However, these phonons pick up asmall mass from two different sources. First, as mentioned before, because the phonons are confined to the Q -ball,they are essentially standing waves with a maximum (Compton) wavelength of λ ≈ R , which implies a minimumrest energy m ψ = 1 /k = 1 / πR . Since in a thick-wall Q -ball ωR = √
3, we have m ψ = ω/ π √ ≈ ω/
22, which issmall, but still a significant fraction of ω ! Second, the baryon-violating term itself introduces a small mass, which wecan see by expanding to second order in ˆ ψ : L lifting ⊃ (cid:20) ( n − m ) ρ n + m − c ( n + m ) / − (Im( g nm ) sin(( m − n ) ωt ) − Re( g nm ) cos(( m − n ) ωt )) (cid:21) ˆ ψ (25)where the quantity in square brackets can be identified with m ψ . Not only is this mass small in magnitude comparedwith the first contribution, but it also has harmonic time dependence, and therefore averages out to zero over timescaleslonger than about | n − m | /ω . Thus, it is safe to assume that the mass from the Compton wavelength is the only massthat contributes. D. Calculation of the decay rate
We now apply equation 15 to calculate the decay rate, focusing on decays to only the Goldstone modes and setting N ρ = 0 and N ≡ N ψ . We take the squared amplitude (which can be simplified because the cross-terms are zero dueto the conflicting delta functions), drop one of the delta functions, and integrate over the final state phase space:Γ Nnm = 4 π n + m − N | g nm | ( C nm N ) R φ D I N (cid:16) | ( n − m ) | ω, R/ √ D, m ψ (cid:17) (26)where I N (Ω , a, m ) = (cid:90) (cid:32) p N (cid:89) p = p d p (2 π ) (cid:112) p + m (cid:33) e − a ( (cid:80) p ) δ (cid:16) Ω − (cid:88) (cid:112) p + m (cid:17) (27)For N = 1 we can get an exact answer: I (Ω , a, m ) = 1(2 π ) e − a (Ω − m ) (cid:112) Ω − m Θ(Ω − m ) (28)However, using the relationships m ψ = ω/ π √ R = √ /ω , we can reduce the integral to something even simpler: I N ( | n − m | ω, R/ √ D, m ψ ) = (cid:90) p N (cid:89) p = p d p (2 π ) (cid:113) p + m ψ e − R D ( (cid:80) p ) δ (cid:16) | ( n − m ) ω | − (cid:88) (cid:113) p + m ψ (cid:17) = ω N − (cid:90) (cid:32) p N (cid:89) p = p d ( p/ω )(2 π ) (cid:112) ( p/ω ) + ( m ψ /ω ) (cid:33) e − π D ( (cid:80) ( p /ω )) δ (cid:18) | n − m | − (cid:88) (cid:113) ( p/ω ) − ( m ψ /ω ) (cid:19) (29)We then change coordinates to ξ ≡ p /ω and substitute the phonon mass so that m ψ /ω ≡ µ = 1 / π √ ω N − (cid:90) ξ N (cid:89) ξ = ξ d ξ (2 π ) (cid:112) ξ + µ e − π n + m − N ( (cid:80) ξ ) δ (cid:16) | n − m | − (cid:88) (cid:112) ξ + µ (cid:17) (30)where we will define the integral in square brackets as J Nnm (note J Nnm = J Nmn and J = 0 if N ≥ n + m or n = m ).Because n, m , and N are integers and J is a dimensionless number, we can simply tabulate all its possible valuesusing numerical integration such as Monte Carlo. However, because of the delta function, we can’t do MC until weintegrate that out. We convert to spherical coordinates and separate the N th coordinate from the rest, then integrateover it to remove the delta function, leaving us with J Nnm = 1(16 π ) N (cid:89) ξ,φ,θ (cid:48) (cid:32)(cid:90) π dφ (cid:90) π dθ sin θ (cid:90) ∞ ξ (cid:112) ξ + µ (cid:33) (cid:90) π dφ N (cid:90) π dθ N sin θ N ξ N ( { ξ } ) × e − π n + m − N (cid:16)(cid:80) (cid:48) i ξ i + (cid:80) (cid:48) i (cid:54) = j ξ i · ξ j + ξ N ( { ξ } )+2 (cid:80) (cid:48) i ξ N ( { ξ } ) ξ i (sin θ N sin θ i cos( φ N − φ i )+cos θ N cos θ i ) (cid:17) (31)where the primed sums/products mean we sum/multiply over all coordinates except the N th, and ξ N ( { ξ } ) ≡ (cid:114)(cid:16) | n − m | − (cid:88) (cid:48) (cid:112) ξ + µ (cid:17) − µ (32) ξ i · ξ j = ξ i ξ j (sin θ i sin θ j cos( φ i − φ j ) + cos θ i cos θ j ) (33)Some sample values of J Nnm (I’ll restrict to n + m > | n − m | = 1 for now) are: N = 1 : J = 5 . × − , J = 1 . × − , J = 3 . × − , J = 1 . × − , J = 7 . × − N = 2 : J = 1 . × − , J = 3 . × − , J = 5 . × − , J = 8 . × − , J . × − N = 3 : J = 2 . × − , J = 5 . × − , J = 1 . × − , J = 2 . × − , J = 7 . × − N = 4 : J = 8 . × − , J = 6 . × − , J = 9 . × − , J = 5 . × − , J = 2 . × − N = 5 : J = 3 . × − , J = 7 . × − , J = 3 . × − , J = 2 . × − Clearly, final states involving more phonons have a smaller amount of phase space volume. The exception is N =1, which gets extra suppression from the fact that any decay involving one final state particle does not conservemomentum. It should be noted that repeated evaluation of the Monte Carlo shows that the uncertainty in theseanswers is quite large; variation in the first digit is common, though the order of magnitude remains consistentover repeated evaluations. It turns out that this is not terribly important for computing the neutron star lifetimes;variations of O (1) in J translate to variations of O (10 − ) in the lifetimes. This is because the dependence of Γ on Q is most important. There is also a small imaginary part attached to some of these numbers which is not shown. Thisis from integrating over a region in phase space which is not kinematically allowed, and it does not contribute to thedecay rate, so we can simply ignore it.If the dimension of the lifting potential is extremely high ( n + m → ∞ ), then the exponential in the integrandbecomes order unity, and we can reduce this even further by transforming to a dimensionless energy coordinate σ = (cid:112) ξ + µ and integrating out all the angles. The integral J Nnm approaches J Nnm → π ) N (cid:32) σ N (cid:89) σ = σ (cid:90) | n − m |− ( N − µµ dσ (cid:112) σ − µ (cid:33) δ (cid:16) | n − m | − (cid:88) σ (cid:17) (34)This can be calculated via Monte Carlo in a similar manner to equation 31. Note that in this limit J Nnm only dependson ∆ Q = | n − m | and N . Now, putting all of this together, we can express the decay rate to N Goldstone bosons asΓ
Nnm = 4 π n + m − N | g nm | ( N !) ( c nm N ) R φ D ω N − J Nnm (35)Writing out R , φ and ω in terms of Q (see equations 8) and lumping all the non-dimensional constants together,Γ Nnm = | g | K Nnm Q (7+2( n + m − N )) M (cid:18) M Λ (cid:19) n + m ) − (36) K Nnm ≡ (5( n + m − N ) − (1 − n + m − N )) π − ( n + m − N/ ( n − m ) N J Nnm (37)where M is the mass scale associated with the potential energy density in the flat direction of the scalar potential( V = M ). We can now simply tabulate the K Nnm and have a semi-analytic expression for the decay rate that will beeasy to use in the analysis of section V.
IV. INTERACTIONS BETWEEN Q -BALLS ANDNEUTRON STARS We would now like to understand how a Q -ball inter-acts with its host star in order to determine the neutronconsumption rate. As discussed in the work of one of us,Loveridge, and Shaposhnikov (KLS) [18], the transportmechanism of neutrons inside a neutron star is compli-cated and is not very well understood. The authors out-line two different possible situations for neutron accre-tion, which we will summarize here for clarity. A. Surface conversion of neutron flux
As a rough estimate, KLS assume the rate of neutronabsorption is simply equal to the flux of neutrons movingacross the surface of the Q -ball. In this scenario, thegrowth rate of the Q -ball is given by˙ Q = b − πR n v = 4 · / n M (4 π ) / Q / ≈ (2 × − GeV) Q / (38)where b = 1 / n ≈ g/cm = 4 × − GeV is the neutron number den-sity at the center of the star, and v ≈ Q -ball. B. Hydrodynamic considerations due to pionproduction
Using a couple different methods, KLS determine thepressure at the center of the star in hydrostatic equilib-rium is approximately P ≈ (0 . . For light degreesof freedom such as pions, electrons and neutrinos, this im-plies a temperature of about 100 MeV from the relation P ≈ gT /π . This temperature cannot be maintainedby thermal effects alone, but can be maintained by thepions produced on the surface of the Q -ball. The rate ofpion loss to decay inside the star is given by˙ N π ≈ π / (cid:114) λ τ n π (0) R (39)where λ ≈ n − / and τ ≈ GeV − are the meanfree path and neutral pion lifetime, respectively. Theyalso assume n π (0) ≈ n in order to maintain pressure. Each neutron only has enough mass and energy to sup-ply about 4-5 pions, so the rate of neutron absorption isabout that much lower, giving us˙ Q = 10 π / n / R b √ τ = 5 πn / / bM τ / ≈ (10 − GeV) Q / (40)This estimate is slightly lower than the raw neutron fluxestimate and is a little more realistic. V. BARYON NUMBER EVOLUTION IN ANINFECTED NEUTRON STAR
Now that have expressions for both the growth rateand decay rate of the Q -ball, we can set up a simpleset of differential equations to model the evolution of thebaryon number in both the Q -ball and the neutron star:˙ B Q = b ˙ Q = − ˙ N n − b | n − m | Γ nm (41)˙ B NS = ˙ N n = − (10 − GeV) Q / (42)where Γ nm ≡ (cid:80) N Γ Nnm . Or, eliminating N n and assum-ing decays are dominated by a specific N (usually either1 or 2), we can put it in a more aesthetically pleasingform: ˙ Q ≈ N Q / − Γ Q α (43)where ˙ N = 10 − GeV, Γ = | g | K Nnm M ( M/ Λ) n + m ) − , and α = (7+2( n + m − N ))( α > N is some ridiculously high number, whichis unlikely). The initial conditions for this system are Q (0) = Q ≈ and N n (0) = B NS = 10 , and thetotal number of neutrons absorbed by the Q -ball is givenby integrating equation 42:∆ N n ( t ) = − (cid:90) t dt (cid:48) ˙ N Q / (44)We can see that equation 43 has late-time attractor so-lutions, whereby setting ˙ Q = 0, we solve for the equi-librium charge: Q eq = (3 ˙ N / Γ ) α − / (see figure 3). Ifthis charge is reached relatively quickly compared to thetotal lifetime of the neutron star, then equation 44 im-plies that the neutron depletion is linear in time, and thelifetime of the star is then τ NS ≈ B NS ˙ N (cid:32) N Γ (cid:33) / / − α (45) × - × - × - t [ yr ] Q n = = = = = = Q - ball charge with Δ Q = FIG. 3. Plot of the evolution of charge within the Q -ball atthe center of a neutron star with decay channels attributedto various ∆ Q = 1 operators, indexed by ( n, m ) = ( n, n + 1).The Q -ball very quickly equilibrates so that the rate of decayis equal to the rate of neutron consumption. Not shown arethe contours for n = 2 and n = 3, which are ruled out becausethe corresponding operators would destabilize the Q -ball infree space on short timescales. In free space, the evolution of the charge of a Q -ball isgiven by equation 43 with ˙ N = 0, which can easily besolved for: Q ( t ) ≈ (cid:20) ( α − (cid:18) Q − α α − t (cid:19)(cid:21) − α (46)The Q -ball lifetime can then be solved for by setting Q ( τ Q ) = 1, which gives us τ Q ≈ − Q − α ( α − (47)If we want to be more exact and take into account decaysfrom all channels (not just the dominant one), we cannumerically solve for τ NS and τ Q by evolving equations41 and 42 until N n = 0 or Q = 1, at which point eitherthe neutron star has been consumed or the Q -ball hasdecayed, and we stop integration (a specific example isgiven in figure 4). This is how we will derive the limitsin the next section. VI. LIMITS ON BARYON-VIOLATING (ANDCONSERVING) OPERATORS
Using equations 41 and 42 and the algorithm pre-scribed in the previous section, we can tabulate the life-times of infected neutron stars and free Q -balls endowedwith the lifting potential of equation 1, indexed by the in-tegers n and m . We will find that baryon-violating termsare necessary if an infected neutron star is to survive topresent day. A. From decay of Q -balls in free space We solve the baryon number evolution equations with˙ N = 0 in order to model the decay of the Q -ball in Q - Δ N n - - - - t [ yr ] Baryon number evolution of Q - ball - infected NS ( n = = ) FIG. 4. Plot of the charge Q contained within a Q -ball andthe number of neutrons consumed by the Q -ball over the lifeof the neutron star. Once − ∆ N n = B NS = 10 , integrationis stopped and the star is gone. This specific example is fora Q -ball with decays mediated by a ( n, m ) = (4 ,
6) operator,resulting in a neutron star lifetime of τ NS = 1 × years. log ( τ [ yr ]) - FIG. 5. Plot of Q -ball lifetimes with an initial charge of Q = 10 as a function of various n , m corresponding tothe terms in the lifting potential. The diagonal n = m isactually completely stable because decays are not permitteddue to restoration of the U (1) B symmetry. free space. The results are plotted in figure 5 and tab-ulated in table I in the appendix. The most strikingfeature is that for n = m , the Q -ball is completely sta-ble because the Goldstone field does not appear in thepotential. We can also see that in general, as the dimen-sion of the operator increases, so does the lifetime of the Q -ball. In fact, all Q -balls with lifting potentials of di-mension 5 or less are unstable and decay in a matter ofhours or less, whereas those with dimension greater than5 are stable on timescales much longer than the age ofthe Universe. This immediately rules out dark matter Q -balls with n + m ≤
5. In the high-dimension limit( n + m → ∞ ), we can calculate J Nnm using equation 31and solve the baryon evolution equations again, thoughthis doesn’t lead to any interesting revelations; the Q -ball lifetime continues to increase as the dimension ofthe operator increases, and is pretty much independent0 log ( τ [ yr ]) FIG. 6. Plot of neutron star lifetimes after being infected bya Q -ball with initial charge Q = 10 as a function of various n , m corresponding to the terms in the lifting potential. Thediagonal n = m is ruled out because the B-violating decaysare forbidden, and the stability of the Q -ball causes it to growwithout bound, quickly consuming the star. of ∆ Q . The largest lifetime calculated (dim = 100) wasover 10 years! B. From lifetime of neutron stars
Solving the baryon number evolution equation with˙ N (cid:54) = 0 and integrating until N n = 0 gives us the lifetimeof an infected neutron star. This information is plottedin figure 6 and table II. As we can see, the diagonal where n = m is ruled out, with a lifetime of about 10 years.This is due to the fact that the Q -ball is absolutely stablein this regime, and therefore grows without bound as iteats away at the neutron star, quickly consuming it. Infact, this is an upper limit on the lifetime; the final chargeof the Q -ball in this situation is 3 × , which is beyondthe critical charge for a flat direction Q -ball to changeinto a curved direction type, which as mentioned before,has an even higher rate of neutron consumption. Thehighest charge for a Q -ball with baryon-violating decaysis only 10 , well below the critical charge. Interestingly,in the regions with operator dimension ≤
4, the Q -balldecays so quickly that it breaks down completely beforethe neutron star is consumed. As mentioned in the previ-ous subsection, Q -balls in this regime aren’t stable in freespace anyway. We can see that as we move away from the n = m diagonal (increasing ∆ Q ), the lifetime of the starbegins to drop, then levels out, with the magnitude ofthe drop decreasing as the operator dimension increases.In order to study the effects of very high-dimension op-erators ( n + m → ∞ ), we once again use equation 31to calculate J Nnm and solve the baryon number evolutionequations. This is plotted in figure 7. What we find isquite interesting: the lifetime appears to approach a lim-iting value around 10 years as the operator dimension Δ Q d i m Neutron star lifetime log ( τ [ yr ]) FIG. 7. Plot of neutron star lifetimes after being infected bya Q -ball with initial charge Q = 10 as a function of thecharge violation per decay ∆ Q >
Q > dim, which is not allowed since itimplies one of either n or m is negative. increases. The lifetime is roughly independent of ∆ Q ,though it does drop slightly near ∆ Q = 0. This appearsto match the trend of figure 6 as the operator dimensionis increased. VII. CONCLUSION
We have shown here that Q -balls can make up darkmatter if baryon-violating terms of dimension n + m > n = m ) are ruled out as well dueto unrestricted Q -ball growth. The baryon number vio-lation is also necessary for the Affleck-Dine mechanismto work. This eliminates the neutron star bounds. Be-yond this, there appears to be no restriction on theseoperators, even at very high dimension. The low level ofbaryon number violation does not affect the experimen-tal limits based on IceCube [13], Super-Kamiokande [24]and other direct detection experiments. However, oneshould keep in mind that Q -balls may carry some electriccharge [8, 24, 25], making them almost invisible to mostdirect-detection searches. (A positively charged Q -ballcannot destabilize nuclei because the Coulomb repulsionprevents any strong interactions between non-relativistic Q -balls and matter nuclei.) This leaves a wide range ofparameters available for dark matter in the form of su-persymmetric Q -balls. VIII. ACKNOWLEDGEMENTS
This work was supported by the U.S. Department ofEnergy Grant No. de-sc0009937. A.K. was also sup-1ported by the World Premier International ResearchCenter Initiative (WPI), MEXT, Japan. A.K. appre- ciates the hospitality of the Aspen Center for Physics,which is supported by National Science Foundation grantPHY-1066293. [1] T. Gherghetta, C. Kolda, S. P. Martin,
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10 2 × × × × × × > > > ∞ TABLE I. Table of Q -ball lifetimes (in years) for various lifting potentials. Lifetimes with an ∞ are absolutely stable due torestoration of the U (1) B symmetry. n \ m × ∞ ∞ × × × × × × × ∞ × × × × × × × × × ∞ × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × ×
10 1 × × × × × × × × × × TABLE II. Table of infected neutron star lifetimes (in years) for various lifting potentials. Lifetimes with an ∞∞