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<ul><li><p>Theoretical and Mathematical Physics, 170(2): 249262 (2012)</p><p>AXIONS AND COSMIC RAYS</p><p>D. Espriu and A. Renau</p><p>We investigate the propagation of a charged particle in a spatially constant but time-dependent pseudo-</p><p>scalar background. Physically, this pseudoscalar background could be provided by a relic axion density.</p><p>The background leads to an explicit breaking of Lorentz invariance; processes such as p p or e eare consequently possible under some kinematic constraints. The phenomenon is described by the QED</p><p>Lagrangian extended with a ChernSimons term that contains a four-vector characterizing the breaking</p><p>of Lorentz invariance induced by the time-dependent background. While the induced radiation (similar to</p><p>the Cherenkov eect) is too small to inuence the propagation of cosmic rays signicantly, the hypothetical</p><p>detection of the photons radiated by high-energy cosmic rays via this mechanism would provide an indirect</p><p>way to verify the cosmological relevance of axions. We discuss the order of magnitude of the eect.</p><p>Keywords: axion, high-energy cosmic ray, galactic magnetic eld</p><p>1. Axions</p><p>Cold relic axions resulting from vacuum misalignment [1], [2] in the early universe is a popular and sofar viable candidate for dark matter. If we assume that cold axions are the only contributors to the matterdensity of the universe apart from ordinary baryonic matter, then its density must be [3]</p><p> 1030 g/cm3 1046 GeV4.Of course, dark matter is not uniformly distributed: its distribution traces that of visible matter (or ratherthe other way around). The galactic halo of dark matter (assumed to consist of axions) would correspondto a typical value for the density [4]</p><p>a 1024 g/cm3 1040 GeV4,extending over a distance of 30 to 100 kpc in a galaxy such as the Milky Way. The precise details of thedensity prole are not so important at this point. The axion background provides a very diuse concen-tration of pseudoscalar particles interacting very weakly with photons and therefore indirectly with cosmicrays. What are the consequences of this diuse axion background on high-energy cosmic ray propagation?Could this inuence cosmic ray propagation similarly to the GreisenZatsepinKuzmin (GZK) cuto [5]?This is the question that we answer here.</p><p>It is quite relevant that the axion is a pseudoscalar, being the pseudo-Goldstone boson of the brokenPecceiQuinn symmetry [6]. Its coupling to photons occurs through the anomaly term; the coecient ishence easily calculable if the axion model is known:</p><p>L = ga 2a</p><p>faFF, (1)</p><p>CERN, Geneva, Switzerland, e-mail: domenec.espriu@cern.ch.Department of Structure and Constituents of Matter, Institute of Cosmos Sciences, University of Barcelona,</p><p>Barcelona, Spain, e-mail: arencer@gmail.com.</p><p>Prepared from an English manuscript submitted by the authors; for the Russian version, see Teoreticheskaya i</p><p>Matematicheskaya Fizika, Vol. 170, No. 2, pp. 304320, February, 2012.</p><p>0040-5779/12/1702-0249</p><p>c</p><p>249</p></li><li><p>Two popular axion models are the model in [7] and the model in [8], in both of which ga 1. We providesome additional details of the theory in Sec. 3.</p><p>2. Cosmic rays</p><p>Cosmic rays consist of particles (such as electrons, protons, helium nuclei, and other nuclei) reachingthe Earth from outside. Primary cosmic rays are those produced by astrophysical sources (e.g., supernovas),while secondary cosmic rays are particles produced by the interaction of primaries with interstellar gas. Westudy the eect of axions on the propagation of these cosmic rays. We separately consider proton andelectron cosmic rays and ignore heavier nuclei because the eect on them is far less important, as becomesclear later (the axion-induced bremsstrahlung depends on the mass of the charged particle).</p><p>2.1. Cosmic ray energy spectrum. The number of protons in cosmic rays is an important char-acteristic. Experimental data indicate that the number of cosmic ray particles with a given energy dependson energy according to a power law J(E) = NiEi , where the spectral index i takes dierent values indierent regions of the spectrum [9]. For protons, we have</p><p>Jp(E) =</p><p>5.87 1019E2.68 eV1m2s1sr1, 109 eV E 4 1015 eV,6.57 1028E3.26 eV1m2s1sr1, 4 1015 eV E 4 1018 eV,2.23 1016E2.59 eV1m2s1sr1, 4 1018 eV E 2.9 1019 eV,4.22 1049E4.3 eV1m2s1sr1, E 2.9 1019 eV,</p><p>(2)</p><p>while the power law for electrons is [10]</p><p>Je(E) =</p><p>5.87 1017E2.68 eV1m2s1sr1, E 5 1010 eV,4.16 1021E3.04 eV1m2s1sr1, E 5 1010 eV,</p><p>(3)</p><p>with the ux typically two orders of magnitude below the proton ux, although it is known less well. Ourignorance about electron cosmic rays is quite regrettable because it has a substantial impact in our estimateof the radiation yield.</p><p>We note that the above values measured locally in the inner solar system. It is known that the intensityof cosmic rays increases with distance from the sun because the modulation due to the solar wind makes itmore dicult for them to reach us, particularly so for electrons. In addition, the hypothesis of homogeneityand isotropy holds for proton cosmic rays but not necessarily for electron cosmic rays. Indeed, becausecosmic rays are deected by magnetic elds they follow a nearly random trajectory within the Galaxy.We know that a hadronic cosmic ray spends an average of about 107 years in the galaxy before escapinginto intergalactic space. This ensures the uniformity of the ux, at least for protons of galactic origin. Incontrast, electron cosmic rays travel for an average of about 1 kpc before being slowed down. But becausel t1/2 for a random walk, 1 kpc corresponds to a typical age of an electron cosmic ray of 105 yr [11]. Inaddition, the lifetime of an electron cosmic ray depends on the energy as</p><p>t(E) 5 105(</p><p>1TeVE</p><p>)</p><p>yr =T0E</p><p>, T0 2.4 1040.</p><p>To complicate matters further, it has been argued that the local interstellar electron ux is not evenrepresentative of the Galactic ux and may reect the electron debris from a nearby supernova 104 yearsago [12].</p><p>250</p></li><li><p>2.2. The GZK cuto. It was stated in [5] that the number of cosmic rays above a certain energythreshold (we call this the GZK limit) should be very small. Cosmic rays particles interact with photons fromthe cosmic microwave background (CMB) to produce pions CMB + p p + 0 or CMB + p n + +.The energy threshold is about 1020 eV. Because of the mean free path associated with these reactions,cosmic rays with energies above the threshold and traveling over distances larger than 50Mpc should notbe observed on the Earth. This is the reason for the rapid fall o of the proton cosmic ray spectrum above1020 eV because there are very few nearby sources capable of providing such tremendous energies.</p><p>We note that the change in the slope of the spectrum at around 1018 eV is believed to be due to theappearance of extragalactic cosmic rays at that energy.</p><p>3. Solving QED in a cold axion background</p><p>In this section, in great detail, we describe the theoretical tools needed for understanding the interac-tions between the high-energy cosmic rays we have just described and the cold axion background describedin Sec. 1.</p><p>The coupling of axions to photons is described by the following part of the Lagrangian:</p><p>La = ga 2a</p><p>faF F , (4)</p><p>whereF =</p><p>12F</p><p>is the dual eld strength tensor.The axion eld is originally in a nonequilibrium state, and in the process of its relaxation to the</p><p>equilibrium conguration, coherent oscillations with q = 0 are produced if the reheating temperature afterination is below the PecceiQuinn transition scale [6]. In late times, the axion eld evolves according toa(t) = a0 cos(mat), where the amplitude a0 is related to the initial misalignment angle. Lagrangian (4)then becomes</p><p>La = ga 21fa</p><p>a0 cos(mat)F F = ga</p><p>faa0 cos(mat)AF .</p><p>Integrating by parts (dropping total derivatives) and taking into account that F = 0, we obtain</p><p>La = ga maa0fa</p><p>sin(mat)ijkAiFjk, (5)</p><p>where the Latin indices run over only the spatial components.A cosmic ray particle (which travels at almost the speed of light) sees regions with quasiconstant values</p><p>of the axion background of a size depending on the axion mass but always many orders of magnitude largerthan its wavelength. Therefore, we can approximate the sine in (5) by a constant (1/2, for example). Itcan then be written as</p><p>La = 14AF , = (, 0, 0, 0), = 4ga</p><p>maa0fa</p><p>. (6)</p><p>The constant changes sign with a period 1/ma.The oscillator has the energy density a = a2max/2 = (maa0)</p><p>2/2, and hence maa0 =</p><p>2a. Theconstant is then</p><p> = ga4</p><p>2afa</p><p> 1020 eV</p><p>for a = 104 eV and fa = 107 GeV = 1016 eV.The extra term in (6) corresponds to the MaxwellChernSimons electrodynamics. Although we can</p><p>then in principle have any four-vector , the axion background provides a purely temporal vector. Weassume that is constant within a time interval 1/ma.</p><p>251</p></li><li><p>3.1. EulerLagrange equations. In the presence of an axion background, the QED Lagrangian is</p><p>L = 14FF + (i e Ame) + 12m</p><p>2AA</p><p> +14A F</p><p> , (7)</p><p>where A = A and similarly for other slashed symbols. Here, we also introduce the eective photon withthe mass m (equivalent to a refractive index [2]). We can write the approximation m2 4ne/me for it.The electron density in the Universe is expected to be at most ne 107 cm3 1021 eV3. This densitycorresponds to m 1015 eV, but we here use the more conservative limit m = 1018 eV (compatiblewith [13]).</p><p>The second term in (7) gives the kinetic and mass term for the fermions and also their interaction withphotons. Dropping it, we obtain the Lagrangian for (free) photons in the axion background (see [14] forfurther details):</p><p>L = 14FF +</p><p>12mAA</p><p> +14A F</p><p> =</p><p>= 12A(A A) + 12m</p><p>2AA</p><p> +14AA .</p><p>The EulerLagrange (EL) equations are</p><p>L</p><p>(A) L</p><p>A= 0,</p><p>where</p><p>L</p><p>(A)= </p><p>(A)</p><p>[</p><p>12</p><p>Agg(A A) + 14</p><p>AA</p><p>]</p><p>=</p><p>= </p><p>{</p><p>12[gg(A A) +</p><p>+ A(gg gg)] + 14 A</p><p>}</p><p>=</p><p>= </p><p>[</p><p>(A A) + 14A</p><p>]</p><p>=</p><p>= A + A + 14 A ,</p><p>LA</p><p>=</p><p>A</p><p>(</p><p>12m2g</p><p>AA +14AA</p><p>)</p><p>=</p><p>=12m2(g</p><p>A + gA) +14A = m2A</p><p> +14A .</p><p>Rearranging the indices, we bring the equations to the form</p><p>A + A m2A 12 A = 0.</p><p>If we choose the Lorenz gauge A = 0, then the second term vanishes. The equations can also be writtenas</p><p>gA gm2A 12 A = 0.</p><p>252</p></li><li><p>We are interested in writing these equations in the momentum space. For this, we dene the Fouriertransform of the eld:</p><p>A(x) =</p><p>d4k</p><p>(2)4eikxA(k).</p><p>The relevant derivatives are written as</p><p>A =</p><p>d4k</p><p>(2)4(ik)eikxA(k), A =</p><p>d4k</p><p>(2)4(k2)eikxA(k).</p><p>The EL equations are then</p><p>d4k</p><p>(2)4</p><p>[</p><p>g(k2 m2) +i</p><p>2k</p><p>]</p><p>eikxA(k) = 0.</p><p>Therefore,[</p><p>g(k2 m2) +i</p><p>2k</p><p>]</p><p>A(k) = 0</p><p>or, equivalently,</p><p>KA(k) = 0, K = g(k2 m2) +i</p><p>2k . (8)</p><p>3.2. Polarization vectors and dispersion relation. We now dene</p><p>S = k</p><p>k.</p><p>This can be more conveniently expressed using the contraction of two Levi-Civita symbols =3![ ] (the minus sign is here because 0123 = 0123 in Minkowski space):</p><p>S = [( k)2 2k2]g ( k)(k + k) + k2 + 2kk .</p><p>It satisesS </p><p> = S k = 0, S = S = 2[( k)2 2k2], SS =</p><p>S</p><p>2S .</p><p>If = (, 0, 0, 0), then we have S = 22k 2 > 0.We now introduce two projectors:</p><p>P =S</p><p>S i</p><p>2Sk .</p><p>These projectors have the properties</p><p>P = P k = 0, gP</p><p> = 1, (P</p><p> )</p><p> = P = P ,</p><p>P P = P , P</p><p> P = 0, P</p><p>+ + P</p><p> = 2</p><p>S</p><p>S.</p><p>With these projectors, we can build a pair of polarization vectors to solve (8). We start from a spacelikeunit vector, for example, = (0, 1, 1, 1)/</p><p>3. We then project it: = P . To obtain a normalized</p><p>vector, we need</p><p>() = P</p><p> P</p><p> = P =</p><p>SS</p><p>=</p><p>=S/2 + 2( k)2</p><p>S= 1</p><p>2+</p><p>( k)22k2</p><p>.</p><p>253</p></li><li><p>(this of course is negative because is spacelike). The polarization vectors are then</p><p> =</p><p>=</p><p>[k2 ( k)22k2</p><p>]1/2P .</p><p>These polarization vectors satisfy</p><p>g </p><p> = 1, g = 0, (9)</p><p> + </p><p> = 2</p><p>S</p><p>S</p><p>= S</p><p>2k2. (10)</p><p>Using the projectors, we can write the tensor in (8) as</p><p>K = g(k2 m2) +</p><p>S</p><p>2(P P+ ).</p><p>For k = (, k), we then have</p><p>K =</p><p>[</p><p>(k2 m2)</p><p>S</p><p>2</p><p>]</p><p> = (k2 m2 |k|) =</p><p>(</p><p>2 k2 m2 |k|)</p><p>.</p><p>Therefore, A = is a solution of (8) i</p><p>(k) =</p><p>m2 |k|+ k2.</p><p>This is the new dispersion relation of photons in the cold axion background in the approximation where is assumed to be piecewise constant.</p><p>4. The process p p4.1. Kinematic constraints. We now consider p(p) p(q)(k) or e(p) e(q)(k). This process</p><p>is forbidden in normal QED by the energy conservation, but it is possible in this background (the cold axionbackground even allows the process e+e [15]). Momentum conservation implies that q = p k. Ifm is interpreted as the mass of the charged particle (proton or electron), then conservation of energy leadsto</p><p>E(q) + (k) = E(p),</p><p>m2 + (p k)2 +</p><p>m2 |k|+ k2 =</p><p>m2 + p 2,</p><p>E2 + k2 2pk cos +</p><p>m2 k + k2 E = 0,(11)</p><p>where we use the expressions</p><p>E = E(p) =</p><p>m2 + p 2, p = |p |, k = |k|, pk cos = p k</p><p>in the last line. As becomes clear in what follows, if is positive or negative, then the process is onlypossible for the respective negative or positive polarization. Therefore, = || in these cases. To takeboth of them into account, we use the minus sign and write instead of ||.</p><p>We square the last equality in (11) twice:</p><p>(4E2 4p2 cos2 + 4p cos 2)k2 2(2E2 + 2m2p cos m2)k + 4E2m2 m4 = 0.</p><p>254</p></li><li><p>Neglecting 2k2, m2k, and m4 in this equation, we obtain</p><p>(E2 p2 cos2 + p cos )k2 (E2 + m2p cos )k + E2m2 = 0.</p><p>This equation has two solutions</p><p>k =E2 + pm2 cos E</p><p>E22 4E2m2 + 4p2m2 cos2 2pm2 cos 2(E2 p2 cos2 + p cos ) , (12)</p><p>which make sense only if the discriminant is positive. In the approximation cos 1 sin2 /2, thecondition 0 becomes</p><p>sin2 p22 2pm2 + m2(2 4m2)</p><p>4p2m2(1 /4p),</p><p>which can be rewritten as</p><p>sin2 2</p><p>4p2m2</p><p>11 /4p(p p+)(p p), (13)</p><p>where</p><p>p =m2 2mm</p><p>1 2</p><p>4m2 2mm</p><p>1 2</p><p>4m2.</p><p>It is clear that p+ > 0 and p < 0. For sin2 to be positive, the condition</p><p>p > p+ = pth =2mm</p><p>1 2</p><p>4m2</p><p>must be satised. This is the threshold below which the process cannot occur kinematically. The energythreshold E2th = m</p><p>2 + p2th is written as</p><p>Eth =2mm</p><p>.</p><p>As 0, the threshold value goes to innity (as expected, the process becomes impossible if vanishes).There is another relevant scale in the problem: m2/. It is many orders of magnitude above the GZK</p><p>cuto. Therefore, we always assume the limit p m2/. The maximum angle of emission for a givenmomentum is given by (13). Its largest value is obtained when p is large (p pth):</p><p>sin2 max =2</p><p>4m2.</p><p>Because this is a small number, photons are emitted in a narrow cone max = /2m . This justies theapproximation for cos .</p><p>At max(p), the square root in (12) vanishes, and we have</p><p>k+[max(p)] = k[max(p)]pthpm2/ 2m</p><p>2</p><p>.</p><p>The minimum value for the angle is = 0:</p><p>k(0) E2 + pm2 (E2 pm2 2m2m2/)</p><p>2(m2 + p).</p><p>255</p></li><li><p>In the framework of the assumption pth p m2/, we hence obtain the maximum and minimum valuesof the photon momentum:</p><p>kmax = k+(0) =E2</p><p>m2, (14)</p><p>kmin = k(0) =m2</p><p>. (15)</p><p>These two values coincide at the energy threshold. Here, we can see that the process is possible for negativeor positive polarization only if respectively > 0 or < 0. Otherwise, the modulus of the photon momentumwould be negative.</p><p>We note that the incoming cosmic ray wavelength ts perfectly within the 1/ma size, and the quantity therefore turns out to be almost perfectly constant. Whether is positive or negative, there is alwaysa state with slightly less energy into which it is possible to decay with a loss of part of its energy (of theorder O()), emitting a soft photon. Th...</p></li></ul>