Hydrogen Atom in the Cosmic Microwave Background
AAPS/123-QED
Hydrogen Atom in the Cosmic Microwave Background
Jose A. Magpantay ∗ Quezon City, Philippines (Dated: February 9, 2021)The cosmic microwave background covers the entire universe which suggests the absence of anyclosed system, except the universe itself. In this paper, I consider the effect of the cosmic microwavebackground on the hydrogen atom, which must be very small otherwise changes in energy levelswould have been measurable. But how small is small? This I compute by considering a system inan environment or bath. I derive the bath’s (in this case the CMB) effect on the hydrogen atom inthe Feynman-Vernon approach to an open system. The effect is small and quantified in terms of acorrection to the hydrogen atom that breaks time-reversal symmetry, as expected of memory effects.This is significant, there are imperceptible changes in the state of the hydrogen atom, which meansthat the pervasive CMB must have similar small effect on other atoms thus breaking time-reversalsymmetry in all physical systems.
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1. Quantum theory started from two outstanding problems in the early 20th century - the blackbody radiationpattern and the spectral lines of the hydrogen atom - see any introductory atomic physics book, for example [1]. Inthis paper, I consider the effect of the cosmic microwave background on the hydrogen atom, which we know fromcosmology started at 380,000 years after the Big Bang when the Universe has cooled to temperature of the order of10 K and has expanded to the size of the order of 10 light years, reducing the energy of the electromagnetic radiationto the order of 10 − ev [2], which is too little to kick out the electron from the clutches of the proton (ionizationenergy is of the order of ev). At 13.8 B years of age, the universe is bathed by a microwave background radiation ofcosmic origin, the electromagnetic radiation that became transparent to the universe at 380,000 years age, now has amuch lower energy of the order of 10 − ev. The microwave background follows extremely well the blackbody radiationpattern derived by Planck at the turn of the century with temperature equal to 2.73 K [3]. About a decade later,Bohr explained the spectral lines of hydrogen atom by considering an isolated proton and electron system by assumingclassical physics - planetary orbits, but proposing quantized angular momentum states. Still another decade later,the hydrogen atom was treated using the new quantum mechanics formulated by Heisenberg and Schroedinger, spinwas later added and the electron made relativistic by Dirac, showing a richer structure with corresponding spectrallines, see for example [4].But up to this time, the treatment of the hydrogen atom neglects the cosmic microwave background. Althougheach individual photon does not have enough energy to alter substantially the energy of the electron, however, thereare around 10 per cm of CMB photons in the universe today, every electron and proton must then be hit by CMBphotons every now and then because the number of particles in the universe today is of the order of 10 − per cm ,thus giving around 10 CMB photons per hydrogen atom [2]. This suggests a situation of a Brownian particle in avery weak and dilute (typically water has 10 molecules per cm that can hit a pollen) ’fluid’.The question now is how to compute the effect of the CMB on the hydrogen atom? The Brownian particle analogy isnot too useful. Fortunately, the quantum formalism for dealing with any system in bath or environment, called an opensystem, is established. The density matrix of such a system, after taking care of the bath degrees of freedom is givenby the Feynman/Vernon method [5] in non-relativistic quantum case. Thus, I will I only consider the non-relativistichydrogen atom. The fully relativistic quantum theory of electron and photon is the realm of QED.2. The CMB is the remnant of the EM radiation that decoupled from matter at 380 K after the Big Bang. It isdescribed then by a free EM field with Lagrangian (SI units) S em = − µ (cid:90) d x ( ∂ µ A ν − ∂ ν A µ )( ∂ µ A ν − ∂ ν A µ ) , (1)where µ is the magnetic permeability of free space. I use the metric with signature (-+++). The use of the Coulombgauge (cid:126)∂ · (cid:126)A = 0 results in S em = 12 µ (cid:90) d x (cid:2) ( ∂ A i ) − ( ∂ i A j ) (cid:3) , (2) ∗ Electronic address: [email protected] a r X i v : . [ c ond - m a t . s t a t - m ec h ] F e b giving the field equation ( ∂ − (cid:126)∂ · (cid:126)∂ ) (cid:126)A = 0 . (3)The solution to equation (3) is given by (cid:126)A ( (cid:126)x, t ) = (cid:88) (cid:126)k,α (cid:104) q (cid:126)k ( t ) (cid:126)ϕ (cid:126)k,α ( (cid:126)x ) + q (cid:126)k ( t ) (cid:126)ϕ ∗ (cid:126)k,α ( (cid:126)x ) (cid:105) , (4a) (cid:126)ϕ (cid:126)kα ( (cid:126)x ) = ( 12 π ) (cid:126)(cid:15) ( (cid:126)k, α ) exp ( i(cid:126)k · (cid:126)x ) , (4b)where (cid:126)(cid:15) ( (cid:126)k, α ) with α = 1 , (cid:126)(cid:15) ( (cid:126)k, ,(cid:126)(cid:15) ( (cid:126)k, , (cid:126)k ) form a right-handedtriad of vectors. Also, the (cid:126)k summation will go into continuum integration when the CMB nature of the EM wavesis accounted for later. For equation (4) to be a solution to equation (3), the q (cid:126)k ( t ) must satisfy d q (cid:126)k dt + ( ck ) q (cid:126)k = 0 , (5)which means q (cid:126)k is a harmonic oscillator of ’mass’ µ c and ’frequency’ ck. It is assumed in equation (4) that theharmonic oscillator solutions are all real (sine and cosine solutions will be used). Substituting equations(4;a,b) inequation (2) results in S em = (cid:90) dt (cid:88) (cid:126)k µ c (cid:34)(cid:18) dq (cid:126)k dt (cid:19) − ( ck ) q (cid:126)k (cid:35) . (6)In this form, the free EM wave is written as a sum of harmonic oscillators.3. A charge particle of charge q interacts with an electromagnetic wave as given in the interaction Lagrangian term L int = q(cid:126)v · (cid:126)A ( (cid:126)x, t ) , (7)A hydrogen atom in a CMB has both proton and electron interacting with the CMB EM wave. Since the proton hasmuch smaller velocity than the electron in the hydrogen atom (typically of the order of 10 − slower), I will neglectthe CMB interaction with the proton and just consider the effect on the electron. For simplicity, I will just put theproton at the origin. The full Lagrangian being considered is now L = L hy + L em + L int , (8a) L hy = 12 m ( d(cid:126)xdt ) + e π(cid:15) x , (8b) L em = (cid:88) (cid:126)k µ c (cid:20) ( dq (cid:126)k dt ) − ( ck ) q (cid:126)k (cid:21) , (8c) L int = − e ˙ (cid:126)x · (cid:88) (cid:126)k,α (cid:104) q (cid:126)k ( t ) (cid:126)ϕ (cid:126)kα ( (cid:126)x ) + q (cid:126)k ( t ) (cid:126)ϕ ∗ (cid:126)kα ( (cid:126)x ) (cid:105) , (8d)where e is the electron charge. Using equation (4b), the interaction of the electron with the EM waves can be writtenas L int = − e ( 1(2 π ) (cid:88) (cid:126)k ˙ (cid:126)x ( t ) · (cid:126)(cid:15) cos( (cid:126)k · (cid:126)x ) q (cid:126)k ( t ) , (9)where (cid:126)(cid:15) = [ (cid:126)(cid:15) ( (cid:126)k,
1) + (cid:126)(cid:15) ( (cid:126)k, (cid:126)k , the calculations done by F/V and C/L are applicable.4. I give the results of the F/V and C/L calculations right up to the point before summing over the different’frequency’ contributions, which I will do in 5. But before I go into that, an admission of error. In a previous paper[7], I argued the difference between my calculation’s equation (25) and C/L’s equation (3.9) and (3.10). First, I misseda factor of in the second line of equation (25) and second, I missed a simple evaluation that would have shown(coth θ + 1sinh θ ) = coth θ ). Taking these corrections, I find the C/L results and there is no basis for claiming anydifference with their result.The C/L quantities, as given in formulas (3.9) to (3.11), translate to this paper’s quantities via m → µ c , (10a) ω i → ck, (10b) c i → e ( 1(2 π ) , (10c) x → ˙ (cid:126)x ( t ) · (cid:126)(cid:15) cos( (cid:126)k · (cid:126)x ) . (10d)The propagator for the density matrix is given by J ( (cid:126)x, (cid:126)y ; (cid:126)x (cid:48) , (cid:126)y (cid:48) ) = (cid:90) endpoints ( d ˜ (cid:126)x )( d ˜ (cid:126)y ) exp i (cid:126) (cid:8) (cid:90) t dt (cid:48) (cid:104) L hy (˜ (cid:126)x ) − L hy (˜ (cid:126)y ) (cid:105) + (cid:90) t dt (cid:48) (cid:90) t (cid:48) dt (cid:48)(cid:48) (cid:88) (cid:126)k µ e π k (cid:110) (cid:104) ˙˜ (cid:126)x ( t (cid:48) ) · (cid:126)(cid:15) cos( (cid:126)k · ˜ (cid:126)x ( t (cid:48) )) − ˙˜ (cid:126)y ( t (cid:48) ) · (cid:126)(cid:15) cos( (cid:126)k · ˜ (cid:126)y ( t (cid:48) ) (cid:105) Xi coth( β (cid:126) ck ck ( t (cid:48) − t (cid:48)(cid:48) )) (cid:104) ˙˜ (cid:126)x ( t (cid:48)(cid:48) ) · (cid:126)(cid:15) cos( (cid:126)k · ˜ (cid:126)x ( t (cid:48)(cid:48) )) − ˙˜ (cid:126)y ( t (cid:48)(cid:48) ) · (cid:126)(cid:15) cos( (cid:126)k · ˜ (cid:126)y ( t (cid:48)(cid:48) )) (cid:105) + (cid:104) ˙˜ (cid:126)x ( t (cid:48) ) · (cid:126)(cid:15) cos( (cid:126)k · ˜ (cid:126)x ( t (cid:48) )) − ˙˜ (cid:126)y ( t (cid:48) ) · (cid:126)(cid:15) cos( (cid:126)k · ˜ (cid:126)y ( t (cid:48) ) (cid:105) sin( ck ( t (cid:48) − t (cid:48)(cid:48) )) X (cid:104) ˙˜ (cid:126)x ( t (cid:48)(cid:48) ) · (cid:126)(cid:15) cos( (cid:126)k · ˜ (cid:126)x ( t (cid:48)(cid:48) )) + ˙˜ (cid:126)y ( t (cid:48)(cid:48) ) · (cid:126)(cid:15) cos( (cid:126)k · ˜ (cid:126)y ( t (cid:48)(cid:48) )) (cid:111)(cid:9) . (11)This propagator is more involved because when we sum over (cid:126)k , the electron coordinate (cid:126)x is entwined with it insidethe cosine term and the electron velocity is entwined with the sum of the two polarization vectors. But the moreimportant question is what weighs the contribution of the different (cid:126)k ?5. Since the electromagnetic waves are supposed to represent the CMB, the sum over (cid:126)k will naturally go to theintegral (cid:88) (cid:126)k → (cid:90) k dkd Ω(1) (12)which means all ’frequencies’ have equal weight because of the weight (1), i.e., the EM waves represent a ’white’ light(note, the k and d Ω are geometrical factors in 3D). This is not representative of CMB. To ensure that the EM wavesin equation (11) is a CMB, it must be weighted on by the blackbody radiation pattern given by Planck. The Planckdistribution given by ρ T ( λ ) dλ = 8 πhcλ βhcλ − dλ, (13)where β = k B T , T = 2 . K and k B is Boltzmann’s constant. The distribution accounts for the number of waves andthe average energy per wave between wavelengths λ and λ + dλ . Since the temperature is 2.7 K, I am not consideringtime starting from 380 K years after the Big Bang (temperature then is of the order of 10 K but only when theUniverse has cooled down to 2.7 K. Current ideas do not know when this exactly happened but present ideas speculatethat the Universe has cooled down to 10 to 20 K after a billion years so maybe a few more billion years will cool downthe Universe to about 2.7 K.In terms of the ’frequency’ k, which is actually the wave number k = πλ , the Planck distribution is given by ρ T ( λ ) dλ = − ρ T ( k ) dk, (14)giving ρ T ( k ) = hc π k β (cid:126) ck − . (15)To have a normalized distribution, so that the contribution of each ’frequency’ k is properly weighted, the distribution ρ T ( k ) must be divided by the norm ρ T = (cid:90) ∞ dkρ T ( k )= ( 2 πβ (cid:126) c ) (cid:126)k in equation(11) should now be replaced by (cid:88) (cid:126)k ( eqn term ) = (cid:90) d Ω (cid:90) ∞ dkOF β (cid:126) ck − k (cid:8)(cid:104) ˙˜ (cid:126)x ( t (cid:48) ) · (cid:126)(cid:15) cos( (cid:126)k · ˜ (cid:126)x ( t (cid:48) )) − ˙˜ (cid:126)y ( t (cid:48) ) · (cid:126)(cid:15) cos( (cid:126)k · ˜ (cid:126)y ( t (cid:48) ) (cid:105) Xi coth( β (cid:126) ck ck ( t (cid:48) − t (cid:48)(cid:48) )) (cid:104) ˙˜ (cid:126)x ( t (cid:48)(cid:48) ) · (cid:126)(cid:15) cos( (cid:126)k · ˜ (cid:126)x ( t (cid:48)(cid:48) )) − ˙˜ (cid:126)y ( t (cid:48)(cid:48) ) · (cid:126)(cid:15) cos( (cid:126)k · ˜ (cid:126)y ( t (cid:48)(cid:48) )) (cid:105) + (cid:104) ˙˜ (cid:126)x ( t (cid:48) ) · (cid:126)(cid:15) cos( (cid:126)k · ˜ (cid:126)x ( t (cid:48) )) − ˙˜ (cid:126)y ( t (cid:48) ) · (cid:126)(cid:15) cos( (cid:126)k · ˜ (cid:126)y ( t (cid:48) ) (cid:105) sin( ck ( t (cid:48) − t (cid:48)(cid:48) )) X (cid:104) ˙˜ (cid:126)x ( t (cid:48)(cid:48) ) · (cid:126)(cid:15) cos( (cid:126)k · ˜ (cid:126)x ( t (cid:48)(cid:48) )) + ˙˜ (cid:126)y ( t (cid:48)(cid:48) ) · (cid:126)(cid:15) cos( (cid:126)k · ˜ (cid:126)y ( t (cid:48)(cid:48) )) (cid:9) , (17)where the constant overall factor OF is given by OF = e µ hc ( β (cid:126) c ) π ∝ − . (18)The SI values of the physical constants and T = 2 . u = ck. (19)Using this in the coth( β (cid:126) ck ) term, the term can be approximated bycoth( β (cid:126) ck ≈ β (cid:126) ( 1 u ) . (20)The reason for this is that substituting the values for the Boltzmann constant k B and Planck constant (cid:126) , the RHS of(20) is ∝
11 1 u and the neglected terms are of the order of (10 − u ) n , where n = 1 , , ... .Another term in equation (17) is cos( (cid:126)k · ˜ (cid:126)x ( t (cid:48) )) = cos( c x ( t (cid:48) ) u cos Θ x ( t (cid:48) )), where Θ x ( t (cid:48) ) is the angle between theelectron’s position vector at t’ and the CMB’s wave number vector (cid:126)k , with an explicit time dependence as shownlater. Since c ≈ − , then the approximation of the cosine term can be used, i.e.,cos( 1 c x ( t (cid:48) ) u cos Θ x ( t (cid:48) )) ≈ −
12 ( x ( t (cid:48) ) cos Θ x ( t (cid:48) ) uc ) . (21)Another angle that appears in equation (17) is in the term ˙˜ (cid:126)x ( t (cid:48) ) · (cid:126)(cid:15) = | ˙˜ (cid:126)x ( t (cid:48) ) | cos Φ x ( t (cid:48) ), the angle between theelectron’s velocity vector at t’ and the CMB’s polarization vector (cid:126)(cid:15) . This is also time dependent and the explicit timedependence will also be shown later.Using equations (19), (20) and (21) in (17) results in Equation (17) = (10 − ) (cid:90) d Ω (cid:90) ∞ du u exp γu − (cid:8) i β (cid:126) cos u ( t (cid:48) − t (cid:48)(cid:48) ) uX (cid:20) | ˙˜ (cid:126)x ( t (cid:48) ) | cos Φ x ( t (cid:48) ) (cid:18) −
12 ( x ( t (cid:48) ) cos Θ x ( t (cid:48) ) uc ) (cid:19) − | ˙˜ (cid:126)y ( t (cid:48) ) | cos Φ y ( t (cid:48) ) (cid:18) −
12 ( y ( t (cid:48) ) cos Θ y ( t (cid:48) ) uc ) (cid:19)(cid:21) X (cid:20) | ˙˜ (cid:126)x ( t (cid:48)(cid:48) ) | cos Φ x ( t (cid:48)(cid:48) ) (cid:18) −
12 ( x ( t (cid:48)(cid:48) ) cos Θ x ( t (cid:48)(cid:48) ) uc ) (cid:19) − | ˙˜ (cid:126)y ( t (cid:48)(cid:48) ) | cos Φ y ( t (cid:48)(cid:48) ) (cid:18) −
12 ( y ( t (cid:48)(cid:48) ) cos Θ y ( t (cid:48)(cid:48) ) uc ) (cid:19)(cid:21) + sin u ( t (cid:48) − t (cid:48)(cid:48) ) (cid:20) | ˙˜ (cid:126)x ( t (cid:48) ) | cos Φ x ( t (cid:48) ) (cid:18) −
12 ( x ( t (cid:48) ) cos Θ x ( t (cid:48) ) uc ) (cid:19) − | ˙˜ (cid:126)y ( t (cid:48) ) | cos Φ y ( t (cid:48) ) (cid:18) −
12 ( y ( t (cid:48) ) cos Θ y ( t (cid:48) ) uc ) (cid:19)(cid:21) X (cid:20) | ˙˜ (cid:126)x ( t (cid:48)(cid:48) ) | cos Φ x ( t (cid:48)(cid:48) ) (cid:18) −
12 ( x ( t (cid:48)(cid:48) ) cos Θ x ( t (cid:48)(cid:48) ) uc ) (cid:19) + | ˙˜ (cid:126)y ( t (cid:48)(cid:48) ) | cos Φ y ( t (cid:48)(cid:48) ) (cid:18) −
12 ( y ( t (cid:48)(cid:48) ) cos Θ y ( t (cid:48)(cid:48) ) uc ) (cid:19)(cid:21) (cid:9) , (22)where γ = (cid:126) k B T and T = 2 . K . Note the considerable reduction of the OF by c ∝ − .From the Table of Integrals [8] (cid:90) ∞ sin au exp γu − du = π γ coth( πaγ ) − a , [ a > , Reγ >
0] (23a) (cid:90) ∞ u cos bu exp γu − du = 12 b − π γ (sinh( bπγ )) − , [ Reγ > , (23b)where a = b = ( t (cid:48) − t (cid:48)(cid:48) ). The u integrals in equation (22) are arrived at by differentiating equations (23) w.r.t. a orb a number of times to arrive at the power of u in each integral. The result of each integration is also simplified bytaking note that γ = (cid:126) k B T ≈ − . With these in mind, I list down the three least small among all the small terms.They are, in decreasing strength, and following equation (11), to include the time integrals: F irst ∝ (10 − ) (cid:90) t dt (cid:48) (cid:90) t (cid:48) dt (cid:48)(cid:48) (cid:90) d Ω (cid:26) t (cid:48) − t (cid:48)(cid:48) ) − π γ exp ( − π ( t (cid:48) − t (cid:48)(cid:48) ) γ ) (cid:27) X (cid:104) | ˙˜ (cid:126)x ( t (cid:48) ) | cos Φ x ( t (cid:48) ) − | ˙˜ (cid:126)y ( t (cid:48) ) | cos Φ y ( t (cid:48) ) (cid:105) (cid:104) | ˙˜ (cid:126)x ( t (cid:48)(cid:48) ) | cos Φ x ( t (cid:48)(cid:48) ) − | ˙˜ (cid:126)y ( t (cid:48)(cid:48) ) | cos Φ y ( t (cid:48)(cid:48) ) (cid:105) ,Second ∝ (10 − ) (cid:90) t dt (cid:48) (cid:90) t (cid:48) dt (cid:48)(cid:48) (cid:90) d Ω (cid:26) − t (cid:48) − t (cid:48)(cid:48) ) + 4( πγ ) exp ( − π ( t (cid:48) − t (cid:48)(cid:48) ) γ ) (cid:27) X (cid:104) | ˙˜ (cid:126)x ( t (cid:48) ) | cos Φ x ( t (cid:48) ) − | ˙˜ (cid:126)y ( t (cid:48) ) | cos Φ y ( t (cid:48) ) (cid:105) (cid:104) | ˙˜ (cid:126)x ( t (cid:48)(cid:48) ) | cos Φ x ( t (cid:48)(cid:48) ) + | ˙˜ (cid:126)y ( t (cid:48)(cid:48) ) | cos Φ y ( t (cid:48)(cid:48) ) (cid:105) ,T hird ∝ (10 − ) (cid:90) t dt (cid:48) (cid:90) t (cid:48) dt (cid:48)(cid:48) (cid:90) d Ω (cid:26) t (cid:48) − t (cid:48)(cid:48) ) − π γ exp ( − π ( t (cid:48) − t (cid:48)(cid:48) ) γ ) (cid:27)(cid:110) (cid:104) | ˙˜ (cid:126)x ( t (cid:48) ) | cos Φ x ( t (cid:48) ) − | ˙˜ (cid:126)y ( t (cid:48) ) | cos Φ y ( t (cid:48) ) (cid:105) (cid:104) | ˙˜ (cid:126)x ( t (cid:48)(cid:48) ) | cos Φ x ( t (cid:48)(cid:48) ) x ( t (cid:48)(cid:48) ) cos Θ x ( t (cid:48)(cid:48) ) − | ˙˜ (cid:126)y ( t (cid:48)(cid:48) ) | cos Φ y ( t (cid:48)(cid:48) ) y ( t (cid:48)(cid:48) ) cos Θ y ( t (cid:48)(cid:48) ) (cid:105) + (cid:104) | ˙˜ (cid:126)x ( t (cid:48) ) | cos Φ x ( t (cid:48) ) x ( t (cid:48) ) cos Θ x ( t (cid:48) ) − | ˙˜ (cid:126)y ( t (cid:48) ) | cos Φ y ( t (cid:48) ) y ( t (cid:48) ) cos Θ y ( t (cid:48) ) (cid:105) (cid:104) | ˙˜ (cid:126)x ( t (cid:48)(cid:48) ) | cos Φ x ( t (cid:48)(cid:48) ) − | ˙˜ (cid:126)y ( t (cid:48)(cid:48) ) | cos Φ y ( t (cid:48)(cid:48) ) (cid:105) (cid:111) . (24a)In the First, the OF is multiplied by β (cid:126) ∝ giving a factor ∝ − . In the Second, there is no modifier to OF.In the Third, the OF is multiplied by β (cid:126) ∝ then by c , which comes from equation (21). resulting in the factor10 − .7. Equation (24) gives the correction to the hydrogen atom action terms in the propagator for the density matrixas shown in equation (11). Evaluating the path integrals in (11) with the time delayed and non-quadratic termsin (24) is clearly not possible. Still, the angular integrations can be done and should be finite (see example in thenext paragraphs) even for an arbitrary ˜ (cid:126)x and ˜ (cid:126)y and time derivatives. Equation (24) then gives memory terms withextremely small factors showing that the time-reversal symmetry violating terms are extremely small.Separating out the x from y contributions to have a clearly L eff ( (cid:126)x, ˙ (cid:126)x ) does not seem to be possible also. However,the structure of the terms of equation (24) shows the separate x, y and cross terms have about the same contribution.This will be shown in this section by considering the effect of the CMB on the Bohr model. This will also makeconcrete how small the effect of the CMB is on the energy levels of the atom.The next problem is to integrate the angles of the CMB EM waves to complete the sum over CMB waves contri-bution. This is done by setting up the hydrogen atom geometry, as treated by Bohr, and superimposing the CMBvector (cid:126)k . It will be clear later in the calculations that I will only provide an estimate of the CMB correction to theBohr energy levels, which is the same as the Schroedinger treatment without spin and relativistic corrections. FIG. 1: The electron’s circular orbit on (x,y) plane superimposed with the CMB propagation vector (cid:126)k , which is along z (cid:48)(cid:48) axisand the polarization vectors along the x (cid:48)(cid:48) and y (cid:48)(cid:48) axes. The electron is doing a circular motion with the Coulomb attraction from the proton providing the centripetalacceleration. The Bohr condition of the quantization of the angular momentum is then added. The relevant equationsare 14 π(cid:15) e r = m v r , (25a) L = mvr = n (cid:126) , (25b)where n is an integer. These equations give the quantized values r = 4 π(cid:15) (cid:126) me n , (26a) v = e π(cid:15) (cid:126) n , (26b) ω = me (4 π(cid:15) ) (cid:126) n , (26c) E = − me )2(4 π(cid:15) ) (cid:126) n = − . n ev, (26d)where ω is the angular velocity and E is the energy of the hydrogen atom.The position and velocity of the electron, assuming it starts at φ = 0 are given by (cid:126)x ( t (cid:48) ) = r cos ωt (cid:48) i + r sin ωt (cid:48) j , (27a)˙ (cid:126)x ( t (cid:48) ) = − rω sin ωt (cid:48) i + rω cos ωt (cid:48) j . (27b)If the y terms are also considered in equation (24), the end point conditions in equation (11) shows different ˜ (cid:126)y endpoints ( ( (cid:126)y (cid:48) , (cid:126)y ) ). In the context of the Bohr atom, it means the y trajectory can have a different quantum number n,say n’.To specify the CMB arbitrary EM wave vector (cid:126)k , start with the unit vectors ( i , j , k ). Then rotate about k by angle ϕ to get the unit vectors ( i’ , j’ , k’ = k ), then follow up with a counter clockwise rotation by angle θ about the i’ axisto get the unit vectors ( i” = i’ , j” , k” ). Now I can identify these RHT of vectors with the RHT of vectors of theCMB EM wave given by ( (cid:126)(cid:15) ( (cid:126)k, ,(cid:126)(cid:15) ( (cid:126)k, , (cid:126)k ). For completeness, the vector (cid:126)k makes an angle θ with the k unit vectorand angle ϕ + π with the i unit vector (if I do the second rotation clockwise, the vector (cid:126)k makes an angle ϕ + π with the i unit vector . This operations will identify the terms in equation (24) as˜ x ( t (cid:48) ) = r, (28a) | ˙˜ (cid:126)x ( t (cid:48) ) | = rω, (28b)cos Φ x ( t (cid:48) ) = cos ωt (cid:48) (sin ϕ + cos ϕ cos θ ) − sin ωt (cid:48) (cos ϕ − sin ϕ cos θ ) , (28c)cos Θ x ( t (cid:48) ) = cos ωt (cid:48) sin ϕ sin θ − sin ωt (cid:48) cos ϕ sin ϕ. (28d)Now, the solid angle averaging can be done and all are finite integrals. The interesting part is the term | ˙˜ (cid:126)x ( t (cid:48) ) || ˙˜ (cid:126)x ( t (cid:48)(cid:48) ) | = v , = − m E n , ≈ +31 ( 13 . n ) ev, (29)This will strengthen the really small terms given by equation (24) by 31 orders of magnitude but still resulting invery weak contributions. The change in action (cid:82) t dt (cid:48) L hy ( t (cid:48) ) from the different contributions are now F irst ∝ − (cid:34)(cid:90) t dt (cid:48) (cid:32)(cid:90) t (cid:48) dτ (cid:26) τ − π γ exp ( − πτγ ) (cid:27) cos ωτ (cid:33)(cid:35) E n , (30a) Second ∝ − (cid:34)(cid:90) t dt (cid:48) (cid:32)(cid:90) t (cid:48) dτ (cid:26) − τ + 4( πγ ) exp ( − πτγ ) (cid:27) cos ωτ (cid:33)(cid:35) E n , (30b) T hird ∝ − (cid:34)(cid:90) t dt (cid:48) (cid:32)(cid:90) t (cid:48) dτ (cid:26) τ − π γ exp ( − πτγ ) (cid:27) cos ωτ (cid:33)(cid:35) E n , (30c)where a change of variables, from t (cid:48)(cid:48) to τ = t (cid:48) − t (cid:48)(cid:48) was made in the integration. The much smaller Third term comesfrom the fact that this term goes with r , which by equation (26a), is ∝ − .All the contributions are small, modulo the time factor given by the two time integrals, which shows memoryeffects. These time integrals are rapidly oscillating integrals because ω ∝ +14 by equation (26c) and modified by thedamping polynomial term in the denominator and also by an exponential decaying term. The exponential dampingterms are straightforward to evaluate. The problematic terms are the polynomial terms in the denominator. Still itis expected that the τ integrations would lead to a finite integral even though the closed form in terms of t (cid:48) is notknown. Then the t (cid:48) integration expression should give an overall effect of the CMB on the hydrogen atom, whichmust be small because of the very small factors that go with each term in equation (30).8. This paper showed that the effect of the CMB on the hydrogen atom is almost negligible, the energy levels arehardly changed, the memory effects are imperceptible but break time-reversal symmetry. Thus, this paper showedthat since everything is made up of atoms, as a matter of principle, time-reversal symmetry is no longer a symmetryof atomic physics because of the CMB effect on these atoms.If the pervasive CMB breaks time-reversal symmetry, surely the other long-range force in the universe, gravity, canalso break time-reversal symmetry. The gravitational wave generated during the Big Bang, although much weakerthan the CMB effect now, must also produce time-reversal symmetry violation. In this sense, time-reversal symmetryis not really present in principle but only ’observed’ because we are not sensitive to the violations. Acknowledgments