aa r X i v : . [ phy s i c s . g e n - ph ] J un Noname manuscript No. (will be inserted by the editor)
Macromechanics and two-body problems
Huai-Yu Wang
Received: date / Accepted: date
Abstract
A wave function can be written in the form of ψ = R e i S/ ¯ h . Weput this form of wave function into quantum mechanics equations and takehydrodynamic limit, i. e., let Planck constant be zero. Then equations of mo-tion (EOM) describing the movement of macroscopic bodies are retrieved.From Schr¨odinger equation, we obtain Newtonian mechanics, including New-tons three laws of motion; from decouple Klein-Gordon equation with positivekinetic energy (PKE), we obtain EOM of special relativity in classical me-chanics. These are for PKE systems. From negative kinetic energy (NKE)Schr¨odinger equation and decoupled Klein-Gordon equation, the EOM de-scribing low momentum and relativistic motions of macroscopic dark bodiesare derived. These are NKE systems, i. e., dark systems. In all cases scalar andvector potentials are also taken into account. The formalism obtained is col-lectively called macromechanics. For an isolated system containing PKE andNKE bodies, both total momentum and total kinetic energy are conserved. Adark ideal gas produces a negative pressure, and its microscopic mechanismis disclosed. Two-body problems, where at least one is of NKE, are investi-gated for both macroscopic bodies and microscopic particles. A NKE protonand a PKE electron can compose a stable PKE atom, and its spectral lineshave blue shifts compared to a hydrogen atom. The author suggests to seekfor these spectral lines in celestial spectra. This provides a way to seek fordark particles in space. Elastic collisions between a body and a dark body areresearched. Keywords hydrodynamic limit of quantum mechanics equations · Newton’slaws of motion · dark matter · negative pressure · blue shift of hydrogenspectral lines · two-body problems Department of Physics, Tsinghua University, Beijing 100084, ChinaTel.: +86-13717938266E-mail: [email protected] Huai-Yu Wang
It is well known that the movement of macroscopic bodies follows classical me-chanics, while that of microscopic particles follow quantum mechanics (QM).Both have the equations of relativistic motion and their low momentum limits.There are certainly relations between classical mechanics and QM. In his-tory, classical physical quantities were replaced by operators to develop Schr¨odingerequation and Klein-Gordon equation from classical mechanics. Usually, oneprefers to obtain quantum mechanics equations (QMEs) starting from classi-cal mechanics by means of quantization [1].Every macroscopic body is composed of microscopic particles, and for theformer, the effect of QM disappears. The disappearance of quantum effectcorresponds to letting Planck constant be zero ¯ h →
0. Therefore, it is logicallycorrect that the QMEs are first principle, and the classical mechanics equationsshould be the derivation of quantum ones under some approximations thatremove quantum effect. We are going to present this procedure in this paper.Madelung [2] first expressed the wave function of Schr¨odinger equation inthe form of R e i S/ ¯ h . This was called hydrodynamic model by Bohm [3,4,5,6,7,8,9,10]. Using this model he tried to explain the implications of Schr¨odingerequation. He also noticed that this model lead to an equation approachingHamilton-Jacobi equation in classical mechanics and he made an attemptionto obtain Newtons second law from this model. A term called “quantum po-tential” by him always remained in his formulism.We find that this model did set a right starting point of a route throughwhich we are able to obtain Newtonian mechanics, or Newtons three laws ofmotion, from Schr¨odinger equation. The key point is to make the approxima-tion ¯ h →
0, to abandon the quantum potential. This approximation eliminatesquantum effect, and is called hydrodynamic approximation or hydrodynamiclimit. In the same way, the equations of motion of special relativity can bederived from Klein-Gordon equation. In doing so, scalar and vector potentialsare also taken into account. Therefore, all the classical equations of motioncan indeed be derived from QMEs. These equations describe the movement ofparticles or bodies with positive kinetic energy (PKE).We [11] have pointed out that for relativistic quantum mechanics equations(RQMEs), the negative kinetic energy (NKE) ought to be treated on an equalfooting as PKE. For Dirac equation, both the PKE and NKE solutions shouldbe considered when solving the motion of a particle. For Klein-Gordon equa-tion, the PKE and NKE branches should be treated separately. For piecewiseconstant potentials, decoupled Klein-Gordon equations were easily obtained.By taking the low momentum approximation, PKE decoupled Klein-Gordonequation leads to usual Schr¨odinger equation, and NKE decoupled one leadsNKE Schr¨odinger equation which was propose by the author. The Schr¨odingerequation and NKE Schr¨odinger equation are generically called low momentumQMEs, and they could also be derived from Dirac equation under low mo-mentum approximation. The NKE Schr¨odinger equation and decoupled NKEKlein-Gordon equation are collectively called NKE QMEs. The NKE solutions acromechanics and two-body problems 3 of Dirac equation and NKE QMEs were believed describing movement of darkparticles.The NKE QMEs can also be taken their hydrodynamic limits with the sameprocedure as mentioned above. The resultant equations are believed describingthe motion of macroscopic dark bodies.Thus, hydrodynamic limits can be taken from QMEs to achieve equa-tions describing macroscopic bodies, including PKE and NKE ones. Presently,we name generically these equations as “macroscopic mechanics”, in short“macromechanics”, which contains PKE one, the classical mechanics we arefamiliar with, and NKE ones describing dark bodies.Few macroscopic bodies can constitute a system with interactions betweenthem. The most obvious example is our solar system. A simplest system com-prises two bodies. The treatment of two-body problem has been sophisticatesince it is equivalent to the following problem: a body with the total mass ofthe two bodies moves freely the position of which is called mass center, anda body with reduced mass moves as if it is acted by the interaction betweenthem but originated from the mass center. We point out that this treatmentis for the systems where both component bodies are of PKE.Since in our universe, there are dark macroscopic bodies as well, we have toinvestigate two-body problems concerning dark bodies either. The two bodiescan be both dark ones or one with PKE and the other with NKE. All thepossible cases are going to be investigated.In the transition from QMEs to macromechanics, analytical mechanics isan important means. Let us here first retrospect the fundamental formulismof analytical mechanics.If the action S , kinetic energy T and potential V of a system are known,the relationship connecting these three quantities is Hamilton-Jacobi equation[12]: − ∂S∂t = T + V. (1.1)The right hand side of Eq. (1.1) is also Hamiltonian [12] H , H = T + V. (1.2)When the system is doing stationary movement, its Hamiltonian is a constant.The value of this constant is just the total energy E of the system, H = E . Inthis case, we have [12] − ∂S∂t = E. (1.3)In terms of the action S , classical momentum can be defined [12] by p = ∇ S. (1.4)Both kinetic and potential energies are functions of coordinates r and momenta p . One of Hamilton formulas is v = ˙ r = ∂H∂ p . (1.5) Huai-Yu Wang
Here a dot on the top of a quantity means taking its time derivative. TheLagrangian can be gained from the Hamiltonian by means of Legendre trans-formation [12], L = p · ˙ r − H. (1.6)The Lagranian in turn is put into Euler-Lagrange equation [12]dd t ∂L∂ ˙ r − ∂L∂ r = 0 (1.7)so as to achieve equations of motion of the system. The resultants are theequations of macromechanics.Our tasks are to derive Eq. (1.1) from QMEs under hydrodynamics limitsfor all possible cases.As long as macromechanics is established, we are able to study all thepossible cases of two-body problems.Here we present the standard procedure achieving Eq. (1.1) from QMEs.In most of cases, a QME can be written in the form ofi¯ h ∂ψ∂t = Hψ, (1.8)where H is the Hamiltonian of the system. The particle density probability isdefined as ρ = ψ ∗ ψ. (1.9)The density probability in this form is definitely nonnegative. From Eqs. (1.8)and (1.9), we construct two equations ∂ρ∂t = ψ ∗ ∂ψ∂t + ψ ∂ψ ∗ ∂t = 1i¯ h ( ψ ∗ Hψ − ψH ∗ ψ ∗ ) (1.10)and i¯ h ( ψ ∗ ∂ψ∂t − ψ ∂ψ ∗ ∂t ) = ψ ∗ Hψ + ψH ∗ ψ ∗ . (1.11)It is apparent that they respectively take the imaginary ψ ∗ Hψ − ψH ∗ ψ ∗ andthe real ψ ∗ Hψ + ψH ∗ ψ ∗ . In view of the left hand side of Eq. (1.10), thisequation ought to be continuity equation, ∂ρ∂t + ∇ · j = 0 . (1.12)That is to say the right hand side of Eq. (1.10) can be written as the form of −∇ · j , where the expression of current probability j depends on Hamiltonian H . The way to take hydrodynamic limit is to let the wave function be theform of ψ = R e i S/ ¯ h , (1.13)where both R and S are real numbers. Plese note that these two quantuitieshave specific physical meanings: the square of R is just the density probability ρ = R (1.14) acromechanics and two-body problems 5 as can easily be seen from Eqs. (1.9) and (1.13). After Eq. (1.13) is put into(1.11) and hydrodynamic limit is taken, Eq. (1.1) can be obtained, which isthe Hamilton-Jacobi equation in classical mechanics [13,14]. S is the action ofthe system, the significance of which is clearly revealed by Eqs. (1.1)-(1.4). h ∂ψ∂t = Hψ = ( − ¯ h m ∇ + V ) ψ. (2.1)When the wave function in the form of Eq. (1.13) is substituted into Eq. (2.1),the real and imaginary parts devote the following formulas. ∂R ∂t = − m ∇ · ( R ∇ S ) (2.2)and − ∂S∂t = 12 m ( ∇ S ) + V. (2.3)Equaiton (2.2) is just continuity equation (1.12), because the density proba-bility is Eq. (1.14) and the current probability is j = − i¯ h m ( ψ ∗ ∇ ψ − ψ ∇ ψ ∗ ) = R ∇ Sm . (2.4)Let us inspect Eq. (2.3) where a term − ¯ h m ∇ RR has been dropped. Thisterm was called quantum potential by Bohm, because it contained a factorof Planck constant ¯ h which embodies quantum mechanics effect. When onewants to ignore the quantum effect he lets ¯ h → − ∂S∂t = 12 m p + V. (2.5)The right hand side is just classical kinetic energy plus potential. By com-parison to Eq. (1.1), Eq. (2.5), as well as (2.3), is the Hamilton-Jacobi equation Huai-Yu Wang of the system and kinetic energy is T = p / m . By use of Eq. (1.5), velocityis calculated to be v = p /m. (2.6)From Eq. (1.6), Lagrangian is L = p / m − V. (2.7)It is just kinetic energy minus potential energy. By use of (2.6), the Lagrangianis reexpressed by L = m v / − V. (2.8)This form can be used in Eq. (1.7) and the result isdd t ( m v ) = −∇ V. (2.9)This is Newtons second law. If the mass is unchanged, Eq. (2.9) can be rewrit-ten as m ˙ v = −∇ V. (2.10)This formula is usually stated as “force is the cause of acceleratio”. However,the original equation should be (2.9) which, by means of (2.6), is rewritten as˙ p = −∇ V = f . (2.11)This is the real Newtons second law: the force felt by a body equals to therate of change over time of the bodys momentum. This statement emphasizesthat the right hand side of Eq. (2.11) is the cause and the left hand side is theconsequence.One can also explain Eq. (2.11) reversely: the left hand side can be a causeand the right hand side is a consequence. Thus Eq. (2.11) has another physicalmeaning: the rate of change of a bodys momentum equals its force exertingon exterior. This meaning has been employed such as to derive the pressureof an ideal gas by means of molecular kinetics.Equation (2.11) has even one more meaning. If a body is not acted byany force, then the right hand side is zero. We have ˙ p = 0. In this case, themomentum is a constant. That is the law of momentum conservation: if notacted by a force, a bodys momentum is conserved.When the mass is fixed, from Eqs. (2.10) and (2.11), f · d r = d( m v )d t · d r = 12 d( m v ) = d K. (2.12)The work done by a force on a body converts to its kinetic energy. This is work-energy theorem. When a body is not subject to a force, its kinetic energy isconserved.Let the potential at infinity be zero. It is possible for a body to do steadymotion only when its potential energy is negative. This is because the particleskinetic energy is positive. A PKE plus a negative potential energy is possibleto make the total energy reach an equilibrium point to meet Virial theorem acromechanics and two-body problems 7 [11]. When a body is acted by a repulsive force, it will move to infinity. Hence,only subject to an attractive force can a body do steady motion.Let us now consider a two-particle system. The Hamiltonian is H = − ¯ h m ∇ − ¯ h m ∇ + V ( r − r ) . (2.13)The two-particle wave function is assumed to be the form of ψ ( r , r ) = R ( r , r )e i S ( r , r ) / ¯ h . (2.14)This wave function is substituted into (1.11), and we proceed the same routineas above. Momenta and velocities of the two particles are respectively definedby p i = ∇ S, i = 1 , r i = v i = ∂H p i = p i m i , i = 1 , . (2.16)Lagrangian is achieved by transformation L = ˙ r · p + ˙ r · p − H. (2.17)The obtained equations of motion of the two bodies are˙ p = −∇ V ( r − r ) = f (2.18a)and ˙ p = ∇ V ( r − r ) = − f , (2.18b)where a force f = −∇ V ( r − r ) has been defined.Equation (2.18) manifests Newtons third law: actions equals minus reac-tions, or, the force the first body acts on to the second one is exactly the sameas that the second one acts on the first one, except in the opposite direction.Now we define the total momentum of the two-body system to be the sumof the constituent bodies. p = p + p . (2.19)Then from Eq. (2.18) it is easily obtained that˙ p = 0 . (2.20)This is the law of the momentum conservation of the system: if there is noexternal force, the total momentum of the system is conserved.The kinetic energies of the two bodies are K i = 12 m i v i , i = 1 , . (2.21)If their masses remain unchanged, by Eq. (2.12) we have f · d r i = 12 d( m v i ) = d K i , i = 1 , . (2.22) Huai-Yu Wang
When there is no exterior force acting on the system, we haved K = d K + d K = 0 . (2.23)This is the law of kinetic energy conservation of an isolated system.For a system containing more particles, the same conclusion can be drawn.Assume that there are N particles. The Hamiltonian of the system is H = − N X i =1 ¯ h m i ∇ i + N X i 12 d( m v − ) ) = d K ( − ) . (2.36)That is to say, the work done by a force on a dark body converts to its NKE!Its NKE is conserved if it is not subject to a force.For a dark body, its kinetic energy is negative. Hence, only a positivepotential is possible to make the total energy reach an equilibrium point tomeet Virial theorem, so that the body can do steady motion. Correspondingly,it is possible for a NKE body to do steady motion only when it is subject toa repulsive force. In one word, the case of a NKE body is contrary to that ofa PKE body. acromechanics and two-body problems 11 Now let us consider two-particle systems. If both particles are of NKE,after the Hamiltonian of the system is put down like Eq. (2.13), the procedureis the same as those Eqs. (2.14)-(2.23).We investigate such a case that the first particle is of PKE and the secondone of NKE. The Hamiltonian is H = − ¯ h m ∇ + ¯ h m ∇ + V ( r − r ) . (2.37)The process (2.14)-(2.23) is still followed. The relations between velocities andmomenta are ˙ r = v = p m , ˙ r = v = − p m . (2.38)The equations of motion of the two bodies are easily put down.˙ p = −∇ V ( r − r ) = f (2.39a)and ˙ p = −∇ V ( r − r ) = − f . (2.39b)Every body obeys Newton’s second law. Apparently, Newton’s third law stillapplies. If we define a total momentum of the system, p = p + p . Then˙ p = 0, which is law of momentum conservation.We put down the kinetic energies of the two bodies. K = 12 m v (2.40a)and K − ) = − m v − ) . (2.40b)Their sum is the total kinetic energy of the system. K = K + K − ) (2.41)It follows from Eqs. (2.36), (2.38) and (2.39) that f · d r = 12 d( m v ) = d K (2.42a)and − f · d r = 12 d( m v ) = d K − ) (2.42b)which results in d K = d K + d K − ) = 0 . (2.43)This is the law of kinetic energy conservation: if an isolated system is notaffected by outside, its total kinetic energy is conserved.Now assume that a system contains N particles among which M are ofPKE and L = N − M are of NKE. The Hamiltonian of the system is H = − M X i =1 ¯ h m i ∇ i + N X i = M +1 ¯ h m i ∇ i + N X i 0. That is to say, the information of spin is eliminated. On the otherhand, Dirac equation manifests spin in terms of the form of spinors. The spinorform representing spin cannot be removed by taking hydrodynamic limit. So,there will be a contradiction when one tries to make macroscopic approxi-mation from Dirac equation. It fails to achieve macromechanics from Diracequation. Substantially, for macroscopic bodies, there is no spin as an intrinsicproperty.3.3 PKE case with scalar and vector potentialsNow we add potentials into Klein-Gordon equation. First, we discuss the casewhere there is only a scalar potential. Klein-Gordon equation is[(i¯ h ∂∂t − V ) − ( m m − c ¯ h ∇ )] ψ = 0 . (3.19)We have stressed that to deal with PKE and NKE branches separately, de-coupled Klein-Gordon equation should be used. The obvious form is that(i¯ h ∂∂t − V + H )(i¯ h ∂∂t − V − H ) ψ = 0 , (3.20)where H is just H (+) in Eq. (3.4). The problem is that the right hand sidesof (3.19) and (3.20) are actually not the same. They differ by H V ψ − V H ψ. (3.21)This is not zero unless the potential V is piecewise constant. H is expandedby Eq. (3.5). In each order, there are many terms, just as (3.8). It is easily seenthat every term contains a factor of powers of Planck constant ¯ h . They cancertainly be dropped when we take hydrodynamic limit ¯ h → 0. The conclusionis that Eq. (3.20) can be discarded.When there is a vector potential, we have to consider PKE and NKE casesseparately.The Hamiltonian of a PKE particle is H (+) = p m c + c ( − i¯ h ∇ − q A ) + V. (3.22)When Hamiltonian (3.21) is substituted into (1.8), this equation was firstsuggested by Scalpter [18], so that it is usually called Scalpter equation [19,20]. Here we call it PKE decoupled Klein-Gordon equation, because we thinkthere is another one, NKE decoupled Klein-Gordon equation the Hamiltonianof which is (3.31) below.The wave function is taken the form of Eq. (2.55). When Eqs. (2.55) and(3.21) are substituted into (1.10), continuity equation is obtained, where thecurrent probability j ( ψ (+) ) is to be determined by − i¯ h ∇ · j ( ψ (+) ) = ψ ∗ (+) p m c + c ( − i¯ h ∇ − q A ) ψ (+) − ψ (+) p m c + c ( − i¯ h ∇ − q A ) ψ ∗ (+) . (3.23) The expression of j ( ψ (+) ) contains infinite terms, each being complicated. Wehave given a method how to calculated each term before [11].After Eqs. (2.55) and (3.22) are put into (1.11) and hydrodynamic limit ismade, the achieved Hamiltonian-Jacobi equation is − ∂S (+) ∂t = q m c + c ( ∇ S (+) − q A ) + V. (3.24)Thus, by Eq. (1.4), the Hamiltonian of a macroscopic body is H (+) = q m c + c ( p (+) − q A ) + V. (3.25)According to Eq. (1.5), the relationship between velocity and momentum is v (+) = c ( p (+) − q A ) p m c + c ( p (+) − q A ) . (3.26)Lagrangian is evaluated by Eq. (1.6): L (+) = − m c + c q ( p (+) − q A ) · A p m c + c ( p (+) − q A ) − V. (3.27)Hamiltonian and Lagrangian are expressed by velocity: H (+) = mc q − v /c + V. (3.28)and L (+) = − mc q − v /c + q v (+) · A − V. (3.29)Finally, the equation of motion is derived from Eqs. (1.7),dd t m v (+) q − v /c = q v (+) × B + q ∂ A ∂t − ∇ V. (3.30)Equations (3.27)-(3.29) have been given in textbooks [17]. Since we have as-sumed that vector potential A is independent of time, the left hand side ofEq. (3.30) can again be written as ˙ p (+) .3.4 NKE case with scalar and vector potentialsThe Hamiltonian of a NKE particle is H ( − ) = p m c + c ( − i ¯ h ∇ − q A ) + V. (3.31)Wave function is taken the form of Eq. (2.66). The process is the same assubsection 3.3. The expression of current probability j ( ψ ( − ) ) of a NKE system acromechanics and two-body problems 21 is just − j ( ψ (+) ) the latter being determined by Eq. (3.23). We merely put downthe final results derived from Eq. (1.11).The relationship between velocity and momentum is v ( − ) = c ( p ( − ) − q A ) p m c + c ( p ( − ) − q A ) . (3.32)Hamiltonian and Lagrangian are respectively H ( − ) = q m c + c ( p ( − ) − q A ) + V = − mc q − v − ) /c + V (3.33)and L ( − ) = m c − c ( p ( − ) − q A ) · A p m c + c ( p ( − ) − q A ) = mc q − v − ) /c + qv ( − ) · A − V. (3.34)The equation of motion is − dd t m v ( − ) q − v − ) /c = q v ( − ) × B + q ∂ A ∂t − ∇ V. (3.35)Since we have assumed that vector potential A is independent of time, the lefthand side of Eq. (3.35) can again be written as ˙ p ( − ) . been aware of that there can be macroscopic NKE bodies, the constituent twobodies can be PKE and NKE ones or both NKE ones.We first briefly retrospect the treatment of PKE two-body problem. Thenwe investigate other cases. Two bodies have masses m and m , respectively. The interaction betweenthem is f ( r − r ). We copy Eqs. (2.18) here: m ¨ r = f ( r − r ) , m ¨ r = − f ( r − r ) . (4.1)After a relative radius vector r − r = r is defined, it follows from Eq. (4.1)that µ ¨ r = f ( r ) , (4.2)where µ = m m m + m . (4.3)is called reduced mass [12]. Its feature is that it is always less than both m and m . Equation (4.2) manifests that a body with the reduced mass, hereaftercalled reduced body in short, moves subject to the force between the twoconstituent bodies, and the original position of the force can be any point inspace although usually set as the origin of coordinates. Please note that thereduced body is of PKE, so that it can be in a steady movement only when itis subject to an attractive force, as discussed below Eq. (2.11).If we define m r + m r = M R , (4.4)Eq. (4.1) can also turn to be M ¨ R = 0 . (4.5)This equation means that a mass M is doing free motion in space. It is em-phasized that the M and R in (4.4) have not been explicitly defined. If oneof them is explicitly defined, the other is either. In customary, M is set asthe sum of the masses of the two constituent bodies, and consequently, R isthe coordinate of the mass center of the two bodies. As a matter of fact, M can also be any other mass, and correspondingly, R can be a shift from masscenter.The position of the mass center is in between the two bodies and on theline connecting them. Taking mass center coordinate as the origin, the so-called center-of-mass frame, is most convenient because in this system thetotal momentum is zero, which facilitates evaluation of physical quantities.For the sake of simplicity, let us assume that the reduced body is doingcircular motion with angular velocity ω around the mass center which is setas the origin. Then the distances between the mass center and the bodies arerespectively r and r , r + r = R . Their ratio is r : r = m : m . This ratiocan also be obtained by setting R = 0 in (4.4). The rotational kinetic energies acromechanics and two-body problems 23 of the two bodies are T = m r ω and T = m r ω , respectively. Thesum of them is that of the reduced body, T + T = µr ω , and their ratio is T : T = m : m .We are aware of that although the ratio r : r is given, the explicit valuesof r and r are not uniquely determined. They are to be determined by initialconditions.We turn to the case where the two constituent bodies are of NKE. Theirequations of motion, following Eq. (2.34), are − m ¨ r = f ( r − r ) , − m ¨ r = − f ( r − r ) . (4.6)After a relative radius vector r − r = r is defined, Eq. (4.6) is combined tobe − µ ¨ r = f ( r ) , (4.7)where the reduced mass (4.3) is used. Comparison of Eq. (4.6) to (4.2) disclosesthat the reduced body is still of NKE. We have pointed out that a NKE bodycan do steady motion only when it is subject to a repulsive force. It is obviousthat as long as we set f = − f ′ in Eqs. (4.6) and (4.7), then their formsbecome exactly the same as (4.1) and (4.2). Therefore, the treatment cancopy the procedure of PKE two-body problem. For example, if it is supposedthat the reduced body is doing circular motion, then the ratio of the rotationalkinetic energies of the two NKE bodies is T : T = m : m . One body with mass m is of PKE and the other with mass m is of NKE.Equations (2.39) are copied here: m ¨ r = V ( r − r ) (4.8)and − m ¨ r = − V ( r − r ) . (4.9)When the two masses are equal, m = m , there is only one equation. In thefollowing, we merely investigate the cases of m = m .After a relative radius vector r − r = r is defined, Eqs. (4.8) and (4.9)combine to be m m m − m ¨ r = f ( r ) . (4.10)We define a “reduced mass” of this system as follows: µ = m m | m − m | . (4.11)If we define m r − m r = M R . (4.12)the sum of Eqs. (4.8) and (4.9) gives M ¨ R = 0 . (4.13) Similar to Eq. (4.5), Eq. (4.13) manifests that a mass M is doing free motionin space. Note that the M and R in (4.13) have not been explicitly defined. Asa simplest case, one can set M = m − m .In the following, we investigate two cases with m < m and m > m separately.(1) m < m In this case, Eq. (4.10) is rewritten as µ ¨ r = f ( r ) . (4.14)This is the equation of motion of a PKE body. That is to say, the reducedbody is of PKE. It can do steady motion if subject to an attractive force.Therefore, in such a system, the net force between the two bodies should beattractive. The primary candidate is universal gravity.The feature of the reduced mass (4.11) is that it is always greater than m : µ > m . Its relation with m depends on the latter: as m < m , µ > m ; as m = 2 m , µ = m ; as m > m , m < µ < m . When m ≪ m , µ ≈ m .For the sake of simplicity, we assume that the force between them is uni-versal gravity, and the “reduced body” is doing circular motion around theorigin with angular velocity ω . By Eq. (4.14), the value of the ω is ω = p G ( m − m ) r / . (4.15)where G is the gravitational constant. The kinetic, potential and total energiesof the reduced body are respectively T = 12 µr ω = Gm m r , (4.16) U = − Gm m r (4.17)and E = T + U = − Gm m r . (4.18)Let us find the position of the “mass center“. Apparently, both bodies shoulddo circular motion around the mass center, and the angular velocity of themis the same as that of the reduced body. The mass center taken as the origin,the distances of the bodies to the origin are r and r , respectively. The forcebetween them determines that m r ω = Gm m r = m r ω . (4.19)Thus, the distance ratio is r r = m m . (4.20)This ratio can also be gained by taking R = 0 in Eq. (4.12). acromechanics and two-body problems 25 Now m is of PKE and m is of NKE. The sum of their kinetic energies isthe PKE of the reduced body. T + T = 12 m r ω − m r ω = 12 µr ω . (4.21)Combination of Eqs. (4.20) and (4.21) results in r = m m − m , r = m m − m . (4.22)Subsequently, r − r = r. (4.23)This reveals that the mass center is still on the line connecting the two bodies,but not in between them. It is on the outside of the dark body.The ratio of the absolute values of their kinetic energies is | T | : | T | = m : m . (4.24)The physical picture of the movement of the two bodies can be outlined.They do circular motion around the mass center in the same radial direction.The radius of body m is larger and that of m , as shown by Eq. (4.20). Theabsolute value of kinetic energy of m is greater and that of m is less, as shownby (4.24). The closer the values of m and m , the closer their radii, and thefarther the mass center position. Consequently, the absolute values of theirkinetic energies and the total kinetic energy become greater. As m → m , r → r . The case of m = m is impossible because they stick together butthe total kinetic energy would become infinite.The line speeds of the two bodies are respectively v = r ω and v = r ω .They move along the same angular direction, i. e., v and v have the samedirection. However, their momenta have opposite directions. According to Eqs.(2.6) and (2.31), their momenta are respectively p = m v , (4.25)and p = − m v = − p . (4.26)In the mass-center frame, the total momentum is zero.Please note again that although the ratio r : r is given, the explicit valuesof r and r are not known, but to be determined by initial conditions.According to this result, if a body or a galaxy moves around a center, therecan be nothing in the center, as long as a dark body or a dark galaxy with agreater mass but smaller radius also moves around the center synchronously.The above discussion takes a simplest model: circular motion, where thedistance between the two bodies is fixed. In reality, the orbitals of celestialbodies are generally elliptical. Then the distance between the two bodies willvary periodically. (2) m > m In this case, Eq. (4.10) is rewritten as − µ ¨ r = f ( r ) . (4.27)The reduced mass is still defined by (4.11). Its feature is that it is alwaysgreater than m , m < m . As m < µ , µ > m ; as m = 2 m , µ = m ; as m > m , µ < m .Equation (4.27) describes the motion of a NKE body. It has been discussedthat a NKE body is in a steady movement only when it is subject to a repulsiveforce. Therefore, in this case the net force between the two bodies should berepulsive. Under a net attractive force the system cannot be stable.The universal gravity always exists. There should be repulsive forces be-tween them, and they are numerically greater than universal gravity. A possiblecase is that they have the same kind of electric charges and the electrostaticrepulsive force is greater than universal gravity. The dark body has to be con-stituted by charged particles with NKE. The body m with PKE should alsobe charged. A possible candidate is a charged black hole.For the time being, we are content with describing some qualitative behav-iors of the systems.The mass center position is on the line connecting the two bodied, but notin between, and it is at the outside of the m side. The sum of the PKE of m and NKE of m is the NKE of the reduce mass. If both bodies do circularmotion around the mass center, the radius of the dark body m is greaterthan that of body m . The absolute value of kinetic energy of m is less thanthat of m . The closer the values of m and m , the closer their radii, and thefarther the mass center position. Consequently, the absolute values of theirkinetic energies and the total kinetic energy become greater. As m → m , r → r . The case of m = m is impossible because they stick together butthe total kinetic energy would become infinite.They move along the same angular direction, i. e., v and v have the samedirection. However, their momenta p and p have opposite directions. In themass-center frame, the total momentum is zero.We list in Table 1 four possible cases of stable two-body systems. We usu-ally see Case I: PKE celestial bodies combine together due to universal gravitybetween them.Table 1. Four possible cases of a stable system composed of a visible bodywith mass m and a dark body with mass m . Two reduced masses are definedby µ ± = m m | m ± m | . acromechanics and two-body problems 27 Case InteractionKE KE Reduced mass between the twoof m of m and its KE bodies that makesthe system stableI PKE PKE µ + , PKE AttractionII PKE NKE µ − , PKE Attraction m < m III PKE NKE µ − , NKE Repulsion m > m IV NKE NKE µ + , NKE RepulsionBy the way let us have some words about the topic related to the hypothe-sized star named Nemesis regarded as a companion of the sun. In 1984, Raupand Sepkoski claimed that they had identified a statistical periodicity in ex-tinction rates of species on the surface of the earth over the last 250 millionyears [21]. The average time interval between extinction events was determinedas 26 million years. The extinction period was further investigated later [22,23, ? ]. It was guessed that the cause of the extinction events were probably relatedto extraterrestrial forces [21]. Possible mechanics were proposed to explain thereasons of the extraterrestrial forces [26,27,28,29,30]. One of them was thatthe sun had a companion [27]. The solar companion star was named as Neme-sis. However, it has not been observed. Some investigations [23,24] thoughtthat the assumed orbital of Nemesis was difficult to explain the periodicity ofthe extinction events.The author thinks that there may be such a companion for the sun. Since wehave known in the present work that dark macroscopic bodies can exist in theuniverse, the solar companion is probably a dark one so that it is not observableto us. For the sake of convenience of discussion, presently, this companion isnamed dark Nemesis. It is conjectured that dark Nemesis is also moving aroundthe center of our galaxy. Because the sun and dark Nemesis connect togetherby universal gravity, it should be Case II in Table 1. According to discussionabove, the mass of dark Nemesis denoted by M Nem , is larger than that ofthe sun M Sun and dark Nemesis is closer to the galaxy center. The distancebetween the sun and the galaxy center is about tens of thousands l.y, and thatbetween the sun and Nemesis is about one l.y or so. Thus, by Eq. (4.20) itis estimated that M Nem − M Sun M Sun ∼ − . That is to say, their masses are quiteclose to each other. Since the distance between the sun and Nemesis varieswith a period about 26 million years, we further conjecture that dark Nemesishas its dark companion. They compose a system, called dark Nemesis system.However, in this system, both objects are dark, which should be Case IV inTable 1. That is to say, the net force between them is repulsive. A possible wayis that both of them carry charges of the same kind. In this way, the distanceof the sun and dark Nemesis can vary with time periodically, and the periodcan be tuned to match the value of 26 million years. Of cause this merely areasonable conjecture for the unseen Nemesis. m and the other is of NKE with mass m . The Hamiltonian waswritten by Eq. (2.37). After r − r = r and m r − m r = M R are defined,Eq. (2.37) is transformed to be H = H µ + H M , (4.28)where H µ = − ¯ h ( m − m )2 m m ∇ r + V ( r ) (4.29)and H M = − ¯ h ( m − m )2 m m ∇ R . (4.30)Apparently, the wave function of (4.28) is in the form of ψ ( r , R ) = χ ( r ) ξ ( R ) . (4.31)The factor ξ ( R ) is the eigenfunction of H M , representing the free motion of aparticle with mass | m − m | M . When m > m , this is a PKE particle, and when m < m a NKE one.The factor χ ( r ) is the eigen function of H µ . The reduced mass µ is definedby Eq. (4.11). When m < m , this reduced mass is of PKE. Only when V ( r )is an attractive potential can make the two-body system be a stable one. When m > m , the reduced mass µ is of NKE. Then only when V ( r ) is a repulsivepotential can make the system be a stable one.In short, if the mass m of the dark body is larger, the two-body systemis a PKE one, and an attractive potential between them can make the systembe stable. While if m is smaller, it is a NKE system, and a repulsive potentialbetween them can make the system stable. This conclusion is in agreementwith that of macroscopic two-body systems. For microscopic particles, therecan indeed be a repulsive interaction between the two particles: they carrysame kind of electric charge.A simplest example is a system composed of a proton and an electron. InTable 2, hydrogen and three possible cases are listed.The dark hydrogen atom and dark combo hydrogen atom cannot be ob-served by us because of their NKE. The combo hydrogen atom is of PKE, sothat its spectrum, if there is, can be observed by us. If the wave number ofa spectral line of hydrogen atom is e ν , then that of the combo hydrogen lineshould be m/M − m/M e ν . There is a slight blue shift. The author suggests to lookfor such lines in celestial spectra. Of cause, when one observes such spectraform a celestial body, the possible red shift due to its going away from theearth should be taken into account.Table 2. Hydrogen and three possible cases of a stable system composedof a proton and an electron. The electric charge q can be either positive or acromechanics and two-body problems 29 negative. The masses of proton and electron are respectively m and M . Tworeduced masses are defined by µ ± = m ± m/M .KE and KE and Reducedcharge charge mass and Nameof proton of electron its KEPKE, q PKE, − q µ + , PKE H (Hydrogen atom)PKE, q NKE, − q µ − , PKE Combo HPKE, q NKE, q µ − , NKE Dark combo HNKE, q NKE, q µ + , NKE Dark H We consider elastic collisions between two bodies.When both bodies are of PKE, the case has been fully researched [12].When both are of NKE, the case can be investigate almost the same as thatof two PKE bodies.Here we study the case that one with mass m is of PKE and the otherwith mass m is of NKE. They move along the x axis before and after thecollision.It is assumed that before collision, m is at x < m is at x > m are respectively v and p , and those of m are respectively v and p . After collision, the quantitiesare denoted by the same symbols with the subscript 0 removed.By the discussion in subsection 2.2, in the process of collision both thetotal kinetic energy and total momentum are conserved. Hence, we have12 m p − m p = 12 m p − m p (5.1)and p + p = p + p . (5.2)The reduced mass is defined by Eq. (4.11). We have to distinguish two casesof m < m and m > m .5.1 The NKE body has a larger mass than the PKE bodyIn this case, m < m . It is solved from Eqs. (5.1) and (5.2) that p = µ [ − m ( p + p ) ± ( p m + p m )] (5.3)and then p is solved from Eq. (5.2). In Eq. (5.3) there are two possible solu-tions. One is always such that p = p and p = p . This means that aftercollision each body goes through the other and continues its moving, whichis unreasonable and is excluded. In each case listed in the following, only the reasonable solution is discussed.(1) Head-on collisionBefore collision, m moves rightward and m moves leftward: v > p > v → − v < 0, and p > 0. In this case, the physically reasonablesolution in Eq. (5.3) is that p = − µ [( 1 m + 1 m ) + 2 p m ] < p = 2 µm p + (1 + 2 µm ) p > p . (5.4b)After collision, m moves leftward with a momentum greater than that beforecollision and m continue to move leftward with a greater momentum. Sincethey move in the same direction, it is required that | v | > | v | . It is so becausethe evaluation is | v | − | v | = v + v .An extreme case is that as m ≪ m and µ ≈ m , v ≈ − v − v , v ≈ v . The larger mass keeps its velocity almost unchanged and the smallermass gains velocity, a slingshot effect.If a celestial body changes its moving direction rapidly, it probably collidesa dark body.(2) Following collision rightwardBefore collision, both m and m move rightward: v > p > v > p → − p < 0. In order for the collision to occur, it is requiredthat v > v . In Eqs. (5.2) and (5.3), p → − p is manipulated, and thephysically reasonable solution is that p = µ [ − ( 1 m + 1 m ) p + 2 p m ] (5.5a)and p = 2 µm p − (1 + 2 µm ) p . (5.5b)What are the moving directions of m and m after collision depends on theirinitial velocities v and v .(3) Following collision leftwardBefore collision, both m and m move leftward: v → − v < p →− p < v → − v < p > 0. In order for the collision to occur, itis required that v < v . In Eqs. (5.2) and (5.3), p → − p is manipulated,and the physically reasonable solution is that p = µ [( 1 m + 1 m ) p − µm p (5.6a)and p = − µm p + (1 + 2 µm ) p . (5.6b) acromechanics and two-body problems 31 What are the moving directions of m and m after collision depends on theirinitial velocities v and v .5.2 The PKE body has a larger mass than the NKE bodyIn this case, m > m . It is solved from Eqs. (5.1) and (5.2) that p = µ [ 1 m ( p + p ) ± ( p m + p m )] . (5.7)and then p is solved from Eq. (5.2). In Eq. (5.7) there are two possible solu-tions. One is always such that p = p and p = p , which is excluded. Ineach case listed below, only the reasonable solution is discussed.(1) Head-on collisionBefore collision, m moves rightward and m moves leftward: v > p > v → − v < p > 0. In this case, the physically reasonablesolution in Eq. (5.7) is that p = µ [( 1 m + 1 m ) p + 2 p m ] > p = − µm p − ( 2 µm − p < − p . (5.8b)After collision, both m and m move rightward, and the velocity of is m greater than that before collision. Since they move in the same direction, it isrequired that v < v . It is so because the evaluation is v − v = − v − v < m ≫ m and µ ≈ m , v ≈ v , v ≈ v +2 v .The larger mass keeps its velocity almost unchanged and the smaller mass gainsvelocity, a slingshot effect.(2) Following collision rightwardBefore collision, both m and m move rightward: v > p > v > p → − p < 0. In order for the collision to occur, it is requiredthat v > v . In Eqs. (5.2) and (5.7), p → − p is manipulated, and thephysically reasonable solution is that p = µ [( 1 m + 1 m ) p − µm p (5.9a)and p = − µm p + ( 2 µm − p . (5.9b)What are the moving directions of m and m after collision depends on theirinitial velocities v and v .(3) Following collision leftward Before collision, both m and m move leftward: v → − v < p →− p < v → − v < p > 0. In order for the collision to occur, itis required that v < v . In Eqs. (5.2) and (5.7), p → − p is manipulated,and the physically reasonable solution is that p = − µ [( 1 m + 1 m ) p + 2 µm p (5.10a)and p = 2 µm p − ( 2 µm − p . (5.10b)What are the moving directions of m and m after collision depends on theirinitial velocities v and v . The equations of motion (EOM) of macroscopic bodies are derived from quan-tum mechanics equations.In doing so, hydrodynamic limits are taken. From Schr¨odinger equation, allthe content of Newtonian mechanics, including Newton’s three laws of motion,are derived. From decoupled Klein-Gordon equation, the EOM of relativisticmotion in special relativity are retrieved. These are for PKE systems. FromNKE Schrodinger equation and NKE decoupled Klein-Gordon equation, theEOM for macroscopic NKE bodies in low momentum and relativistic motionsare obtained.NKE bodies are believed dark bodies. A unique feature is that for a darkbody, its velocity and momentum have opposite directions. For dark bodies,Newton’s three laws of motion still apply.The laws of momentum conservation and total kinetic energy conservationare still valid. Explicitly, suppose a system is composed of N interactive bodies,among which a part is of PKE and the remainings are of NKE. If this systemis not acted by forces from outside, the total momentum and total kineticenergy of the system are conserved.In one word, all the known laws in physics remain unchanged.The achieved mechanical EOM are collectively called macroscopic mechan-ics, in short macromechanics.For a dark ideal gas, the pressure is negative, which is proved in viewpointof molecular kinetics. The key point is that for a dark molecule, its velocityvector and momentum vector are antiparallel to each other. The velocity playsa role to determine its position in space, while its momentum plays a roleyielding physical effect. Here the physical effect is pressure.Two-body problems are studied. If both bodies are dark, the system canbe investigated by mimicking that of a system composed of two PKE bodies.If one with mass m is of PKE and the other with mass m is of NKE, thereduced mass is defined by µ = m m | m − m | , and the mass center is on the lineconnecting the two bodies but not in between. If m < m ( m > m ), the acromechanics and two-body problems 33 reduced body is of PKE (NKE), and only a net attractive (repulsive) forcebetween them can make a stable system.The two-body system containing one PKE and one NKE particles is alsoresearched by means of low momentum quantum mechanics. A NKE protonand a PKE electron can constitute a stable system by the attractive electricpotential between them, which is called combo hydrogen atom. Its spectrallines are those of a hydrogen atom multiplied by a factor m/M − m/M , where M and m are the masses of the proton and electron, respectively. That is tosay, they have blue shifts compared to hydrogen spectral lines. The authorsuggests to seek for such lines in celestial spectra. Therefore, we provide away to detect possible dark matter. Following this thinking, it is possible tocalculate the spectra of more PKE systems composed of some PKE and someNKE particles, and then to seek for corresponding lines in celestial spectra.Elastic collisions between two bodies are studied. The results provide usat least one clue to seek for the evidence of celestial dark bodies: if a celestialbody changes its moving direction rapidly, it probably collides a dark body.The author thinks that there are two possible ways to detect dark matter.One is to observe PKE systems composed of PKE and NKE particles. Thecalculated combo hydrogen atom is a first example. The other way is to probethe interaction between PKE bodies (particles) and NKE bodies (particles).The two-body elastic collision give an simplest example. The author believesthat there will be more examples in both ways. Acknowledgements This research is supported by the National Key Research and Devel-opment Program of China [Grant Nos. 2018YFB0704304-3 and 2016YFB0700102]. References 1. Gon¸calves, L. A., Olavo L. S. F.: Foundations of Quantum Mechanics: Derivation of adissipative Schr¨odinger equation from first principles. Ann. Phys. 59 (2017)2. Madelung Von E.: Quantentheorie in hydrodynamischer Form. Z. Physik (3-4) 322(1926)3. Bohm David: A Suggested Interpretation of the Quantum Theory in Terms of ”Hidden”Variables. I Phys. Rev. (2) 166 (1952)4. Bohm David: A Suggested Interpretation of the Quantum Theory in Terms of ”Hidden”Variables. II Phys. Rev. (2) 180 (1952)5. Bohm David: Reply to a Criticism of a causal Re-Interpretation of the Quantum Theory.Phys. Rev. (2) 389 (1952)6. Bohm David: Proof That Probability Density Approaches | ψ | in Causal Interpretationof the Quantum Theory. Phys. Rev. (2) 458 (1953)7. Bohm D., Vigier J. P.: Model of the Causal Interpretation of Quantum Theory in Teliiisof a Fluid with Irregular Fluctuations. Phys. Rev. (1) 208 (1954)8. Bohm D. J., Hiley B. J.: On the Intuitive Understanding of Nonlocality as Implied byQuantum Theory. Foundations of Physics (1) 93 (1975)9. Bohm D. J., Dewdney C., Hiley B. H.: A quantum potential approach to the Wheelerdelayed-choice experiment. Nature 23 (1985)10. Bohm D. J., Hiley B. H., Kaloyerou P. N.: 1987 AN ONTOLOGICAL BASIS FOR THEQUANTUM THEORY. Phys. Rep. (6) 321 (1987)11. Wang Huai-Yu: The negative kinetic theory of dark matter. unpublished (2020)4 Huai-Yu Wang12. Landau L. D., Lifshitz E. M.: Mechanics, Vol. 1 of course of Theoretical Physics 3rd ed.Pergmon Press: New York 2-3, 29, 44-45, 138-149 (1976)13. Husimi K.: Miscellanea in Elementary Quantum Mechanics. 2. Prog. Theor. Phys. 381 (1953)14. Delos J. B.: Semiclassical Calculation of Quantum Mechanical Wavefunctions. Adv.Chem. Phys. 161 (1986)15. Beringer J. et al. (Particle Data Group): Review of Particle Physics. Phys. Rev. D 328 (1952)19. Kowalski K., Rembieliski J.: Salpeter equation and probability current in the relativisticHamiltonian quantum mechanics. Phys. Rev. A 204 (2018)21. Raup David M., Sepkoski J. John Jr.: Periodicity of extinctions in the geologic past.Proc. Nati. Acad. Sci. USA 801 (1984)22. Alvarez Walter, Mullert Richard A.: Evidence from crater ages for periodic impacts onthe Earth. Nature (5961) 718 (1984)23. Melott Adrian L., Bambach Richard K.: Nemesis reconsidered. Mon. Not. R. Astron.Soc. L99 (2010)24. Melott Adrian L., Bambach Richard K.: DO PERIODICITIES IN EXTINCTIONWITHPOSSIBLE ASTRONOMICAL CONNECTIONSSURVIVE A REVISION OF THE GE-OLOGICAL TIMESCALE? The Astrophysical Journal (5961) 685 (1984)26. Rampino Michael R., Stothers Richard B.: Terrestrial mass extinctions, cometary im-pacts and the Sun’s motion perpendicular to the galactic plane. Nature (5961) 709(1984)27. Schwartz Richard D., James Philip B.: Periodic mass extinctions and the Sun’s oscilla-tion about the galactic plane. Nature (5961) 712 (1984)28. Whitmire Daniel P., and Jackson IV Albert A.: Are periodic mass extinctions drivenby a distant solar companion? Nature (5961) 713 (1984)29. Davis Marc Hutt, and Muller Richard A.: Extinction of species by periodic comet show-ers. Nature (5961) 715 (1984)30. Kenyon Scott J., Bromley Benjamin C.: Stellar encounters as the origin of distant SolarSystem objects in highly eccentric orbits. Nature432