CESKE VYSOKE UCENI TECHNICKE V PRAZE
FAKULTA JADERNA A FYZIK ALN E INZENYRSKA
Vyzkumny ukol
Studium merenı doby letu atomuantivodıku v experimentu AEGIS
Praha 2010 Michal Spacek
Abstrakt
Nazev prace:Studium merenı doby letu atomu antivodıku v experimentu AEGIS
Autor: Michal Spacek
Obor: Jaderne inzenyrstvı
Druh prace: Vyzkumny ukol
Vedoucı prace: doc. RNDr. Vojtech Petracek, CSc., Katedra fyziky, Fakulta jaderna a fyzikalne inze-nyrska,Ceske vysoke ucenı technicke v Praze
Abstrakt: Tato prace se zabyva pohybem atomu antivodıku v obecnych elektrickych a magnetickychpolıch a v gravitacnım poli Zeme. V uvodu autor objasnuje, ktere casti experimentu AEGIS se pracedotyka a jaky vyznam pro experiment ma rozresenı t´eto problematiky. Podrobne jsou odvozeny kine-matika nabojove vymeny a kinematika deexcitace urychleneho antivodıku - vysledky jsou overeny neza-vislymi postupy a demonstrovany na jednoduchych prıkladech. V ramci kvaziklasickeho modelu, kteryje postupne budovan a analyzovan, je nalezena pohybovarovnice antivodıku ve vnejsım poli a jsourozebrana jejı omezenı. Prubezne jsou zduraznovany dulezite vlastnosti antivodıku.
Klıcova slova: AEGIS, antivodık, positronium, nabojova vymena, deexcitace, vlastnı elektricky dipol,indukovany elektricky dipol, vlastnı magneticky dipol, indukovany magneticky dipol, pohybova rovnicedipolu, elektricke pole, magneticke pole, gravitacnipole, starkovske urychlenı rydbergovskych atomu.
Abstract
Title:Study of the antihydrogen atoms’ time of flight in the AEGIS experiment
Author: Michal Spacek
Branch of study: Nuclear engineering
Type of work: Research project
Supervisor: doc. RNDr. Vojtech Petracek, CSc., Department of Physics, Faculty of Nuclear Sciencesand Physical Engineering, Czech Technical University in Prague
Abstract: This work deals with the motion of antihydrogen atoms in general electric and magneticfields and in the Earth’s gravitational field. The Author firstclarifies which part of the AEGIS experimentis concerned and what is the importance of solving the problematics. Charge exchange kinematics andantihydrogen deexcitation kinematics are derived in detail - the outcomes are verified by independentderivation methods and illustrated with simple examples. Within a quasiclassical model which is con-structed and analysed progressively, the antihydrogen’s equation of motion in an external field is foundand its limits are analysed. The important qualities of the antihydrogen are stressed continuously.
Key words: AEGIS, antihydrogen, positronium, charge exchange, deexcitation, intrinsic electric di-pole, induced electric dipole, intrinsic magnetic dipole,induced magnetic dipole, dipole equation ofmotion, electric field, magnetic field, gravitational field,Stark acceleration of Rydberg atoms.
4
Obsah
1 Introduction 5
2 Charge exchange kinematics 7
3 Antihydrogen deexcitation kinematics 15
4 Antihydrogen dynamics 21
5 The bibliography 30
5
1 Introduction
This text’s aim is to provide an analysis how to deal with the antihydrogen kinematics and dynamicsin the AEGIS experiment at CERN.
The first scientific goal of AEGIS (Antimatter Experiment: Gravity, Interferometry, Spectroscopy) isto study the gravitational interaction of antimatter, for the first time in history. For this purpose, a devicehas been designed for antihydrogen production and experimental use [1].
Although antimatter has been known since 1930’s, the scientists haven’t got the opportunity to test itsgravity properties experimentally yet. There are many difficulties: The antimatter itself, whether naturalor artificially prouduced, is extremelly rare; much of antimatter are charged particles (positrons, antipro-tons, ...) therefore a presence of any weak electric or magnetic field makes gravity studies unfeasible;and much of antimatter propagates almost at the speed of light (antineutrinos).
Gravity is the weakest of all the fundamental interactions.If its repulsion magnitude is comparedto Coulombic repulsion of two protons at Bohr radius, the ratio gives 10−36. Gravity prevails if thedistances are enormous or if the two interacting bodies are huge. However, at AEGIS, any long distancemeasurements can not be afforded and even though one of the interacting body (the Earth) is hugeenough, the second (an antimatter particle) is always microscopic.
Fortunately, there is no need the gravity prevails in this sense - the interaction’s another characteristicis that it can not be ”cancelled”1. If an area is prepared with both electric and magnetic fieldsscreened out(”cancelled”), the gravity would remain weak but also the only effective interaction - this is the generalway of gravity measurements on antimatter at AEGIS.
There are three major theories of gravity interaction of antimatter [2]. The first theory predicts forantimatter exactly the same behaviour as it is known for a classical matter; the second theory proposesanalogy to Coulombic force in which antimatter particle would fall up in the Earth’s gravitational field;and the third theory suggests existence of two more exotic potentials so antimatter would fall down but abit faster thana classical matter.
No matter which of these theories (or another one) is valid, due to small dimensions of the AEGISexperiment device it is clearly reasonable to treat the gravitational acceleration (positive or negative) asconstant in the device’s volume (that is the gravitational field would be uniform).
Because of the gravity’s relative weakness, for studying gravity behaviour on charged antimatterparticles one would need both electric and magnetic fields tobe screened out with a precision10−36 orbetter. This is completely impossible so the only promisingway now is to work with electrically neutralantimatter particles. Another problem resulting from the gravity’s weakness is that for particles withlarge initial velocities almost nothing could be judged by their trajectories.
Antimatter is not only so rare on Earth, it also can not survive a touch with a classical matter - thisalways leads to a complete annihilation of all concerned particles. On account of that, one has to keepantimatter in a space of very hight vacuum and slow the antiparticles down so that they stay in theirpreparation volume until the experimental device is ready for measurements.
For AEGIS, antihydrogen was found to be the only possible particle for gravity studies and measure-ments. It is (when not ionised) electrically neutral2 and it can be handled at velocities small enough tomake gravity measurements possible.
1According to one of the theories of antimatter’s gravitational interaction, the gravity could be ”cancelled”in the same way,as the electric interaction of the Earth in space is - the bothcharge types are of the same number and space distribution. However,antimatter amount is still easily overwhelmed by the mass ofthe Earth. So the statement about gravity’s uncancellationremainsvalid.
2Even though a particle is electrically neutral as a whole, itcan still exhibit electrical qualities, especially if it has a non-uniform charge distribution in space. Then electric and magnetic fields play roles and may affect the gravity measurements. Onthe other hand, the fields’ outscreening does not have to be asmuch as10−36. This will be discussed later.
6
An antihydrogen atom is a compound state of one antiproton (negative charge) and one positron(positive charge) in a general quantum stateΨn,l,m. Its mass is equal to the hydrogen atom in the samestate. Antihydrogen is likely not to exist in nature close toEarth at all so one has to produce it artificallythrough so called charge exchange reaction.
In charge exchange, a positronium (a compound state of one positron and one electron) hits an an-tiproton which displaces the electron in the positronium. One antihydrogen and one electron are obtainedas products. Note that the quantum state of the gained antihydrogen depends on that of the positronium.
In AEGIS, it is impossible to keep antihydrogen in the vacuumvolume at small velocities for a longtime and the experimental device can not work continuously.However, one is able to handle its element- antiproton - very well with a system of electrods (Malmberg-Penning trap) inside a strong magneticfield; the antiprotons follow closed curves at given small average velocities.
The philosophy of the experiment is following: Antiprotonsare being trapped untill their numberis large enough. Then (still inside the trap) they undergo the charge exchange reactions which resultin antihydrogen production. Since all the antihydrogen rise almost simultaneously, the trap electrodeschange their mode in this moment and become an accelerator (so called Stark accelerator of Rydbergatoms) suddenly; the magnetic field is not switched off - it istime-independent. The electric field ofthe accelerator makes a gradient which forces the antiatomsto move in the field even though they areelectrically neutral3. The accelerator gives each suitable antihydrogen an impulse and the antiatoms beginto fly just about horizontally. The electrodes are turned offafter the antiatoms leave their space and nosignificant electric field is in the presence for the time. Nowit is necessary that the antiatoms reach a spacewhere the gravitational field would be the only effective one(see above). Accordingly, the antiatoms haveto fly through a zone of a magnetic field gradient from a strong field area, where they rose, to almost zerofield volume. There (as other fields are succesfully screenedout) the antiatoms follow a parabola typicalfor a uniform gravitational field and some of them hit a position sensitive detector. Unfortunately, thiscoordinate itslef can not provide any information about theantimatter’s gravitational acceleration. Tofit a parabola, one need together with the final point coordinates another information - starting pointcoordinates and starting velocity vector4.
It is impossible to measure these starting quantities directly. Instead, one knows the coordinates ofthe antihydrogen production zone, the time of starting and switching the accelerator off, the time ofantihydrogen hitting the position sensitive detector and the electric and magnetic field properties. Thecrucial then is to know how much of the total time of flight is spent in the final (gravity-only) area andwhether the antihydrogen keeps to fly horizontally or is elevated or dropped a bit by the electric andmagnetic fields.
This text tries to find the dynamics of antihydrogen moving incombined electric, magnetic andgravitational field to answer the questions in the previous paragraph. Much of the brief analysis in thisintroduction is discussed in detail further as well as all the important characteristics of antihydrogen.
3In an electric field with a gradient, the antihydrogen gains induced electric dipole moment which is sensitive to that gradientretroactively and forces the antihydrogen to accelerate.
4These starting quantities may be given at any point of the antihydrogen’s trajectory which is in the area with electric andmagnetic field screened out.
7
2 Charge exchange kinematics
In charge exchange reaction, an antiproton replaces an electron in a positronium and as a result aneutral antihydrogen is obtained. It is the only mechanism of antihydrogen production in AEGIS. In thischapter, the quantities linked to an antiproton and a positronium are denoted as initial, whereas those con-cerning antihydrogen or electron as final. Only the velocityvectors infinitesimally before (respectivellyafter) the reaction are considered. Symbolically:
Ps + p → H + e− (2.1)
In this chapter, the focus is on the charge exchange kinematics. The main goal is to derive an identityputting all the proper quantities together, namely: initial velocities of an antiproton and of a positronium(vp andvPs), the antihydrogen’s final velocity (vH), angles between these velocity vectors (θPs a θH)and the difference in the binding energy of the initial positronium and the final antihydrogen (Q). Thederivation will be performed in a more general relativisticway.
Obrazek 2.1: The setting of the charge exchange reaction. In the picture, theθe angle is missing,however, it is well-defined in the text and its placing is intuitive. The excitations of the positronium andthe antihydrogen are reminded by asterisk signs but not usedin the text anymore. Taken from [1].
It is necessary to define the setting closer. The fact that only the instantenous initial velocities matter1
supports an inertial frame of reference implementation. The frame of reference is attached to the labora-tory, does not rotate and a classical Cartesian coordinate system is used. The initial velocity vectors of anantiproton and a positronium unambiguously form a plane in which the whole reaction takes place. Thex-axis is chosen in that way, the vector of an antiproton points in the axis’ positive direction; they-axisis perpendicular tox-axis and lays fully in the plane given by the initial vectors; and thez-axis is per-pendicual to both previous axes and plays no role in the kinematics; the origin is put in the point wherethe charge exchange takes place. These assumption make it clear that for every single charge exchange
1The kinematics does not depend on the choice of the coordinate system’s origin and also not on a position itself. If theantiproton undergoes nonzero acceleration in the moment ofthe charge exchange reaction, it is caused by the present externalforce field (generally the combined electric, magnetic and gravitational field). Since the incipient antihydrogen differs in properdynamical qualities from both positronium and antiproton,the antihydrogen gains acceleration from the field ”suddenly”(socalled jump change). Therefore only the instantaneous velocities of initial and final particles matter. The antihydrogens’ position(known at least approximately from the experimental geometry) and instantaneous velocity (given by the kinematic identity)serve as initial conditions for accelerated motion in the external fields.
8
reaction event, a different coordinate system is separately materialized. However, since the kinematicidentity will be formulated only in the terms of scalars, there is no loss of generality.
Every velocity vector is given by three independent real numbers. In the described coordinate system,the z coordinate is zero for every vector of the charge exchange reaction so two coordinates are non-trivial. These could be the two remaining Cartesian coordinates as well as two polar coordinates - themagnitude of the vector (never negative) and its polar angle. Let this polar angle be measured contra-clockwise from the positivex half-axis.
Following from how the coordinate system was defined, the angle is always zero for the antiproton.Therefore only its velocity magnitude is non-trivial and isdenoted asvp. The three remaining velocityvectors need two coordinates to be fully described - a pair ofa velocity magnitude and a polar angle.For the positronium these arevPs andθPs, for the antihydrogenvH andθH and for the electronve andθe. The meaning of the three angle quantities are according to the coordinate system definition theirorientation in reference to the velocity vector of the initial antiproton.
The last which is necessary to define is theQ quantity. It is the difference between the bindingenergies of the positronium and the antihydrogen (both can be in a general quantum state). There are twopossible ways of definition:
Q = ((mp + mPs) − (mH + me)) c2 (2.2a)
Q = RH
(
1
n2H
−1
2n2Ps
)
(2.2b)
TheR quantity is the Rydberg constant of antihydrogen [3]:
R =q4e
8h2ǫ20
·mpme
mp + me(2.3)
If the mass of the antiproton in the second fraction is replaced by the mass of electron, the Rydbergconstant of positronium is obtained.
The energy of a relativistic body (an indiceα = p, P s, H, e denotes the particular particle) and itsTaylor power series expansion to the fourth order inv give:
Eα =mαc2
√
1 − v2α
c2
≈ mαc2 +1
2mαv2
α +3
8mα
v4α
c4(2.4)
Similarly for the particle’s linear momentum:
pα =mαvα√
1 − v2α
c2
≈ mαvα +1
2mα
v3α
c2(2.5)
Besides kinematic derivation further, this formula is important for one more reason: It shows thatlinear momentum of a body and its velocity are always parallel. Therefore all the reasoning about aframe of reference and a coordinate system in the beginning of this chapter applies to linear momentumsas well as to velocities2.
With no doubt, the energy conservation holds in charge exchange:
Ep + EPs = EH + Ee (2.6)
2The velocity model for the definition of the frame of reference and of the coordinate system was introduced because itillustrates the motivation of the kinematics. However, thelinear momentum is much more suitable as the starting point ofidentity’s derivation since it is very often conserved in a reaction in contrast to velocity.
9
Because of the outlined relativistic approximations and the way theQ quantity has been defined, theprevious equation could be written as:
1
2mev
2e +
3
8me
v4e
c4=
1
2mpv
2p +
3
8mp
v4p
c4+
1
2mPsv
2Ps +
3
8mPs
v4Ps
c4−
1
2mHv2
H−
3
8mH
v4H
c4+ Q (2.7)
If both terms containing the electron velocity are exclusively moved on the left-hand side of theequation and the whole right-side is denoted asP1, one gets a biquadratic equation for the electronvelocity ve:
v4e +
(
4
3c2
)
v2e −
(
8c2P1
3me
)
= 0 (2.8)
There is only one physical solution (that is a real one) to theequation:
v2e,1 =
2
3c2
(
√
1 +6P1
mec2− 1
)
(2.9)
Now one can apply the same procedure to the linear momentum conservation law which is of coursealso valid. However, this conservation law is 3-dimensional. From how the coordinate system was de-fined, two equations are sufficient since the third (linked tothe z-axis) says trivially0 = 0. The pairis:
pp + pPs cos θPs = pH cos θH + pe cos θe (2.10a)
pPs sin θPs = pH sin θH + pe sin θe (2.10b)
The terms includingpH (andθH as well) can be easily transfered onto the other sides in eachof theequations. Then, by squaring the both equations, summing them and by using the Pythagorean identity,one gets a single equation instead of two, with theθe angle completely eliminated.
p2p + p2
Ps + p2H− 2pppH cos θH + 2pppPs cos θPs − 2pPspH cos(θPs − θH) = p2
e (2.11)
Although the proper Taylor series expansion of linear momentum has been already introduced, a bitmore is necessary to know in this phase of derivation: expansion of a product of two momentums. Usingthe equation(2.5) and keeping the terms to fourth order inv, this is obtained:
pαpβ ≈ mαmβvαvβ +1
2mαmβ
vαvβ
c2
(
v2α + v2
β
)
(2.12)
Forα = β simply:
p2α ≈ m2
αv2α +
m2αv4
α
c2(2.13)
Plugging these into the(2.11), another biquadratic equation forve is obtained, this time within thelinear momentum conservation law.
v4e +
(
c2)
v2e −
(
c2P2
m2e
)
= 0 (2.14)
10
P2 refers to:
P2 = P2a + P2b + P2c + P2d (2.15)
P2a = m2pv
2p + m2
p
v4p
c2+ m2
Psv2Ps + m2
Ps
v4Ps
c2+ m2
Hv2H
+ m2H
v4H
c2(2.15a)
P2b = −2mpmHvpvH
(
1 +v2p + v2
p
2c2
)
cos θH (2.15b)
P2c = 2mpmPsvpvPs
(
1 +v2p + v2
Ps
2c2
)
cos θPs (2.15c)
P2d = −2mHmPsvHvPs
(
1 +v2H
+ v2Ps
2c2
)
cos(θPs − θH) (2.15d)
Like the first one (following from the energy conservation),this biquadratic equation has only onephysical solution, too:
v2e,2 =
1
2c2
(√
1 +4P2
m2ec
2− 1
)
(2.16)
Setting the two solutionsve,1 andve,2 equal, the whole kinematics is expressed by a single iden-tity. However, any information about the electron (that isve andθe) has been lost. On the other hand,mathematically, one can get it by return if necessary and thelittle mass of the electron compared to theantihydrogen suggests its negligable role in the reaction kinematics. The identity in the terms ofP1 andP2 is:
64P 21
m2ec
4+
9P 22
m4ec
4+
8P1
mec2−
48P1P2
m3ec
4−
4P2
m2ec
2= 0 (2.17)
Both P1 andP2 have been defined and the last step of the derivation is to plugthe two definitionsinto (2.17) but now keeping only the terms to second order in velocities3. The algebraic process is verylong and error prone but it leads to the final result:
A1v2H
+ A2v2p + A3v
2Ps + A4vpvH cos θH + A5vpvPs cos θPs + A6vPsvH + A7Q = 0 (2.18)
3The fourth order terms served in the biquadratic equation finding and are no more required.
11
TheA1, A2, A3, A4, A5, A6, A7 symbols’ meanings are:
A1 = 1 +me
mH
+ 16Q
mHc2+ 12
Q
mec2(2.18a)
A2 =
(
1 −me
mp− 16
Q
mpc2+ 12
Q
mec2
)(
mp
mH
)2
(2.18b)
A3 =
(
1 −me
mPs− 16
Q
mPsc2+ 12
Q
mec2
)(
mPs
mH
)2
(2.18c)
A4 = –2
(
1 + 12Q
mec2
)(
mp
mH
)
(2.18d)
A5 = 2
(
1 + 12Q
mec2
)
(
mpmPs
m2H
)
(2.18e)
A6 = −2
(
1 + 12Q
mec2
)(
mPs
mH
)
(2.18f)
A7 = –2
(
1 + 8Q
mec2
)
(
me
m2H
)
(2.18g)
It is reasonable now to use these approximations:
Q
mHc2,
Q
mpc2,
Q
mPsc2,
Q
mec2,
me
mH
,me
mp≪ 1 (2.19a)
mPs ≃ 2me,mH ≃ mp (2.19b)
The last two approximations do not apply to theQ quantity. After implementation of all these ap-proximations, the final identity can be written and understood in the form of a quadratic equation for theantihydrogen velocity:
v2H
+ B1vH + B2 = 0 (2.20)
B1 = −2vp cos θH–4
(
me
mp
)
vPs cos(θPs–θH) (2.20a)
B2 = v2H
+ 2
(
me
mp
)2
v2Ps + 4
(
me
mp
)
vpvPs cos θPs–2meQ
m2p
(2.20b)
The formal solution is:
vH =1
2
(
−B1 ±√
B21–4B2
)
(2.21)
12
Exactly the same formula is the result of a fully non-relativistic approach, too. The conservation lawsare:
1
2mPsv
2Ps +
1
2mpv
2p =
1
2mev
2e +
1
2mHv2
H− Q (2.22a)
mpvp + mPsvPs cos θPs = meve cos θe + mHvH cos θH (2.22b)
mPsvPs sin θPs = meve sin θe + mHvH sin θH (2.22c)
The derivation is very similar, only the solutions of two quadratic equations are set equal instead oftwo biquadratic equations. The result is the same. This could serve as a control.
Note that the relation in the AEGIS Proposal [1] is differentfrom the one just derived, namely bothB1 andB2 are different (if the former identity takes the same form of quadratic equation) - they aredenoted now asB1 andB2:
B1 = −4(me
mp)vPs cos(θPs − θH) (2.23a)
B2 = v2p + 2(
me
mp)v2
Ps–2vpvPs cos θPs–2meQ
m2p
(2.23b)
In a special case when the velocities of the entering particles (antiproton and positronium) are zero,the relations are the same and give the same results. Otherwise, the difference is easy to see. The identityis usefull for instance if one need to know the maximum velocity of an antihydrogen on the exit of thereaction.
To demonstrate the difference of the former and the new identity on an example, let the antiprotonand the positronium fly on the same line (the faster positronium following the antiproton) with samplevelocites and principal quantum numbers4:
vp = 45 m/s, vPs = 300 m/s, θPs = 0, θH = 0, nPs = 29, nH = 41 (2.24)
Then, according to the new formula, the antihydrogen velocity just after the reaction is45.3 ±0.2 m/s, whereas the former identity gives158.4 m/s.
4For the chosen pairnPs = 29 andnH = 41, the absolute value ofQ quantity is the smallest up tonPs = 43 andnH = 60.Therefore the difference in the two indentities should be most noticeable.
13
Furthermore, the vector of the final antihydrogen can not be known precisely since all the informationabout the electron has been eliminated - there is a class of possible pairsvH - θH solving the identity.However, the effect of the electron is very small. In the picture below, it is shown (still within the previousvalues) how the class of solutions looks like (the green curve) and how it looks like if the angle of thepositronium is changed to perpendicularθPs = π
2 (the red curve):
Obrazek 2.2: Two classes of solutions of the identity - the green curve forθPs = 0 and the red curve forθPs = π
2 . The antihydrigen velocity magnitude is on the vertical axis and the angle of the antihydrogenon the horizontal axis.
The maximum relative difference is1.3 %. For the same values, if one is interested what is thefunctional dependence of the deviation of the curve centres(from the previous picture) on theθPs angle,this is obtained:
Obrazek 2.3: The dependence of the deviation of the antihydrogen velocity solutions class centre on theangle of the positronium. The horizontal axis refers to thatangle and the vertical means the velocitydeviation.
14
The deviation (still the effect of the elimination of the electron’s information) is negligable.
15
3 Antihydrogen deexcitation kinematics
In this chapter, the focus is on this reaction:
H∗
n → H∗
n′ + γ (3.1)
1 ≤ n′ < n (3.1a)
n, n′ ∈ N (3.1b)
In words: An antihydrogen atom in an excited staten switches to a lower energy leveln′ (”deexcites”)which leads to a single photon radiation.
It is the antihydrogen’s nature that it prefers to get rid of any extra internal energy and for the AEGISexperiment it has to be taken into account.
All the possible energy states are discrete and they are all unambigiously labeled with a naturalnumber (n or n′) called the principal quantum number. The final state is not necessary the ground state.
In this chapter, the term deexcitation photon will be used torefer to the photon in(3.1) and inaccordance with this notation, the antihydrogen on the left-hand side is called the initial antihydrogenand the one on the right-hand side is called the final antihydrogen1. The quantities linked to the finalantihydrogen have usualy a prime aside.
The rest energy of an antihydrogen is given by the summ of the rest energies of both its components(an antiproton and a positron) reduced by an amount of the system binding energy:
E(n) = mpc2 + mec
2 −R
n2(3.2a)
E(n′) = mpc2 + mec
2 −R
n′2(3.2b)
In this sense, the rest masses of antihydrogen are simply:
m0 = mp + me −R
n2c2(3.3a)
m′
0 = mp + me −R
n′2c2(3.3b)
The total antihydrogen energy in a frame of reference is thengiven by a product of its rest energyand its Lorentz factor (”gamma-factor”):
E = E(v, n) = E(n)γ(v) =mpc
2 + mec2 − R
n2
√
1 − v2
c2
=m0c
2
√
1 − v2
c2
(3.4a)
E′ = E(v′, n′) = E(n′)γ(v′) =mpc
2 + mec2 − R
n′2
√
1 − v′2
c2
=m′
0c2
√
1 − v′2
c2
(3.4b)
For sure, there is a frame of reference in which the initial antihydrogen stays at rest.
1In the chapter of Charge exchange kinematics, the antihydrogen was always refered to be final because its definition wasdifferent even though it was not necessary to attribute anything to the antihydrogen. Here, the term initial is used as well andthe two ensure the definiteness.
16
One can write the conservation laws of the deexcitation in this frame of reference quite easily; infact, the conservation equations take their simpliest forms there.
m0c2 = E′ + Eγ (3.5)
p′ cos θH′ = pγ cos θγ (3.6a)
p′ sin θH′ = pγ sin θγ (3.6b)
The meanings of the symbols in the first equation (energy conservation) was shown in the beginningof this chapter. Like in any other double-particle decays, the three linear momentum vectors form aplane so only two equations for linear momentum conservation are necessary. If the polar coordinatesare chosen, thep′ andpγ quantities refer to linear momentum magnitudes of the final antihydrogen andthe deexcitation photon respectively (both never negative). TheθH′ quantity is the polar angle of the finalantihydrogen in the polar coordinates system andθγ is defined for the deexcitation photon in the sameway. The inertial frame of reference, where the initial antihydrogen stays at rest and where the polarcoordinates are defined, will be used for most of this chapter.
Anyway, from that the chosen frame of reference is the centreof mass one, then it should be clear thatp′ = pγ : Since the linear momentum of the initial antihydrogen is always zero, the linear momentumsof the two final particles have to be equal in magnitudes (and opposite in directions). The proof is thatthep′ = pγ information follows from the(3.6a) and(3.6b) laws - squaring them, summing and using ofPythagorean identity results in the same equation for linear momentum conservation.
Any photons’ energy and linear momentum satisfy:
pγ =Eγ
c(3.7)
Setting this and the previous momentum information equal, one gets:
Eγ
c= pγ = p′ (3.8)
So the energy of the deexcitation photon can be expressed as amultiple of the final antihydrogenlinear momentum magnitude:
Eγ = p′c (3.9)
This can be plugged into the energy conservation equation:
m0c2 = E′ + p′c (3.10)
Now, the final antihydrogen’s energy is tranfered onte the opposite side of the equation:
p′c = m0c2 − E′ (3.11)
Squaring of this formula leads to:
(
p′c)2
=(
m0c2 − E′
)2(3.12)
Now the(p′c)2 is expressed. Furthermore, there is one more general way of how the quantity can befound - the Lorentz invariant of the final antihydrogen. It says:
(
p′c)2
= E′2 −(
m0c2)2
(3.13)
17
Setting the two expressions equal:
m20c
4 − 2m0c2E′ + E′2 = E′2 − m′2
0c4 (3.14)
Eliminating theE′2 terms:
2m0c2E′ = m′2
0c4 + m2
0c4 (3.15)
Now it is easy to findE′. Moreover, an independent definition ofE′ is what the(3.4b) says. Nowthese two expressions of theE′ quantity can be set equal:
m′20c
4 + m20c
4
2m0c2= E′ =
m′
0c2
√
1 − v′2
c2
(3.16)
If both sides are multiplied with 1m′
0c2, the Lorentz factor of the final antihydrogen stands on its side
of the equation alone:
1√
1 − v′2
c2
=m′2
0c4 + m2
0c4
2m0m′0c4
(3.17)
Setting the both fractions reciprocal:
√
1 −v′2
c2=
2m0m′
0c4
m′20c
4 + m20c
4(3.18)
Squaring:
1 −v′2
c2=
(
2m0m′
0c4
m′20c
4 + m20c
4
)2
(3.19)
Now the goal is to find the value ofv′ or v′
c.
v′2
c2= 1 −
2m0m′
0c4
m′20c
4 + m20c
4
2
=m′4
0c8 + 2m2
0m′20c
8 + m40c
8 − 4m20m
′20c
8
(
m′20c
4 + m20c
4)2 =
=m′4
0c8 − 2m2
0m′20c
8 + m40c
8
(
m′20c
4 + m20c
4)2 =
(
m20c
4 − m′20c
4
m′20c
4 + m20c
4
)2
(3.20)
So the result is:
v′ =m2
0c4 − m′2
0c4
m′20c
4 + m20c
4c (3.21)
The just derived velocity of the final antihydrogen is linkedto the final antihydrogen’s linear momen-tum and possesses analogical qualities - the velocity, likethe momentum, is one of the polar coordinatesand therefore is never negative.
The fraction is appropriate to abbreviate even though the reason is just aesthetic.
v′ =m2
0 − m′20
m′20 + m2
0
c (3.22)
The rest masses of the initial and final antihydrogen atoms has been already introduced and now theywill be replaced with their definitions in the expression forv′:
18
v′ =
(
mp + me −R
n2c2
)2−(
mp + me −R
n′2c2
)2
(
mp + me −R
n2c2
)2+(
mp + me −R
n′2c2
)2 c =
=(mp + me)
2 − 2 (mp + me)R
n2c2+ R2
n4c4− (mp + me)
2 + 2 (mp + me)R
n′2c2− R2
n′4c4
(mp + me)2 − 2 (mp + me)
Rn2c2
+ R2
n4c4+ (mp + me)
2 − 2 (mp + me)R
n′2c2+ R2
n′4c4
c =
=−2 (mp + me)
Rc2
(
1n2 − 1
n′2
)
+ R2
c4
(
1n4 − 1
n′4
)
2 (mp + me)2 − 2 (mp + me)
Rc2
(
1n2 + 1
n′2
)
+ R2
c4
(
1n4 + 1
n′4
)c (3.23)
It follows that the velocity of the final antihydrogen depends on the initial and final states’ principalnumbers only - the rest in(3.23) are constants, namely: the rest mass of an antiproton, the rest massof a positron, the Rydberg constant of an antihydrogen and the speed of light in vacuum. They are allknown with a very high accurancy. A brief analysis of the(3.23) equation indicates that the fraction isdimensionless since the dimension of the rightmost speed oflight corresponds with the final antihydro-gen velocity’s dimension on the left-hand side of the equation. Therefore both the numerator and thedenominator are of the same dimension and for the purpose of afurther analysis, no matter whether SIunits or atomic units are chosen.
For instance, if one prefers to work in the atomic units, thenthese values will be substituted for thesymbols of the constants [3]:
mp = 938272013 eV
me = 510999 eV
R = 13.6 eV
c = 1
Only the principal numbers and the rightmostc constant will be untouched:
v =−2.553 · 1010
(
1n2 − 1
n′2
)
+ 1.85 · 102(
1n4 − 1
n′4
)
1.763 · 1018 − 2.553 · 1010(
1n2 + 1
n′2
)
+ 1.85 · 102(
1n4 + 1
n′4
)c (3.24)
This should help to simplify the expression of the final velocity a bit.In the denominator, both the second and the third term are limited because( 1
n2 + 1n′2
) ≤ 2 and
( 1n4 + 1
n′4) ≤ 2 for all possible pairs of natural numbersn andn′. So there is no need to discuss the
particular values of the second and the third term, since they form less then one nth of the denominator.In the numerator, one can see that0 ≤ 1
n2 ≤ 1, ≤ 1n′2
≤ 1, ≤ 1n4 ≤ 1 and≤ 1
n′4≤ 1 which leads to
−1 ≤ 1n2 − 1
n′2≤ 1 and−1 ≤ 1
n4 − 1n′4
≤ 1. Furthermore,1n4 − 1
n′4= n′4
−n4
n4n′4= (n′2+n2)
n2n′2
(n′2−n2)
n2n′2=
( 1n2 + 1
n′2)( 1
n2 −1
n′2). It was mentioned in the analysis of the denominator that( 1
n2 + 1n′2
) ≤ 2 and now it
follows that the term1n4 −
1n′4
can not be more than twice as higher than1n2 −
1n′2
. Therefore the secondterm in the numerator can be neglected, too.
19
These sensible approximations simplify the formula for thefinal antihydrogen velocity:
v′ =−2 (mp + me)
Rc2
(
1n2 − 1
n′2
)
2 (mp + me)2 c =
R
mpc2 + mec2
(
1
n′2−
1
n2
)
c (3.25)
The fully non-relativistis approach can serve as a control.The proper energy and linear momentumconservation laws are:
−R
2n2= −
R
2n′2+
1
2(mp + me)
2 v′2+ Eγ (3.26)
0 = (mp + me) v′ −Eγ
c→ Eγ = (mp + me) cv′ (3.27)
Plugging the second equation into the first one, the result isa quadratic equation for the final antihy-drogen velocity.
v′2+ 2cv′ −
2R
mp + me
(
1
n′2−
1
n2
)
= 0 (3.28)
Its positive solution is:
v′ =
−2c +
√
4c2 + 4c2 2Rmpc2+mec2
(
1n′2
− 1n2
)
2(3.29)
The Taylor series of the to the first order:
v′ =
(
−1 + 1 +1
2· 2 ·
Rc2
mpc2 + me
(
1
n′2−
1
n2
)
)
c =R
mpc2 + mec2
(
1
n′2−
1
n2
)
c (3.30)
The results of two different approaches are the same.Some of the characteristics of the result are simple to find. For 1 ≤ n′ ≤ n, thev′ value is positive,
naturally, and forn′ = n (nothing happens to the antihydrogen), the velocity is zero, as expected - notethe frame of reference where the initial antihydrogen staysat rest has not been abandoned.
The result depends on four constants:R, mp, me andc. If the values are used instead of symbols,this is obtained:
v′ = 4.343046
(
1
n′2−
1
n2
)
[m
s
]
(3.31)
The number 4.343046 (with its proper SI unit metre per second) also means the theoretically maxi-mum velocity which the final antihydrogen can gain in the frame of reference where the initial antihy-drogen stays at rest.
If one would like now to work with quantities in the laboratory frame of reference, they had totransfer thev′ result into that frame of reference. According to the fact that the values of the initialantihydrogen velocity in the laboratory frame of reference2 would not be greater than10000 metres persecend, the classical Galilean velocity transformation can be used - the possible difference to the ”slowLorentz transformation” [4] would be approximately10−11.
If the Cartesian coordinates of the initial antihydrogen velocity in the laboratory frame of referenceare denoted asvx, vy and vz and the Cartesian coordinates of the final antihydrogen velocity in the
2This is the same as the relative velocity of the two frames of reference - the laboratory one and the one in which the initialantihydrogen stays at rest.
20
laboratory frame of referencevx′, vy
′ andvz′, then (using polar coordinates for the final antihydrogen
velocity) this follows:
vx′ = vx +
R
mpc2 + mec2
(
1
n′2−
1
n2
)
c cos φ cos θ (3.32a)
vy′ = vy +
R
mpc2 + mec2
(
1
n′2−
1
n2
)
c sin φ cos θ (3.32b)
vz′ = vz +
R
mpc2 + mec2
(
1
n′2−
1
n2
)
c sin θ (3.32c)
Finally, one can evaluatev′ for chosen pairs ofn andn′:
n = 35, n′ = 34 → v′ = 2.116 · 10−4[m
s
]
(3.33a)
n = 35, n′ = 1 → v′ = 4.340[m
s
]
(3.33b)
Evidently, the effect of the deexcitation can reach the order of metres per second and the smaller thedeexcitaion difference in principal numbers is, the smaller is the velocity change effect.
21
4 Antihydrogen dynamics
This chapter presents the derivation of a simple antihydrogen dynamics in electric (~E(~R, t)), mag-netic (~B(~R, t)) and gravitational field~g. The limits of the approach will be stressed continuously. Themain phenomenon which affects the dynamics is the deexcitation of the antihydrogen - the possiblechange of kinematic qualities in such an event was discussedin the chapter Antihydrogen deexcitationkinematics. The outcomes of the chapter Charge exchange kinematics serve as the initial conditions foran equation of motion which this chapter aims to find.
Before starting, it is useful to remind the dynamics of a classical charged particle in an electromag-netic field. The equation of motion is:
~P (~R) =M ~R
√
1 −~R2
c2
= q(
~E(~R, t) + ~R × ~B(~R, t))
(4.1)
whereP means the linear momentum of the charged particle,M its mass and~R the current position
of the particle (then~R refers to the velocity);c is the speed of light and~g the gravitational acceleration,
as usual. The ”dot”sign always refers to the first total time derivative (like ~R = ddt
~R)In this chapter, a general chargeq (in the Lorentz force, for example) will be distinguished from a
particular chargeQ (the charge of an antiproton, for instance).This equation includes the so called Lorentz force which is nonzero if the particle posses an intrinsic
electric charge - this charge has to be constant and globallynonzero. However, this does not imply thata particle with a zero charge is insensive to electromagnetic field and experiences only the gravity. Anantihydrogen atom is such a particle. Even though its globalcharge is zero (if not ionized), the spacedistribution of the charge need not be uniform. This kind of space distribution is possible to represent inso called multiple expansion whose first order term is the electric dipole.
In a given frame of reference, this is the definition of the electric dipole~p located at~R:
~p(~R) =
∫
V
ρ(~r)(
~r − ~R)
d3~r (4.2a)
for the continuous charge distributionρ(~R); and for a set of N point charges (qj):
~p(~R) =
N∑
j=1
qj
(
~rj − ~R)
(4.2b)
There are two different types of an electric dipoles - an intrinsic electric dipole and an induced electricdipole. The intrinsic dipole moment is possessed by a molecule of water, for instance, its absolute valueis constant and only the space orientation of the vector is influenced by the field. Although this is notwhat an antihydrogen atom represents, the derivation of such dipole’s dynamics in an electric field willhelp to find the more complicated dynamics of the induced dipole in electromagnetic field.
Electric charge is positive or negative. Assume the two kinds are possible to treat separately (possibleto summarize them and to find the centres of their distributions - q+ = q(~r+) and q− = q(~r−)). Inaddition, if these charges are equal in the absolute value (|q+| = |q−| = Q), the definition of the electricdipole becomes:
~p(~R) = Q~l (4.3)
where the~l quantity stands for the relative position of the two charges:
22
~l = ~r+ − ~r− (4.4)
A charged particle in an electric field experiences a force:
~F (~r) = q ~E(~r) (4.5)
For the two charges forming the intrinsic dipole it follows:
~F+ = ~F (~r+) = q+~E(~r+) = Q~E(~r+) (4.6a)
~F− = ~F (~r−) = q− ~E(~r−) = −Q~E(~r−) (4.6b)
It is possible to expand one of the previous equation (say thefirst one) in Taylor series to the firstorder inl:
~F+ = Q~E(~r+) = Q~E(~r− +~l) ≈ Q(
~E(~r−) +(
~l∇r−
)
~E(~r−))
(4.7)
The definition of a special operator was used:
~a∇b =3∑
j=1
aj∂
∂bj(4.8)
Thej indices just the Cartesian coordiante. The meaning of the Taylor expansion is that the distancel between the two charges is very small with respect to the molecule’s displacement as it moves and thechange of the electric field intensity alongl the first order effect.
The total force on the intrinsic dipole is given by the sum of the two forces:
~F = ~F+ + ~F− ≈ Q(
~E(~r−) +(
~l∇r−
)
~E(~r−))
− Q~E(~r−) =(
~l∇r−
)
~E(~r−) (4.9)
There is no need to keep the subscribe of the negative charge anymore - for an obsrever, the differencebetween~r+ and~r− is undetectable (~r+, ~r− ≫ ~l) and~r ≈ ~r− can be used instead:
~F =(
~l∇r
)
~E(~r) (4.10)
In (4.10), the definition of the operator provides:
~l∇r =
3∑
j=1
lj∂
∂xi(4.11)
wherexi are the Cartesian coordinates of~r.The equation(4.10), however, does not provide the complete information about the dynamics since
the two coupled charges have more than three degrees of freedom (namely five). The constraint here isthe constant distance between the two charges and as a whole,the molecule can be treated as a rigidbody. The corresponding theory suggests the second (independent) equation of motion which is based ontorque [5] and [4]. Its derivation follows from the(4.6a) and(4.6b) like the force equation - it is theiranother ”combination”. In [4], this is proposed:
~M = (~r+ − ~r−) × ~F+ = Q(~r+ − ~r−) × ~E(~r+) = Q~l × ~E( ~r+) = ~p × ~E(~r+) (4.12)
Like before, the~r+ ≈ ~r approximation can be implemented and the torque equation becomes:
~M = ~p × ~E(~r) (4.13)
23
The configuration of the intrinsic dipole is now described bythree Cartesian coordinates~r and bytwo independent coordinates of its orientation~p. Therefore the intrinsic dipole is said to be point-like inthis model.
To sum up, the intrinsic electric dipole has five degrees of freedom and just a one equation of motionwould be not sufficient to know its dynamics completely. Beside the force equation, the sedond torqueequation is introduced. The pair of equation of its dynamicsis:
~F =(
~l∇r
)
~E(~r) (4.14a)
~M = ~p × ~E(~r) (4.14b)
This outcome is very often found in textbooks, however, the antihydrogen atom can not be countedas the intrinsic electric dipole. It is reasonable now to remind some of its qualities. The antihydrogenis a coupled state of two charges: the positive one (qe = Q) is linked to the positron (its mass isme)and the negative one (qp = −Q) is linked to the antiproton (whose mass ismp). If the antihydrogen isset free (outside any electric or magnetic field), the centres of the two charges are situated in the sameplace - the antihydrogen possess zero global charge (anytime if not ionized) and, moreover, there is noprefered direction of its charge distribution orientation. In other words, its electric dipole is zero so, inthe first order, one can neglect its electric qualities. On the other hand, if the same antihydrogen atom isput into nonzero electric field, both of the charges will tendto move in opposite directions - even in ahomogeneous electric field, the antihydrogen stays at rest,however, it gains the induced electric dipole.
Evidently, the induced electric dipole is very different from the intrinsic one: outside the field theinduced electric dipole was zero, whereas inside a field, theelectric dipole was gained by the particle.Therefore the condition that the intrinsic dipole has to be constant in magnitude does not hold. Theconstant magnitude of a dipole is equivalent to the presenceof a holonomic constraint. In the case ofantihydrogen, just a different holonomic constraint applies, thus in this sense the two kinds of dipoles arenot extremely distinct.
The mechanism is that the antihydrogen gains an induced electric dipole in an electric field and thedipole is sensitive to that field retroactively.
If the positions of the centres of the two charges are denotedas rp and re, the variability of thedistance between them has to be taken into account:
~l = ~l(~R) = ~re − ~rp (4.15)
The ~R quantity is the position of centre of the antihydrogen mass:
~R =mp~rp + me~re
mp + me(4.16)
In this model, the induced dipole is treated as point-like (in the sense introduced above), too. The~Rquantity will refer to the position of the whole antihydrogen.
The~l quantity’s dependence on~R implicitly says that the distance between the centres of thetwocharges is given by the electric intensity at~R:~l = ~l( ~E) = ~l(~E(~R)) = ~l(~R). Note that the intrisic dipole’sdefinition formula~p(~R) = Q~l = Q~l(~R) remains valid for the induced dipole, therefore~p = ~p(~E(~R)) =~p(~R). According to the source [6], the dependance is linear:
~p(~R) = α ~E(~R) (4.17)
The constant of proportionalityα will be discussed later.
24
The equations(4.15) and(4.16) together form simultaneous equations forrp andre. The solution issimple to find:
~rp = ~R −me
mp + me
~l(~R) (4.18a)
~re = ~R +mp
mp + me
~l(~R) (4.18b)
The two charges experience the Lorentz force introduced in the beginning of this chapter. First, thesame kind of expansions like in the intrinsic dipole’s section will be implemented for the electric forces:
~FEp = ~FE(~rp) = −Q~E(~rp) = −Q~E(~R −me
mp + me
~l(~R)) ≈
≈ −Q~E(~R) +me
mp + meQ(
~l(~R)∇R
)
~E(~R) (4.19a)
~FEe = ~FE(~re) = +Q~E(~re) = +Q~E(~R +mp
mp + me
~l(~R)) ≈
≈ +Q~E(~R) +mp
mp + meQ(
~l(~R)∇R
)
~E(~R) (4.19b)
The total electric force on the induced electric dipole is the sum of the previous two:
~FE = ~FEp + ~FEe =(
~d(~R)∇R
)
~E(~R) (4.20)
Now, the same expansion for the magnetic forces on the charges:
~FBp = ~FB(~rp) = −Q~rp × ~B(~rp) =
= −Q
(
~R −me
mp + me
~l(~R)
)
× ~B(~R −me
mp + me
~l(~R)) ≈
≈ −Q
(
~R −me
mp + me
~l(~R)
)
×
(
~B(~R) −me
mp + me
(
~l(~R)∇R
)
~B(~R)
)
=
= −Q~R × ~B(~R) +me
mp + meQ~R ×
((
~l(~R)∇R
)
~B(~R))
+
+me
mp + meQ~l(~R) × ~B(~R) −
m2e
(mp + me)2 Q~l(~R) ×
((
~l(~R)∇R
)
~B(~R))
(4.21a)
~FBe = ~FB(~re) = +Q~re × ~B(~re) =
= +Q
(
~R +mp
mp + me
~l(~R)
)
× ~B(~R +mp
mp + me
~l(~R)) ≈
≈ +Q
(
~R +mp
mp + me
~l(~R)
)
×
(
~B(~R) +mp
mp + me
(
~l(~R)∇R
)
~B(~R)
)
=
= +Q~R × ~B(~R) +mp
mp + meQ~R ×
((
~l(~R)∇R
)
~B(~R))
+
+mp
mp + meQ~l(~R) × ~B(~R) +
m2p
(mp + me)2 Q~l(~R) ×
((
~l(~R)∇R
)
~B(~R))
(4.21b)
25
The sum of the previous two forces is the total magnetic forceon the induced electric dipole:
~FB = ~FBp + ~FBe =
= Q~R ×((
~l(~R)∇R
)
~B(~R))
+ Q~l(~R) × ~B(~R)+
+(mp + me) (mp − me)
(mp + me)2 Q~l(~R) ×
((
~l(~R)∇R
)
~B(~R))
=
= ~d(~R) × ~B(~R) +
(
~R +1
q
mp − me
mp + me
~d(~R)
)
×((
~d(~R)∇R
)
~B(~R))
(4.22)
Summing the forces~FE and ~FB , one gets the total force which the antihydrogen would experience ifthere were no more antihydrogen dynamic qualities (like magnetic dipole, for instance). The force wouldbe:
~FEB = ~FE + ~FB =
=(
~d(~R)∇R
)
~E(~R) + ~d(~R) × ~B(~R) +
(
~R +1
q
mp − me
mp + me
~d(~R)
)
×((
~d(~R)∇R
)
~B(~R))
(4.23)
The already discussed retroactivity of the electric field tothe induced electric dipole will be nowfully implemented into the force from the information~p(~R) = α ~E(~R):
~FEB = α(
~E(~R)∇R
)
~E(~R) + α ~E(~R) × ~B(~R)+
+α
(
~R +1
q
mp − me
mp + meα ~E(~R)
)
×((
~E(~R)∇R
)
~B(~R))
(4.24)
Following the derivation of intrinsic dipole’s dynamics, one would try to find the torque equation.However, the introduced formula~p(~R) = α ~E(~R) makes the situation easier. If the torque on the inducedelectric dipole is similar to that of the intrinsic electricdipole, the cross product of two identical vectorsgives zero:
~M = ~p(~R) × ~E(~R) = α ~E(~R) × ~E(~R) = 0 (4.25)
A different proof supports this estimation: The constraint~p(~R) = α ~E(~R) actually not only assertsthe magnitude of the induced electric dipole1 but also its orientation. While the constraint of the constantmagnitude reduced the number of degrees of freedom of the intrinsic electric dipole from six to five,induced dipole’s constraint reduces the number of degrees of freedom from six to three. Therefore theforce equation (with just the electric dipole implemented so far) fully describes the dynamics of theantihydrogen.
However, this is not everything because the antihydrogen has a magnetic dipole, too: in fact, twokinds which are both sensitive to the magnetic field. The firstkind is the induced magnetic dipole whichis very similar to the induced electric dipole. For any magnetic dipole ~m, the source [4] states the forceequation:
~FM (~R) =(
~m(~R)∇R
)
~B(~R) (4.26)
1The magnitude of the intrinsic electric dipole is explicitely given, too, albeit trivially.
26
And the same source provides the dependence of~m(~R) on ~B(~R):
~m(~R) = β ~B(~R) (4.27)
Because of this similarity, the induced magnetic dipole’s contribution to the total force of antihydro-gen will be:
~FMind= β
(
~B(~R)∇R
)
~B(~R) (4.28)
The two quantitiesα and β are independent of the fields, however, they depend on the particularinner state of the antihydrogen. The source [6] provides estimations for both of them (in the case of ahydrogen atom):
α = f(n, l,m)4πǫ0r3B (4.29a)
β = −q2e < r2
0 >
6me(4.29b)
The senses of symbols are:ǫ0 is the vacuum permitivity,rB is the Bohr radius;r0 is the root meansquare of the electron radius and has to be treated by quantummechanics tools as well asf(n, l,m)where the three parameters are the principal, azimuthal andmagnetic quantum numbers. Fortunately, thesource [6] states that it shouldf(1, 0, 0) = 9
2 - this perfectly agrees with something solved in [7]:
f(n, l,m) =1
8
(
17n2 − 9m2 + 19)
(4.30)
There is no torque equation for the introduced induced magnetic dipole, too. The two induced dipolespoint exactly in the same direction like the proper fields’ lines. In other words, both of them answer tothe change of the field immediately. The causality is not the problem since the reaction takes place in thesame place as the action - the(4.17) and(4.27) could be both just ”adiabatic”approximations. However,in this model the both equations are treated as valid.
The intrinsic magnetic dipole~µ experience torque but to keep within the simple model, its orientationis prescribed at every instant:
~µ(~R) =
∣
∣
∣~µ(~R)
∣
∣
∣
∣
∣
∣
~B(~R)∣
∣
∣
~B(~R) (4.31)
The intrinsic magnetic dipole experiences the force:
~FMint=
∣
∣
∣~µ(~R)∣
∣
∣
∣
∣
∣
~B(~R)∣
∣
∣
(
~B(~R)∇R
)
~B(~R) (4.32)
Finally, the ordinary gravitational force is implemented:
~FG = (mp + me)~g (4.33)
27
If all the intermediate outcomes are summarized, the total force on the antihydrogen atom in thecombined fields within the model is:
~F = α(
~E(~R)∇R
)
~E(~R) + α ~E(~R) × ~B(~R)+
+α
(
~R +1
q
mp − me
mp + meα ~E(~R)
)
×((
~E(~R)∇R
)
~B(~R))
+
+
|~µ|∣
∣
∣
~B(~R)∣
∣
∣
+ β
(
~B(~R)∇R
)
~B(~R) + (mp + me)~g (4.34)
And the equation of motion is:
(mp + me) ~R = α(
~E(~R)∇R
)
~E(~R) + α ~E(~R) × ~B(~R)+
+α
(
~R +1
q
mp − me
mp + meα ~E(~R)
)
×((
~E(~R)∇R
)
~B(~R))
+
|~µ|∣
∣
∣
~B(~R)∣
∣
∣
+ β
(
~B(~R)∇R
)
~B(~R) + (mp + me)~g (4.35)
The Newtonian formulation of mechanics is chosen since the number of the degrees of freedom isthree and since for all the forces, a simple expression in Cartesian coordinate system is known. The non-relativistic approach agrees with the fact that the velocities of antihydrogen will be definitely smallerthan a few tenths kilometres per second. The result and its derivation are ”quasi-classical”- they treat theantihydrogen as a classical point-like particle and its qualities like electric dipole and magnetic dipoleare prescribed from the outcomes of quantum mechanics or experiments.
The initial values are~R(t0) and ~R(t0) wheret0 refers to the time of the antihydrogen creation.The final equation of motion can be written with the Cartesianindices rigorously emphasized. Let~E(~R, t) = (E1(~R, t), E2(~R, t), E3(~R, t)) = (E1, E2, E3),~B(~R, t) = (B1(~R, t), B2(~R, t), B3(~R, t)) = (B1, B2, B3),
~g = (g1, g2, g3), ~R = (X1,X2,X3) and ~R = (X1, X2, X3); ǫijk is the Levi-Civita symbol. Thedefinition of the total time derivative is used. In this form,there are three equations of motion (i = 1, 2, 3):
(mp + me) Xi = α
(
3∑
l=1
El∂Ei
∂Xl
)
+ ǫijkα
(
∂Ej
∂t+
3∑
l=1
∂Ej
∂Xl
Xl
)
Bk+
+ǫijkα
(
Xj +1
q
mp − me
mp + meα
(
∂Ej
∂t+
3∑
l=1
∂Ej
∂Xl
Xl
))(
3∑
l=1
El∂Bk
∂Xl
)
+
+
(√
∑3l=1 µ2
l∑3
l=1 B3l
+ β
)(
3∑
l=1
Bl∂Bi
∂Xl
)
+ (mp + me) gi (4.36)
If the number of degrees of freedom were higher than three andtherefore the force equation werenot enough, to find the equations of motion, the Lagrange formulation of mechanics would be easier toimplement (any problems concerning number of degrees of freedom, constraints or correct identificationof the forces are solved automatically). If~Fl denotes to coordiantes of all the Cartesian forces experiencedby the system or by its components, if parametresqi (generalized coordinates) are used instead of the
28
Cartesian coordinatesxl and ifT = T (q1, ...) stands for the kinetic energy, then the equations of motionfor all possiblei are [8]:
d
dt
∂T
∂qi+
∂T
∂qj=∑
l
Fl∂xl
∂qi(4.37)
Then the particular form for antihydrogen depends on how theequations(4.17) and(4.27) are mod-ified.
29
Podekovanı
Chtel bych podekovat vedoucımu sveho vyzkumneho ukolu, doc. RNDr. Vojtechu Petrackovi, CSc.,za vytrvalou a nevsednı pomoc, trpelivost a cas, kteremi nezistne venoval po celou dobu vytvarenı tetoprace a dıky kterym jsem zıskal dostatecne znalosti,vedomosti a orientaci. Dekuji mu take za vsechnypodnety, kterymi me usmernoval a dıky kterym se uroven textu stale zvysovala. Pod jeho vedenım meprace velmi tesila.
Take dekuji RNDr. Davidu Brenovi, Ph.D. za poskytnutıkonzultacı a prom.fyz. Jirımu Adamovi,CSc. za jeho zajem, pravidelne diskuze a odbornou pomoc s neobvyklymi okruhy fyziky, na ktere sivyhrazoval cas po cely rok.
Konecne bych chtel podekovat sve rodine a pratelum, dıky jejichz podpore jsem mel to spravnepracovnı zazemı.
30
5 The bibliography
[1] AEGIS collaboration.Proposal for the AEGIS Experiment at CERN Antiproton Decelerator :Antimatter Experiment: Gravity, Interferometry, Spectroscopy. [s.l.] : [s.n.], 2007. 125 s.Dostupny z WWW:<http://doc.cern.ch//archive/electronic/cern/preprints/spsc/public/spsc-2007-017.pdf>.
[2] Goldman, T.; Hughes, R. J.; Nieto, M. M.Phenomenological Aspects Of New GravitationalForces: IV. New Terrestrial Experiments. Phys. Rev. D36, 1254(1987).
[3] Particle Physics Booklet. [s.l.] : Elsevier, 2008. 296 s. Dostupne z WWW:<http://pdg.lbl.gov/pdgmail>.
[4] SEDLAK, Bedrich; STOLL, Ivan.Elektrina a magnetismus. Vydanı 2., opravene a rozsırene.Praha : Academia, 2002. 636 s. ISBN 80-200-1004-1.
[5] STOLL, Ivan.Mechanika. Vydanı druhe. Praha : VydavatelstvıCVUT, 2003. 210 s. ISBN80-01-02692-2.
[6] STOLL, Ivan.Elektrina a magnetismus. Vydanı druhe. Praha : VydavatelstvıCVUT, 2003. 216s. ISBN 80-01-02693-0.
[7] LANDAU, Lev Davidovich; LIFSHITZ, Evgeny Mikhailovich. Quantum mechanics :Non-relativistic theory. Third edition, revised and enlarged. [Oxford] : Pergamon Press, 1977.677 s. ISBN 0-08-020940-8.
[8] HORSKY, Jan; NOVOTNY, Jan;STEFANIK, Milan. Mechanika ve fyzice. Vydanı 1. Praha :Academia, 2001. 416 s. ISBN 80-200-0208-1.
