Klein-Gordon and Dirac particles in non-constant scalar-curvature background

0.2 MB·English

by  M. Alimohammadi

Klein-Gordon and Dirac particles in non-constant
scalar-curvature background
∗
M. Alimohammadi and A. A. Baghjary
Department of Physics, University of Tehran,
North Karegar Ave., Tehran, Iran.
8
0
0
2
Abstract

n
a TheKlein-GordonandDiracequationsareconsideredinasemi-infinite
J lab (x>0) in thepresenceof backgroundmetrics ds2 =u2(x)ηµνdxµdxν
0 and ds2 = dt2+u2(x)ηijdxidxj with u(x)=e±gx. These metrics have
−
1 non-constantscalar-curvatures. Variousaspectsofthesolutionsarestud-
ied. For the first metric with u(x)=egx, it is shown that the spectrums

] arediscrete,with thegroundstateenergyE2 =p2c2+g2c2~2 forspin-
c min
q 0 particles. For u(x) = e−gx, the spectrums are found to be continuous.
- For the second metric with u(x) = e−gx, each particle, depends on its
r transverse-momentum, can have continuous or discrete spectrum. For
g
[ Klein-Gordon particles, this threshold transverse-momentum is √3g/2,
while for Dirac particles it is g/2. There is no solution for u(x) = egx

2 case. Some geometrical properties of these metrics are also discussed.
v
6
6 1 Introduction
3
1
Studying the quantum mechanical effects of gravity is an important and inter-
.
1
estingbranchofphysicswhichhasbeenstartedfromthe earlydaysofquantum
0
mechanics. Thesimplestexampleoftheseeffectsisthe behaviorofthe nonrela-
8
0 tivisticspinlessquantumparticle,i.e. theSchrodingerequation,inthepresence
: of constant gravity [1]. This phenomenon has been experimentally verified by
v
the famous experiment of Collela et al. [2]. The latest of these experiments is
i
X onereportedbyNesvizhevskyetal.,inwhichthequantumenergylevelsofneu-
r tronsintheEarth’sgravitationalfieldhavebeenmeasured[3,4]. Otheraspects
a
of gravitational effects in quantum physics are appeared, for example, in neu-
trino oscillation in gravitational background [5, 6, 7], Berry phase of spin-1/2
particles moving in a space-time with torsion [8, 9], etc.
Another branch of researches in this area is the study of the behaviors of
Dirac and Klein-Gordon particles in the curved background and distinguishes
their physicalcharacteristics. This is aninteresting subject since it makesclear
theimportanceofthespinoftheparticles,whichisapurelyquantummechanical
property, in the gravitational interactions. Chandrasekhar, for example, has
consideredthe Dirac equationin aKerr-geometrybackground[10],with results
which have been followed by others [11, 12].
∗[email protected]
1
A quick review in the literature of this field shows that the number of in-
tegrable models is very few. For example in the case of Schwarzschild metric,
where the metric’s components depend only on one spatial coordinate r, the
problem is too complicated to be solvedanalytically. So trying to exactly solve
somerelativisticquantummechanicalexamplesincurvedbackground,mayshed
light on this important topic and can help us to achieve more insight into the
realistic problems.
A class of background metrics which can be considered in this area, is one
whichdepends onlyononespatialvariablex. In[13],a semi-infinite laboratory
(x>0) has been considered in background metric
ds2 =u2(x)( dt2+dx2)+dy2+dz2. (1)
−
Infact,aninfinitebarrierhasbeenassumedinx<0region. TheKlein-Gordon
and Dirac equations have been studied for the constant-gravity approximation
of this background, i.e. u(x) 1+gx, and some interesting features of this
≃
problemhavebeendiscussed. Forexampleithasbeenshownthatinthecaseof
zerotransverse-momentum,thereexistsanexactrelationbetweenthesquaresof
energy eigenvalues of spin-1/2 and spin-0 particles with same masses: E2 =
Dirac
E2 +mg~c. So the gravity clearly distinguishes between the Fermions and
KG
Bosons. The scalar-curvature of metric (1) is R=0.
In [14], another member of this class, that is the metric
ds2 = dt2+dx2+u2(x)(dy2+dz2), (2)
−
has been considered. For the case u(x) = e gx, the Klein-Gordon and Dirac
−
equations have been studied and their eigenfunctions have been obtained. As
anexactresult,it hasbeenshownthatthe spin-0particleshavespecificground
state energy : E m2c4+g2c2~2, while the spin-1/2 particles have the
KG
≥
natural rest-mass energy ground state : E mc2. The scalar curvature
p Dirac ≥
of metric (2) is R = 6g2. So both metrics (1) and (2) have constant scalar
curvatures.
In this paper, we are going to study the behaviors of relativistic spin-0 and
spin-1/2 particles in one-variable-dependent-metrics with non-constant scalar-
curvature. For a semi-infinite labwith an infinite potential barrierin x<0, we
consider the following two metrics:
ds2 =u2(x)( dt2+dx2+dy2+dz2), (3)
−
ds2 = dt2+u2(x)(dx2+dy2+dz2). (4)
−
It is worth nothing that in metrics (1) and (2), only two of the metrics’ com-
ponents are nontrivial, while here we study the next steps and take three ( in
eq.(4)) and four ( in eq.(3)) of the components nontrivial.
To make our problems more solvable, we assume u(x) = e gx, which are
±
the similar choices that have been considered in the preceding cases. As the
result,bothmetrics(3)and(4)gainthex-dependentscalarcurvatures,andthe
Dirac and Klein-Gordon equations show interesting properties with significant
different behaviors. It is worth mentioning that changing the variable from x
to X, defined by X = u(x)dx, transforms the metric (4) to (2), but with
new u(x), i.e. u(x) f(X). Now if one decides to study both metrics (2)
→ R
2
and (4) with the same u(x), as we do in this paper, then these two metrics are
independent.
The plan of the paper is as follows: In section 2, after fixing our notations,
wediscusstheKlein-GordonandDiracequationsinbackgroundmetric(3)with
u(x) = e gx. It is shown that the spectrum of energy eigenvalues are discrete
±
for u(x) = egx and continuous for u(x) = e gx. The discrete spectrums are
−
compared numerically and the geometrical properties of the metrics, including
their geodesics, are discussed. The importance of geodesic in this problem is
that it determines how much it is possible to consider the x = 0-plane as the
floorofthelaboratory. Notingthattheendpointsofallclassicalfallingparticles
are the floor, then only if the classical trajectories finally intersect x = 0, this
plane can be considered as floor,otherwise not. We see that this is the case for
u(x)=e gx.
−
In section 3, the same is done for metric (4) and it is shown that only the
caseu(x)=e gx isconsistentwiththedesiredboundaryconditions. Forthisu,
−
it is shown that in both cases, i.e. spin-0 and spin-1/2 particles, the spectrums
have interesting properties. They are continuous for p < p and discrete for
0
p > p . p is the transverse-momentum of the particles and the value of p is
0 0
different for Dirac and Klein-Gordon particles. The geometrical properties of
metric (4) are also discussed. Finally in section 4, we review our main results
and bring some comments on metric
ds2 =u2(x)( dt2+dx2)+v2(x)(dy2+dz2), (5)
−
which is somehow a combination of two metrics (1) and (2).
2 Conformally-flat metric ds2 = u2(x)η dxµdxν
µν
In a space-time with metric g , the Klein-Gordon equation in c = ~ = 1 unit
µν
is
1 ∂ ∂
detg gµν m2 ψ =0. (6)
" −detgµν ∂xµ (cid:18) − µν ∂xν(cid:19)− # KG
p
The Dirac eqpuation in curved background is
[γa(∂ +Γ ) m]ψ =0. (7)
a a D
−
γas are the Dirac matrices and Γ s are spin connections which can be obtained
a
from tetrads ea through
dea+Γa eb =0,
b
∧
Γa =Γa ec,
b cb
1
Γ = [γ ,γ ]Γc b. (8)
a b c a
− 8
The boundary conditions of the semi-infinite lab (x > 0) with an infinite po-
tential barrier at x=0 are as follows. For Schrodinger equation, the boundary
condition is lim ψ = 0, which comes from the fact that the Schrodinger
x 0 sch
→
equation is second order in x, so ψ must be continuous at x = 0. The same
sch
is true for Klein-Gordon equation, so the same boundary condition exists:
ψ (0)=0. (9)
KG
3
But the Dirac equation is of first order in x and therefore ψ can be discon-
D
tinuous at x = 0, if the potential goes to infinity there. In this case, it can be
shown that the desired boundary condition is [13]
γ1 1 ψ (0)=0. (10)
D
−
The square-integrability of w(cid:0)avefunc(cid:1)tions ψ and ψ in (0, ) region also
KG D
∞
leads to
lim detg ψ 2 0. (11)
µν
x | | →
→∞
p
2.1 The Klein-Gordon equation
For the metric (3), the Klein-Gordon equation (6) becomes
∂2 ∂2 ∂2 ∂2 u ∂
+ + + +2 ′ m2u2 ψ =0. (12)
−∂t2 ∂x2 ∂y2 ∂z2 u ∂x − KG
(cid:18) (cid:19)
Since u depends only on x, one may seek a solution whose functional form is
ψ (t,x,y,z)=exp( iEt+ip y+ip z)ψ (x). Then ψ satisfies
KG 2 3 KG KG
−
d2 u d
E2 p2 p2+ +2 ′ m2u2 ψ (x)=0. (13)
− 2− 3 dx2 u dx − KG
(cid:18) (cid:19)
The above equation becomes solvable if we assume u/u=constant. So we take
′
u(x)=e gx. (14)
±
Let us first consider u(x)=egx case.
Substitute u(x)=egx into eq.(13), and defining φ(x) through
ψ =e gxφ(x), (15)
KG −
one finds
d2φ(x)
+ g2 λ2 1 m2e2gx φ(x)=0, (16)
dx2 − −
in which (cid:2) (cid:0) (cid:1) (cid:3)
p2 :=p2+p2,
2 3
E2 p2
λ2 := − . (17)
g2
In terms of new variable X =(m/g)egx, eq.(16) reduces to the modified Bessel
equation
d2φ(X) dφ(X)
X2 +X X2+1 λ2 φ(X)=0 (18)
dX2 dX − −
with solutions K (X) and I (X) (ν =√(cid:0)1 λ2). Usin(cid:1)g the asymptotic behav-
ν ν
−
iors of Bessel functions
1
I (z) zν , K (z) (ν =0), (19)
ν ∼ ν ∼ zν 6
in the limit z 0 ( with similar relations for Bessel functions J (z) and Y (z),
ν ν
→
respectively), and
ez e z
−
I (z) , K (z) , (20)
ν ν
∼ √z ∼ √z
4
in the limit z , it can be easily seen that the boundary condition (11)
→ ∞
discards I (X):
ν
lim detg ψ 2 e2(ν+1)gx . (21)
µν KG
x | | ∼ →∞
→∞
p
So the wavefunction becomes
m
ψKG =Ce−gxKν( egx). (22)
g
The boundary condition at x = 0, eq.(9), forces us to take ν a pure imaginary
number, since K becomes zero only when ν is pure imaginary[15]. So one has
ν
to take λ2 > 1, which results the ground state energy of the spin-0 particles in
background metric (3) with u(x)=egx to be:
E2 E2 =p2c2+g2c2~2. (23)
≥ min
Other energy eigenvalues can be obtained by equation:
m
K ( )=0, (24)
i√λ2 1 g
−
which clearly results in a discrete spectrum.
In u(x)=e gx case, a similar argument leads to
−
m m
ψKG =egx C1Iν( e−gx)+C2Kν( e−gx) , (25)
g g
(cid:20) (cid:21)
with ν =√1 λ2 and λ defined through eq.(17). Since detg =e 4gx, the
µν −
−
functionI (me gx)satisfiestheboundarycondition(11)forallpositiveνs. But
ν g − p
one has detg ψ 2 e2(ν 1)gx for K (me gx), which satisfies (11) only
µν| KG| ∼ − ν g −
if (ν)<1. For λ2 <1, ν =√1 λ2 <1 is a real number and for λ2 >1, ν is
p
ℜ −
pure-imaginarynumber (so (ν)=0). Therefore in all cases,one has (ν)<1
ℜ ℜ
whichimplies thatC andC ineq.(25) arearbitrarynon-zeroconstants,upto
1 2
the normalization condition of wavefunction. The boundary condition (9) then
gives
m m
C I ( )+C K ( )=0, (26)
1 ν 2 ν
g g
which results in a continuous energy spectrum.
2.2 The Dirac equation
For metric (3), the nonvanishing Γa s are Γ0 = Γ1 = (u/u2)e0, Γ2 =
b 1 0 ′ 1
Γ1 = (u/u2)e2 and Γ3 = Γ1 = (u/u2)e3, from which Γ0 1 = Γ1 0 =
2 ′ 1 3 ′ 0 0
− − −
Γ2 1 = Γ1 2 = Γ3 1 = Γ1 3 = u/u2. Therefore one finds Γ = 0 and
2 2 3 3 ′ 1
− −
Γ = (u/2u2)γ γ (a = 0,2,3). Noting that γa∂ = γaeµ∂ , in which
a − ′ 1 a a a µ
eµ =diag(u 1,u 1,u 1,u 1), the Dirac equation (7) leads to:
a − − − −
1 ∂ ∂ 3 u
(γ0 +γk )+ ′γ1 m ψ =0. (27)
u ∂t ∂xk 2u2 − D
(cid:20) (cid:21)
Sinceu=u(x),itisnaturaltotakeψ (t,x,y,z)=exp( iEt+ip y+ip z)ψ (x).
D 2 3 D
Defining ψ˜ (x) through −
D
ψ (x)=u 3/2ψ˜ (x), (28)
D − D
5
eq.(27) then results in
(O +O )φ (x)=0. (29)
1 2 D
In above equation, O , O and φ (x) are:
1 2 D
d
O = iEγ1+γ0 muγ1γ0, (30)
1
− dx −
O =i(p γ2+p γ3)γ1γ0, (31)
2 2 3
φ (x)=γ1γ0ψ˜ (x). (32)
D D
Using the fact that [O ,O ] = 0, it may be possible to choose the common
1 2
eigenspinors for O and O . The eigenvalues of O are ip, with p defined
1 2 2
±
in eq.(17) and each eigenvalues are two-fold degenerate. For ip-eigenvalue, the
eigenspinors are
i(p p)/p 0
2 3
−
1 0
χ1 = 0  , χ2 =i(p +p)/p . (33)
2 3
 0   1 
   
   
So one can choose φ (x) = φ (x)χ +φ (x)χ and determines the unknown
D ′1 1 2 2
functions φ (x) and φ (x) such that φ (x) satisfies (29). If we write φ as
′1 2 D D
φ
1
φ
φD =φ′1, (34)
′2
φ2
 
 
then φ and φ relate to φ and φ as following
1 ′2 ′1 2
i(p p)
2
φ1 = p− φ′1,
3
i(p +p)
2
φ = φ . (35)
′2 p 2
3
Now it is sufficient to obtain two functions φ (x) and φ (x), from which φ (x)
1 2 D
and therefore ψ (x) will be determined. Noting that O φ = ipφ , eqs.(30)
D 1 D D
−
and (34) result in:
dφ
1
=pφ (E+mu)φ ,
1 2
dx −
dφ
2
= pφ +(E mu)φ . (36)
2 1
dx − −
It can be also easily shown that the boundary condition (10) for ψ (0) reduces
D
to the following boundary condition on φ and φ :
1 2
(φ (x)+φ (x)) =0. (37)
1 2 x=0
|
Defining ψ and ψ through:
1 2
ψ =φ +φ ,
1 1 2
ψ =φ φ , (38)
2 1 2
−
6
the differential equations (36) then lead to:
dψ
1
=(p+E)ψ muψ ,
2 1
dx −
dψ
2
=(p E)ψ +muψ . (39)
1 2
dx −
We first consider u(x) = egx case. Introducing the new variable X =
(2m/g)egx, the differential equation of ψ becomes
1
d2ψ dψ 1 1
X2 1 +X 1 + λ2+ X X2 ψ =0, (40)
dX2 dX 2 − 4 1
(cid:18) (cid:19)
where λ is defined in eq.(17). Defining ψ˜ through
1
ψ =X 1/2ψ˜ , (41)
1 − 1
eq.(40) leads to:
d2ψ˜ 1 1/2 1/4+λ2
1 + + + ψ˜ =0, (42)
dX2 −4 X X2 1
(cid:18) (cid:19)
which is Whittaker differential equation with solution
ψ˜ =e X/2Xiλ+1/2[C M(iλ,1+2iλ,X)+C U(iλ,1+2iλ,X)]. (43)
1 − 1 2
M(a,c,x) and U(a,c,x) are confluent hypergeometric functions. The bound-
ary condition (11) implies ψ (x ) = X 1/2ψ˜ (X) = 0. But the
1 − 1 X
asymptotic behaviorof M(a,c,x)→is e∞x/xc a, so C =0. T|he→s∞econdboundary
− 1
condition (37) results in ψ (x = 0) = X 1/2ψ˜ (X) = 0, which leads
1 − 1 X=(2m/g)
|
to:
iλ
2m 2m
U(iλ,1+2iλ, )=0. (44)
g g
(cid:18) (cid:19)
ThisequationdeterminesthediscreteenergyeigenvaluesofDiracparticleswhen
they are in background metric (3) with u(x)=egx.
For u(x)=e gx, the same procedure leads to:
−
ψ =e Y/2Yiλ[C M(1+iλ,1+2iλ,Y)+C U(1+iλ,1+2iλ,Y)], (45)
1 − 1 2
whereλisdefinedineq.(17)andY (2m/g)e gx. Heretheboundarycondition
−
≡
(11) implies ψ (Y 0) 0 which can not discard none of the constants C
1 1
→ →
and C . The energy eigenvalues can be obtained by (37), which results in
2
m m
C M 1+iλ,1+2iλ,2 +C U 1+iλ,1+2iλ,2 =0. (46)
1 2
g g
(cid:18) (cid:19) (cid:18) (cid:19)
Since C and C are arbitrary constants, up to the normalization condition,
1 2
eq.(46) results in a continuous energy spectrum.
It may be worth noting that wavefunctions of scalar and spin-1/2 particles
in two background metrics which relate by a conformal transformation, can
be obtained from each other if the particles are massless. If we consider the
7
conformal transformation g (x) g¯ (x) = Ω2(x)g (x), then if R = 0, one
µν µν µν
can show that (cid:3)φ=0 (cid:3)¯φ¯=0→in which [16]
→
φ¯(x)=Ω(x)(2 n)/2φ(x). (47)
−
n is the dimension of space-time. For massless fermions, one also has
ψ(x) ψ¯(x)=Ω(x)(1 n)/2ψ(x). (48)
−
→
In our problem, R is zero ( for flat metric ds2 = η dxµdxν). Now if we take
µν
m=0 in eq.(16), the eqs.(15) and (16) result in
ψ (x)=u 1ψflat(x), (49)
KG − KG
which is consistent with eq.(47). Also if we put m=0 in eq.(27) and inserting
eq.(28) into eq.(27), we obtain
∂
γµ ψ˜ (x)=0, (50)
∂xµ D
which is the free Dirac equation in flat space-time, i.e. ψ˜ =ψflat. This shows
D D
ψDnon−flat =u−3/2ψDflat which is again consistent with eq.(48).
2.3 Comparing the spectrums
For u(x)=egx, the spectrums of spin-0 and spin-1/2 particles can be obtained
byeqs.(24)and(44),respectively. Noneoftheseequationscanbesolvedanalyt-
ically and only in mc/g~>>1 limit, an approximate solution can be obtained
for eq.(24) [15]. The main difference between two spectrums is that the Klein-
Gordoneigenvalueshasagroundstate,i.e. λ2 >1,butforDiracparticleboth
KG
λ2 <1 and λ2 >1 cases are possible. To obtain the numerical values of
Dirac Dirac
energies ( λs ), one must fix the values of m and g and then finds the roots of
twoequations(24)and(44). Forexampleformc/g~=0.1,Table 1showssome
of the lowest energy levels of Dirac and Klein-Gordon particles. The energies
can be found by using eq.(17): E = p2c2+(λgc~)2.
q
Table 1: The lowest ten values of λ and λ for (mc/g~)=0.1
KG D
λ 1.52 2.27 3.02 3.74 4.44 5.12 5.78 6.42 7.04 7.68
KG
λ 0.85 1.82 2.65 3.42 4.14 4.84 5.51 6.17 6.82 7.46
D
2.4 The metric properties
The scalar curvature of metric (3) is
R=6u 3u . (51)
− ′′
So for u = egx, R = 6g2e 2gx and for u = e gx one has R = 6g2e2gx. Both
− −
Rs are x-dependent, and in u = e gx case, the scalar-curvature R and the
−
Kretschmann-invariantK =R Rµναβ =6g4e4gx diverge at x .
µναβ
→∞
8
The classical trajectories of particles in these backgroundsare also interest-
ing. For u=e gx, one can show that
±
1
x(t)=x(0) lncosh 1 v2g(t t ) ,
0
∓ g − −
(cid:20)q ⊥ (cid:21)
y(t)=y(0)+v t, (52)
0y
z(t)=z(0)+v t,
0z
in which v and v are arbitrary constants and v2 = v2 + v2 . The x-
0y 0z 0y 0z
component of velocity is ⊥
v (t)= 1 v2 tanh 1 v2g(t t ) . (53)
x 0
∓ − − −
q ⊥ (cid:20)q ⊥ (cid:21)
Note that in~=c 1 unit, v2+v2 <1. Eq.(53) showsthatfor u=egx, v (t)
is always negative (≡the particxles fa⊥ll in ( x)-direction ) and in u= e gx cxase,
−
−
v (t)>0 and the particles fall in (+x)-direction ( towardthe singular region).
x
3 The ds2 = dt2 + u2(x)η dxidxj metric
ij
−
In this section we study the Klein-Gordon and Dirac particles in the presence
of the metric (4):
ds2 = dt2+u2(x) dx2+dy2+dz2 (4)
−
(cid:0) (cid:1)
.
3.1 The spin-0 particles
For metric (4), the Klein-Gordon equation (6) becomes
∂2 1 ∂2 ∂2 ∂2 u ∂
+ + + + ′ m2 ψ =0. (54)
−∂t2 u2 ∂x2 ∂y2 ∂z2 u3∂x − KG
(cid:20) (cid:18) (cid:19) (cid:21)
To make the above equation solvable, we assume that 1/u2 u/u3 or u(x) =
′
∝
e gx. Let us first consider u(x)=e gx case.
± −
Following the same steps of Sec.2.1, that is taking ψ (t,r) = exp( iEt+
KG
−
ip y+ip z)ψ (x), defining ψ (x)=egxφ(x) and X =(k/g)e gx, where
2 3 KG KG −
k2 =E2 m2, (55)
−
eq.(54) then reduces to
d2
X2 +X2 η2 φ(X)=0, (56)
dX2 − KG
(cid:18) (cid:19)
in which η = p/g (p2 = p2 +p2). The solutions of the above differential
KG 2 3
equation are √XJ (X) and √XY (X). µ is defined through
µ µ
1
µ= η2+ . (57)
4
r
9
The limitx correspondstoX 0. Since detg =e 3gx andY (X)
µν − µ
→∞ → ∼
1/Xµ (for µ = 0), the boundary condition (11) leads to detg ψ 2
6 p µν| KG| ∼
X2(1 µ). So for µ µ = 1, i.e. p p = (√3/2)g, both Bessel functions
− ≤ 0 ≤ 0 p
J and Y are acceptable and the Klein-Gordon wavefunction in background
µ µ
metric (4) with u=e gx becomes
−
k k √3
ψ (x)=e(1/2)gx C J e gx +C Y e gx , p g . (58)
KG 1 µ − 2 µ −
(cid:20) (cid:18)g (cid:19) (cid:18)g (cid:19)(cid:21) ≤ 2 !
The eigenvalues are determined by eq.(9) which results in a continuous spec-
trum. Forµ>1,Y (X)doesnotsatisfytheboundarycondition(11). Therefore
µ
the wavefunction is
k √3
ψKG(x)=Ce(1/2)gxJµ e−gx , p> g , (59)
g 2
(cid:18) (cid:19) !
and its corresponding eigenvalues can be obtained by
k √3
J =0, p> g . (60)
µ
g 2
(cid:18) (cid:19) !
If we consider u=egx, instead of eq.(56), one obtains
d2
Y2 +Y2 η2 φ(Y)=0, (61)
dY2 − KG
(cid:18) (cid:19)
whereψ (x)=e gxφ(x)andY (k/g)egx. Againthesolutionsare√YJ (Y)
KG − µ
≡
and √YY (Y), but now the x limit corresponds to Y . Since the
µ
→ ∞ → ∞
asymptotic behaviors of J (Y) and Y (Y) in the limit Y are:
µ µ
→∞
2 π π
J (Y) cos Y µ ,
µ
→ πY − 2 − 4
r
(cid:16) (cid:17)
2 π π
Y (Y) sin Y µ , (62)
µ
→ πY − 2 − 4
r
(cid:16) (cid:17)
and detg = e3gx Y3, the boundary condition (11) does not satisfy. So
µν
∼
the Klein-Gordon equation has no solution in this case.
p
3.2 The spin-1/2 particles
ForDirac particlesinthe presenceofmetric (4), the procedureis similarto one
introduced in Sec.2.2, but with some differences. Here, instead of eq.(28), one
has ψ (x) = u 1ψ˜ (x), and instead of (30), it is O = iEuγ1+γ0d/dx
D − D 1
− −
muγ1γ0, which again [O ,O ] = 0. Finally one arrives at, instead of eq.(36),
1 2
the following equations for φ and φ :
1 2
dφ
1
=pφ (E+m)uφ ,
1 2
dx −
dφ
2
= pφ +(E m)uφ . (63)
2 1
dx − −
10