Some properties of a $Δ-$string

7 Pages·0.14 MB·

by  V. Dzhunushaliev

Some properties of a ∆ string
−
V. Dzhunushaliev ∗
February 7, 2008
Dept. Phys. and Microel. Engineer., KRSU, Bishkek,
Kievskaya Str. 44, 720000, Kyrgyz Republic
Abstract
3 The properties of 5D gravitational flux tubes are considered. With the cross section and 5th dimension
0 in the Planck region such tubes can be considered as string-like objects, namely ∆−strings. A model of
0 attachment of ∆−string to a spacetime is offered. It is shown that the attachment point is a model of an
2 electric charge for an observer living in thespacetime. Magnetic charges are forbidden in this model.

n
a
1 Introduction
J

9
2 Recently [1] it was assumed that some kind of gravitationalflux tube solutions in the 5D Kaluza-Kleingravity
canbe consideredasstring-likeobjects,namely∆ strings. The idea is thatthe gravitationalflux tubes canbe

1 −
very thin (approximately l ) and arbitrary long. Evidently that such object looks like to a string attached
Pl
v ≈
totwoUniversesortoremotepartsofasingleUniverse. Thethicknessofthisobjectissosmallthatneartothe
8
1 point of attachment to an external Universe the handles of a spacetime foam will appear between this object
1 and the Universe. This is like a delta ofthe riverflowing into the sea. We callsuchobjects as a ∆ string. The
−
1 ∆ string has interesting properties : it can transfer electromagnetic waves and consequently the energy; the
0 mo−uth is similar to a spread charge; it is a realization of Einstein-Wheeler’s idea about “mass without mass”
3
and “charge without charge”; it has wormhole and string properties; in some sense it is some realization of
0
Einstein’s idea that all should be constructed from vacuum (in this case the ∆ string is the vacuum solution
/
c −
of 5D gravitational equations).
q
At first these gravitational flux tubes solutions were obtained as 4D Levi-Civita - Robertson - Bertotti
-
r solutions [2] in 4D Einstein-Maxwell theory. It is an infinite flux tube filled with parallel electric F and and
g 01
magnetic F fields
: 23
v
Xi ds2 = a2 cosh2ζdt2 dζ2 dθ2 sin2θdϕ2 , (1)
− − −
r F01 = ρ1/(cid:0)2cosα, F23 =ρ1/2sinα, (cid:1) (2)
a
where
G1/2aρ1/2 =1. (3)
α is an arbitrary constant angle; a and ρ are constants defined by Eq. (3); a determines the cross section of
the tube and ρ the magnitude of the electric and magnetic fields; G is Newton’s constant (c = 1, the speed of
light); F is the electromagnetic field tensor. For cosα=1 (sinα=1) one has a purely electric (or magnetic)
µν
field.
In Ref. [3] was obtained similar 5D solutions. The properties of these solutions depend on the relation
δ = 1 E/H between electric E and magnetic H fields. The purpose of this paper is to investigate more
EH
−
careful the properties of the gravitationalflux tubes with 0<δ 1.
EH
≪
2 Numerical calculations
In this section we will investigate the solutions of 5D Einstein equations
1
R η R=0 (4)
AB AB
− 2
here A,B are 5-bein indices; R and R are 5D Ricci tensor and the scalar curvature respectively; η =
AB AB
diag 1, 1, 1, 1, 1 . The investigated metric is
{ − − − − }
a(r) ∆(r)
ds2 = dt2 dr2 a(r) dθ2+sin2θdϕ2 e2ψ(r)(dχ+ω(r)dt+Qcosθdϕ)2 (5)
∆(r) − − − a(r)
∗E-mail: [email protected] (cid:0) (cid:1)
thefunctionsa(r),∆(r)andψ(r)aretheevenfunctions;Qisthemagneticcharge. Thecorrespondingequations
are
a ∆
′ ′
R15 =ω′′+ω′ +2 +3ψ′ = 0, (6)
−a ∆
(cid:18) (cid:19)
a aψ 2 Q2∆e2ψ
′′ ′ ′
R = + + = 0, (7)
33 a a − a a3
aψ Q2∆e2ψ
2 ′ ′
R R =ψ +ψ + = 0, (8)
11− 55 ′′ ′ a − 2a3
∆ ∆a ∆ψ 2 aψ
′′ ′ ′ ′ ′ ′ ′
2R +R R R = +3 + 6 = 0, (9)
11 22 33 55
− − ∆ − ∆a ∆ a − a
∆2 4 ∆2 Q2∆e2ψ aψ ∆a ∆ψ
R55−R11−R22+2R33 = ∆′2 + a − a2 e2ψω′2− a3 −6 ′a ′ −2 ∆′a′ +2 ∆′ ′ = 0. (10)
The solution of Maxwell equation (6) is
qae 3ψ
−
ω = (11)
′
∆2
here q is the electric charge. It is interesting that equations (6)-(9) have not any contribution from the electric
field (11). Neverthelessthe electric field hasaninfluence onthe solutiondue the equationfor initial values(10)
at the point r=0
q2+Q2 =4a(0) (12)
as a(0) = ∆(0) = ψ (0) = 0. The first analysis was done in Ref. [3] : the conclusion is that the solution
′ ′ ′
depends on the magnitude of the magnetic charge Q. It is found that the solutions to the metric in Eq. (5)
evolve in the following way :
1. Q=0. The solution is a finite flux tube located between two surfaces at r where ds2( r )=0. The
H H
± ±
throat between the r surfaces is filled with electric flux.
H
±
2. 0<Q<Q = 2a(0). The solution is again a finite flux tube. The throat between the surfaces at r
0 H
±
is filled with both electric and magnetic fields.
p
3. Q=Q . Inthiscasethesolutionisan infinite flux tube filledwithconstantelectricalandmagneticfields.
0
The cross-sectionalsize of this solution is constant (a=const.).
4. Q <Q<Q =2 a(0). In this case we have a singular finite flux tube located between two (+) and
0 max
(-) electrical and magnetic charges located at r . At r = r this solution has real singularities
p ± sing ± sing
which we interpret as the locations of the charges.
5. Q=Q . This solution is again a singular finite flux tube only with magnetic field filling the flux tube.
max
In this solution the two opposite magnetic charges are confined to a spacetime of fixed volume.
The results of numerical calculations of equations (6)-(9) are presented on Fig. 1, 2, 4 and 4. These numerical
calculations allow us to suppose that there is the following relation between functions a(r),∆(r) and ψ(r)
a(r)+∆(r)e2ψ(r) =2a(0). (13)
In the next sections we will demonstrate the correctness of this relation by some approximate analytical calcu-
lations.
8 1.2
6 0.8
a(r) 4 D(r) 0.4
2 0
0 -0.4
0 4 8 12 16 20 0 4 8 12 16 20
Figure 1: The function a(r). The parameterδ = Figure2: Thefunction∆(r). Theparameterδ =
Q Q
1 Q =10 10. 1 Q =10 10.
− Q0 − − Q0 −
16 2.2
12 (r)) 2.1
y(r) 8 y(r) exp(2 2
D+
4 a(r) 1.9
0 1.8
0 4 8 12 16 20 0 4 8 12 16 20
Figure3: Thefunctionψ(r). TheparameterδQ = Figure 4: The suma(r)+∆(r)e2ψ(r). The param-
1 Q =10 10. eter δ =1 Q =10 10.
− Q0 − Q − Q0 −
3 Approximate solution near to Q = 0 solution
In this section we will consider the metric in the form
a(r) = a (r)+δa(r), (14)
0
∆(r) = ∆ (r)+δ∆(r), (15)
0
ψ(r) = ψ (r)+δψ(r) (16)
0
and Q is small parameter Q2/a(0) 1.
≪
a (r) = r2+r2, (17)
0 0
∆ (r) = r2 r2, (18)
0 0−
ψ (r) = 0 (19)
0
the functions a (r),∆ (r) and ψ (r) are the solutions of Einstein equations with Q = 0. The variation of
0 0 0
equations (6)-(10) with respect to small perturbations δa(r),δ∆(r) and δψ(r) give us the following equations
set
a Q2∆
δψ+ ′0δψ 0 = 0, (20)
a ′− 2a3
0 0
Q2∆
0
δa +a δψ + = 0, (21)
′′ ′0 ′ a2
0
a ∆ δa ∆ a δ∆ ∆
δ∆ ′0δ∆ ′0 ′ + ′0 ′0δa+3∆ δψ +2 2 0δa = 0. (22)
′′− a ′− a a2 ′0 ′ a − a2
0 0 0 0 0
The solution of this system is
1 Q2
δψ(r) = , (23)
−4r2+r2
0
Q2 r r Q2
δa(r) = arctan + , (24)
− 2 r r 4
0 (cid:18) 0(cid:19)
Q2 r r r2 r2 1
δ∆(r) = arctan − 0 + . (25)
2 r r − r2+r2 2
(cid:20) 0 (cid:18) 0(cid:19) 0 (cid:21)
(26)
Substituting into equation (13) gives us
Q2
a(r)+∆(r)e2ψ(r) =2r2+ (27)
0 2
but
Q2 r r Q2
a(r) r2+r2 arctan + (28)
≈ 0 − 2 r r 4
0 (cid:18) 0(cid:19)
and
Q2
a(0) r2+ (29)
≈ 0 4
therefore
a(r)+∆(r)e2ψ(r) =2a(0) (30)
with an accuracy of Q2.
4 Expansion into a series
In this section we will derive the solution by expansion in terms of r. Using the MAPLE package we have
obtained the solution with accuracy of r12
1 5 1 7 41 1
ψ(x) = Q˜2x2+ Q˜4 Q˜2 x4+ Q˜4+ Q˜6+ Q˜2 x6
4 48 − 4 −30 720 4
(cid:18) (cid:19) (cid:18) (cid:19)
1 213 111 281
+ Q˜2+ Q˜4 Q˜6+ Q˜8 x8+
−4 560 − 560 8064
(cid:18) (cid:19)
341 3749 1553 5147 1
Q˜4+ Q˜6 Q˜8+ Q˜10+ Q˜2 x10+O x12 , (31)
−630 8400 − 9450 226800 4
(cid:18) (cid:19)
5 1 9 47 1 (cid:0) (cid:1)
∆(x) = 1 x2+ Q˜2 Q˜4 x4+ Q˜2+ Q˜4 Q˜6 x6+
− 6 − 4 −10 90 − 12
(cid:18) (cid:19) (cid:18) (cid:19)
6 373 3 13
Q˜4+ Q˜6 Q˜8+ Q˜2 x8+
−7 1260 − 80 14
(cid:18) (cid:19)
218 2893 5219 17 37
Q˜4 Q˜6+ Q˜8 Q˜2 Q˜10 x10+O x12 , (32)
175 − 4200 28350 − 18 − 1890
(cid:18) (cid:19)
1 1 1 11 (cid:0) (cid:1)
a(x) = 1+ 1 Q˜2 x2+ Q˜2x4+ Q˜2+ Q˜4 x6+
− 2 6 −10 180
(cid:18) (cid:19)(cid:20) (cid:18) (cid:19)
13 73 1
Q˜4+ Q˜6+ Q˜2 x8+
−140 2520 14
(cid:18) (cid:19)
239 887 1 13
Q˜4+ Q˜8 Q˜2 Q˜6 x10 +O x12 (33)
2100 56700 − 18 − 175
(cid:18) (cid:19) (cid:21)
(cid:0) (cid:1)
here we have introduced the following dimensionless parameters : x = r/ a(0) and Q˜ = Q/ a(0). The
substitution in eq. (13) shows that it is valid.
p p
One ofthe mostimportantquestionhere is : howlong is the ∆ string? Let us determine the lengthof the
−
∆ stringas2r wherer istheplacewhereds2( r )=0(atthesepoints∆( r )=0anda( r )=2a(0)).
H H H H H
− ± ± ±
This is very complicated problem because we have not the analytical solution and the numerical calculations
indicatethatrH dependsveryweaklyfromδQ−1 =(1−Q/Q0)−1 parameter: thebigmagnitudeofthisparameter
leads to the relative small magnitude of r . The expansion (33) allows us to assume that
H
a(r)=a(0)+a (r,Q)δ (34)
1
here δ =1 Q2/Q2; Q2 =2a(0). If Q Q = 2a(0) then
− 0 0 ≈ 0
p
a(r) a(0)+a (r,Q )δ. (35)
1 0
≈
Using the MAPLE package we have obtained a (r) with accuracy of r20
1
1 2 1 2
a (x)=x2+ x4+ x6+ x8+ x10+
1
3 45 315 14175
(36)
2 4 1 2
x12+ x14+ x16+ x18+O x20 .
467775 42567525 638512875 97692469875
(cid:0) (cid:1)
It is easy to see that this series coincides with cosh2(x) 1. Therefore we can assume that in the first rough
−
approximation
Q2 r
a(r) a(0) 1+ 1 cosh2 1 . (37)
≈ ( (cid:18) − Q20(cid:19)" a(0)!− #)
It allows us to estimate the length L=2r of the ∆ string frompequation (13) as
H
−
r
a(r )=a(0) 1+δ cosh2 H 1 =2a(0). (38)
H
( " a(0)!− #)
p
For r a(0) we have
H
≫
1
p L a(0)ln . (39)
≈ δ
p
One can compare this result with the numerical calculations : δ = 10 10, r / a(0) 15, 1ln1/δ 15;
− H ≈ 2 ≈
δ =10 9, r / a(0) 11, 1ln1/δ 10; δ =10 8, r / a(0) 10, 1ln1/δ 9. We see that r / a(0) and
− H ≈ 2 ≈ − H ≈ 2 ≈ p H
ln1/δ have the same order. This evaluation should be checked up (in future investigations) more carefully as
p p p
the convergence radius of our expansion (31)-(33) is unknown.
5 The model of the ∆ string
−
The numerical calculations presented on Fig’s (1)-(4) show that the ∆ string approximately can be presented
−
as a finite tube with the constant cross section and a big length joint at the ends with two short tubes which
have variable cross section, see Fig. (5).
1
3
3
2
Figure5: The approximatemodelofthe ∆ string. 1 isthe longtube whichis the partoftheQ=Q solution.
0
2 are two cones which are the Q=0 soluti−ons, 3 are two hypersurfaces where ds2 =0.
For the model of the central tube we take the part of the infinite flux tube solution with Q=Q [3]
0
Q2
a = 0 =const, (40)
∞ 2
a r
eψ = ∞ =cosh , (41)
∆ √a
r∞
ω =√2sinh (42)
√a
∞
here we have parallel electric E and magnetic H fields with equal electric q and magnetic Q charges.
For a model of two peripheral ends (cones) we take the solution with Q=0 (equations (17)-(19)). Here we
have only the electric field E - (11). At the ends of the ∆ string (r = r ) the function ∆( r )=0 and the
H H
− ± ±
term Q2∆e2ψ/a3 in equations (7)-(10) is zero that allows us to set this term as zero at the peripheral cones.
We have to join the components of metric on the r =r =r a(0) (here a(0)=a ).
1 H
− ∞
r a˜(r ) p a˜˜(r )
cosh2 1 = 1 =g˜ (r )=g˜˜ (r )= 1 (43)
a(0)! ∆˜(r1) tt 1 tt 1 ∆˜˜(r1)
here˜and˜˜mean that the correspondping quantities are the metric components for Q=Q and Q=0 solutions
0
respectively. According to equations (17)-(19)
a˜˜(r) = a(0)+(r r )2, (44)
1
−
∆˜˜(r) = α a(0) (r r )2 (45)
1
− −
a˜˜(r ) = a˜(hr )=a(0) i (46)
1 1
here r2 from equation (17)-(19) is replacedwith a˜˜(0) and some coefficient α is introduced as equations (7)-(10)
0
with Q=0 have the terms like ∆ /∆ and ∆/∆ only. It gives us
′′ ′
1 −2r1
α= 4e√a(0) (47)
≈
cosh2 r1
√a(0)
(cid:18) (cid:19)
since for the long ∆ string r / a(0) 1. The next joining is for
1
− ≫
1=p∆˜(r1)e2ψ˜(r1) =g˜ (r )=g˜˜ (r )= ∆˜˜(r1)e2ψ˜˜(r1) (48)
a˜(r1) 55 1 55 1 a˜˜(r1)
here the constant term 2ψ˜˜ is added to solution (19) 2ψ = 0 since again equations (7)-(10) have ψ and ψ
1 0 ′′ ′
terms only. Consequently
e2ψ˜˜1 = 1 e2r1/√a(0) or ψ r1 ln4. (49)
1
α ≈ 4 ≈ a(0) −
p
The last component of the metric is
a(0)=a =g˜ (r )=g˜˜ (r )=a˜˜(r ). (50)
θθ 1 θθ 1 1
∞
For the electric field we have
E˜(r ) = qeψ˜(r1) = q cosh r1 = q 1 (51)
1
a(0) a(0) a(0)! a(0)√α
E˜˜(r1) = q∆˜˜a˜˜2((rr1))e−3ψ˜˜(r1) = a(q0p)√1α (52)
1
and consequently
E˜(r )=E˜˜(r ). (53)
1 1
Thus our final result for this section is that at the first rough approximation the ∆ string looks like to a tube
−
with constant cross section and two cones attached to its ends.
6 ∆ string as a model of electric charge
−
In this section we want to present a model of attachment of the ∆ string to a spacetime.
−
Maybe the most important question for the ∆ string is : what will see an external observer living in the
−
Universe to which the ∆ string is attached. Whether he will see a dyon with electric and magnetic fields or
−
an electric charge ? This question is not very simple as it is not very clear what is it the electric and magnetic
fields on the ∆ string. Are they the tensor F and fields E and H
µν i i
−
F = ∂ A ∂ A , µ,ν =0,1,2,3, (54)
µν µ ν ν µ
−
E = F , i,j,k =1,2,3, (55)
i 0i
H = e Fjk (56)
i ijk
or something another that is like to an electric displacement D = εE and H = µB ? The answer on
i i i i
this question depends on the way how we will continue the solution behind the hypersurface r = r (where
H
±
ds2 =0). Wehavetwopossibilities: thefirstwayisthesimplecontinuationoursolutionto r >r ,thesecond
H
| |
way is a string approach- attachment the ∆ string to a spacetime. In the first case ∆(r)>0 by r <r and
H
− | |
∆(r)<0 by r >r it means that by r >r the time and 5th coordinate χ becomes respectively space-like
H H
| | | |
andtime-likedimensions. Itisnotsogoodforus. ThesecondapproachisliketoastringattachedtoaD-brane.
We believe that physically the second way is more interesting.
Let us consider Maxwell equations in the 5D Kaluza-Kleintheory
1
R = ∂ √ gg Fµν =0 (57)
5µ ν 55
√ g −
−
(cid:0) (cid:1)
hereg =detg =a4sin2θe2ψ. These5DMaxwellequationsaresimilartoMaxwellequationsinthecontinuous
AB
media where the factor √ gg is like to a permittivity for R equation and a permeability for R equations.
55 50 5i
−
It allows us offer the following conditions for matching the electric and magnetic fields at the attachment point
of ∆ string to the spacetime
−
√ g g F0i = √ gF0i (58)
− ∆ ∆55 ∆ −
√ g g Fij = √ gFij (59)
− ∆ ∆55 ∆ −
here the subscript ∆ means that this quantity is given on the ∆ string and on l.h.s there are the quantities
−
belonging to the ∆ string and on the r.h.s. the corresponding quantities belong to the spacetime to which we
−
want to attach the ∆ string. As an example one can join the ∆ string to the Reissner-Nordstro¨m solution
− −
with the metric
ds2 =e2νdt2 e 2νdr2 r2 dθ2+sin2θdϕ2 . (60)
−
− −
Then on the ∆ string (cid:0) (cid:1)
− ∆2e3ψ q
√−g∆g∆55F∆tr =− a ω′ =−aa (61)
For the Reissner-Nordstro¨msolution
√ gFtr = r2E . (62)
r(RN)
− −
Therefore after joining on the event horizon r =r (a=a ) we have
H H
q
E = (63)
r(RN) a
H
here we took into account that a =r2 .
H H
For equation (59) we have
Q∆e3ψ
√ g g Fθϕ = . (64)
− ∆ ∆55 ∆ a2
On the surface of joining ∆(r ) = 0 consequently we have very unexpected result : the continuation of Fθϕ
H
(and magnetic field H ) from the ∆ string to the spacetime (D-brane) gives us zero magnetic field H .
r r
−
Finally our result for this section is that the ∆ string can be attached to a spacetime by such a way that
−
it looks like to an electric charge but not magnetic one. Such approach to the geometrical interpretation of
electric charges suddenly explain why we do not observe magnetic charges in the world.
7 Discussion and conclusions
In this paper we have consideredsome interesting properties ofthe ∆ string. We have shownthat there is the
−
relation (13) between the metric components. Physically this relationshows that the cross section of ∆ string
−
is the same order on the interval ( r ,+r ). In fact we can choose the cross section as l that guarantees
H H Pl
− ≈
us that the cross section of ∆ string is in Planck region everywhere between points r where the ∆ string
H
− ± −
can be attached to D-branes.
It is shown that in some rough approximationthe length of ∆ string depends logarithmically from a small
−
number which is connected with the difference between 1 Q/Q .
0
−
Thecalculationspresentedheregiveusanadditionalcertaintythatthestringcanhaveaninnerstructureand
in gravity such string is the ∆ string. This approach is similar to Einstein point of view that the electron has
−
aninner structure whichcanbe describedinside of a gravitationaltheoryas a vacuum topologicallynon-trivial
solution.
Another interesting peculiarity is that the gravitational flux tube solutions appear in a relatively simple
matterintheclassical5Dgravity. Thisisincontrastwithquantumchromodynamicwherereceivingafluxtube
stretchedbetweenquarkandantiquarkisanunresolvedproblem. Inbothcasesstringsaresomeapproximation
for the tubes. It is possible that it is a manifestation of some nontrivial mapping between a classical gravity
and a quantum field theory.
It is necessary to note that similar idea was presented in Ref. [4] : the matching of two remote regions was
done using 4D infinite flux tube which is the Levi-Civita - Bertotti - Robinson solution filled with the electric
and magnetic fields.
8 Acknowledgment
I am very grateful to the ISTC grant KR-677 for the financial support.
References
[1] V.Dzhunushaliev,Class.Quant.Grav.,19,4817(2002);V.Dzhunushaliev,Phys.Lett.B.,553,289(2003);
V. Dzhunushaliev, “∆ string - a hybrid between wormhole and string”, gr-qc/0301046.
−
[2] T. Levi-Civita, Atti Acad. Naz. Lincei, 26, 519 (1917) ; B. Bertotti, Phys. Rev., 116, 1331 (1959); I.
Robinson, Bull. Akad. Pol., 7, 351 (1959).
[3] V. Dzhunushaliev and D. Singleton, Phys. Rev. D59, 064018(1999).
[4] E. I. Guendelman, Gen. Relat. Grav., 23, 1415 (1991).