On nonstationary solutions to Yang–Mills equations Текст научной статьи по специальности «Математика»

UDC 530.182:537.813

On Nonstationary Solutions to Yang—Mills Equations

A. S. Rabinowitch

Moscow State University of Instrument Construction and Information Sciences 20, Stromynka str., Moscow, 107996, Russia

We study Yang-Mills fields with SU(2) symmetry generated by classical field sources. It is shown that in this case the Yang-Mills equations can be regarded as a reasonable nonlinear generalization of the equations of Maxwell's electrodynamics. We seek new classes of solutions to the examined Yang-Mills equations and find their nontrivial solutions in the case of nonstationary spherically symmetric sources and a wide class of their non-Abelian wave solutions.

Key words and phrases: Yang-Mills equations, SU(2) symmetry, classical field sources, nonstationary spherically symmetric solutions, non-Abelian wave solutions.

1. Introduction

Let us study the Yang-Mills equations with SU(2) symmetry. They can be represented as [1,2]

D^F= d^F+ geklmF ^ A™ = (4n/c) Jk'v, (1)

F= d^Ak'u - d"- gekimAl'^Am'u, (2)

where ¡i, v = 0, 1, 2, 3, k, I, m = 1, 2, 3, D^ is the Yang-Mills covariant derivative, A1''1, Fk''AW are potentials and strengths of a Yang-Mills field, respectively, Skim is the antisymmetric tensor, e123 = 1, g is the constant of electroweak interactions, Jk'w are three four-dimensional vectors of current densities, and d^ = d/dx^, where x^ are orthogonal space-time coordinates of the Minkowsky geometry.

Consider Eqs. (1)-(2) in the case of the following field sources:

J 1'v = r, J2'u = J3'u = 0, (3)

where Jv is a classical four-dimensional vector of current densities.

Then the Yang-Mills equations (1)-(2) have trivial solutions in which A2'v = A3'" = 0, F2'^u = F3'^u = 0 and the potentials A1'" and strengths F satisfy the Maxwell equations with the sources Jw. Besides, the expressions for the La-grangian and energy-momentum tensor of the Yang-Mills field are similar to those of the Maxwell field. That is why the considered Yang-Mills equations with the classical field sources (3) can be regarded as a reasonable nonlinear generalization of the Maxwell equations. This nonlinear theory was studied in our works [3-5], where several classes of exact solutions to Eqs. (1)-(3) were found. In our monograph [6] these solutions are applied to a number of anomalous phenomena that remain still unexplained within the framework of the linear Maxwell theory.

It should be noted that the Yang-Mills equations (1)-(2) with the field sources (3) are not independent. Namely, from (3) and the well-known identities for the Yang-Mills covariant derivative D^ [1,2] we have that when S1 = 1, 62 = 53 = 0,

4[D^Fk- (4n/c)Jk'»] = 0. (4)

That is why in Refs. [3-6] one more equation to the Yang-Mills equations (1)-(2) with the field sources of the form (3) was proposed to uniquely determine the field

Received 11th June, 2012.

Rabinowitch A. S. On Nonstationary Solutions to Yang-Mills Equations 275

strengths Fk,,JlW. This equation has the form

Ik>"h,„ = Jk'" Jk,u, Ik>" = Jk'" - (gc/4n)eklmFl>>"A™, (5)

where, as follows from Eq. (1), the components Ik,w satisfy the charge conservation equations duIk,w = 0 and can be regarded as four-dimensional densities of full currents which include not only source current densities Jk,w but also current densities of field virtual particles.

The additional equation (5) implies the conservation of the intrinsic energy in a small part of a field source when charged particles are created inside the source [3-6].

Using Eq. (1), we can represent Eq. (5) in the form

daF k,a» d? Fk^ = (4K/c)2Jk,» JktV. (6)

Further we will seek exact solutions to the Yang-Mills equations (1)-(2) with the field sources (3) that also satisfy the additional equation (6).

2. Nonstationary solutions to the Yang—Mills equations with spherically symmetric sources

Consider the Yang-Mills equations (1)-(2) with the following spherically symmetric sources:

(4n/c)J1,° = j °(r,r), (4n/c)J 1,n = xnj (T,r), n = 1,2,3, (7)

J2," = J3,» = 0, t = x°, r = V(x1)2 + (x2)2 + (x3)2,

where t = ct, t is time, and r is distance from the source center. Let us seek the field potentials Ak,w in the form

Ak,° = pk(T, r), Ak,n = xnak(T, r). (8)

Then from Eq. (2) we find

pk,°u = xnuk(T,r), Fk,in = 0, k,i,n = 1, 2, 3, (9)

where

uk = dak/dr + (1/r)dpk/dr + gekimalpm. (10)

Substituting expressions (7)-(9) into the Yang-Mills equations (1), we derive

r-duk/dr- + 3uk - gr2eklmulam = -j°8k, Si = 1, ¿2 = ¿3 = 0, (11)

duk/Or + geklmulpm = j5k. (12)

As is well-known, the Yang-Mills equations (1)-(2) have the following consequence [1,2]:

Du Jk," = 0. (13)

From (3) and (13) we find

dj °/dr + rdj/dr + 3j = 0, (14)

j°Pk - r2jak = 0,

(15)

where k = 2, 3. However, since J2'w = J3'u = 0, we can impose one gauge condition on the field potentials and choose the gauge so as to have Eq. (15) satisfied for k = 1. That is why we will further consider Eq. (15) fulfilled for k = 1,2, 3.

Let us now multiply Eq. (11) by j and Eq. (12) by j0 and then add the products. Then using (15), we derive

j0duk/dr + j (rduk/dr + 3uk ) = 0. Multiplying Eq. (11) by uk and summing over k, we find

3

^ uk (rduk/dr + 3uk ) = -j0u1.

k=1

Eq. (11) also gives two equations for ak.

Let us turn to Eq. (16). To solve it we introduce the function

r

q(r,r) = J r2j°(r,r) dr.

0

From (18) we find, using equality (14),

dq/dr = f r2 ^j— dr = — j r2 ( r ^- + 3j ) dr = —

dq/dr = r2j0.

From (19) and (20) we have

j0 dq/dr + rjdq/dr = 0.

Using now the equaliy (21), we obtain the following solution of Eq. (16):

uk = Pk (q)/r3,

where Pk(q) are arbitrary differentiable functions of the argument q. Indeed, from (21) and (22) we derive

1 d Pk

fduk/dr + j(r duk/dr + 3uk) = (j0 dq/dr + rjdq/dr) = 0.

rA d^

(16)

(17)

(18)

Jr2{rfr +3i)dr = — 1 dr = —r3j, (19)

(20) (21)

(22)

(23)

Hence, formula (22) gives solutions to Eq. (16). This formula describes general solutions to Eq. (16) since it contains three arbitrary differentiable functions Pk (q) and Eq. (16) presents three partial differential equations of the first order.

Substituting formula (22) into Eq. (17), we obtain

k=1

k dP^ dq d q dr

= —r 2fP \

(24)

7

Using formula (20), from (24) we find

3

^P kdP k/dq = -P1. (25)

k=1

Let us turn now to Eq. (6). From it and formulas (7) and (9) we derive

3

£[(rduk/Or + 3 uk)2 - r2(duk/Ot)2] = (j0)2 - r2j2. (26)

k=l

Substituting formula (22) into Eq. (26), we obtain

3

Y, (d Pk/dq)2 [(dq/dr)2 - (Oq/Or)2] = r4[(j0)2 - r2j2]. (27)

k=l

Using formulas (19) and (20), from Eq. (27) we find

3

^(dPfc /dq)2 = 1. (28)

k=l

Let us seek solutions to Eqs. (25) and (28) in the following form, taking into account that J2'w = J3'u = 0 and hence the equivalence of the second and third gauge axes:

P1 = -P cos P2 = P3 = -2-1/2P sin ^ P = P(q), £ = £(q). (29)

Then from (25) and (28) we derive

d P/d q = cos (d P/d q)2 + P2 (d£/dq)2 = 1. (30)

From these equations we obtain

dP = cos £dq, Pd£ = ± sin ^d^. (31)

Eqs. (31) give

d P/P = ± cot^. (32)

Integrating this equation and choosing the sign '+' to have no singularity at £ = 0, we find

P = K0 sin K0 = const. (33)

Substituting this formula into Eqs. (31), we obtain

d£/dq = 1/Ko, C = q/Ko + Ki, Ki = const. (34)

As follows from formulas (18), (22), and (29), q(0) = 0 and P(0) = 0. That is why we choose K1 = 0 in order to satisfy formula (33) atr = 0. Then from (33) and (34) we find

P = Ko sin(9/Ko), £ = q/Ko. (35)

Substituting these expressions for P and £ into formulas (29) and then (22) and (9), we come to the following formulas for the field strengths F:

Fi,n0 = K sin(%^) ^^, K = K0 = const, K 3 2

F2,u0 = F3,n0 = ^K j _ COs(^A^

(36)

Fk'in = 0, k,i,n = 1, 2, 3.

As follows from (7) and (18), the function q(r, r) presents the charge of the part of the field source situated in the spherical region of radius r at time t = t/c. From (36) we find

xn , f q(T, r)

F 1'n0 = qeff(t, r)Xg , qeff(t, r) = K si^^K^). (37)

Here qeff(r, r) can be regarded as an effective charge at the time t = t/c in the spherical region of the radius r which includes not only the source charge q(r, r) but also charged quanta of the Yang-Mills field.

The constant K should be considered as some sufficiently large charge. Then when |q/K| ^ 1 we have qeff = q and formula (37) describes the classical electric field. Therefore, this formula can be regarded as a nonlinear generalization of the classical formula in the cases of spherical sources with sufficiently large charges.

Formula (37) was applied in Refs. [3,6] to explain the phenomenon of ball lightning, where a relation between its maximum diameter and the constant K was found. Using the known estimate of the maximum diameter of the ball lightning which is about 100 cm [7], from this relation we obtain that the constant K ~ 107 coul.

It should be noted that in Ref. [8] a nonlinear model of the Earth ionosphere is proposed in which strong electric fields are taken into account and described by formula (37). Besides, as shown in [8], just the obtained esimate of the constant K ~ 107 coul provides good agreement of density distributions in the ionosphere computed by means of the proposed model with experimental data derived from artificial satellites.

Let us now study another class of exact solutions to the examined Yang-Mills equations.

3. Non-Abelian expanding waves

Consider the Yang-Mills equations (1)-(2) in the region outside field sources where

Jk'u = 0. (38)

Let us seek their wave solutions in the form

Ak'° = uk(yo, yi, V2, y3), Ak'n = ^rAk'°1 yo = x0 - r, yn = xn, (39)

k,n = 1, 2, 3, r = V(x1 )2 + (x2)2 + (x3)2,

where uk are some functions of the wave phase y0 = x°-r and of the spatial coordinates yn = xn.

Substituting these expressions for the potentials Ak'w into formula (2) for the strengths F, we readily find

Fk'0n = ^, Fk>™ = jf Vi^ _ yn , M,n = 1, 2, 3. (40) oyn r \ -yn -yi

As will be shown below, these field strengths satisfy Eq. (6) in the considered case (38).

Let us now substitute formulas (38)-(40) for Jk'", Ak>", and Fk>'lv into the Yang-Mills equations (1).

xn

Then when the index v = 0 we obtain

yi d2uk y,

WdJL - vi pL- + !l Vi^=o (41)

v dVi r dyooyi r dyi )

i= 1

and when the index v = n = 1, 2, 3 we derive after reductions

Dn ( d2uk d2uk yiduk ,dum\

— }. Vi^—ß.--r -K-2 + --9£ kimyiu

r i=1 \ dyodyi dy22 r dyi dyi J

n d u d u i d u

Vi^—ö--r +------ge kimyiu —— +

/A duk\

IS

+i=°(42)

It should be noted that Eqs. (41) and (42) can be represented in the form

d»Fk>»0 = g^ y-Skimul , d,F^n = g^ £ * Skmu1 ^ ■ (43)

d i d i

From (43) we readily find that the field strengths of the form (40) satisfy

Eq. (6) in the considered case (38). Let us denote

k ^ duk k ^d2uk

pk = £y^, = £-T^. (44)

i=i dyi dy

Then from (41) and (42) we find

1 / dp

(

^ = r\dP^ - 9£klmulpmj , r = y/y2 + y2^ + yi , (45)

yn[ - dpk - Qk + 4 - -ZkimulpA + ^ = 0, n = 1, 2, 3. (46)

d o 2 d n

As follows from (40) and (44), in the case pk = 0 the considered expanding waves are transverse. At the same time when pk = 0, these waves also have longitudinal components.

Let us substitute expression (45) for qk into Eqs. (46). Then we easily obtain

^ + = 0, n = 1, 2, 3. (47)

r2 dyn

As can be readily verified, these equations have the following solution:

pk = ^M, (48)

where sk are arbitrary differentiable functions of the argument y0. From (44), (45), and (48) we get

£ i-u = IM,

-yi r

^Q2uk 1 \dsk(yo) i mf '

x - -ge k imulsm( yo)

jL^t Qy.2 r2

i=1 %

dy 0

As can be readily verified, Eq. (49) has the following solution:

(50)

uk (yo,yu V2, V3) = - ^^ + fk (yo ,£i, 6,6), di = f, i = 1,2,3, (51)

where r = 1/y\ + y\ + y2 and fk are arbitrary differentiable functions. Actually, from (51) we derive

duk sk(y0)yi 1 dfk yi dfk duk sk(y0) , ^

— = ( 30) y +- - -T, Y, yi = ■ (52)

J yi r3 r J& r3 J£n - yi r

Consider Eq. (50) using formula (51). For the functions fk(y0,£i,£3) we have

JH = 1^ JH (<5..-£.£.) ¿=123 £• = m

-yi = r 2-/ J^. ^), 4 = 12,3, ^ = r,

Su = 1, 5ij = 0 when j = i,

QUk

Q y2

r2

j,n=\

Q 2fk

= Q^jdin

( Sa - tej )( in Ci^n)

-iv

Q/

k

fe(1 - + ]. (53)

T j=i Qlij

Let us substitute expression (51) for the functions uk into Eq. (50) and take into account that the function 1/r is harmonic. Then using (53) and the evident equality & + $ + d = 1, we obtain

3

£ =1

Q2jk

Qt 2

(1 - 42)~èr - 26

df

1 Q Ù

2 k

, =1

=

Q ÙQ

dsk(yo) dy o

-g£k irnf lsm( yo). (54)

The arguments = y%/r of the functions fk are not independent, since Ç2 + + Ç2 = 1. That is why instead of £ 1,£2, £3 we can choose two independent arguments. Let us choose the following two arguments 6 and a:

fk (y o,ii, 6,6) = hk( yo,e,a),

•=2*0+

(1+1 )

a = arctan —

(55)

Then we have

Qfk nQhk

—— = 0-,

Qi 1 0 Qd ■

Q2 k

Q

2 =0

Q k Qh k

—— = 7f3-,

3 Qa,

2 fQ^hk + Qhk

Qe2 1 Q6

Q k Qh k

—— = -7/£2—,

Q6 K 2 Qa ,

0 =

1 - £2;

& + £

) ■

2 k

Q£ 2

2 k

= 7 % ù

Q2 h ~d02

-2 2

Qhk

1

1

)

1

2 k

2

7 26 6

d 2hk dhkN HO! + 2e

2 k

-2

- 1 - 2

/37 6

d2hk dOda'

d2fk ddid£3

-37 6

d2hk d2fk

d6 da d &d 6

- 72 2 3

d2hk

dh k

8a* + «2 - ^

and as can be readily verified, the left-hand side of (54) acquires the form

(56)

£

=1

(1- £2) 2C dtk (1 4 i )d£i2 24 id£i

- E ^

ij = 1

=

d2fk id£j

1 d2hk 1 d2hk . .

+ ~~o o~ . (57)

1 - £2 de2 a + m da2

Since the variables ^ = yi/r satisfy the equality + £3, = 1 — £1, from (54), (55), and (57) we come to the following equation:

d2hk d^h^ 2 ~862 + ~d02 =( - ^

d sk (y0) d yo

-g £ k lm hl sm( yo)

€i = tanh 6. (58)

Let us put

hk = vk(yo, 0, a) + k(yo)sk(yo) ln(cosh (9) + dk(yo),

(59)

where vk(yo,d,a), k(yo), and dk(yo) are some functions.

Then substituting (59) into (58) and taking into account that ekim are antisymmetric, we get

d2 k d2

+

d 2 d a2

= (1 - tanh2 )

ds k (yo) d yo

- k(yo)sk(yo) - gekim (vl + dl(yo)) sm(yo)

. (60)

Let us require that the four functions k(yo) and dk(yo) should satisfy the following system of three algebraic equations which are linear with respect to them:

- k(yo)sk(yo) - gekimdl(yo)sm(yo) = 0.

(61)

dsk(yo) d yo

Then from (60) we get

d2 k d2 k

-¿T + = -9(1 - tanh2 d)£kimvlsm(yo), vk = vk(yo,9,a). (62)

After multiplying (61) by sk(yo) and summing it over the index k, we derive the following simple formula for the function k(yo):

K(yo) = ^^, (s(yo))2 = £ (,k (yo))2.

( o) d o

(63)

k=1

k

Let us turn to Eq. (62). We will seek its solution in the following form:

N

vk( yo, o, a) = R ^Vk(yo, 0) exp(-n(6 + ia)), (64)

n=o

where Vt(yo, 0) are some complex functions and n,N are nonnegative integers. Then

substituting (64) into Eq. (62), we come to the equations d2 V k d V k

-r^t - 2n-vn- = -g(1 - tanh2 d)ek^s™(yo). (65)

Let us choose the variable p = tanh 0 instead of 6 and put

Vt = Vt (y o,p), P = tanh d = a. (66)

Then Eq. (65) acquires the following form, since d p/d9 = 1 - tanh2 6 = 1 - p2: d2 V k d V k

ipt- 2<"+p> up

(1 -^2)—r - 2(n + ^ = -gtkirnVlsm(yo), M < 1. (67)

Putting

r] = (1 -p)/2, 0 1, -1 <p < 1, (68)

from (67) and (68) we get

d2 V k d V k

^ - 1) -frf - (n + 1 - 2v) - gEkirnVlsm(yo) = 0, Vt = Vt (yo, v). (69)

Let us seek solutions Vt(yo,v) to Eq. (69) in the form

Vt = Y,Xin(yoyrf, (70)

=o

where Xkt(yo) are some complex functions. Then substituting (70) into (69), we obtain the recurrence relation for X!l t, X!k t, ^k t,....

A^m = ^ + 1)^ £kl:Xjt°m(yo), j = 0,1, 2,..., (71)

j+i ,t (j + 1)(j + 1 + n) ' J ' ' ' ' v ;

^kt = ^kt( yo)

where the complex functions Ak t = Xk t(yo) may be assigned arbitrarily.

From (71) we can easily derive that the sequence \Xkt\ is bounded for any yo. Actually, let us denote

L(yo) = ,max lgskimsm(yo^ (72)

and consider (71) when j > L(yo) - 1 for an arbitrary yo. Then we find

max \ \k \ < /(j +J + L(yo\ max \\k\ < max \xk\ . (73)

j + 11 (j + 1)(j + 1 + n) jl jl

This formula precisely proves that that the sequence \ Xlk \ is bounded for any yo.

From (73) we also get that the values max^k^3 |Ak|, 0 < j < to, are bounded by their maximum when 0 ^ j ^ L(y0).

Consider the case in which the source of non-Abelian waves under examination is situated along the axis x1. Then for the waves outside the source we have —to < 6 < to and from (66) and (68) we find

0 <q< 1. (74)

Since, as shown above, the sequence | Ak | is bounded for any y0, the considered power series (70) is absolutely convergent when 0 < r] < 1. Therefore, in the case |6>| < to under eximination the functions Vnk(y0, r]) can be determined by formula (70). After that we can find the functions vk(y0,d,a) and hk(y0,d,a) using formulas (64) and (59). Then applying formulas (55) and (51) we determine the functions uk(yo, yi, y2, y3) = u (x0 v , x , x , x 3 ) describing non-Abelian expanding waves radiated from the considered source situated along the axis xi .

As indicated above, the considered wave solutions to the Yang-Mills equations can have longitudinal components when the functions pk of the form (48) are non-zero. This property of the found non-Abelian waves can be used to detect cosmic sources of Yang-Mills fields.

References

1. Ryder L. H. Quantum Field Theory. — Cambridge: Cambridge University Press, 1987.

2. Faddeev L. D., Slavnov A. A. Gauge Fields: Introduction to Quantum Theory. — London: Benjamin, 1990.

3. Rabinowitch A. S. On Nonlinear Electrodynamics with Yang-Mills Equations // Bulletin of PFUR, ser. Physics. — 2005. — Vol. 13, No 1. — Pp. 68-77.

4. Rabinowitch A. S. Yang-Mills Fields of Nonstationary Spherical Objects with Big Charges // Russian Journal of Mathematical Physics. — 2008. — Vol. 15, No 3. — Pp. 389-394.

5. Rabinowitch A. S. On a New Class of Non-Abelian Expanding Waves // Physics Letters B. — 2008. — Vol. 664, No 4-5. — Pp. 295-300.

6. Rabinowitch A. S. Nonlinear Physical Fields and Anomalous Phenomena. — New York: Nova Science Publishers, 2009.

7. Singer S. The Nature of Ball Lightning. — New York: Plenum, 1971.

8. Rabinowitch A. S., Abakumov S. Y. Nonlinear Model of the Earth Atmosphere with Regard to Rocket and Satellite Data // Bulletin of PFUR, ser. Mathematics, Information Sciences, Physics. — 2012. — No 3. — Pp. 44-52.

УДК 530.182:537.813

О нестационарных решениях уравнений Янга—Миллса

А. С. Рабинович

Московский государственный университет приборостроения и информатики Россия, 107996, Москва, ул. Стромынка, 20

Исследуются поля Янга—Миллса с Би(2) симметрией, создаваемые классическими полевыми источниками. Показывается, что в данном случае уравнения Янга—Миллса могут быть рассмотрены как естественное нелинейное обобщение уравнений максвеллов-ской электродинамики. Ищутся новые классы решений исследуемых уравнений Янга— Миллса и находятся их нетривиальные решения в случае нестационарных сферически-симметричных источников и широкий класс их неабелевых волновых решений.

Ключевые слова: уравнения Янга—Миллса, Би(2) симметрия, классические источники поля, нестационарные сферически-симметричные решения, неабелевые волновые решения.