Rev . Acad . Cana r. Ci~ n c . , I V ( núms . y 2) ,1 27 - 1 38 ( 1992)
THE D1A1 SYMMETRY ON THE CLASSICAL ELECTROMAGNETIC GAUGE.
APPLICATION TO SPHERICAL ANO STATIC BLACK HOLES
Juan Méndez-Rodríguez.
Oto. Informática y Sistemas.
Universidad de Las Palmas de Gran Canaria.
Po. Box 322, Las Palmas 35080, Spain.
Abstract: A gauge to electromagnetic field is considered. This gauge is most restrictive as the
Lorentz gauge which is verified by the proposed. lt is based on several symmetric expressions
narned DmSAn lorms. lt is included a application to a spherical and static geornetry as is
generated by a big gravitational object, as a black hale, with central symmetry and electrically
charged. The solution is the classica1 on far space but has strong discrepancy with the
classical result near ol the event horizon of the black hole.
Key words: General Relativity, Field Theory, Gravitation. Black Hale.
1. INTRODUCTION
On classical electrornagnetic field theory in 4-dimensional riemannian spaces, (4] (6] . the potential vector A is
defined from the electromagnetic field F as
F;¡=A¡;i~A ;j
From this definition can be concluded that the potential is not well defined. Several polential transforrnations
with field invariance are possible. A transformation of this type, called A-invariance. is valid:
A 1=A +'VA
These invariances are dueto the presence of several freedom degrees in the potential definition. To reduce
sorne freedom degrees sorne gauge conditions are imposed. The most common is know as Lorentz gauge,
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which is useful but do not fix the potential. In this paper all functions are assumed anatytic, based in this, a
identification of these freedom degrees is show. Let a local trame with cartesian coordinates. This trame is
locally plane and the Chistoffel symbols are nulls. Let the electromagnetic field F and potential A as:
F¡j(x)-4 ;¡+6;¡,,;x" +O(x2)
A;(x) "' «¡ +~¡fi +~Y¡j,¡xix" +O(x3)
Based on the antisymmetric property of the electromagnetic field, its coefficients mus! verifie:
Then, mus! be:
~¡¡- ~¡; 4¡¡ Y;¡11 - Yjik"'6jik
Let the following symmetric-antisymmetric decomposition:
Then must be:
And the potential is expressed as:
The potential in this local frame has three terms. The first is an arbitrary constant, the secorx::I is a term well
correlated with the field coefficients, arx::I the third, which has symmetric coefficients, can be arbitrarily defined.
From this is ctear that the antlsymmetric expression A¡:rA¡ :i is well defined, but the symmetric expression
A¡:¡+ A¡:i which do not includes the field coefficients, can be arbitrarily fixed:
This symmetric expresslon has 1 O components which are called in this paper the freedom degrees of the
potential definition. The Lorentz gauge partially defines these components. This gauge fix the addition of the
components in this way:
This work is related to a gauge based on the previous symmetric expression.
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The SAn tensor symmetries are a generalization of symmetric and antisymmetric tensors. Let all tensors on C.,.
Oelinltion 2.1 : A Tensor, X¡ 1 ¡2 ... in , has the Sn symmetry if all its components in a index cyclic rotatlon are
equal:
Oefinttion' 2.2: A Tensor, X¡ 1 ¡2 .. in , has the ~ symmetry if the addition ot all its components In a index cyclic
rotation is null:
From these definitions is founded that the second degree symmetric and antisymmetric tensors have the S2
and A2 symmetries respectively.
Oefinttion 2.3: The expression of an arbitrary tensor X as the sum of Sn and An tensors is callad its SAn Form:
X;, i2 .. .in"'S¡1 i2 .. .in+A¡1 i2 .. .in
Where S and A are Sn and An tensors, and are the Sn y~ parts of X respectively.
Theorem 2.1: The parts of a SAn form are univocaly determinad
Prool: Let X an arbitrary tensor where S and A are its Sn and An parts, as done on Definition 2.3. Adding a
cyclic index rotation:
X¡, i2 ... in +X¡z ... in ;1+ ... +X;n i l .. .in-1=(S;1 i2 ... in +S;z ... in if+ .. +S;n íl ... in-1)+ "4;1 i2 .. .in+A;2 .. .in it+ ... +A¡n i1...in-1 l ,.nSu i2 .. .in
can be obtained that:
X·, ·2 · +X·2 · ·1+ .+X;n il ... in-1
Aif i2 ... in=X¡¡ i2 ... in- 1 1 ... m 1 ... m ~ .
The parts are well defined trom X. •
Oefinition 2.4: An arbitrary tensor has the DmSn symmetry, if its m-th derivate tensor has the Sm +n symmetry.
Delinition 2.5: An arbitrary tensor has the DmAn symmetry, if its m-th derivate tensor has the ~+n symmetry.
Oefinition 2.6: The OmSAn Form of an arbitrary tensor is its expression as sum ot DmSn y DmAn tensors.
Theorem 2.2: Atl tensors with the D1S1 symmetry can be expressed as a scalar gradient. That is, if is verified:
Then:
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Proof: The 0 1$ 1 definition is exaclly the integrability condition for the next expression, (5], a particular case
of Poincare Lemma, {2) (3] :
dS(K')•Xpx1 •
Theorem 2.3: All tensors with the D1A1 symmetry have nutl divergence. That is, if is verified:
Then:
Proof: Reducing the i and j index in the D1A1 definition:
g ii(X¡;1+X¡;;) • 2X¡rO •
Lemma 2.1: lf a tensor X¡ and its D1SA1 parts are analytics, and is known one of the its D1A1 or 0 1$ 1 parts
in a point, then the D1SA1 parts are well defined.
Proof: Let a local frame centered at point p where a part is known. Then:
X,=X¡+X¡
aj,.a;x¡
a¡.,.a/¡=O
X1(p)·Z1
In this case is known the D1A1 part. Let:
Then. must be:
And the solution is:
X¡(x) =8.¡+b¡¡Xj +~C¡¡~ixk + O(x3)
X;(x)='a;+b;f¡ t~C;¡irix"+ O(x3)
X;(x) =Z;+6;¡X¡ +~é;¡irixk + O(x3)
b¡¡=bji C¡¡k=Cpk
Í>¡¡+6¡;=0 é¡¡k+é¡;k" º
ª;"'ii¡+Z¡ b;¡"'b;¡+6;¡ C¡¡k=C;¡k+C;¡k
X;(x):(a¡-Z;) t~(b;¡•b¡;)xi +~(c;¡k +c¡;k)xix" + O(x3)
X;(x) .,z;+~(b;¡-b¡;)xl +~(C;¡vcjik)xfxk + O(x3)
Ali the parts are well defined. •
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3. THE DmSA0 PARTS OF ELECTROMAGNETIC FIELD. KERNEL POTENTIAL
The electromagnetlc fleld F has the A2 and D1A2 symmetrles. lt Is verified that:
F¡¡+F¡;:•O
F;¡;k+F¡k;1•FkiJªO
The electromagnetic potential A has not these symmetries. Let the 0 1SA1 form of A as:
Where S is a scalar and Á is a D1A1 tensor called in this paper the kernef potential. lf in a physical problem the
potentlal has solutlon, then, basad on the Lemma 2.1 and sorne contour conditlons the kernel potential has also
solution. The primitiva relation between F and A is transformad on a relation between F and Á as follow:
F;¡><A¡;;-A¡JzÁj;l-Ái'J
Á¡J+Á¡;;=afr+-ªÍir2r k;fk=O
In the previously considerad local trame, the kernel potential and scalar S are:
[3.1)
(3.2]
Definilion 3.1: In this context, the freedom degrees of a tensor X¡ is the number of a¡X¡ components which can
be arbitrarily defined.
Theorem 3.1: The maximum freedom degrees of a tensor with 0 1A1 symmetry is 6.
Proof: The number of a¡X¡ is 16, but the D1A1 symmetrydefinition provides 10 linear equations about a¡X¡- Then
is concluded that only 6 components can be arbi1rarily defined. •
Theorem 3.2: 11 the electromagnetic field F is known, then the kernel potential has not freedom degrees. that
is, all the 3iÁ¡ are fixed.
Proof: The linear equations system (3.1] and (3.2] has 6 equations from the field definition and 10 equations
from the D1A1 definition. Then. there are 16 tinearly independent equations about the 16 3i.Á;- •
Theorem 3.3: lf in a Space the Cristoffel symbols are independents from a coordinate, -/", then the freedom
degrees of tensor X with 0 1A1 symmelry is reduced to 2.
Proof: lf the Cristoffel symbols are independents of a coordinate, -/", then the lineal equations system (3.2) has
this solution:
lt is possible to define 4 addi1ional equations related to the 3iX¡ as:
Ll i
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The more general solutlon Is:
Where Yu are the eigenvalues. Basad on Theorem 3. 1 an joining these 4 additional equatlons, then, only is
possible to arbitrarily define 2 components 3iX¡ . •
Theorem 3.4: lf in a Space the Cristoffel symbols are independer.ts from a coordinate, x"', then only 2
components of the electromagnetic field can be arbitrarily defined.
Proof: In this case the freedom degrees of the kernel potential is reduced to 2, dueto Theorem 3.3, and only
can be defined 2 equations including the electromagnetic field, because any more equations can be generate
an incompatible system. •
Theorem 3.5: lf in a Space the Cristoffel symbols are independents from two coordinates, x"' and x9, then can
not be defined any arbitrary component of the electromagnetic field.
Proof: In this case, there are 10 equations about the D1A1 symmetry definition and 8 equations about the
independent coordinates. lf this system with 18 equations is incompatible then a null solution is provided to the
kernel potentiat and electromagnetic field. lf the system is compatible, then a solution is obtained for the
electromagnetic field, and thls solution is the only solution. Any more compatible equations are lineal
combinations of the previous, and the solutions are the same. In this case the electromagnetic field is an
intrinsic property of the space. •
Based on these considerations is proposed this Lemma:
Lemma 3. 1: ff an electromagnetic field F is physically possible in a space, then there is solution for íts kernel
potential Á Thar is, there is solution to the equatíons sysrem {3.1/[3.2}.
This Lemma is called the 0 1A1 gauge in this paper.
4. ELECTROMAGNETIC FIELD BASED ON D1A1 GAUGE
A physical case with two independent coordinates exist in a spherical, static and charged black hote. The
solution to a D1A1 tensor fer this case is obtained in the Appendix.
Theorem 4. 1: In a Spherical and Static Geometry asymptotically plane, the asymptotically null solution to
electromagnetic field based on the 0 1A1 gauge is:
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The metric is [6):
Proot: Based on the Appendix results, if the metric is asymptotically plane n:ust be:
Lim«i (r) ;LimA(r) ;Q ,_ ,_
Then:
Lim dll>(r) •Lim dA(r) •O
,...... dr ,...... dr
lf the next condition is imposed:
LimF .. o
Then must be K2=K3: 0, and the solution obtained is:
From which is obtained the Theorem result •
The results provides by the classlcal electromagnetic field theory to the metric and field for a sphericat and
static geometry is known as the Reissner-Nordstrom solution (7)[8] which is:
2
e 2' (r);e -2h(r),.1-~+~
r ,2
This result are obtained from the Einstein equation G¡¡"' kT¡¡ , which determines both the metric an field. The
work here shown only provides the field if the metric is defined. The obtained solution based on Lemma 3.1
using the metric of classical solution with Ko=r,/rm is:
The complete solution with the classical metric and the field based on Lemma 3.1 is:
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This result has discrepancy with the exact solution provided by the classical theory. The discrepancy is weak,
about 0(1 ;r\ on the far space. Near the horizon of a spherical static and charged black hale the field
discrepancy is strong as is shown the Figure l ., where O:s:2lr8 l :s:rm. The solution is similar to a space
polarization in a dielectric with a relative permittivitie done as:
t \ E.(r)
\
\
E(r)
·-•.•.. __
Figure. 1. The metric. g00(r), and the field, Ec(r), solutions to Reissner-Nordstrom geometry.
E(r) is the solution based on Lemma 3.1.
5. CONCLUSION
A sludy about the freedom degrees in the computation of the electromagnetic potential in riemannian spaces
is presented. A proposal to reduce these freedom degrees is presented, based on several tensor symmetries
which define a new gauge type. For several space metrics types, this gauge determines the field from the
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intrinsic space properties. A study case is resolved, showing similarities and differences with the classsically
accepted solution. The study case concerning with a electrically charged black hole can not be successfutly
resolved basad only on the paper thesis, because the presented framework determines. far this case, the field
from the space metric, and In order to obtain the resolution of the problem is need to determine also the metric
from the field, that is performed by the classlcat theory.
6. APPENDIX: THE D1A1 TENSORS IN SPHERICAL ANO STATIC GEOMETRIES
A sotution to the next equations system is provided in a spherical and static 4-dimensionat riemannian
geometry:
x1r x;r a,x,.a;x¡-2r •,,x,.o
The metric with the (t,r,0,$) Schwarzschild coordinates is:
ds2=-e2t (r)dt2 +e2AVldr2+r2(de2+sin2ed$2)
Using this (· + + +) metric, the non null Cristoffel symbols are [4] 1:
r\lO"'e-2• ·2A ~ r°o1 "'~ r1 11 =~
r 122•-re -2A r323:C016 r 133•-8 -2Arsin26
r212 "r31 3=~
r2 33 .. - sinecose
(6.1(
(6.2(
The (6.1) equations system has 10 homogeneous differential equations and the Chistoffel symbols are
independents from the (t,$) coordinates, then the solutions have this form:
X¡(t,r,0,$):eªt•Uy;(r,0) (6.3)
From (6.1). [6.2] and (6.3] the next equations system is obtained:
a:Y0-r\lO(r)Y1 (6.4)
a,Y0+a:Y1 .. 2r 001 (r)Y0 (6.5)
a6Y0+a:Y2=0 (6.6)
PY0-.a:Y3"'o (6.7)
a,y1,.,r 111 (r)Y1 (6.8)
a6Y, +a,y2 .. 2r 212(r)Y2 (6.9)
\ andau·Lilshitz use a melric (+ • • ·).
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From [6.8} is obtained:
From [6.11} and [6.14J is deduced:
pY, +a,Y3=2I' 313(r)Y3
a8Y2 .. r \ 2(r)Y1
PY2+a6Y3 .. 2cotaY3
Y2(r,6) •-re-AVlz2(6) •U2(r) z;(6) •Z1(6)
From [6.9) and (6.15] is obtalned:
[6.10)
[6.11)
[6.12)
[6.13)
[6.14)
[6.15)
There are two possible solutions based on factoring variables in the previous equation. First solution is obtained
when factoring the r variable. lt is verifled that:
Where w is a constant. This solution is pos~ible only when the metric is restricted to a particular case. The
second solution, the only here considered, is obtained when factoring with the e variable. lt is verified:
The solution for this case must be:
Then the partial solution is:
Y1(r,6)•0
From [6.4), (6.5] and (6.6] is obtained :
Case1 : a =O
Y0(r,6) - Koe 21 VI
Case 2: ª'"º· From [6.4], [6.6], [6.7] and [6.16] is obtained :
Y0(r,0) • Y1(r,0)•Y2(r,6) •Y 3(r,0) ·O
There is a sotution only if « =0. In this case from [6.10] is obtained:
Y3(r,6) · r 2Z3(6) •U3(6)
From (6.13] is obtained:
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[6.16[
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p (r2Z3(0) •U3(0)) • - sin0cosex,r2
From the previous equatlon two cases are consldered :
Case 3: l} =O, in this case K2::0. From [6.10), [6.12] and (6.16) is obtained:
Y2(r,0)·0
Case 4: 1J "'º· must be:
u3(0) =O pZ3(0) --sinecos0K2
For lhis case, from [6.7], [6.12], [6.16] and [6.17) is deduced lhat:
Y0(r,0) · 0 P2•-1
The final solutions are:
Solution 1: ª"º:
X;(t,r,0,<!>)=0
Solution 2: ª"'º· 13 .. 0, 13 .. ± i
X;(l,r,0,<l>) ·O
Solution3: o: =O, 1} =±1
X0(t,r,0,<I>) · X1(t,r,0,<I>) =0 X2(t,r,0,<I>) =K,r2e;,
l(f+.!.)
X3(t,r,0,$) · K,t2sin0cos6e 2
Solution4: a =O, 1}=0
X0(t,r,0,<!>)=K,e21 Vl X1(t,r,0,<!>) · X2(t,r,0,<!>) · 0 X3(t,r,0,<l>)• K:f2sin20
The most general solution is:
X0(t,r,0,$):K0e2• (r)
X1(t,r ,0,<l>) •O
X2(t,r,0,<I>) · K,r2e•
7. ACKNOWLEDGEMENTS
[6.17]
The author like to express its acknowledgment to the Oepartment of Informática y Sistemas and to sorne of
its menbers, including the department leader R. Moreno-Dfaz, the research members A Falcón Martel and J.
Cabrera Gámez which provide suggestion and specially to F.M. Hernández Tejera far its paper corrections.
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8. REFERENCES
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(4) LANOAU LO. and LIFSHITZ E.M. (1962), The Classica/ Theoryof Field, Addison-Westey, ReaQing Mass.
[5} LEVl-CIVITA T. (1977), The Aosolute Differential Calculus, Dover Pub., New York.
[6J MISNER C.W., THORNE K.S. and WHEELEA J.A. (1973), Gravitation, Freeman & Co., New York.
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[8} REISSNER H. (1916), Uber die Eigengravitation de elektrischen Feldes nach Einsteinschen Theorie,
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(9] TONNELAT M.A. (1965), Les Théc-ries Uniraires de L'électro-magnétisme et de la Gravitation,
Guathier-Viflar Ed., Paris.
Rec i bido : 20 de Ene r o de ! 992
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