Three function decomposition theorems in Clifford analysis with applications in electromagnetism
Ghislain R. Franssens
Belgian Institute for Space Aeronomy Ringlaan 3, B-1180 Brussels, Belgium
Abstract. Three theorems are presented which give the, generally non-unique, representation of ak-vector function in terms of ak−1-vector function and ak+1-vector function, in ultrahyperbolic Clifford analysis with general signature(p,q). When specialized to the case(p,q) = (1,3)andk=2, these theorems have a direct application in the domain of electromagnetism.
The specialized theorems characterize: (i) the decomposition of an electromagnetic field in terms of an electric field and a magnetic field, (ii) the representation of an electromagnetic field in terms of potential fields and (iii) the representation of an electromagnetic field in terms of source fields.
Keywords: Clifford analysis, Electromagnetism, Potentials, Sources PACS: 02.90.+p, 02.30.Em, 41.90.+e, 41.20.Jb.
INTRODUCTION
Letp,q∈N:np+q≥2 andΩ⊆Rp,q. We consider in ultrahyperbolic Clifford analysis, [1], [5], [6], the rep- resentation of a k-vector function F :Ω→Clkp,q, with 0<k<n, in terms of functions with neighboring grades k−1 andk+1, hence relative to a given 1-vector object.
Three theorems are presented, which state the represen- tation ofk-vector functionsF left relative to: (i) a given 1-vector fieldv, (ii) the Dirac operator∂and (iii) the right inverse (Teodorescu) operator∂−1of the Dirac operator.
The first theorem is purely algebraic and only uses the real Clifford algebraClp,q, [2], [3], while the other two make use of ultrahyperbolic Clifford analysis. A typical property of the considered representations is the non- uniqueness of the neighboring grade functions. This is a consequence of the fact that the sum of the number of components of the two neighboring grade functions is al- ways higher than the number of components of the given function, whenn>2.
When specializing to the case:(p,q) = (1,3)andk= 2, these theorems have a direct application in the domain of electromagnetism. LetF:Ω⊆R1,3→Cl12,3,S:Ω→ Cl11,3⊕Cl13,3 with suppS⊂Ω and ∂ ∈Cl11,3 the Dirac operator. In this case, the equation
∂F=S (1)
is the Clifford algebra model for electromagnetism, [10], [7], [11]. The specialized theorems then characterize: (i) the decomposition of an electromagnetic field in terms of an electric field and a magnetic field, (ii) the represen- tation of an electromagnetic field in terms of potential fields and (iii) the representation of an electromagnetic field in terms of source fields.
DECOMPOSITION WITH RESPECT TO A VECTOR FUNCTION
Theorem 1 (i) Any F∈
Ω,Clkp,q
, with0<k<n, can be represented, with respect to a v∈
Ω,Cl1p,q
:v2=0 inΩ, as
F=v∧E+v·B, (2) with E∈
Ω,Clk−p,q1
and B∈
Ω,Clk+p,q1 .
(ii) The representation (2) is not unique. The non- uniqueness in E and B is of the form
E+B
−(E+B) =v∧φ+v·Φ, (3) whereinφ∈
Ω,Clk−p,q2
, for2≤k, andΦ∈
Ω,Clk+p,q2 , for k≤n−2, are arbitrary. If k=1,φ=0or if k=n−1, Φ=0.
(iii) Iff
v·E+v∧B=0, (4) then F has the unique decomposition given by
E+B=v−1F. (5) Proof.(i) Sincev2=0, existv−1=v/v2such thatv−1v= 1=vv−1. Using associativity, we have
F =
vv−1 F,
= v v−1F
,
= v·
v−1·F+v−1∧F +v∧
v−1·F+v−1∧F ,
= v∧ v−1·F
+v·
v−1∧F .
Hererin isv−1·F of gradek−1 and v−1∧F of grade k+1. Hence, anyF:Ω→Clkp,q, with 0<k<n, can be represented in the form (2).
9th International Conference on Mathematical Problems in Engineering, Aerospace and Sciences
(ii) Applying (i) to gradesk−1 andk+1, anyE−E and anyB−Bcan be represented with respect tovas
E−E = v∧φ+v·ωE, B−B = v∧ωB+v·Φ,
withφ :Ω→Clk−2p,q, for 2≤k,ωE,ωB:Ω→Clkp,qand Φ:Ω →Clk+p,q2, for k≤n−2. If k=1, φ =0 or if k=n−1,Φ=0. Invariance ofFrequires that
0 = F−F,
= v∧ E−E
+v· B−B
,
= v∧(v∧φ+v·ωE) +v·(v·Φ+v∧ωB),
= v∧(v·ωE) +v·(v∧ωB),
= v·(v·ωE) +v∧(v·ωE) +v·(v∧ωB) +v∧(v∧ωB),
= v(v·ωE+v∧ωB).
Multiplying this equation on the left withv−1and using associativity, we find that invariance ofFrequires that
v·ωE+v∧ωB=0,
or, after splitting in gradesk−1 andk+1, that v·ωE = 0,
v∧ωB = 0.
Hence, any non-uniqueness inE andB, such that F remains invariant, can be represented as (2), whereinφ andΦare arbitrary.
(iii) The representation (2) can be written as v·E+F+v∧B=v(E+B). If (4) holds, then
v−1F = v−1(v(E+B)),
= v−1v
(E+B),
= E+B,
soFhas the unique decomposition given by (5).
IfFhas the unique decomposition given by (5), then v·E+F+v∧B = v(E+B),
= v v−1F
,
= vv−1
F,
= F, which implies (4).
The internal degree of freedom in the representation (2) can be called an (algebraic) gauge transformation.
One could call the particular gauge (4) the orthogonal
or uniqueness gauge. Notice that iff n=2, then the representation (2) is unique.
From Theorem 1 follows that any functionE and any functionBgiven by
E = v∧φ+v−1·F, (6) B = v−1∧F+v·Φ, (7) and whereinφ andΦ are arbitrary, reproduce a given functionFby the representation (2).
Let E,B andE,B be two neighboring grade fields satisfying the conditions (4). Then, sincev2=0,
0 = v·E−v·E
= v·(v∧φ) =v(v∧φ)
⇔ v∧φ=0, 0 = v∧B−v∧B
= v∧(v·Φ) =v(v·Φ)
⇔ v·Φ=0,
which, after substitution in (3), shows that E = E,
B = B.
When specializing to(p,q) = (1,3)andk=2, The- orem 1 gives the decomposition of an electromagnetic fieldF in terms of an electric fieldE and magnetic field B, typically relative to the time-like unit vectorv( i.e., v2=1) tangent to a sub-luminal observer’s world line.
For an observer in the luminal frame (e.g., a photon) is v2=0 and then v−1does not exist. Hence, Theorem 1 shows that a decomposition of an electromagnetic field F in an electric fieldE and a magnetic fieldBonly ap- plies to non-luminal observers. One can show that for observers in a super-luminal frame the roles of electric and magnetic fields are reversed.
PRELIMINARIES FROM CLIFFORD ANALYSIS
LetΣ⊂Ω⊆Rp,qbe a compact simply connected domain (satisfying a technical condition, see [6]),n:δΣ→Cl1p,q the outward normal function on the coherently oriented boundaryδΣofΣ,Cx∈D⊗Cl1p,q, ∀x∈Ω, a Cauchy kernel for the Dirac operator ∂, |δΣ restriction to δΣ and,S Schwartz pairing between a continuous linear functional and a continuous function, [8], [9], [4].
Definition 2 The (Teodorescu) volume operator ∂−1: Ω,C0⊗Clp,q
→
Ω,C1⊗Clp,q
is defined as
−∂−1Cx,...S,∀x∈Ω. (8)
Definition 3 The (Cauchy-type) boundary operatorβ : Ω,C1⊗Clp,q
→
Ω,C1⊗Clp,q
is defined as β
Cx|δΣ,n...|δΣ
S,∀x∈Ω. (9) First we need the following generalized Cauchy inte- gral theorem from ultrahyperbolic Clifford analysis, [6].
Theorem 4 For any F:Ω→C1⊗Clp,q:∂F=S with S:Ω→Cc0⊗Clp,qandsuppS⊂Σholds inΣthat
F=∂−1S+βF. (10) From Theorem 4 follows
∂∂−1 = Id and∂β =0, (11)
∂−1∂ = Id−β. (12) Hence,∂−1is a right inverse of the Dirac operator∂.
REPRESENTATION WITH RESPECT TO THE DIRAC OPERATOR
Theorem 5 (i) Any F∈
Ω,Cc0⊗Clkp,q
, with0<k<n, can be represented inΣ, with respect to∂, as
F=∂∧A+∂·B, (13) with A∈
Ω,C1⊗Clk−p,q1
and B∈
Ω,C1⊗Clk+p,q1 . (ii) The representation (13) is not unique. The non- uniqueness in A and B is of the form
A+B
−(A+B) =∂∧φ+∂·ωA+∂∧ωB+∂·Φ, (14) wherein φ ∈
Ω,C1⊗Clk−2p,q
, for 2 ≤ k, and Φ∈
Ω,C1⊗Clk+p,q2
, for k≤n−2, are arbitrary and ωA,ωB∈
Ω,C2⊗Clkp,q
satisfy∂(∂·ωA+∂∧ωB) =0.
If k=1,φ=0or if k=n−1,Φ=0.
(iii) Iff
∂·A+∂∧B=0, (15) then the representation (13) is unique, with A and B given by
A+B=∂−1F+β(A+B). (16) Proof.(i) Using Theorem 4 (withFhere playing the role ofS), we have
F = ∂
∂−1F ,
= ∂·
∂−1·F+∂−1∧F +∂∧
∂−1·F+∂−1∧F ,
= ∂·
∂−1·F +∂·
∂−1∧F +∂∧
∂−1·F +∂∧
∂−1∧F .
Equating equal grades yields 0 = ∂·
∂−1·F , F = ∂∧
∂−1·F +∂·
∂−1∧F , 0 = ∂∧
∂−1∧F .
Herein is∂−1·F of grade k−1 and ∂−1∧F of grade k+1. Hence, anyF:Ω→C0⊗Clkp,q, with 0<k<n, can be represented in the form (13).
(ii) Applying (i) to gradesk−1 andk+1, anyA−A and anyB−Bcan be represented, with respect to∂, as
A−A = ∂∧φ+∂·ωA, B−B = ∂∧ωB+∂·Φ,
with φ : Ω→C1⊗Clk−p,q2, for 2≤k, ωA,ωB :Ω → C2⊗Clkp,q andΦ:Ω→C1⊗Clk+2p,q , fork≤n−2. If k=1, φ =0 or if k=n−1, Φ=0. Invariance of F requires that
0 = F−F,
= ∂∧ A−A
+∂· B−B
,
= ∂∧(∂∧φ+∂·ωA) +∂·(∂·Φ+∂∧ωB),
= ∂∧(∂·ωA) +∂·(∂∧ωB),
= ∂·(∂·ωA) +∂∧(∂·ωA) +∂·(∂∧ωB) +∂∧(∂∧ωB),
= ∂(∂·ωA+∂∧ωB).
Clearly,φ andΦ are arbitrary and the invariance ofF requires that∂(∂·ωA+∂∧ωB) =0.
(iii) The representation (13) is equivalent with
∂·A+∂∧B+F=∂(A+B). (17) If (15) holds, then (17) becomes
∂(A+B) =F.
We can solve forA+B, using Theorem 4, and get inΣ, A+B=∂−1F+β(A+B),
andFhas the unique representation (13) withA+Bgiven by (16).
If (16) holds inΣ, then by operating on the left of (16) with∂ we get
∂(A+B) =F. Then, (17) becomes
∂·A+∂∧B+F = ∂(A+B),
= F, so (15) is necessary.
The internal degree of freedom in the representation (13), and characterized by (14), is called a gauge trans- formation in Physics. Notice that, without the bound- ary condition for the potentialsβ(A+B) =0, the multi- vector potentialA+Bis defined up to a left holomorphic multi-vector function.
From Theorem 5 also follows that the non-unique k−1-vector potentialAand the non-uniquek+1-vector potentialB, which reproduce a given functionF by the representation (13), is given by
A = ∂∧φ+∂·ωA+∂−1·F, (18) B = ∂∧ωB+∂·Φ+∂−1∧F, (19) and whereinφandΦare arbitrary and∂·ωA+∂∧ωBis left holomorphic.
Let A,B and A,B be two neighboring grade fields satisfying the conditions (15). Then, from (14) follows
0 = ∂·A−∂·A
= ∂·(∂∧φ) =∂(∂∧φ)
⇔ ∂∧φ=0, 0 = ∂∧B−∂∧B
= ∂∧(∂·Φ) =∂(∂·Φ)
⇔ ∂·Φ=0,
which, after substitution in (14) again, shows that A−A = ∂·ωA,
B−B = ∂∧ωB. Adding yields
A+B
−(A+B) =∂·ωA+∂∧ωB,
which is a left holomorphic multi-vector function. Iff A,BandA,Bin addition satisfy the boundary condition β(A+B) =0, holds that
0 = β
A+B
−β(A+B)
= β(∂·ωA+∂∧ωB).
This result combined with the fact that∂·ωA+∂∧ωBis left holomorphic gives
∂·ωA+∂∧ωB=0,
and then we have uniqueness of the representation (13).
Theorem 5 generalizes Helmholtz’ theorem for three dimensional vector fields.
Theorem 5 is also somewhat related to Hodge’s de- composition theorem for differential forms. Hodge’s de- composition however is formulated with respect to a globalL2-inner product and requires that the differential form coefficient functions either have compact support
or belong to a suitable Sobolev space, so that the bound- ary contribution (at infinity) is zero. Theorem 5 differs from Hodge’s decomposition in the sense that (i) it is not a decomposition in orthogonal complements with repect to a global inner product, (ii) it is formulated over a com- pact domainΣand uses the boundary contribution onδΣ and (iii) it holds in the (ultra-)hyperbolic setting, while Hodge’s theorem is restricted to elliptic complexes.
When specializing to(p,q) = (1,3)andk=2, The- orem 5 gives the representation of the electromagnetic field in terms of a 1-vector potentialAand a 3-vector po- tentialB.
REPRESENTATION WITH RESPECT TO THE RIGHT INVERSE DIRAC
OPERATOR
Theorem 6 (i) Any F∈
Ω,C1⊗Clkp,q
can be repre- sented inΣ, with respect to∂−1, as
F=∂−1∧J+∂−1·K+βF, (20) with J ∈
Ω,C0⊗Clk−p,q1
, K ∈
Ω,C0⊗Clk+p,q1 and supp(J+K)⊂Σ.
(ii) The representation (20) is not unique. The non- uniqueness in J and K is of the form
J+K
−(J+K) =∂−1∧φ+∂−1·Φ, (21) wherein φ ∈
Ω,C0⊗Clk−2p,q
, for 2 ≤k, and Φ ∈ Ω,C0⊗Clk+2p,q
, for k≤n−2, are arbitrary. If k=1, φ=0or if k=n−1,Φ=0.
(iii) Iff
∂−1·J+∂−1∧K=0, (22) then the representation (20) is unique, with J and K given by
J+K=∂F. (23)
Proof.(i) Using Theorem 4, we have F = ∂−1(∂F) +βF,
= ∂−1·(∂·F+∂∧F) +∂−1∧(∂·F+∂∧F) +βF,
= ∂−1·(∂·F)
+∂−1·(∂∧F) +∂−1∧(∂·F) +βF +∂−1∧(∂∧F).
Equating equal grades yields 0 = ∂−1·(∂·F),
F = ∂−1∧(∂·F) +∂−1·(∂∧F) +βF, 0 = ∂−1∧(∂∧F).
Hererin is∂−1·F of gradek−1 and∂−1∧F of grade k+1. Hence, anyF:Ω→C0⊗Clkp,q, with 0<k<n, can be represented in the form (20).
(ii) Applying (i) to gradesk−1 andk+1, anyJ−J andK−Kcan be represented, with respect to∂−1, as
J−J = ∂−1∧φ+∂−1·ωJ, K−K = ∂−1∧ωK+∂−1·Φ,
with φ :Ω →C0⊗Clk−2p,q , for 2≤k, ωJ,ωK : Ω → C0⊗Clkp,q andΦ:Ω→C0⊗Clk+p,q2, fork≤n−2. If k=1,φ=0 or ifk=n−1,Φ=0. The boundary terms β(J−J)andβ(K−K)are zero, since it is given that supp(J+K)⊂Σ. Invariance ofFrequires that
0 = F−F,
= ∂−1∧ J−J
+∂−1· K−K
+β F−F
,
= ∂−1∧
∂−1∧φ+∂−1·ωJ
+∂−1·
∂−1·Φ+∂−1∧ωK
+β
F−F ,
= ∂−1∧
∂−1·ωJ
+∂−1·
∂−1∧ωK
+β
F−F ,
= ∂−1·
∂−1·ωJ
+∂−1∧
∂−1·ωJ
+∂−1·
∂−1∧ωK
+∂−1∧
∂−1∧ωK
+β
F−F ,
= ∂−1
∂−1·ωJ+∂−1∧ωK
+β
F−F . Clearly,φ andΦ are arbitrary. Acting with ∂ from the left on this equation yields
0=∂−1·ωJ+∂−1∧ωK,
which is a necessary condition for F to be invariant.
Hence, the non-uniqueness inJandKis as stated in (21).
(iii) The representation (20) is equivalent with
∂−1·J+∂−1∧K+F=∂−1(J+K) +βF. (24) (iii.1) If (22) holds, (24) becomes
F=∂−1(J+K) +βF.
Acting with∂ from the left on this equation, we get J+K=∂F,
soFhas the unique representation (20), withJ+Kgiven by (23). Hence, (22) are sufficient conditions.
(iii.2) If (23) holds, we can solve forFusing Theorem 4, and get
∂−1(J+K) +βF=F. Then (24) becomes
∂−1·J+∂−1∧K+F = ∂−1(J+K) +βF,
= F,
so (22) is necessary.
The internal degree of freedom in the representation (20), and characterized by (21), could be called a source gauge transformation, to distinguish it from the usual potential gauge transformation.
Obviously, (20) takes the form of Cauchy’s integral representation for a functionF, which is holomorphic except at the compact support ofJ+K. Where Cauchy’s integral representation yieldsF for a given source func- tionJ+K, Theorem 6 yieldsJ+Kfor a given function F.
From Theorem 6 also follows that the non-unique k−1-vector sourceJ and the non-unique k+1-vector source K, which reproduce a given function F by the representation (20), is given by
J = ∂−1∧φ+∂·F, (25) K = ∂−1·Φ+∂∧F, (26) and whereinφandΦare arbitrary.
Let J,K and J,K be two neighboring grade fields satisfying the conditions (22). Then, from (21) follows
0 = ∂−1·J−∂−1·J
= ∂−1·
∂−1∧φ
=∂−1
∂−1∧φ
⇔ ∂−1∧φ=0, 0 = ∂−1∧K−∂−1∧K
= ∂−1∧
∂−1·Φ
=∂−1
∂−1·Φ
⇔ ∂−1·Φ=0,
which, after substitution in (21) again, shows that J = J,
K = K.
When specializing to(p,q) = (1,3)andk=2, The- orem 6 gives the representation of the electromagnetic fieldF in terms of a 1-vector electric monopole charge- current density J and a 3-vector magnetic monopole charge-current densityK. Theorem 6 shows that a given electromagnetic fieldF can in general be generated by different sources inΣ. To assure that the inverse source problem has a unique solution, additional conditions need to be imposed. These conditions are given by (22), which are the necessary and sufficient conditions under which the inverse source problem acquires a unique so- lution.
ACKNOWLEDGMENTS
The author acknowledges partial financial support from the Belgian Federal Science Policy Office (BELSPO) ALTIUS project.
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