Two Theorems by Helmholtz
H.Fleming
Instituto deFsia,UniversidadedeS~aoPaulo
Reebidoem22/01/2001. Aeitoem21/02/2001
Why isitthat divandurlropup everywhereinthestudyofvetorelds? These theoremsby
Helmholtzexplain.
Por que os operadores div erotapareem emtodaparte, noestudo deampos vetoriais? Estes
teoremasdeHelmholtzexpliam.
I Introdution
WhatisusuallyalledHelmholtztheorem[1℄isthefat
that avetoreld ~
V whihvanishesatinnityis
om-pletely determined by giving, everywhere, div ~
V and
url ~
V. Helmholtzarrivedat itin hisstudyofvorties
in uids[4℄. Nowadaysit is mostly used in
Eletrody-namis, where the veryMaxwell equations express its
importanebybeingnothingelsethanthespeiation
of what the div and url of theeletri and magneti
elds are.
II The First Theorem
Atually,theHelmholtztheorem 1
provesaslightly
dif-ferent thing, from whih the statement above follows
immediately: a vetor eld ~
V whih vanishes at the
boundaries an be written as the sum of two terms,
one of whih is irrotational and the other, solenoidal
(thatis,divergeneless)[2℄. Considerthefollowing
well-knownidentityforanarbitraryvetoreld ~
Z(~r):
~
r 2
~
Z= ~
r( ~
r : ~
Z)+ ~
r ~
r ~
Z : (1)
Ifwenowtakeourvetoreld tobe
~
V = ~
r 2
~
Z (2)
thenitfollowsthat
~
V = ~
rU+ ~
r ~
W (3)
with
U = ~
r: ~
Z (4)
and
~
W = ~
r ~
Z : (5)
Equation(3) is Helmholtz's theorem, as ~
rU is
irrota-tionaland ~
r ~
W issolenoidal.
But,is itgeneral? It assumesthat ourvetoreld
anbewrittenastheLaplaianofsomeotherone. This
onstitutes,however,noproblemif ~
V vanishesat
inn-ityfastenough,for,then, theequation
~
r 2
~
Z = ~
V ; (6)
whihisPoisson'sequation, hasalwaysthesolution
~
Z(~r)= 1
4 Z
d 3
~
r 0
~
V( ~
r 0
)
j~r ~
r 0
j
: (7)
It is now a simplematter to prove,from Eq.(3),that
~
V isdeterminedfromitsdiv andurl. Taking,infat,
thedivergeneofEq.(3),wehave
div ~
V = ~
r 2
U ; (8)
whihis,again,Poisson'sequation,and,so,determines
U as
U(~r)= 1
4 Z
d 3
~
r 0
~
r 0
: ~
V( ~
r 0
)
j~r ~
r 0
j
: (9)
Takenowtheurlof Eq.(3). Wehave
~
r ~
V =
~
r ~
r ~
W
= ~
r( ~
r : ~
W) ~
r 2
~
W : (10)
Now, ~
r: ~
W = 0, as ~
W = ~
r ~
Z, so another Poisson
equationdetermines ~
W. UsingU and ~
W sodetermined
inEq.(3)provesourontention.
III Appliations
Westart witha verysimpleone. Considera
homoge-neous,ohmianondutor,andlettheurrentdensity ~
j
vanishatthelosedsurfaewhihenirlesit. Wewant
toinquireastothepossibilitythatastationaryurrent
mayexist inthisondutor. Wethenhave ~
r: ~
j=0,as
theurrentissupposed to bestationary, and ~
j = ~
E,
whihis Ohm's law. Also, ~
r
~
E =0,for astati ~
E.
Butthen,
~
r ~
j= ~
r ~
E=0 (11)
andsowehaveboth ~
r : ~
j =0and ~
r ~
j =0. Besides,
~
j vanishes at the boundaries. It follows, then, from
Helmholtz'stheorem,that ~
j=0. So,nostationary
ur-rentanrunundertheseonditions. Notiethatthisis
truealsoforatorusanditsontinuousdeformations,so
thatitappliestoanylosediruit,provingthe
nees-sitythattheondition ~
j = ~
E bebrokensomewherein
theiruit(wheretheeletromotiveforeisloated).
Considernow the eletromagneti potentials. The
Maxwellequation
~
r : ~
B=0 (12)
statesthat thereexistsavetoreld ~
Asuhthat
~
B= ~
r ~
A; (13)
this ~
A being alled the vetor potential. We may
as-sume that ~
Avanishes atinnity. Now, what weknow
about ~
Aisjustitsurl. Therefore,byHelmholtz's
the-orem,wearefreetohoosethevalueofitsdivergene.
Forinstane, wemaytake ~
r: ~
A=0,determining
om-pletely ~
A. Thisistheso-alledCoulombgauge.Butwe
analsoput ~
r: ~
A= 1
t
,beingthesalarpotential.
ThisistheLorentz gauge.
IV The Seond Theorem
Thereis anotherimportantresultbyHelmholtzinthe
samepaperwementionedabove. Itonernsthemost
general innitesimal deformationof a plasti (that is,
non-rigid)body. 2
Consider a small volume element of a deformable
body (a uid, say). Putting the origin of the
oordi-nates insidethis volumeelement, let~rbetheposition
ofageneripointP. TheoriginisdenotedbyO. After
a deformation, the material point P has a new
posi-tionvetor,~r+~s. Also thepoint ofthe body loated
at the origin moved, its position now being given by
~s
0
. Weassume, as physially reasonable, that~s is, as
afuntionof position,ontinuousanddierentiableto
any order. In what follows, in order to get results of
enoughgenerality, weassume P verylose to O, that
is to say, we onsider an innitesimal volume of the
body. Notie that ~s itself does have to be small in
whatfollows,though,inonsideringdeformations,this
is somewhat aademi, as nite deformations an
al-waysbeobtainedfrominnitesimalones(forinstane,
byLie groupmethods). This notwithstanding,the
ge-ometrial interpretation we will get is only lear for
innitesimal~s. Ontheotherhand,theappliationfor
theeletrield whih isdonebelowwould be
impov-erishedbyonning~stobeinnitesimal.
ATaylorexpansionfor~sgives
~s(~r)=~s
0 +(~r:
~
r )~s (14)
negleting further orretions, what is allowed by the
restrition to innitesimal volume. Here we wrote~s
0
for~s( ~
0). Inmoredetail, ifs
i
isthe i-thomponentof
~
s,one has
s
i
(~r)=(s
0 )
i +
X
l x
l (
s
i
x
l )
~r=0
: (15)
Thisanbeabbreviatedto
s
i =(s
0 )
i +x
l
l s
i
(16)
or,
s
i =(s
0 )
i +x
l T
li
: (17)
where,obviously,
T
li =(
l s
i )
~ r=0
: (18)
In order to analysethis deformationin termsof more
basiones,letusdeomposeT
li
inthefollowingway:
T
li =
1
2 (T
li +T
il )+
1
2 (T
li T
il
) (19)
and onsider separately the symmetri and the
anti-symmetriparts. Theantisymmetripartis
1
2 (
l s
i
i s
l )=
1
2
lik
(url~s)
k
: (20)
The symmetri part is S
li
1
2 (T
li +T
il
). It an be
deomposed asfollows:
S
li =
1
3 Æ
li S+(S
li 1
3 Æ
li
S) (21)
whereSTr(S
li )=S
ii
:Now,S
li =
1
2 (
l s
i +
i s
l ),so
that
S =
i s
i =
~
r :~s: (22)
WeanthereforewriteEq.(21)as
S
li =
1
3 Æ
li ~
r:~s+S 0
li
; (23)
where S 0
li
is a traeless symmetritensor. Going now
toEq.(17),weanwrite
s
i =(s
0 )
i +x
l 1
2
lik
(url~s)
k +
1
3 x
i
div~s+x
l S
0
li (24)
2
or,
~ s=~s
0 +
1
2
(url~s~r)+ 1
3 ~
rdiv~s+~r:S 0
(25)
wherethelasttermisavetorwhosei-thomponentis
x
l S
0
li
. AnobjetlikeS 0
issometimesalled adyadi.It
is aseond-ordertensor. Letus examinethe meaning
of the several terms of Eq.(25). Therst term of the
seond member is atranslation. Theseond is an
in-nitesimalrotationaroundtheaxis(url~s)
~r=0
,andthe
remainingtermsdesribeadilatationofthevolume
ele-ment. Tounderstandthembetter,letussuppose
tem-porarily that the translation and the rotation vanish,
sothatwehave,foraomponentof~s:
s
i =
1
3 div~sx
i +x l S 0 li : (26) Now,S 0 li
isasymmetrimatrix,soit anbe
diagonal-ized. Thismeansthatweanhangeoordinateaxesin
suhawaythat,in thenewones,thematrixelements
of S 0
havetheform S 0 li =Æ li S 0 ii
, whereno sumis
im-pliedinthislastexpression. So,ifthenewoordinates
aredenoted byX
i
,wehave
X l x l S 0 li = X l X l ÆliS 0 ii =X i S 0 ii
nosumoni; (27)
sothat s i = 1 3 div~sX
i +S 0 ii X i
(nosum): (28)
Consider theinnitesimal \ube" entered at O, with
P as a vertex. (The \ube" may have urvilinear
edges, at least after the deformation). Its volume
before the deformation was, in the new oordinates,
V = X
1 X
2 X
3
. After the deformation, it is V 0 = (X 1 +s 1 )(X 2 +s 2 )(X 3 +s 3
). As we have,
onsider-ingonlythedilatation,
s
i =
1
3 div~sX
i +S 0 ii X i =X i ( 1 3
div~s+S 0 ii ) (29) then X i +s i =X i (1+ 1 3
div~s+S 0
ii )X
i (1+
i
) (30)
and, forthevolume,
V 0 =X 1 X 2 X 3 (1+ 1 )(1+ 2 )(1+ 3 ) (31)
whih,to rstorder is
V 0
=V(1+
1 + 2 + 3 ) (32) that is, V 0
=V(1+div~s+S 0 11 +S 0 22 +S 0 33
): (33)
But,asS 0
ij
is traeless,
V 0
=V(1+div~s): (34)
So,theS 0
ij
donotontributeto thehangeofvolume,
butdo ontributetothehangeofthe formofthe
lit-tle ube. Referring again to Eq.(25), we see that ~s
is written as a sum of a translation, plus a rotation,
plusan isotropi dilatation, plus a volume-onserving
deformation. 3
Notiethat Eq.(25)wasobtainedusingonly
Calu-lus. So,itshould applyto anyvetoreldwhatsoever.
Toshed morelightontherole ofeahterm ofthe
ex-pansion, let us use it for the eletri eld. We then
have:
~
E(~r)= ~ E( ~ 0)+ ~r 3 (div ~ E) 0 + 1 2 [(url ~ E) 0
~r℄+~r:S 0 : (35) As(div ~ E) 0 =4( ~
0),theseondtermreads
~r 3 (div ~ E) 0 = ~r 3 4( ~ 0)= 4 3 r 3 ( ~ 0) ~ r r 3 : (36)
Now,this is theeld ofauniformly harged sphereof
radiusrat apointofitssurfae. It isaradialeld to
beaddedto ~
E
0 .
The seond term vanishes if the elds are stati.
Otherwise, (url ~ E) 0 = 1 ( ~ B t ) 0 (37) and 1 2 [(url ~ E) 0 ~r℄=
1 2 t [ ~ B( ~
0)~r℄: (38)
Consider a uniform magneti eld of value ~
B( ~
0). A
vetorpotentialorrespondingtoitis
~
A(~r)= 1
2 ~
B( ~
0)~r (39)
sothatwehave
1 2 [(url ~ E) 0 ~r℄=
1 ~ A t : (40)
Thismeans that thethird term of Eq.(35)is the
on-tributionofthemagnetield attheorigin,treatedas
follows: extendthevalueof ~
B at ~
0to auniformeld.
Compute its vetor potential and then add the term
1 ~ A t to ~ E( ~
0). All therestof theeld at~r(andthat
ouldbealot!) omesfrom thetermS 0
.
3
Thisis,infat,thedeomposition,followingWeyl[3℄,ofthetensorT
li
intoitsirreduibleomponentsunderrotations. This
V Conlusion
TherstHelmholtztheoremexplainswhythediv and
urlaresoubiquitousin vetorelds. Italsoexplains
thestruture of Maxwellequations, usually written in
suhawaythattherstmemberofeahiseitheradiv
or aurl. The seond explains the meanings, and so
also the names, of those operators. Considering that
HermannvonHelmholtzgraduatedinMediine,notin
Physis,thatwas notabadjobatall!
I wishtothankWalterF.Wreszinskifor
enlighten-ingomments.
Referenes
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andMagnetism,NewYork,1970.
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[4℄ H.v. Helmholtz,
Uber Integrale der
hydrodynamis-hen Gleihungen, welhe den Wirbelbewegungen
entsprehen,Crelles J.55,25(1858).
[5℄ A.Sommerfeld,Mehanis of Deformable BodiesNew
York(1959).
[6℄ R. Aldrovandi, J.G. Pereira, An Introdution to
