[PDF] Differential geometric formulation of Maxwells equations

Differential geometric formulation of Maxwells equations

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Dierential geometric formulation of

Maxwell's equations

Maris Ozols

January 16, 2012

Abstract

Maxwell's equations in the dierentialgeometricformulation are as follows:dF =dF = 0. The goal of these notes is to introduce the necessary notation and to derive these equations from the stan- dard dierential formulation. Only basic knowledge of linear algebra is assumed.

1 Introduction

Here are Maxwell's equations (in a charge-free vacuum) in their full glory:

8>>>>>>><

>>>>>>:@B x@t =@Ey@z @Ez@y @B y@t =@Ez@x @Ex@z @B z@t =@Ex@y @Ey@x ;@B x@x +@By@y +@Bz@z = 0;(1) 8 >>>>>>:@E x@t =@Bz@y @By@z @E y@t =@Bx@z @Bz@x @E z@t =@By@x @Bx@y ;@E x@x +@Ey@y +@Ez@z = 0:(2) HereEx(t;x;y;z) denotes the strength of the electric eld alongx-axis at timetand at point (x;y;z); similarly,Bx(t;x;y;z) denotes the strength of 1 the magnetic induction in the same direction and at the same time and same coordinates. It turns out that using a more modern notation we can rewrite the same equations in a very concise form: dF = 0; dF = 0:(3) These notes explain the meaning of these two expressions and why they are equivalent to Equations ( 1 ) and ( 2 ), respectively.

2 Maxwell's equations in the dierential form

LetE= (Ex;Ey;Ez) andB= (Bx;By;Bz) be vectors that represent the two elds. Then we can rewrite Equations ( 1 ) and ( 2 ) using vector notation: @B@t =r E;r B= 0;(4) @E@t =r B;r E= 0:(5) Note that these equations are invariant under the following substitution:

E7!B;B7! E:(6)

In these equationsr=@@x

;@@y ;@@z is a formal vector callednabla. The inner productandcross productwithrare dened as follows: r A=@Ax@x +@Ay@y +@Az@z ;(7) r A=@Az@y @Ay@z ;@Ax@z @Az@x ;@Ay@x @Ax@y :(8) These two operations can also be expressed using matrix multiplication: r A=@@x @@y @@z 0 @A x A y A z1 A ;(9) r A=0 B @0@@z @@y @@z 0@@x @@y @@x 01 C A0 @A x A y A z1 A :(10) 2

3 Dierential geometric formulation

3.1 Electromagnetic tensor

Let us combine the vectorsEandBinto a single matrix called theelectro- magnetic tensor: F=0 B

B@0ExEyEz

Ex0BzBy

EyBz0Bx

EzByBx01

C

CA:(11)

Note thatFis skew-symmetric and its upper right 13 block is the matrix corresponding to the inner product withEas in Equation (9); similarly, the lower right 33 block corresponds to the cross product withBas in

Equation (

10

3.2 Electromagnetic tensor as a2-form

We can label the rows and columns of matrixFby (t;x;y;z) and represent it as a 2-form, i.e., a formal linear combination of elementary 2-forms (each elementary 2-form represents one matrix element by the exterior product of the labels of the corresponding row and column). In particular, let

F = E + B (12)

where E and B are dened as follows:

E =Exdt^dx+Eydt^dy+Ezdt^dz;(13)

B =Bxdz^dy+Bydx^dz+Bzdy^dx:(14)

Here the 2-forms E and B encode those entries of matrixFthat correspond to the electric and magnetic eld, respectively. Note that the matrix representation of vectorsEandBin Equation (11) is redundant, since each entry appears twice (in particular,Fis skew- symmetric). However, Equations ( 13 ) and ( 14 ) only contain half of the o-diagonal entries ofF(those with positive signs); the remaining entries are represented implicitly, since the exterior product is anti-commutative (e.g.,dy^dz=dz^dy). 3

3.3 Hodge dual

Let us introduce an operation known asHodge starwhich establishes a duality betweenk-forms and (nk)-forms. Roughly speaking, it replaces exterior product ofkvariables by exterior product of the complementary set ofnkvariables (up to a constant factor, which depends on the metric tensor and the order of the variables in the two products). More precisely, let= (i1;i2;:::;in) be a permutation of (1;2;:::;n); then for anyk2 f0;1;:::;ngtheHodge dualof the corresponding elemen- taryk-form is dxi1^dxi2^^dxik) = sgn()"i1"i2:::"ikdxik+1^dxik+2^^dxin(15) where sgn() is the sign ofand ("1;"2;:::;"n)2 f+1;1gnis the signature of the metric tensor. Once dened for the standard basis, the Hodge dual is extended by linearity to the rest of the exterior algebra. 1 If we live in a Minkowski spacetime with signature \+" thenn= 4 and"t="x="y="z= 1. For example, we have:

1 =dt^dx^dy^dz;(dt^dx^dy^dz) = 1(1)3=1:(16)

In particular, \" isnotan involution (i.e., in generalf6=f). As an exercise, one can check that the Hodge duals of 1-forms are (dt) =dx^dy^dz;(17) (dx) =dt^dy^dz;(18) (dy) =dt^dx^dz;(19) (dz) =dt^dx^dy:(20) In fact, since the electromagnetic tensor is described by a 2-form, we are only interested in duals of 2-forms. The duals of the elementary 2-forms are summarized in these two columns of equations: (dt^dx) =dz^dy;(dz^dy) =dt^dx;(21) (dt^dy) =dx^dz;(dx^dz) =dt^dy;(22) (dt^dz) =dy^dx;(dy^dx) =dt^dz:(23)1 Note that the order of terms in the exterior product on the right-hand side of Equa- tion ( 15 ) can be chosen arbitrarily. However, the exterior product is anti-commutative, so this will be compensated by the sign of the permutation. Thus the denition of the

Hodge dual is consistent.

4 From this we immediately see that for anyv2R3it holds that

E(v) = B(v);B(v) =E(v);(24)

where E(v) and B(v) denote the 2-forms dened in Equations (13) and (14) with the coecients given by the components of vectorv. Notice that this is the same duality that we observed in Equation ( 6

3.4 Exterior derivative

Let us dene one more operation on the exterior algebra, known as the exterior derivative. It is dened onk-forms as d(f dxi1^dxi2^ ^dxik) =nX j=1@f@x jdxj^(dxi1^dxi2^ ^dxik) (25) and extended by linearity. Note that it mapsk-forms to (k+ 1)-forms.

Let us compute the derivative of E(v):

dE(v) =d(vxdx+vydy+vzdz)^dt(26) @vx@y dy+@vx@z dz ^dx @vy@x dx+@vy@z dz ^dy(27) @vz@x dx+@vz@y dy ^dz# ^dt:

After rearranging terms we get:

dE(v) =dt^" @vz@y @vy@z dz^dy @vx@z @vz@x dx^dz(28) @vy@x @vx@y dy^dx# =dt^B(r v):(29) 5

Similarly, for B we have:

dB(v) =d(vxdz^dy+vydx^dz+vzdy^dx)(30) @vx@t dt+@vx@x dx ^dz^dy @vy@t dt+@vy@y dy ^dx^dz(31) @vz@t dt+@vz@z dz ^dy^dx#

After rearranging terms we get:

dB(v) =dt^ @v x@t dz^dy+@vy@t dx^dz+@vz@t dy^dx! (32) @v x@x +@vy@y +@vz@z dx^dy^dz(33) =dt^B@v@t (r v)dx^dy^dz:(34)

3.5 Resulting equations

Let us verify thatdF =dF = 0 is equivalent to Equations (4) and (5).

First, let us computedF:

dF =dE(E) + B(B)(35) =dt^B(r E) +dt^B@B@t (r B)dx^dy^dz(36) =dt^B@B@t +r E (r B)dx^dy^dz:(37) By settingdF = 0 we recover Equation (4). Similarly, using Equation (24) we can computedF: dF =dB(E)E(B)(38) =dt^B@E@t (r E)dx^dy^dzdt^B(r B) (39) =dt^B@E@t r B (r E)dx^dy^dz:(40) By settingdF = 0 we recover Equation (5). ThusdF = 0 anddF = 0 are equivalent to Equations ( 4 ) and ( 5 ), respectively. 6quotesdbs_dbs47.pdfusesText_47