Naber G.L. The Geometry of Minkowski Spacetime. An Introduction to the Mathematics of the Special Theory of Relativity

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104 2 Skew-Symmetric Linear Transformations and Electromagnetic Fields

Next we claim that ker F  ker F

.kerF ⊆ ker F

is obvious. Now, suppose

ker F =kerF

, i.e., Fx =0iff F

x =0.Then

x = F

(Fx)=0 =⇒ Fx ∈ ker F

=kerF

=⇒ F(Fx )=0

=⇒ F

x =0

=⇒ Fx = 0 by assumption,

so F

x =0=⇒ Fx = 0 and we conclude that ker F

=kerF . Repeating the

argument gives ker F

=kerF .Butby(2.3.6), ker F

= M so ker F = M

and F is identically zero, contrary to hypothesis. Thus, ker F  ker F

and

we may select a nonzero v ∈Msuch that F

v = 0, but Fv =0.Thus,

Fv ∈ rng F ∩ ker F as required. 

Exercise 2.3.2 Show that if F : M→Mis a nonzero, null, skew-symmetric

linear transformation on M,thenF

v is null (perhaps 0) for every v ∈M.

Hint:Beginwith(2.3.6).

All t

hat

remains is to show that if F is null, then every 2-dimensional invari-

ant subspace U satisﬁes U∩U

⊥

= {0}.

Lemma 2.3.6 Let F : M→Mbe a nonzero, skew-symmetric linear trans-

formation on M. If there exists a 2-dimensional invariant subspace U for F

with U∩U

⊥

= {0},thenU

⊥

is also a 2-dimensional invariant subspace for

F and there exists a real number α such that F

u = αu for every u ∈U.

Proof: U

⊥

is a subspace of M (Exercise 1.1.2) and is invariant under F

(Proposition 2.2.2(b)). Notice that the restriction of the Lorentz inner prod-

uct to U cannot be degenerate since this would contradict U∩U

⊥

= {0}.

Thus, by Theorem 1.1.1, we may select an orthonormal basis {u

} for U.

Now, let x be an arbitrary element of M.Ifu

and u

are both spacelike,

then v = x −[(x ·u

+(x ·u

] ∈U

⊥

and x = v +[(x ·u

+(x ·u

]

so x ∈U+ U

⊥

Exercise 2.3.3 Argue similarly that if {u

} contains one spacelike and

one timelike vector, then any x ∈Mis in U + U

⊥

and explain why this is

the only remaining possibility for the basis {u

Since U∩U

⊥

= {0} we conclude that M = U

⊥

so dim U

⊥

=2.

Now we let {u

} be an orthonormal basis for U and write Fu

= au

and Fu

= cu

+ du

. Then, since neither u

nor u

is null, we have

0=Fu

· u

=(au

+ bu

) · u

= ±a so a = 0 and, similarly, d =0.

Thus, Fu

= bu

and Fu

= cu

,soF

= F (bu

)=bFu

= bcu

and

= bcu

.Letα = bc. Then, for any u = βu

+ γu

∈Uwe have

u = βF

+ γF

= β(αu

)+γ(αu

)=α(βu

+ γu

)=αu as required.



2.4 Canonical Forms 105

With this we can show that if F is null and nonzero and U is a 2-

dimensional invariant subspace for F ,thenU∩U

⊥

= {0}. Suppose, to the

contrary, that U∩U

⊥

= {0}. Lemma 2.3.6 implies the existence of an α ∈ R

such that F

u = αu for all u in U.Thus,F

u = F

u)=F

(αu)=

αF

u = α

u for all u ∈U.But,by(2.3.6), F

u =0forallu ∈Uso α =0

and F

=0onU. Again by Lemma 2.3.6 we may apply the same argument

to U

⊥

to obtain F

=0onU

⊥

.SinceU and U

⊥

are 2-dimensional and

U∩U

⊥

= {0}, M = U

⊥

so F

= 0 on all of M. But then, for every

u ∈M,F

u ·u =0soFu ·Fu = 0, i.e., rng F contains only null vectors. But

then dim(rng F ) = 1 and this contradicts Proposition 2.2.1(c) and we have

proved:

Theorem 2.3.7 Let F : M→Mbe a nonzero, skew-symmetric linear

transformation on M. If F is null, then F has 2-dimensional invariant sub-

spaces and every such subspace U satisﬁes U∩U

⊥

= {0}.

Combining this with Theorem 2.3.4 gives:

Corollary 2.3.8 Let F : M→Mbe a nonzero, skew-symmetric linear

transformation on M. Then F has 2-dimensional invariant subspaces and F

is regular i f f there exists such a subspace U such that U∩U

⊥

= {0} (so F is

null i f f U∩U

⊥

= {0} for every such subspace).

2.4 Canonical Forms

We now propose to use the results of the preceding section to prove that, for

any skew-symmetric linear transformation F : M→M, there exists a basis

for M relative to which the matrix of F has one of the two forms

⎡

⎢

⎣

0 δ 00

−δ 000

000

00 0

⎤

⎥

⎦

⎡

⎢

⎣

0 000

00α 0

0 −α 0 α

00α 0

⎤

⎥

⎦

depending on whether F is regular or null respectively. We begin with the

regular case.

Thus, we suppose F : M→Mis a nonzero, skew-symmetric linear trans-

formation and that (Corollary 2.3.8) there exists a 2-dimensional subspace U

of M which satisﬁes FU⊆Uand U∩U

⊥

= {0}. Then (Lemma 2.3.6) U

⊥

also a 2-dimensional invariant subspace for F and there exist real numbers α

and β such that

u = αu for all u ∈Uand (2.4.1)

v = βv for all v ∈U

⊥

. (2.4.2)

106 2 Skew-Symmetric Linear Transformations and Electromagnetic Fields

Since M = U

⊥

and U

⊥⊥

= U we may assume, without loss of generality,

that the restriction of the Lorentz inner product to U has index 1 and its

restriction to U

⊥

has index 0.

We claim now that α ≥ 0andβ ≤ 0. Indeed, dotting both sides of (2.4.1)

with

itself

giv

es F

u ·u = α(u ·u), or −Fu ·Fu = α(u ·u) for any u in U.Now

if u ∈Uis timelike, then Fu is spacelike or zero so u · u<0andFu · Fu ≥ 0

and this implies α ≥ 0. Thus, we may write α = 

with  ≥ 0so(2.4.1)

becomes

u = 

u for all u ∈U. (2.4.3)

Exercise 2.4.1 Show that, for some δ ≥ 0,

v = −δ

v for all v ∈U

⊥

. (2.4.4)

Now, select a future-directed unit timelike vector e

in U.ThenFe

spacelike or zero and in U so we may select a unit spacelike vector e

in U

with Fe

= ke

for some k ≥ 0. Observe that 

= F

= F (Fe

F (ke

)=kFe

.Thus,

kFe

· e

= 

·e

k(−e

·Fe

)=

(−1),

k(e

·(ke

)) = 

· e

= 

so k

= 

and k =  (since k ≥ 0and ≥ 0). Thus, we have

= e

. (2.4.5)

Notice that {e

} is an orthonormal basis for U.

Exercise 2.4.2 Show that, in addition,

= e

. (2.4.6)

Now, let e

be an arbitrary unit spacelike vector in U

⊥

.ThenFe

is space-

like or zero and in U

⊥

so we may select another unit spacelike vector e

⊥

and orthogonal to e

with Fe

= ke

for some k ≥ 0(ife

×e

·e

is −1

rather than 1, then relabel e

and e

Exercise 2.4.3 Show that k = δ.

Thus,

= δe

(2.4.7)

and, as for (2.4.6),

= −δe

. (2.4.8)

2.4 Canonical Forms 107

Now, {e

} is an orthonormal basis for M and any basis constructed

in this way is called a canonical basis for F.From(2.4.5)–(2.4.8) we ﬁnd that

the

ix of F relative to such a basis is

⎡

⎢

⎣

0 δ 00

−δ 000

000

00 0

⎤

⎥

⎦

. (2.4.9)

This is just the matrix of the F

deﬁned in Section 2.2. In a canonical basis

an observer measures electric and magnetic ﬁelds in the x

-direction and of

magnitudes  and δ respectively. We have shown that such a frame exists

for any regular F .Observethat|



−|



= δ

− 

and



= δ.

Since these two quantities are invariants, δ and  can be calculated from the

electric and magnetic 3-vectors in any frame. The canonical form (2.4.9)of

F is

icularly convenient for calculations. For example, the fourth power

of the matrix (2.4.9) is easily computed and found to be

⎡

⎢

⎣

000

0 δ

00

000

⎤

⎥

⎦

so that, unlike the null case, F

= 0. The eigenvalues of F are of some interest

and are also easy to calculate since the characteristic equation (2.3.4) becomes

+(δ

−

)λ

−δ



= 0 i.e., (λ

−

)(λ

+δ

) = 0 whose only real solutions

are λ = ±. The eigenspace corresponding to λ =  is obtained by solving

⎡

⎢

⎣

0 δ 00

−δ 000

000

00 0

⎤

⎥

⎦

⎡

⎢

⎣

⎤

⎥

⎦

⎡

⎢

⎣

v

⎤

⎥

⎦

i.e.,

⎡

⎢

⎣

δv

−δv

v

⎤

⎥

⎦

⎡

⎢

⎣

v

⎤

⎥

⎦

If  =0andδ =0,thenv

= v

=0,whereasv

and v

are arbitrary. Thus,

the eigenspace is Span{e

}. Similarly, if  =0andδ =0,v

= v

and v

= v

so the eigenspace is Span{e

+ e

}.Ifδ =0,δv

= v

and

−δv

= v

again imply v

= v

= 0; in addition, v

= v

so the eigenspace

is Span{e

+ e

}. In the ﬁrst case the eigenspace contains two independent

null directions (those of e

+ e

and e

− e

), whereas in the last two cases,

there is only one (e

+ e

). For λ = −, the result is obviously the same in

108 2 Skew-Symmetric Linear Transformations and Electromagnetic Fields

the ﬁrst case, while in the second and third the eigenspace is spanned by

− e

. The null directions corresponding to e

± e

are called the principal

null directions of F.

Now we turn to the case of a nonzero, null, skew-symmetric linear trans-

formation F : M→Mand construct an analogous “canonical basis”. Begin

with an arbitrary future-directed unit timelike vector e

in M.

Exercise 2.4.4 Show that Fe

is spacelike. Hint: Fe

=0wouldimply

∈ (rng F )

⊥

Thus, we may select a unit spacelike vector e

in M such that e

·e

=0and

= αe

(2.4.10)

for some α>0. Observe that e

= F





∈ rng F . Next we claim that

is a nonzero vector in rng F ∩ ker F . Fe

= 0 is clear since Fe

=0=⇒

∈ rng F ∩ker F , but e

is spacelike and this contradicts Proposition 2.3.5.

∈ rng F is obvious. Now, by Exercise 2.3.2, F

is either zero or null and

nonzero. F

= 0 implies F (Fe

)=0soFe

∈ ker F as required. Suppose,

on the other hand, that F

is null and nonzero. Fe

·F

= 0 implies that

is not timelike. Fe

·e

= 0 implies that Fe

is not spacelike since then rng

F would contain a null and two orthogonal spacelike vectors, contradicting

Proposition 2.3.5. Thus, Fe

is null and nonzero. But then {e

, Fe

} is a

basis for rng F and Fe

is orthogonal to both so Fe

∈ (rng F )

⊥

=kerF as

required.

Now we wish to choose a unit spacelike vector e

such that e

· e

=0,

·e

= 0 and Span{e

+ e

} =rngF ∩ker F . To see how this is done select

any null vector N spanning rng F ∩ker F such that N · e

= −1. Then let

= N − e

. It follows that e

·e

=(N −e

) ·(N −e

)=N ·N −2N ·e

·e

=0−2(−1)−1=1soe

is unit spacelike. Moreover, e

= N spans

rng F ∩ker F .Also,e

·e

=(N −e

) ·e

= N ·e

−e

·e

= −1 −(−1) = 0.

Finally, e

+ e

∈ (rng F)

⊥

implies 0 = (e

+ e

) ·e

= e

·e

+ e

·e

= e

·e

and the construction is complete. Now, there exists an α



> 0 such that

= α



+ e

). But α = e

· (αe

)=e

· Fe

= −e

· [α



+ e

)] =

−α



·e

+ e

·e

]=−α



[0 − 1] = α



= α(e

+ e

). (2.4.11)

Next we compute Fe

= F(N − e

)=FN − Fe

=0− αe

= −αe

. (2.4.12)

Finally, we select a unit spacelike vector e

which is orthogonal to e

and

and satisﬁes e

× e

·e

= 1 to obtain an admissible basis {e

}

a=1

Exercise 2.4.5 Show that

=0. (2.4.13)

Hint: Show that Fe

· e

=0fora =1, 2, 3, 4.

2.5 The Energy-Momentum Transformation 109

AbasisforM constructed in the manner just described is called a canon-

ical basis for F .ThematrixofF relative to such a basis (read oﬀ from

(2.4.10)–(2.4.13)) is

⎡

⎢

⎣

00α 0

0 −α 0 α

00α 0

⎤

⎥

⎦

(2.4

.14)

andiscalledacanonical form for F . This is, of course, just the matrix of the

transformation F

introduced in Section 2.2 and we now know that every

null F takes this form in some basis. An observer in the corresponding frame

sees electric



= αe

and magnetic



= αe

3-vectors that are perpendicular

and have the same magnitude α.

Exercise 2.4.6 Calculate the third power of the matrix (2.4.14)andim-

pro

(2.3.5) by showing

=0(F null). (2.4.15)

For any two vectors u and v in M deﬁne a linear transformation u∧v : M→

M by u ∧ v(x)=u(v · x) − v(u · x).

Exercise 2.4.7 Show that, if F is null, then, relative to a canonical basis

}

a=1

F = Fe

∧ e

. (2.4.16)

The only eigenvalue of a null F is, of course, λ =0.

Exercise 2.4.8 Show that, relative to a canonical basis {e

}

a=1

,the

eigenspace of F corresponding to λ = 0, i.e., ker F , is Span{e

+ e

}

and so contains precisely one null direction (which is called the principal null

direction of F ).

2.5 The Energy-Momentum Transformation

Let F : M→Mbe a nonzero, skew-symmetric linear transformation on M.

The linear transformation T : M→Mdeﬁned by

T =

4π

tr(F

)I −F

, (2.5.1)

where F

= F ◦ F, I is the identity transformation I(x)=x for every x

in M and tr(F

)isthetraceofF

, i.e., the sum of the diagonal entries

in the matrix of F

relative to any basis, is called the energy-momentum

110 2 Skew-Symmetric Linear Transformations and Electromagnetic Fields

transformation associated with F .ObservethatT is symmetric with respect

to the Lorentz inner product, i.e.,

Tx · y = x · Ty (2.5.2)

for all x and y in M.

Exercise 2.5.1 Prove (2.5.2).

Moreov

er, s

ince tr(I)=4,Tis trace-free, i.e.,

tr T =0. (2.5.3)

Relative to any admissible basis for M the matrix [T

]ofT has entries

given by

4π



− F



,a,b=1, 2, 3, 4. (2.5.4)

Although not immediately apparent from the deﬁnition, T contains all of

the information relevant to describing the classical “energy” and “momen-

tum” content of the electromagnetic ﬁeld represented by F in each admissible

frame. To see this we need the matrix of T in terms of the electric and mag-

netic 3-vectors



and



Exercise 2.5.2 With the matrix of F relative to {e

} written in the form

(2.2.3), calculate the matrix of F

relative to {e

} and show that it can be

written as

⎡

⎢

⎣

)

− (B

)

− (B

)

+ B

− E

+ B

)

− (B

)

− (B

)

+ B

− E

+ B

)

− (B

)

− (B

)

− E



⎤

⎥

⎦

(

2.5.5)

Now,

4π

times the oﬀ-diagonal entries in (2.5.5) are the oﬀ-diagonal entries

in [T

]. Adding the diagonal entries in (2.5.5)givestr(F

)=2(|



−|



)

tr(F

((E

)

+(E

)

+(E

)

−(B

)

−(B

)

−(B

)

). Subtracting

the diagonal entries in (2.5.5) from the corresponding diagonal entries in

tr(F

)I gives 4π times the diagonal entries in [T

]. Thus,

8π

[−(E

)

+(E

)

+(E

)

− (B

)

+(B

)

+(B

)

8π

[(E

)

− (E

)

+(E

)

+(B

)

− (B

)

+(B

)

8π

[(E

)

+(E

)

− (E

)

+(B

)

+(B

)

− (B

)

], (2.5.6)

= −

8π





+ |





Notice once again that the nonzero index of the Lorentz inner product has

the unfortunate consequence that the matrix of a symmetric linear transfor-

mation on M is not (quite) a symmetric matrix.

2.5 The Energy-Momentum Transformation 111

In classical electromagnetic theory the quantity

8π



+ |



]



= −T



is called the energy density measured in the given frame of reference for the

electromagnetic ﬁeld with electric and magnetic 3-vectors



and



.The

3-vector

4π



=(E

− E

+(E

− E

+(E

−

= T

+ T

= −



+ T



is called

the Poynting 3-vector and describes the energy density ﬂux of the ﬁeld. Fi-

nally, the 3 × 3matrix





i,j=1,2,3

is known as the Maxwell stress tensor

of the ﬁeld in the given frame. Thus, the entries in the matrix of T relative

to an admissible basis all have something to say about the energy content of

the ﬁeld F measured in the corresponding frame.

Notice that the (4, 4)-entry in the matrix [T

]ofT relative to {e

} is

= −Te

·e

= −

8π



]. Thus, we deﬁne, for every future-directed

unit timelike vector U,theenergy density of F in any admissible basis with

= U to be TU ·U. In the sense of the following result, the energy density

completely determines the energy-momentum transformation.

Theorem 2.5.1 Let S and T be two nonzero linear transformations on M

which are symmetric with respect to the Lorentz inner product, i.e., satisfy

(2.5.2). If SU · U = TU · U for

every

ture-directed unit timelike vector U,

then S = T .

Proof: Observe ﬁrst that the hypothesis, together with the linearity of S

and T imply that SV · V = TV · V for all timelike vectors V . Now select a

basis {U

}

a=1

for M, consisting exclusively of future-directed unit timelike

vectors (convince yourself that such things exist). Thus, SU

·U

= TU

·U

for each a =1, 2, 3, 4. Next observe that, for all a, b =1, 2, 3, 4, Lemma 1.4.3

implies that U

+ U

is timelike and future-directed so that

S(U

+ U

) · (U

+ U

)=T (U

+ U

) ·(U

+ U

· U

+2SU

· U

+ SU

· U

= TU

· U

+2TU

· U

+ TU

·U

· U

= TU

·U

Exercise 2.5.3 Show that

Sx ·y = Tx · y (2.5.7)

for all x and y in M.

Now, let {e

}

a=1

be an orthonormal basis for M.Then(2.5.7)gives

·e

= Te

· e

(2.5.8)

for all a, b =1, 2, 3, 4. But (2.5.8) shows that the matrices of S and T relat

to {e

} are identical so S = T . 

We investigate the eigenvalues and eigenvectors of T by working in a canon-

ical basis for F. First suppose F is regular and {e

} is a canonical basis for

F . Then the matrix [F

]ofF relative to {e

} has the form (2.4.9)anda

simple calculation gives

112 2 Skew-Symmetric Linear Transformations and Electromagnetic Fields

]

⎡

⎢

⎣

−δ

000

0 −δ

00

000

⎤

⎥

⎦

so tr(F

)=2(

−δ

) and therefore [T

4π

[

tr(F

)I −[F

]

]isgivenby

8π

(

+ δ

)

⎡

⎢

⎣

10 0 0

01 0 0

00−10

00 0−1

⎤

⎥

⎦

det(T −λI) = 0 therefore gives (λ+



+δ

8π

)

(λ−



+δ

8π

)

=0soλ = ±



+δ

8π

∓T

(the energy density). The eigenvectors corresponding to λ =



+δ

8π

are

obtained by solving



+ δ

8π

⎡

⎢

⎣

10 0 0

01 0 0

00−10

00 0−1

⎤

⎥

⎦

⎡

⎢

⎣

⎤

⎥

⎦



+ δ

8π

⎡

⎢

⎣

⎤

⎥

⎦

i.e.,

⎡

⎢

⎣

−v

⎤

⎥

⎦

⎡

⎢

⎣

⎤

⎥

⎦

so v

= v

=0,whereasv

and v

are arbitrary. Thus, the eigenspace is

Span{e

} which contains only spacelike vectors. Similarly, the eigenspace

corresponding to λ = −



+δ

8π

is Span{e

} which contains two independent

null directions (e

± e

) called the principal null directions of T .

If F is null and {e

} is a canonical basis, then [F

] has the form (2.4.14)so

]

⎡

⎢

⎣

0000

0 −α

0 α

0000

0 −α

0 α

⎤

⎥

⎦

and therefore tr F

=0so

]=−

4π

]

4π

⎡

⎢

⎣

000 0

010−1

000 0

010−1

⎤

⎥

⎦

2.6 Motion in Constant Fields 113

Exercise 2.5.4 Show that λ = 0 is the only eigenvalue of T and that the

corresponding eigenspace is Span{e

+ e

}, which contains only one

null direction (that of e

+ e

), again called the principal null direction of T.

Exercise 2.5.5 Show that every eigenvector of F is also an eigenvector of

T (corresponding to a diﬀerent eigenvalue, in general).

Exercise 2.5.6 Show that the energy-momentum transformation T satisﬁes

the dominant energy condition, i.e., has the property that if u and v are

timelike or null and both are future-directed, then

Tu · v ≥ 0. (2.5.9)

Hint: Work in canonical coordinates for the corresponding F.

2.6 Motion in Constant Fields

Thus far we have concentrated our attention on the formal mathematical

structure of the object we have chosen to model an electromagnetic ﬁeld at

a ﬁxed point of M , that is, a skew-symmetric linear transformation. In or-

der to reestablish contact with the physics of relativistic electrodynamics we

must address the issue of how a given collection of charged particles gives

rise to these linear transformations at each point of M and then study how

the worldline of another charge introduced into the system will respond to

the presence of the ﬁeld. The ﬁrst problem we defer to Section 2.7. In this

section we consider the motion of a charged particle in the simplest of all

electromagnetic ﬁelds, i.e., those that are constant. Thus, we presume the

existence of a system of particles that determines a single skew-symmetric

linear transformation F : M→Mwith the property that any charged par-

ticle (α, m, e) introduced into the system will experience changes in world

momentum at every point on its worldline described by (2.1.1). More partic-

ula

ly,

we have in mind ﬁelds with the property that there exists a frame of

reference in which the ﬁeld is constant and either purely magnetic (



)

or purely electric (



). To a reasonable degree of approximation such

ﬁelds exist in nature and are of considerable practical importance. Such a

ﬁeld, however, can obviously not be null (without being identically zero) so

we shall restrict our attention to the regular case and will work exclusively

in a canonical basis.

Suppose then that F : M→Mis nonzero, skew-symmetric and regular.

Then there exists an admissible basis {e

}

a=1

for M and two real numbers

 ≥ 0andδ ≥ 0sothatthematrixofF in {e

} is