5 Four Vectors

Home , Gravitation (book), Invariant (physics), Line element, Principle of covariance, Rest frame, Unit vector

Physics 139 Relativity

Relativity Notes 2002

G. F. SMOOT

Oce 398 Le Conte

DepartmentofPhysics,

University of California, Berkeley, USA 94720

Notes to b e found at

http://aether.lbl.gov/www/classes/p139/homework/homework.html

5 Four Vectors

A natural extension of the Minkowski geometrical interpretation of Sp ecial Relativity

is the concept of four dimensional vectors. One could also arrive at the concept by

lo oking at the transformation prop erties of vectors and noticing they do not transform

as vectors unless another comp onent is added. We de ne a four-dimensional vector

or four-vector for short as a collection of four comp onents that transforms according

to the Lorentz transformation. The vector magnitude is invariant under the Lorentz

transform.

5.1 Co ordinate Transformations in 3+1-D Space

One can consider co ordinate transformations manyways: If x ;x ;x ;x = x; y ; z ; ict,

1 2 3 4

then ordinary rotations in x x plane around x

1 2 3

x = x cos + x sin cos sin

1 2

x = x sin + x cos sin cos

1 2

But in x x plane:

1 4

x = x cos + x sin cos sin

1 4

x = x sin + x cos sin cos

1 4

where the angle is de ned by

2 2 2

cos =1= 1v =c =1= 1+tan =

tan v=c

= sin = i

2 2

1+tan

1 v =c

tan = iv =c = i :

And thus one has the trignometic identity:

2 2 2 2

cos + sin = 1 =1 1

x = [x +icti ] = [x ct]

1 1

x = [x i x ]

4 1

ict = [ict i x ]

ct = [ct x ]

t = [t x =c]

So the extension to 3+1-D includes Lorentz transformations, if angles are

imaginary.

Really,we are considering the set of all 4 4 orthogonal transformations

matrices in which one angle may b e pure imaginary.

In general all angles may b e complex, combining real rotations in 2-space with

imaginary rotations relativeto t.

An alternate way of writing this is

x = xcosh ctsinh

ct = xsinh + ctcosh

where = cosh .

x = xcosi+ictsini

ict = xsini+ ictcosi

and

= i = icosh ; tan = i = iv =c

Still another notation is with x = ict

x = x + i x

1 4

x = x i x

4 1

The transformation matrix is then

0 1

0 0 i

B C

0 1 0 0

B C

@ A

0 0 1 0

i 0 0

Still yet another notation is with x = ct

x = x i x

0 1

x = x + i x

1 0

The transformation matrix is then

0 1 2 3

0 1

0 0 0

B C

1 0 0

B C

@ A

2 0 0 1 0

3 0 0 0 1 2

5.1.1 Generalized Lorentz Transformation

For spatial co ordinates the Lorentz transform ts the linear form

x = x 1

sub ject to the condition that the prop er length

X X

2 2 0 2 2

cd = ds = x = x =ct j~xj 2

is an invariant. This condition requires that the co ecients form an orthogonal

matrix:

= 3

where the Kronecker delta is de ned by = = = 1 when = and 0

otherwise.

The invariance group can b e enlarged to b e the Poincare0 group by the addition

of translations:

x = x + a 4

The full group includes: translations, 3-D space rotations, and the Lorentz b o osts.

5.2 The Inner Pro duct of 3+1-D Vectors

The de nition of the inner pro duct dot pro duct must b e mo di ed in 3+1 dimensions.

~ ~

A B = A B + A B + A B + A B

1 1 2 2 3 3 4 4

if x = ict. But with our usual convention

~ ~

A B = A B A B A B A B

0 0 1 1 2 2 3 3

or with the opp osite signature metric one has

~ ~

A B = A B + A B + A B + A B

0 0 1 1 2 2 3 3

~ ~

A B = A B + A B + A B A B

1 1 2 2 3 3 4 4

if x = ct which is often the convention for the opp osite sign convention. It is an

exercise to show that the inner pro duct is unchanged under a Lorentz transformation.

Can b e done simply by substitution. This can b e extended to the general class of

Lorentz transformations. 3

5.3 Four Velo city

So wehave the p osition 4-vectorx ~ =x ;x ;x ;x and the displacement 4-vector

0 1 2 3

dx =dx ;dx ;dx ;dx . What other 4-vectors are there? That is what other 4-

0 1 2 3

vectors are natural to construct? What we mean by a four-vector is a four-dimensional

quantity that transforms from one inertial frame to another by the Lorentz transform

which will then leave its length norm invariant.

Consider generalizing the 3-vector velo cityv ;v ;v =dx=dt; dy =dt; dz =dt

x y z

what can wedotomake this into a 4-vector naturally? One clear problem is that we

are dividing by a comp onent dt of a vector so that the ratio is clearly going to Lorentz

transform in a complicated way.We need to take the derivative with resp ect to a

quantity that will b e the same in all reference frames, e.g. d the di erential of the

prop er time, and add a fourth comp onent to make the 4-vector. It is clear that the

derivative of the 4-vector p osition ct; x; y ; z with resp ect to the prop er time will

b e a 4-vector for Lorentz transformations since ct; x; y ; z transform prop erly and d

is an invariant. So we can de ne the 4-velo cityas

dx dx dy dz dct

u = ; u~ = ; ; ; 5

d d d d d

Note that

2 2 2

dx dy dy

2 2 2 2 2 2 2 2 2

c c d = c dt dx dy dz = dt

dt dt dt

2 2 2 2 2 2 2 2

c v c v v v = dt = dt

x y z

or the time dilation formula we got b efore

d 1 dt

2 2

= = = 1 v =c ; and

2 2

dt d

1 v =c

So we can now explicitly write out the 4-velo city using the chain derivative rule:

dx dx dt

u = =

d dt d

u~ =u ;u ;u ;u = c; v ; v ; v = c; v ;v ;v

0 1 2 3 x y z x y z

Thus three comp onents of the 4-velo city are the three comp onents of the 3-vector

velo city times .

Note also that the norm - the magnitude or vector invariant length - of the

four-velo city is not only unchanged but it is the same for all physical ob jects matter

plus energy. For 3+1 dimensions the norm or magnitude is found from the inner

pro duct or dot pro duct which has the same signature as the metric see just ab ove

so that

2 2

1 v =c

2 2 2 2 2 2 2 2 2 2 2

u~ u~ = u u u u = c v v v = c = c

0 1 2 3 x z z

2 2

1 v =c 4

Thus every physical thing, including light, moves with a 4-velo city magnitude of c

and the only thing that Lorentz transformations do is change the direction of motion.

A particle at rest is moving down its time axis at sp eed c. When it is b o osted to a

xed velo city, it still travels through space-time at sp eed c but more slowly down the

time axis as it is also moving in the spatial directions.

One should also note that as the spatial sp eed three-velo city approaches c,

all comp onents of the 4-velo city u are unb ounded as !1. One cannot then de ne

a Lorentz transformation that moves to the rest frame. Thus all massless particles

will have no rest frame.

5.3.1 LawofTransformation of a 4-Vector

We can write the transformation in our standard algebraic Lorentz notation

A = A A =1= 1

0 1

A = A A

1 0

0 0

A = A ; A = A

2 3

where and refer to the relativevelo city V of the frames.

5.3.2 LawofTransformation of a 4-Velo city

u = u u

1 0

where and are for the relativevelo city of the frames and not of the particle. But

in the formula for the 4-velo city

u~ =u ;u ;u ;u = c; v ; v ; v = c; v ;v ;v

0 1 2 3 x y z x y z

The is for the particle! So we should have lab eled it and the and for the

frame transform and . Then wehave

f f

0 0

v = v c

f p x f p

p x

So we can get out a formula for v

f p

q q

v V = v V v =

x x

1 1

p f

2 2

This is our old friend on the law of transformation of 1 u =c

q q

0 2 2 2 2

1 u =c 1 V =c

2 2

1 u =c =

0 2

1+u V=c

x 5

and

q q

2 2 2 2

1 u =c 1 V =c

0 2 2

1 u =c =

1+u V=c

which is simply

1 1

0 2

1 u V=c

p f x

v V

v =

1 u V=c

as derived earlier by the di erential route.

Continuing onward

0 0 0

u = u ; or v = v

2 p y

so that

v = v

0 2

1 u V=c

f x

and

2 2

1 V =c

v = v

1 u V=c

which is the same relationship as b efore from the di erential Lorentz transform.

0 0

Similarly for v and v :

z t

u = u u

f 0 f 1

Explicitly this is

0 2

c = c u V=c= c 1u V=c

f p p x p f x

0 2

= 1 u V=c

p f x

which is our relation from the transformation of 's and its recipro cal used ab ove.

5.4 Four Momentum

What is the natural extension of the 3-vector momentum to 4-momentum. The answer

is clear from dimensional/transform analysis and from our exp erimental approachon

how masses transformed. The 4-momentum is simply:

p = m u ;~p=p ;p ;p ;p = m c; v ;v ;v 6

o 0 1 2 3 o x y z 6

The three spatial comp onents are just the Newtonian 3-momentum with the mass of

the particle replaced by m .

We can see that the 4-momentum also has an invariant norm by making use

of our results for the 4-velo city:

2 2 2 2 2 2 2 2 2 2

p~ p~ p p p p = E =c p = m u~ u~ = m c

0 1 2 3 o o

Thus the invariant length of the 4-momentum vector is just the rest mass of the

particle times c.

5.5 The Acceleration Four-Vector

In a similar way one may derive the acceleration four-vector. Again we di erentiate

with resp ect to the prop er time .

a = 7

The four-vector acceleration will have a part parallel to the acceleration three-vector

and a part parallel to the velo city three-vector.

Prove that the inner pro duct of the 4-acceleration and the 4-velo city are Exercise:

zero; ~a u~ = 0 as they must b e if the norm of the four-velo city is to remain constant

Wehave also constructed the 4-acceleration to b e a 4-vector so that ~a ~a is an

invariant. Evaluate it in the rest frame ~a a~ = ja~j

~a a~ = ja~ j

rest frame

in any frame. This can b e very useful in various calculations and we will use it later

to treat radiation from and accelerating charged particle.

Acceleration 4-vector transforms by the relations:

0 0

a = a a ; a = a ;

f 0 f 1 2

0 2

0 0

a = a a ; a = a ;

f 1 f 0 3

1 3

This is the b est starting place from which to derive the detailed Lorentz

transformation equations for acceleration.

5.6 The Four Vector Force

Wenow consider the four-vector force, whichwe de ne the following way:

dp~

8 F

dp~ dp~ dt dp~

F = =

d dt d dt 7

~ ~

9 F F ;F ;F ;F = W =c; F ;F ;F ! F ; F ;F ;F

0 1 2 3 N 1 N2 N3 N N 1 N2 N3

~ ~

where F is the three-dimensional Newtonian force, e.g. F =F ;F ;F

N N N1 N2 N3

Note that the four force can b e space-like, time-like and null. If a frame can

b e found where the three-force on an ob ject is zero but the ob ject is exchanging

internal energy with the environment, then the four-force is time-like. The converse

is space-like.

Then the 4-vector force F has the same transformation law as all 4-vectors:

~ ~

F F F =

0 f 1 f

~ ~

F F F =

1 f 0 f

0 0

~ ~ ~ ~

F = F ; F = F

2 3

So we can now conveniently transform any of the familiar vectors used in mechanics,

but not electric and magnetic elds, and pseudovectors obtained from cross-pro ducts,

such as angular momentum and angular velo city.We will treat these later.

The 4-vector force transforms are much easier than the 3-D fource transforms

whichinvolvea . See the homework problem for the transformaiton of acceleration

to grasp howmuch more complicated it is.

5.7 4-D Potential

It is convenienttodophysics in terms of p otential and nd the resulting force as the

derivative, e.g. the gradient, of the p otential. Classical physics examples are:

F = mr Newtonian Gravitation

G G

F = q r Electrostatics 10

E E

Once wehave a 4-D p otential, then we need to learn how to take derivatives in 4-D

spaces.

One approach is to make the simplest p ossible frame-indep endent scalar

estimate of the interaction of two particles. This manner of thinking eventually leads

one to the interaction Lagrangian as a the pro duct of the two currents electrical,

matter, strong, weak, gravitational.

~ ~

L = j j 11

1 2

where is the coupling constant and the next term is the inner 4-D dot pro duct of

the current of particle 1 and the current of particle 2. When the two currents are in

contact zero prop er distance separation, there is an interaction. When they are not

in prop er distance contact, there is no interaction. This means that all interaction

is on the prop er distance null the light cone. Thus there is no action at a prop er

distance. It is manifestly invariant as the inner pro duct of two 4-D vectors. 8

From this Lagrangian we can generate the 4-D p otential of the e ect of all

~ ~

other currents or a single current j on our test particle which has current j .

2 1

ZZZZ ZZZZ

2 2 2 2

~ ~

A~x = f s j ~x dV dt = f c t t r jdV dt 12

1 2 2 2 2 1 2

12 12

2 2 2 2 2

where s jx~ x~j =ct t r is the invariant separation b etweenx ~

1 2 1 2 1

12 12

andx ~ dV is the 3-D spatial volume and dt is the time. f s is a function whichis

zero every where but p eaks when the square of the 4-vector distance s between the

source 2 and the p ointofinterest 1 is very small. The integral over f s is also

normalized to unity. The Dirac delta function is the limiting case for f s . Thus

f s is nite only for

2 2 2 2 2

s = c t t r 13

1 2

12 12

Rearranging and taking the square ro ot

ct t r 1 r r 1 14

1 2 12 12

2 2

r 2r

12 12

15 t t

1 2

c 2cr

whichsays that the only times t that are imp ortant in the integral of A are those

which di er from the time t , for which one is calculating the 4-p otential, by the

delay r =c ! { with negligible correction as long as r .Thus the Bopp theory

12 12

approaches the Maxwell theory as long as one is far away from any particular charge.

By p erforming the integral over time one can nd the approximate 3-D volume

2 2

integral by noting that f s has a nite value only for t =2 =2r c, centered

2 12

at t r =c. Assume that f s =0=K, then

1 12

Z Z

jt r =c; ~x K

12 2

dV 16 jt ;~x fs dV dt A~x =

2 2 2 2 2 1

c r

which is exactly the 3-D version, if we pick K so that K =1.

5.8 Derivative in 4-Space

The 3-D vector gradient op erator is DEL:

@ @ @

; ; 17 r =

@x @y @z

which b ehaves as a 3-D vector.

This can b e generalized to 4-D:

@ @ @ @

2 = ; ; ; 18

@x @x @x @x

0 1 2 3 9

How do es it transform?

@ @ @ @

2 = ; ; ; 19

0 0 0 0

@x @x @x @x

0 1 2 3

Op erate rst on a scalar function x ;x ;x ;x

0 1 2 3

X X

@x ;x ;x ;x @ @x @

0 1 2 3

R 20 = =

0 0

@x @x @x @x

where R is the rotation matrix/tensor de ned by

x = a x

1 0

x = a x 21

1 y y

A = a means transp ose, if a is orthogonal.

x = a x 22

@ @

= a

@x @x

x = a x 23

so that

~ ~

2 = a 2 24

and 2 is a Lorentz 4-vector.

5.9 Op erate with 2

Op erate with 2 on a Lorentz 4-vector, to get the dot inner pro duct:

@ct @x @y @z

2 x~ = + + +

@ct @x @y @z

= 1 +1+1 +1=4=invariant 25

Now op erate on velo city 4-vectoru ~ :

@ v @ v @ v @ c

x y z

+ + + 2 u~ =

@ct @x @y @z

@ 1 @ v @ v @ v

x y z

p p p p

= + + +

2 2 2 2

@t @x @y @z

1 1 1 1

~v @ 1

p p

+r 26 =

2 2

1 1

This equation is an expression related to continuity. 10

5.9.1 Hydro dynamics

Conservation of uid matter is expressed by the equation:

+ r ~v=0 27

If one integrates this equation over a xed volume containing mass M

Z Z

dxdy dz + r ~vdxdy dz = M 28

vol vol

The rst term is the mass contained in the volume and the second part is the

divergence theorem and yields:

+ ~v ndS^ =0 29

sur f ace

= - outward transp ort of mass and equals the inward transp ort of mass.

Since our expression for 2 u~ is

1 ~v @

p p

+r 30 2 u~ =

2 2

1 1

the role of densityisplayed by =1= 1 .

5.10 The Metric Tensor

Now b efore moving to make electromagnetism consistent with our relativistic

mechanics, we need to generalize the concepts of the distance, vectors, vector algebra

and tensors as they work in 3+1 D space.

The metric tensor de nes the measurement prop erties of space-time. Metric

means measure { Greek: metron = a measure.

Cartesian { at space

2 ij

ds = g dx dx 31

i j

i;j

by de nition g = g since the measure must b e symmetric under interchange of

ij ji

co ordinate multiplication order.

In the general case: Cartesian { at space

2 i j

ds = g dx dx = scalar invariant 32

i;j

Note the sup erscripts. Section of covariant and contravariantvectors explains this.

If g is diagonal, the co ordinates are orthogonal.

ij 11

Physical interpretation: g = h , where h is de ned by the comp onents of the

ii i

vector line element, ds = h dx . An example of this is spherical p olar co ordinates:

i i i

2 2 2 2 2 2 2

ds = dr + r d + r sin d 33

2 3

1 0 0

6 7

g = 0 r 0 34

4 5

2 2

0 0 r sin

For the 3 + 1 dimension Minkowski space-time

2 2 2 2 2 2

ds = dc = dct dx dy dz 35

3 2

1 0 0 0

7 6

0 1 0 0

7 6

7 36 g = 6

5 4

0 0 1 0

0 0 0 1

In general the symbol is used to denote the Minkowski metric. Usually it

is displayed in rectangular co ordinates ct; x; y ; z orx;x ;x ;x but could b e

0 1 2 3

expressed in spherical ct; r; ; or cylindrical ct; r; ;z equally well.

~ ~

The o -diagonal g = h h ds ds for i 6= j . An example is skew

ij i j i j

co ordinates in two dimensions.

x^ x^

2 2 -

- -

O O

x^ x^

1 1

By the law of cosines

2 2 2

ds = dx + dx +2dx dx cos

1 2

2 2

+ g dx dx + g dx dx 37 + g dx = g dx

12 1 2 21 1 2 22 11

2 1

ds = dx ; ds = dx

1 1 2 2

2 2

g = h =1; g h =1

11 22

1 2

h h cos = cos 39 g = g =

1 2 12 21

1 cos

40 g =

cos 1 12

5.11 Contra & CovariantVectors

First we consider a simple example to illustrate the signi cance of contravariant and

covariantvectors. Consider two non-parallel unit vectors ^a anda ^ in a plane with

1 2

a^ a^ = cos 6=1.

1 2

^a a^

2 2

- -

O O

a^ a^

1 1

A displacement from O to P can b e represented bya vector, S. Its comp onents

1 2

in the directions ofa ^ anda ^ can b e denoted S and S :

1 2

S = S a^ + S a^ 41

1 2

Another set of basis vectors ~a and ~a , resp ectively,may b e de ned, b eing

p erp endicular to ^a anda ^ and having lengths found the following way: Let ^a be a

1 2 3

unit vector normal to the plane, prop ortional to ^a a^ . Then

1 2

e^ ^a ^a

2 3

= 42 ~a =

a^ a^ a^ sin

1 2 3

^a ^a e^

3 1

~a = = 43

a^ a^ a^ sin

1 2 3

We denote the triple scalar pro duct by[ ] .

123 13

1 H

j~a S j

H 1

j~a S j

Hj ?

The displacementvector, S may also b e expressed by its comp onents S and

S as follows:

1 2

S = S ~a + S ~a : 44

1 2

The relations among S , S , S , and S may b e found by elementary geometry:

1 2

They are:

1 2

v = v + v cos 45

1 2

v = v cos + v 46

1 2

v =v v cos =sin 47

1 2

2 2

v =v cos + v =sin : 48

1 2

Using the original pair of unit vectors,

2 1 2 2 2 1 2

S = S +S +2S S cos

i j

= g S S 49

i;j =1

with the metric tensor

1 cos

50 g =

cos 1

De ned to b e symmetric.

The tensor g is de ned by

ij i

g g = : 51

j =1

It is easy to nd that

1 cos

: 52 g =

cos 1

sin 14

From this relation one nds that

S = g S 53

i i;j

and

i i;j

S = g S : 54

The comp onents S are contravariant and the comp onents S are covariant.

The square of the length of S is as given ab ove

i;j

X X

2 i j ij

jSj = g S S = g S S ; 55

ij i j

i;j

but is given more compactly by

2 j

S = S S 56

Other relations of interest are:

Signed Minor of g Cofactor of g

ij ij

g = = 57

Det g g

2 i+j

For this example Det g = g = sin ; the cofactor of g is 1 g =

ij ij ji

i+j ij

1 g b ecause g and g are symmetric.

ij ij

Returning to the original sets of basis vectors

a^ a^ a^ a^

2 3 2 3

~a = = 58

a^ ^a a^ []

1 2 3

123

and others by cycling indices, by substitution one has:

2 3 2 3

a^ a^ a^ a^

= 59 ^a =

123

1 2 3

a^ ^a a^

[]

1 1

123

[] = = 60

[] sin

123

Also one has

1 1

= : 61 Detg =

Detg sin

ij 15

5.12 Electric Charge

Wenow consider the implications for electric charge. We de ne electric charge density

as the charge p er volume, .Wehavealaw of conservation of charge: Charge cannot

b e created or destroyed. Thus

+ r~v=0: 62

So the charge-current density Lorentz 4-vector

j ~ =c; v ;v ;v =j ;j ;j ;j 63

x y z 0 1 2 3

where = and

2j =0 64

is the equation for the conservation of charge. j is the 4-vector charge current.

Now consider the vector and scalar p otentials of the electromagnetic elds.

ZZ Z

1 jdV

~ ~ ~ ~

B = rA where A =

c r

ZZZ

dV 1@A

~ ~

where = 65 E = r

c @t r

The Lorentz 4-vector p otential is

ZZZ

1 j dV

A =;A ;A ;A =A ;A ;A ;A where A = 66

x y z 0 1 2 3

c r

Then the inner pro duct gives

2 A = 2 A

@ @A @A @A

x y z

= + + +

@ct @x @y @z

1 @

~ ~

+ rA =0 67 =

c @t

This is the equation of Lorentz gauge invariance.

5.12.1 BoxonAis a four vector

It is clear that j = u~ is a four vector sinceu ~ was constructed to b e one and we

constructed j as a scalar rest frame charge density times that four vector. However,

I merely asserted that A was a four vector. That is true only if dV =r is invariant

under Lorentz transforms. Wehave this as an exercise for the student to show

that is true. The following are hints: Show that dV =1+ cos dV and that

0 0 0

= r 1 + cos and thus dV =r = dV =r . r 16

5.13 Lorentz Force Law

The 3-D vector form of the force lawis

~ ~ ~

F =q E+~vB 68

We need to write this in 4-D vector form to show that it is Lorentz invariant. The

relativistic force laymust involve the particle velo city and the simplest form is linear

in the 4-D velo city. The 4-D vector form then would b e

q q

~ ~

F = F u;~ F = F u 69

c c

To obtain the 4-D expression for the electromagnetic elds we need second

rank tensors, i.e. F .

Since wewant the force F to b e rest-mass preserving, wehave the requirement

that F u = 0 and thus F u u = 0. Since this must hold for all u , the F must

be antisymmetric.

A cartesian at-space second rank tensor has comp onents C . The tensor is

the sum of a symmetric tensor S and an antisymmetric tensor A :

ij ij

1 1

C = C + C + C C

ij ij ji ij ji

2 2

= S + A 70

ij ij

S = S ; A = A 71

ij ji ij ji

The prop erty of b eing symmetric or of b eing antisymmetric is preserved under

orthogonal transformations.

Now construct the antisymmetric tensor in a generalized curl

F = 2 A 2 A = @ A @ A = A A 72

; ;

Note that

F = F = F = F =0

00 11 22 33

@A @A

2 3

~ ~

=rA =B F =

x x 23

@x @x

2 3

Similarly, F = B , F = B .

31 y 10 z

@A @ @A @A

1 x 0

= = E F =

x 10

@x @x @x @ct

1 0

and similarly F = E and F = E . So the full tensor is

20 y 30 z

2 3

0 E E E

x y z

6 7

E 0 B B

6 7

x z y

F = 6 7 73

4 5

E B 0 B

y z x

E B B 0

z y x 17

F is the electromagnetic eld tensor.

The contravariant form of the electromagnetic eld tensor is

3 2

0 E E E

x y z

7 6

E 0 B B

7 6

x z y

7 74 F = 6

5 4

E B 0 B

y z x

E B B 0

z y x

One can raise and lower indices by use of the metric tensor.

X X

F = g F g 75

In 3-D Maxwell's equations are:

1 @ E ~v j

~ ~

rB = =

c @t c c

~ ~

r E =

1@B

~ ~

= 0 rE +

c @t

~ ~

rB = 0 76

Nowwe take the 4-D divergence of the electromagnetic eld tensor

2 F = j=c 77

which reduces to the rst two Maxwell equations. The continuity equation is simply

j =0: 78

Since there were actually two p ossible ways to unify the electric and magnetic

elds into a single entity,wenow de ne the dual electromagnetic eld tensor:

2 3

0 B B B

x y z

6 7

B 0 E E

6 7

x z y

G = 6 7 79

4 5

B E 0 E

y z x

B E E 0

z y x

The second set of Maxwell's equations can b e simply written as

=0 80

Or, if one do es not wish to resort to the dual electromagnetic eld tensor, then the

second set of Maxwell's equations can b e simply written as

@ F + @ F + @ F =0 81

a generalized curl. 18

5.14 Transformation of the EM Fields

One can derive the transformation of the electromagnetic eld by using the Lorentz

~ ~ ~ ~ ~ ~

force law F = q E + V B as a de nition of the E and B and by the transformation

~ ~

of second rank tensors as shown b elow. To derive the E and B requires using three

reference frames in order to see how b oth transform.

Do use the Lorentz force lawwe need a test electron or charge to prob e the

force and thus how the elds must transform. We consider the eld acting on an

electron lo cated at the origin of three reference frames in relative motion.

0 0 0

~ ~ ~ ~ ~ ~

~ ~ ~

F , E , B F , E , B

F , E , B

o o o

6 6 6

~u relativetoS

V relativetoS

~v relativetoS

- - - -

q -

Electron velo city ~v in S

Electron velo city ~u in S

Electron at rest in S

The electron is at rest relative to reference frame S ,moving with velo city ~v

with resp ect to reference frame S, and moving with velo city ~u with resp ect to reference

frame S .We arrange the co ordinate systems so that the velo cities all lie along the

x axes. Thus the relativevelo city V of the frames S and S is given by the velo city

addition formula as

u + v

V =

1+uv =c

We can write simple expression for the Lorentz force comp onents in frames S,

S , and S , resp ectively:

S S S

0 0

F = eE F = eE F = eE

x x ox ox

x x

0 0 0

F = eE vB F = eE uB F = eE

y y z oy oy

y y z

0 0 0

F = eE + vB F = eE + uB F = eE

z z y oz oz

z z y

Note that in S the electron is not moving so that the magnetic eld do es not pro duce

a force.

The equations for the transformation of force for u = 0 give

F = F F = F

x ox ox

q q

2 2 2 2

1 v =c F = F 1 u =c F = F

oy y oy

q q

2 2 2 2

F = F 1 v =c F = F 1 u =c

y oy oz

Then wehave

E =E E = E

x ox ox

q q

0 0

2 2 2 2

E vB = E 1 v =c E uB = E 1 u =c

y z oy oy

y z

q q

0 0

2 2 2 2

1 v =c E uB = E 1 u =c E + vB = E

oz z y oz

z y 19

We can see at once that E = E .From the velo city addition lawwehave

v u=c + V=c

c 1+u=cV=c

and thus

V u

c c

q q q

2 2 2 2 2 2

1 v =c 1u =c 1 V =c

Thus

0 0

E uB

E vB

y z

q q

= E =

2 2 2 2

1 v =c 1 u =c

so that

2 3

0 0 uV

E uB

u + V

y z

c c

4 5

q q q

E = B

y z

2 2 2 2 2 2

1+uV =c

1u =c 1 V =c 1 u =c

If these equations are to hold true for all values of u, then since the terms which

contain u must b e equal and those that do not must also b e equal:

E VB

y z

E =

2 2

1 V =c

V=cE + B

y z

B =

2 2

1 V =c

Similarly by equating the expression for E one nds

E + VB

z y

E =

2 2

1V =c

V=cE + B

z y

B =

2 2

1 V =c

This gives the transformation law for 5 of the six comp onents of the

electromagnetic eld. We are missing B since we started with a stationary electron

in frame S . This can b e found by considering an electron moving at right angles to

B and recalling that the force is unchanged in the x direction. Thus B = B .

x x

Now do the derivation of eld transformation from the transformation of a

second rank tensor and apply that to F .

X X

a a F 82 F =

applied to either the electromagnetic eld tensor F or its dual gives

0 0

E = E B = B

x x

0 0

E = E B B = B + E 83

y z y z

y y

0 0

E = E + B B = B E

z y z y

z z 20

5.15 The Equations of Motion for a Charge Particle

The 3-D Lorentz force law

d~p ~v

~ ~ ~

F = = q E + B 84

dt c

We can turn this into 4-D vector equation by rst replacing dt = d and 3-vector

velo city ~v by the 4-vector velo city~u.

F = = qF u 85

5.16 The Energy-Momentum Tensor

First a brief review to provide motivation for the study and understanding of tensors:

1 Electromagnetism describ ed by a tensor eld 4 by4

2 Gravity represented by a tensor eld 4 by4

3 elastic phenomena in continuous media mechanics classical 3 x 3

4 metric tensor for generalized co ordinates

First we found a 4-vector equation of motion for a single particle:

dp~ dp dp~

~ ~

= F = F = F 86

d d d

Next we found the equation of motion for a single particle in an electromagnetic eld

as:

dp du

= m = F u 87

d d

Later we will nd that the equation of motion for a single particle in a weak gravitation

eld is

dp du 1

= m = h m u u 88

0 ; o

d d 2

The last equation the second rank tensor h is obvious but there is another simple

second rank tensor there m u u . This is an imp ortant tensor. The next paragraph

supplies a little more motivation to study this imp ortant and one of the simplest that

one could think to form.

In classical mechanics one has the concept that the integral of the force times

distance is the work done energy gained and that the gradient of the p otential is

the force.

~ ~ ~

W =E = F d~x F =rV 89

All this p oints to the need to develop the same concept in 4-D.

Z Z Z Z

d~p d~x

E = F d~x = d~x = d~p = ~u d~p 90

dt dt 21

From the last part of the equality one nds that the integral to get the \4-p otential"

will involve p u . The tensor p u is lab eled the energy-momentum tensor. We can

write out explicitly the tensor for a particle.

T = p u = m u u

3 2

x y z

7 6

x x y x z

2 2

7 91 = m c 6

5 4

y x y y z

z x z y z

since u = c1; ; ; .

x y z

2 2

The quantity, m c = E, seems a bit strange but not so when we consider

a collection of particles or a continuum in density of material, . = since one

factor of comes from the mass increase and another factor of comes from the

volume contraction due to length contraction along the direction of motion.

2 3

x y z

6 7

x x y x z

T = c 6 7 92

4 5

y x y y z

z x z y z

and nowwe see that the energy-momentum tensor comp onents are the transp ort of

energy-momentum-comp onentin -direction into the -direction.

Consider an interesting case: a large ensemble of non-interacting elastic

scattering only particles { an ideal gas. For an ideal gas, < >= 0 and < >=0,

i i j

2 2 2

for i 6= j , and ==, so that the energy-momentum tensor is

x y z

diagonal

2 3 2 3

c 0 0 0

0 0 0

6 7 6 7

0 P

6 7 6 7

T = 6 7=6 7 93

ideal gas

4 5 4 5

0 P

where is the full energy density due to the mass density, and P = which

is easily derived for an ideal gas PV = nk T = nm.

We can write a simple formula for the energy-momentum tensor for a p erfect

uid in a general reference frame in which the uid moves with 4-D velo city u as

T = + p=c u u pg 94

which reduces to the equation ab ove in its rest frame.

5.17 The Stress Tensor

Nowwe can consider the case of a medium or eld that can have non-zero o -diagonal

comp onents. First it is go o d to review the concept of stress. Stress is de ned as force

p er unit area, same a pressure which is a particularly simple stress, 22

Imagine a distorted elastic solid or a viscous uid such as molasses in motion.

Imagine a surface conceptual/mathematical in the medium The surface can and

will b e curved or distorted. with a plus and a minus side and unit normal vector for

every p oint on it. A di erential area element dA, with normaln ^ will exert forces on

each of its sides. The forces are equal and opp osite by Newton's second law, since

~ ~

the mass of the element is zero. F = m~a =0,so F +F =0

total +on on +

The force p er unit area on the small element of the surface is the stress. It is

avector, not necessarily known. It underlies the dynamics of continuous media.

Consider a small piece of material at the surface

~ ~ 6

F = T dxdz

2 2

~ ~

F = T dy dz

1 1

~ ~

F = T dxdy

3 3

We de ne stress which stretches as p ositive and stress which compresses as

negative.

Clearly each of the three axes has a vector force asso ciated with it so that we

have a second rank tensor eld asso ciated with the stress. We de ne the stress tensor,

is when the vector T is co-directional with the normal E T . Normal Stress

i ij ij

n^ .

If E = C , C is the hydrostatic pressure, if C<0.

ij ij

Simple Tension Consider E = C n^ n^ , then T = E n^ = C n^ n^ n^ = C n^

ij i j ij i i j i j

thus is co-directional with pmn^ . If m^ has directional orthogonal to n , then

i i i

T = C n^ n^ m^ =0.

i j i

If C is negativeC<0, the stress is simple compression.

is sp eci ed by E = C ^n m^ +^n m^ Shearing Stress

ij i j j i

We will see by example the following generalization: A simple tension

in one direction and a single compression along an orthogonal direction

is equivalent to a shearing stress along along shearing stress along the

direction bisecting the angle b etween the two directions.

In anticipation of later integration to 4-D we can call the stress tensor

E = T Force p er area on the surface along the i-axis along the surface with

ij ij

normal in the j -direction by the material on the side with smaller x . Since action

must equal reaction T = force by material on the side of larger x .

ij j

Now return to our in nitesimal cub e of the medium, with sides lined up along

the cartesian co ordinate planes: 23

x+dx

Consider the front face: F is exerted on it toward inside in the x-direction is

dy dz 95 dx F = T x + dxdxdy = T x+

x xx xx

The force on the back face is

F =+T xdxdy = T xdy dz 96

x xx xx

The net force on the cub e is F is exerted on it toward inside in the x-direction is

F = dxdy dz 97

If T > 0, inside pushes on the outside, pressure: compressive stress. If

T < 0, inside pulls on the outside, tension: tensile stress.

T and T are shear stresses.

xy yx

x+dx

Similarly to the treatmentabove the net force in the y -direction, F ,onthe

front and back face is

F = dxdy dz 98

@x 24

and

F = dxdy dz 99

Thus the total F on the material inside is

@T @T @T

xx yx zx

F total = + + dxdy dz

@x @x @x

F = dV 100

i=1

Now consider F on the two faces p erp endicular to x and F on the two faces

y x

p erp endicular to y as exerted from the outside.

F = T dA

y xy yz

F = T dA

F =+T dA

y yx yz

F =+T dA

y xy xz

The sign changes b ecause from the surface the force is toward the inside. Now

calculate the net torque. The two x faces have a counter-clo ck-wise torque:

torque from x face = Force moment arm = T dy dz dx=2 101

torque from y face = T dxdz dy =2 102

To the net torque is

103 =T T dxdy dz =2=I

xy yx

2 2

where I / mr dxdy dz r is the moment of inertia and d! =dt is the angular

acceleration so that

T T / r 104

xy yx

as we consider an in nitesimal cub e, r ! 0 so that

T = T 105

xy yx

which means the stress tensor must b e symmetric. The stress tensor is symmetric, so

only six indep endent comp onents. 25

5.18 Consideration of Shear

Simple shear displacement is like sliding a deck of cards.

A pure shear displacementkeeps the center at the same place and is what our

four forces try to do:

If the little cub e is cut di erently, e.g. cut at 45 to the previous cub e, a

di erent e ect o ccurs:

@ @

Thus pure shear is a sup erp osition of tensile and compressive stresses of equal

size at right angles to each other.

Let us follow our example of shear a little further:

3 2

0 T 0

7 6

T 0 0 106 T =

5 4

xy ij

0 0 0

We can lo ok at the transformation prop erties by considering on the 2 2

p ortion. Now rotate the axes 45 .How do the tensor comp onents change?

X X

S = a a S 107

ik jl kl

k l 26

where a is the matrix for the co ordinate transformation, rotation:

x a a x cos sin x

11 12

= = 108

y a a y sin cos y

21 22

For 45 , the rotation matrix is:

1 1

p p

2 2

[A ]= 109

1 1

p p

2 2

so that

T = a T + a a T + a a T + a a T

11 11 11 12 11 21 11 12 12 12 21 22

= T + T + T + T =T = T 110

11 21 12 22 21 12

T = a a T + a a T + a a T + a a T

12 11 21 11 11 22 12 12 21 21 12 22 22

T + T T + T =0 111 =

11 12 21 22

T = a a T + a a T + a a T + a a T

22 21 21 11 21 22 12 22 21 21 22 22 22

= T T T + T =T = T 112

11 12 21 22 21 12

So that for the 45 rotation wehave

T 0

T = 113

0 T

Thus wehave shown that a pure shear stress rotated by45 is equivalent to equal

amounts of tension and compression stress at right angles to each other with the pure

shear bisecting the angle they make.

5.19 Electric and Magnetic Stress

In this section we see that using the Faraday lines of force concept that b oth the

electric and magnetic eld lines can b e under tension or compression and thus by the

argument just ab ove under shear stress.

First consider two opp osite charges, magnitude q , a distance 2d apart, lo cated

symmetrically opp osite the origin on the x-axis. The force b etween them is F =

2 2

q =4d according to the Coloumblaw. We can imagine putting a metal plate p erfect

conductor in the y z plane and know that an image charge will form and have the

same force on it and thus the plate. This makes sense in terms of the Faraday lines

of force. We can calculate the total integrated mean square value of the electric eld

in the y z plane. 27

s s

s -s

d d -q d

2 3 2 2 2

The only non-zero comp onentisE =2q cos =r =2q d=r where r = d .

Z Z Z

2 2 2

1 1

d[ + d ] 1 2q 2 d

2 2 2 2 =0 2 2 2

=4q d =4q d j = E dA =4q d

=1 x

2 2 3 2 2 3 2 2 2 2

+ d + d 2 + d d

=0 0

114

2 2

The actual force b etween the charges is q =4d , so that the force p er unit

area in eld must b e which is a tensile stress and is along the lines of electric eld.

Now consider the same situation but with b oth charges having the same sign.

In this case the lines b end and b ecome tangent to the y z plane and are clearly in

compression. By symmetry the only non-zero comp onent of the electric eld is that

that go es radially in the ^ direction.

2q 2q 2q

sin = = E =

2 2 3

d d r r

2 2 2

where E = E + E . Again we can compute the total integrated mean square electric

y z

eld strength in the y z plane:

Z Z Z

2 2 2 2 2

1 1

r d 1 d 2q

2 2 2 2 2 d

E dA =4q 2 d =4q dr =4q j =

6 2 3 2 2 2

r r r 2r d

0 d

115

Thus again we nd the compressive stress p erp endicular to the electric eld lines is

E =8 .

Consider another simple case of tension along the lines of electric eld, which

is the familar simple capacitor. 28

- +

Clearly the lines of force, electric eld lines are under tension. We can consider

the charge on each of the capacitor faces to have a surface charge density equal to

. Then by Gauss's lawwe can construct the usual pill b ox which has a uniform

electric eld passing though the face with area A and not on the sides or outside face.

Thus in Gaussian units 4 = E in Heavyside-Lorentz units, = E and the force

between the plates p er unit area is

F E E

= = 116

A 2 8

or in Heavyside-Lorentz units, E =2.

Nowwe turn to magnetic stress. First consider a very long solenoid or a current

sheet.

Integral Path

B =

c `

The magnetic eld is parallel to the solenoid and

B d` = j

so that B =4 j =c`. The Lorentz force on the currentis

~ ~ ~

F =q~vB=jB

The force p er unit area is equal to the average of the magnetic eld at each edge of

the solenoid or for an ideal solenoid this is half the internal magnetic eld. We then 29

have pressure stress

P = 117

magnetic

The factor c dep ends up on the units one uses. Thus we see that like the electric eld,

the magnetic eld can have compression p erp endicular to the magnetic eld lines.

Nowwe observe tension along magnetic eld lines. Consider two magnets

placed with p oles near each other. If the p oles are opp osite, the magnets are attracted

{ tension in the direction of the lines. If the p oles are the same, the magnets are

repulsed { compression p erp endicular to the lines.

We can see that this reduces to exactly the same case as for the charges

calculated ab oveby considering two long magnets.

S N N S

N S S

As the magnets get longer and longer, each p ole acts exactly as if it is an

isolated charge and the math is the same.

Nowwe see that we need to have a momentum-energy tensor or more prop erly

stress-energy tensor for electromagnetism.

5.20 Stress-Energy Tensor

We need to generalize this to 4-vectors and Lorentz invariance. This will require the

use of second rank tensor - the stress-energy tensor.

In relativistic mechanics for continuous media the energy-momentum or stress-

energy tensor, T , is usually de ned as:

ij i j ij i0 0i i 00

T = u u E ; T = T = u ; T = 118

where is the density and E is the Cartesian stress tensor usually de ned as

the tensor that describ es the surface forces on a di erential cub e around the p oint

in question. The normal surface force is pressure but there can b e terms for

tension/compression and shearing stress.

Then the equations of motion of a continuous medium is

T = f 119

;

where f is the 4-force density. That is the net force on material in a volume V is

ZZZ ZZ

F = f dV= T dA 120

V sur f ace

where the last equality comes from invoking Stoke's theorem. 30

In the case of electromagnetism in the 3-dimensional form the parallel

equations are

ZZ Z ZZZ

3 3

~ ~ ~ ~ ~

F = E+~vB d V = E+jB dV 121

V V

Thus the force density f is

~ ~

f = E + j B 122

Nowwewant to replace and j by the elds via Maxwell's equations.

@E 1

~ ~ ~ ~

= r E; j = rB

c @t

Thus

1 @ E

~ ~ ~ ~ ~ ~

f =r EE +rB B

c @t

Through suitable use of Maxwell's equations this can b e recast to

1 1 @

2 2

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

EB f =r EE ErE+r BB BrB r E + B

2 c @t

~ ~

This is not a particularly elegant expression but is symmetrical in E and B . The

approach can b e simpli ed byintro ducing the Maxwell Stress Tensor,

1 1

2 2

+ B B 123 T = E E E B

i j ij i j ij ij

2 2

For example the indices i and j can refer to the co ordinates x, y , and z , so that the

Maxwell Stress Tensor has a total of nine comp onents 3 3. E.g. with and

0 0

explicitly stated instead of the units we usually use with c

1 1 1

2 2 2 2 2 2

B B B E E + E E E + B B

0 x y 0 x y

x y z x y z

2 2

0 0

1 1 1

2 2 2 2 2 2

T = 6 E E + B B B B B E E E +

ij 0 x y x y 0

y z x y z x

2 2

0 0

1 1 1

E E + B B E E + B B E

0 x z x z 0 y z y z 0

0 0

And thus the force p er unit volume is then

@S 1

~ ~ ~

124 f = r T

c @t

And by Stoke's Law

ZZ ZZ Z

d 1

~ ~ ~ ~

F = Sd V 125 T dA

c dt

sur f ace V

This turns out to b e a much more compact equation in 4-D vector notation. 31

For 4-dimensions the force lawisf =F j .

Wewant the full generalized relation b etween the energy-momentum tensor,

T , and the 4-force to b e:

~ ~

F = 2 T 126

X X

f = T T 127

; ;

where the last term represents the rep eated indices summation convention. One

uses indicates partial derivative with resp ect to x and rep eated index to

;index r mindex

indicate summation on that index to make the equations easier to write and view.

For example,

f = T + T + T

x xx;x xy ;y xz ;z

force = pressure + shear stress 128

For electromagnetism the force equation is

f = F j = F F 129

;

since F = j .Thus wehave

;

T = F F 130

; ;

A tensor satisfying this equation is

1 1

T = 131 F F F F

4 4

1 1

T = F F F F 132

4 4

1 1

F F g F F 133 T =

4 4

First consider the Maxwell stress tensor,

1 1 1

2 2

T = E E + 134 E B B B

ij 0 i j ij i j ij

2 2

2 2 2 2 2 2

T = E E E + B B B 135

x y z x y z

2 2

B B 136 T = E E +

x y xy 0 x y

and so on. Bear in mind that the stress tensor is symmetric. It is also p ossible to

add some additional terms.

1 1

00 2 2

~ ~

T = E + B + rE 137

8 4 32

1 1

~ ~ ~ ~

T = E B + rA E 138

4 4

1 1 @

~ ~ ~ ~

T = E 139 E B + rB

4 4 @x

The added terms uses the free eld j = 0 Maxwell equations and included for

completeness. If the elds are reasonably lo calized, then T is the eld energy

0i i

density, and the T = cP is the comp onents of the eld momentum densityor

f ield

the Poynting vector S .Thus a simpli ed form is

2 2

E + B

T = 140

S Maxwell Stress Tensor

2 3

2 2

E +B

E B E B E B E B E B E B

y z z y z x x z x y y x

2 2 2 2 2 2

6 7

E E E +B B B

x y z x y z

6 7

E B E B E E + B B E E + B B

y z z y x y x y x z x z

6 7

T = 6 7

2 2 2 2 2 2

E E E +B B B

y x z y x z

6 7

E B E B E E + B B E E + B B

z x x z x y x y y z y z

4 5

2 2 2 2 2 2

E E E +B B B

z x y z x y

E B E B E E + B B E E + B B

x y y z x z x z y z y z

141 33

5.21 Bopp Theory

In classical electromagnetic theory there are two additional factors that must b e taken

into account: 1 the nite sp eed of light which means that the charge distribution

can change and the change only propagates at the sp eed of light and 2 the 1/r form

of the p otential means that any p ointcharge has in nite energy.To takeinto account

the motion of charges one must end up using retarded p otentials. In 3-D one has:

1 j tr =c; ~x

i 12 2

A t; ~x = dV 142

i 1 2

c r

Bopp suggested a simpler form of the 4-vector p otential which he thought

might handle b oth problems:

ZZZZ

A ~x = jt;~x fs dV dt 143

1 2 2 2 2

Where f s is a function which is zero every where but p eaks when the square of

the 4-vector distance s between the source 2 and the p ointofinterest 1 is very

small. The integral over f s is also normalized to unity. The Dirac delta function

2 2

is the limiting case for f s . Thus f s is nite only for

12 12

2 2 2 2 2

s = c t t r 144

1 2

12 12

Rearranging and taking the square ro ot

ct t r 1 r 1 145 r

1 2 12 12

2 2

r 2r

12 12

t t 146

1 2

c 2cr

whichsays that the only times t that are imp ortant in the integral of A are those

which di er from the time t , for which one is calculating the 4-p otential, by the

delay r =c ! { with negligible correction as long as r .Thus the Bopp theory

12 12

approaches the Maxwell theory as long as one is far away from any particular charge.

By p erforming the integral over time one can nd the approximate 3-D volume

2 2

integral by noting that f s has a nite value only for t =2 =2r c, centered

2 12

at t r =c. Assume that f s =0=K, then

1 12

Z Z

j t r =c; ~x K

12 2

dV 147 A ~x = j t ;~x fs dV dt

2 1 2 2 2 2

c r

which is exactly the 3-D version shown ab oveifwe pick K so that K =1.

This manner of thinking eventually leads one to the interaction Lagrangian as

a the pro duct of the two currents electrical, matter, strong, weak, gravitational. 34

5.22 The Principle of Covariance

The laws of physics are indep endent of the choice of space-time co ordinates.

General Relativity applies this to all conceivable space-time co ordinates:

rotating, accelerating, distorting, non-Euclidean, non-orthogonal, etc.

Sp ecial Relativity applies this only to the choices of Euclidean pseudo-

Euclidean, non-rotating co ordinate moving with constantvelo cities with resp ect to

each other.

Einstein said that this principle is an inescapable axiom , since co ordinates are

intro duced only by thought and cannot a ect the workings of Nature.

Therefore the Principle of Covariance cannot have Physical Content to

determine the laws of any part or eld of physics.

Tensors are essential b ecause all tensor equations of prop er form are manifestly

covariant; their functional form do es not change when co ordinates are changed.

Prop er form means that b oth sides of the equation result in tensors of the same

rank and, if the equation matches the classical limit formula, then it is the only

correct form. get the stu in these parentheses, precisely right.

The form of a tensor equation provides no guide for selecting a particular

\ xed" or \at rest" co ordinate system. However, its contentmay provide this.

Covariance Language has heuristic invariance:

1 It guides in pro ceeding, without telling where to go.

2 It helps to prevent errors from staying with particular co ordinates through

oversight or error.

3 One should take as a rst approximation to physical laws those which are simple

in tensor language, but not necessarily simple in a particular co ordinate system. 35