PH4506: Notes for Part 2: Lorentz Covariant Formulation

Dr Leek Meng Lee

Version: 5th February 2015

AbstractThe logic flow for this part is:

1. We derive Lorentz transformations based on Einsteins 2 axioms of Special Relativity.

2. With Lorentz transformations, we can get all the standard consequences of Special Relativity.

3. Then we formalize Special Relativity by setting up the vector space for it (i.e. by defining theMinkowski metric) and discuss about vectors and scalars in that space.

4. Then we attempt to rewrite electrodynamics into the language of Special Relativity by firstdeducing how the fields transform. Turns out they transform like a tensor in Minkowski space.

5. Then all the important equations in electrodynamics are written into a Lorentz covariant form(i.e. every inertial frame sees the same form of equations) with perfect consistency. Lorentzforce equation has the upgrade where the momentum is now the relativistic version.

6. Using the new language for electrodynamics, we revisit the point charge case and (i) rederiveLienard-Wiechert potentials in covariant form, (ii) resolve the coincidence why the electric fieldpoints from the present position and not the retarded position, (iii) realise that power is aLorentz scalar and deduce a suitable generalization of Larmors formula.

7. Finally, we recast the Lorentz covariant form of electrodynamics into the framework of Hamil-tons variational principle and Lagrangian dynamics.

Contents

1 Special Relativity 21.1 Derivation of Lorentz transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.2 Basic consequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.2.1 Time Dilation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.2.2 Length Contraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.2.3 Velocity Addition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.3 Further consequences (Minkowski spacetime diagram, invariant interval and causality) . . 61.4 Further Developments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

1.4.1 Minkowski metric, 4-vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.4.2 Momentum 4-vector and Einstein relation . . . . . . . . . . . . . . . . . . . . . . . 91.4.3 Derivative 4-vectors and dAlemberts operator . . . . . . . . . . . . . . . . . . . . 111.4.4 Relativistic Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

2 Electrodynamics recast into a manifestly Lorentz covariant form 132.1 Motivation: Magnetism as a relativistic phenomena . . . . . . . . . . . . . . . . . . . . . . 132.2 Transformations of ~E and ~B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152.3 Rewriting electrodynamics or unification of ~E and ~B . . . . . . . . . . . . . . . . . . . . . 18

2.3.1 Maxwells equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182.3.2 Conservation Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212.3.3 Lorentz invariants in electrodynamics . . . . . . . . . . . . . . . . . . . . . . . . . 232.3.4 Point charge revisited: Lienard-Wiechert Potentials . . . . . . . . . . . . . . . . . 232.3.5 Point charge revisited: Fields of a charged particle in constant velocity . . . . . . . 252.3.6 Point charge revisited: Lienards generalization of Larmors formula . . . . . . . . 27

3 Lagrangian description of a classical relativistic U(1) gauge theory 313.1 Review of Lagrangian mechanics and Noethers theorem . . . . . . . . . . . . . . . . . . . 313.2 Relativistic Lagrangian of a charged particle interacting with external EM field . . . . . . 323.3 Lagrangian of the EM field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

1

1 Special Relativity

1.1 Derivation of Lorentz transformation

Einsteins axioms for special relativity are:

1. Laws of nature and the results of all experiments are equivalent in all inertial frames. Inertialframes are reference frames at relative constant velocities to one another.

2. Speed of light has the same value in all inertial frames and is independent of the motion of thelight source. Speed of light is also the upper limit of physical entities.

Note that from axiom 1, we can say that space is isotropic (all spatial directions are equivalent)and spacetime is homogeneous (origin of spacetime can be chosen arbitrarily). This will mean that thetransformation is linear.

In fact, without axiom 2, we have the familiar Galilean transformation in Newtonian mechanics. Itis a linear transformation.

Figure 1: Two inertial frames with S-frame at speed u relative to S-frame.

Assuming the origins coincide at t = 0, we have x = x + ut inverse x = x ut (1)= dx

dt=dx

dt+ u (2)

= v = v + u (3)which is the usual relative velocity expression.Now we start with a more general linear transformation and incorporate axiom 2 at the end.

x = (u)(x ut) (4)y = (u)y (5)z = (u)z (6)t = (u)t+ (u)x (7)

The coefficients of y and z are the same due to isotropy. Again due to isotropy, flipping x, x and uwill result in no change (we flipped y and y so that we maintain a right handed coordinate system).

x = (u)(x+ ut) (8)y = (u)y (9)z = (u)z (10)t = (u)t (u)x (11)

We compare to get (u) = (u), (u) = (u), (u) = (u) and (u) = (u) so only is an oddfunction. We write (u) into this form where (u) is even.

(u) = u(u)

(u) (12)

so the set of transformation equations are

x = (u)(x ut) (13)

2

y = (u)y (14)z = (u)z (15)

t = (u)(t u

(u)x

)(16)

We can jump into S-frame and by axiom 1, the equations must have the same form. We interchangeprimed and unprimed coordinates and reverse the sign of u.

x = (u)(x + ut) (17)y = (u)y (18)z = (u)z (19)

t = (u)

(t +

u

(u)x)

(20)

Substitute y = (u)y to get y = (u)2y so we choose = 1 (21)

Then we proceed further with the x and t equations,

x = (x ut) x = x ut (22)t =

(t u

x

)u ut = ut u

2

x (23)

Eliminate t = x+ ut = x u2

x (24)

x =x+ ut

u2 (25)

=x

(

1 u2) + ut

(

1 u2) (26)

Compare with x = (x + ut) = x + ut to get =1

(

1 u2) and = 1

(

1 u2) so that =

and =1

1 u2(u). The positive root is chosen so that when u 0, 1 and x x.

The transformation equations are now,

x = (x ut) (27)y = y (28)z = z (29)

t = (t u

x

)(30)

We determine by using axiom 2 where every inertial frame sees speed of light as c. So we need thevelocity addition formula. Take infinitesimal intervals,

dx = (dx udt) (31)dt =

(dt u

dx

)(32)

divide one equation by the other,dx

dt=

dx udtdt u dx

(33)

=dxdt u

1 u dxdt(34)

v =v u1 uv

(35)

3

where v is the velocity of the object measured in S-frame and v is the velocity of the object measuredin S-frame. Now consider the object to be light, then v = c and v = c, so

c =c u

1 uc(36)

= c2 (37)

Finally with =1

1 u2c2, the Lorentz transformation equations are,

x = (x ut) (38)y = y (39)z = z (40)

t = (t u

c2x)

(41)

We generalise this for arbitrary relative velocities. We denote ~u to be the relative velocity in an arbitrarydirection. We resolve the position vector into a component parallel to ~u and a component perpendicularto ~u.

~x ~u = ux cos = ux (42)x =

~x ~uu

(43)

~x =~x ~uu

u (44)

=~x ~uu2

~u (45)

hence, ~x = ~x ~x = ~x ~x ~uu2

~u (46)

The transformations of the parallel component and the perpendicular component has been derived earlier.We note it again,

~x = (~x ~ut) (47)~x = ~x (48)

t = (t ~u ~x

c2

)(49)

and,

~x = ~x + ~x (50)

= (~x ~ut) + ~x (51)

=

(~x ~uu2

~u ~ut)

+ ~x ~x ~uu2

~u (52)

= ~x+ ( 1)~x ~uu2

~u ~ut (53)

Since ~x ~u = 0, we can also write ~x ~u = ~x ~u, so

t = (t ~u ~x

c2

)(54)

1.2 Basic consequences

1.2.1 Time Dilation

We recall the 2 inertial frames setup where S-frame is moving at speed u with respect to the x-axis ofthe S-frame. Let a clock be in the S-frame (or S-frame is comoving with a clock that is moving withspeed u along the x-axis of the S-frame).

4

Let event 1 be the second-hand of the clock at the 12 oclock-mark. Let event 2 be the second-hand of the clock at the 1 oclock-mark.Recall the inverse transformation: t =

(t + uc2x

) and we want to find a relationship between timeintervals as measured from each frame.

t1 = (t1 +

u

c2x1)

and t2 = (t2 +

u

c2x2)

(55)

Take difference, t2 t1 = (t2 t1 +

u

c2(x2 x1)

)(56)

t = (

t +u

c2x

)(57)

| since the clock is in the same x coordinate, x = 0t = t (58)

Thus if t = a 5 second interval (measured in S-frame), then t > 5 seconds as measured in S-frame.In other words, if there is a similar clock in S-frame and both clocks are set to start ticking from the12 oclock-mark together, when the S-clock is ticking at the 1 oclock-mark, the S-clock is seen to beticking at the -mark. 1

1.2.2 Length Contraction

Let a stick be on the x-axis in the S-frame.

Let event 1 be the left end of the stick. Let event 2 be the right end of the stick.

Recall the inverse Lorentz transformation: x = (x + ut) and we want to find a relationship betweenspatial intervals as measured from each frame.

x1 = (x1 + ut

1) and x

2 = (x

2 + ut

2) (59)

Take difference, x2 x1 = (x2 x1 + u(t2 t1)) (60)= (x + ut) (61)| seeing the stick means both ends are observed simultaneously so t = 0| then recall t =

(t +

u

c2x

)so t = u

c2x

=

(x u

2

c2x

)(62)

| recall = 11 u2c2

=x

(63)

So, x is the length of the stick measured in S-frame where the stick has no relative motion and xis the length of the stick measured in S-frame and x < x. This is due to relative simultaneity: werequire both ends of the stick to be observed simultaneously in S-frame but actually these 2 events arenot simultaneous in S-frame. The opposite case is true also. This fact can be illustrated clearly on theMinkowski spacetime diagram which we will learn later.

1.2.3 Velocity Addition

We have already derived the velocity addition formula which is v =v u1 uvc2

but lets state the situation

more precisely.Basically, there are 3 inertial frames: S, S and S. S-frame is moving at speed u relative to S-frame

and S-frame is moving at speed v relative to S-frame. Then v is the speed of S-frame relative toS-frame.

1Assuming < 12 for simplicity.

5

Figure 2: A 3 inertial frame setup.

We shall go further and relate the various factors: (v) =1

1 v2c2, (u) =

11 u2c2

and

(v) =1

1 v2c2,

(v) =1

1 v2c2(64)

=

(1 1

c2

(v u1 uvc2

)2)1/2(65)

=

((1 uvc2

)2 (vu)2c2(1 uvc2

)2)1/2

(66)

| expand the numerator

=

(1 2uv

c2+u2v2

c4 v

2 2uv + u2c2

)1/2 (1 uv

c2

)(67)

=

(1 v

2

c2 u

2

c2+u2v2

c4

)1/2 (1 uv

c2

)(68)

=

((1 u

2

c2

)(1 v

2

c2

))1/2 (1 uv

c2

)(69)

= (u)(v)(

1 uvc2

)(70)

1.3 Further consequences (Minkowski spacetime diagram, invariant intervaland causality)

Minkowski spacetime diagram: The Minkowski spacetime diagram is one way of graphically rep-resenting both S and S-frames together so that by referring to the S-frame axes, we get the coordinatesin S-frame and by referring to the S-frame axes, we get the coordinates in S-frame.

We will construct S-frame in the usual way where the ct vs x axis are 90 degrees to each other. TheS-frame shall be constructed based on Lorentz transformations.

By definition, ct axis means x = 0,

x = (x ut) x=0 0 = x ut (71)ct =

c

ux (72)

By definition, x axis means t = 0,

t = (t ux

c2

)t=0 0 = t ux

c2(73)

6

ct =u

cx (74)

Note that the spacings are not the same on the axes of different frames: take coordinate (x, ct) = (1, 0)and using Lorentz transformations

1 = (x ut) and 0 = t uxc2

= x = and t = uc (75)

Figure 3: Summary of the main features in Minkowski spacetime diagram.

Invariant interval: Consider this strange Pythagoras theorem for length in S-frame

c2t2 + x2 = c22(t ux

c2

)2+ 2(x ut)2 (76)

= 2(c2t2 + 2tuxc

2

c2 u

2x2

c2+ x2 2xut+ u2t2

)(77)

= 2(c2t2 + x2 u

2x2

c2+ u2t2

)(78)

=1

1 u2c2

(c2t2

(1 u

2

c2

)+ x2

(1 u

2

c2

))(79)

= c2t2 + x2 (80)

so this length is the same in S-frame and therefore this length is called the invariant interval becauseit is the same (numerical value) regardless of the inertial frame.

If we choose a certain inertial frame where a clock is stationary, the time measured (in this comovingframe) is called the proper time and is denoted by . As the clock is always at the origin in this frame,x2 = 0 and so the invariant interval in this frame is c22. We can write the following forms: c22 = c2t2 + x2

Full 4D: c22 = c2t2 + x2 + y2 + z2

Define infinitesimal invariant interval ds: (ds)2 = c2(d)2 = c2(dt)2 + (dx)2 + (dy)2 + (dz)2

7

Conventional writing: ds2 = c2d2 = c2dt2 + dx2 + dy2 + dz2

Since there are 3 positive terms and 1 negative term in the invariant interval, it can be positive, zero ornegative. We illustrate this on the Minkowski diagram.

Figure 4: 3 separate regions in the Minkowski spacetime diagram.

There should be a z-axis also but I cant draw it. Thus the light cones should really be hypercones.The 3 regions are named as follows:

For interval < 0, it is called the timelike region. This is the region where particles with massmove because particles with mass can only move slower than c.

For interval = 0, it is called the lightlike region. This is the region where photons travelbecause photons only travel with speed c.

For interval > 0, it is called the spacelike region. This is the region where particles movefaster than c.

1.4 Further Developments

1.4.1 Minkowski metric, 4-vectors

We will now carry out a more formal treatment so that we can handle 4D vectors (in Minkowski space).This is an upgrade of the treatment of 3D vectors.

3D ~x =i

xiei where ei = ex, ey, ez are the basis vectors (previously denoted as x, y, z) (81)

4D x =

xe where e = ect, ex, ey, ez are the basis vectors (82)

Now we introduce the Einstein summation convention: whenever 2 identical indices appear (one in thesuperscript, one in the subscript), a summation is implied. Greek indices run from 0 to 3 and Latinindices run from 1 to 3. Index 0 = ct, 1 = x, 2 = y and 3 = z.

Now consider the dot product of 3D and 4D vectors with itself which gives the squared length:

3D ~x ~x = xiei xj ej = xixj(ei ej) (83)4D x x = xe x e = xx(e e) (84)

We shall define e e as the Minkowski metric tensor since the basis vectors characterise the space.Note that because the dot product commutes, therefore = . Note that the 4D dot product issupposed to give the squared length so this 4D dot product should be equal to the invariant interval.

c2t2 + x2 + y2 + z2 = x x = xx(e e) = xx (85)

with x, x = (x0, x1, x2, x3) = (ct, x, y, z) we require =

1

11

1

(86)8

We can take another perspective where the dot product is formed by a vector from a space with anothervector from the conjugate space. This is like in quantum mechanics where the dot product is formedbetween a vector in ket-space and another vector in bra-space and the 2 spaces are hermitian conjugatesof each other.

Define the contravariant position 4-vector: x = (x0, x1, x2, x3) (87)

= (ct, x, y, z) (88)

Define the covariant position 4-vector: x = x (89)

=

1

11

1

ctxyz

(90)

=

ctxyz

(91)= (ct, x, y, z) (92)

The dot product: xx =

( ct x y z )

ctxyz

(93)= c2t2 + x2 + y2 + z2 (94)

and the metric tensor maps a contravariant vector to a covariant vector. This is casually described asthe metric tensor lowering an index.

To map from a covariant vector to a contravariant vector (or raising an index), we need the inversemetric tensor.

x = x = x

= x (95)

which means is the inverse of .

Explicitly, =

1

11

1

since =

11

11

.1.4.2 Momentum 4-vector and Einstein relation

We want to define a velocity 4-vector and proceed to construct a momentum 4-vector from there. Aquantity is a 4-vector if it transforms like a 4-vector under Lorentz transformations so we can definitequantities as we like it but we have to check its transformation to make sure that it is a 4-vector.

Define the contravariant velocity 4-vector: u =dx

d(96)

This guess is made by requiring a time derivative on the position 4-vector and choosing the time to bethe proper time because the numerator is already a 4-vector and so we do not want the denominator totransform also. Thus we can

Define the contravariant momentum 4-vector: p = mu = mdx

d(97)

where m is called the rest mass which is the mass measured in the comoving frame of the particle.Now we need to check that the objects are indeed 4-vectors. First we work out the time component

(or 0-component),

p0 = mdx0

d= m

cdt

d(98)

| from time dilation, dt = (u1)d , (u1) = 11 u21c2

9

= mc(u1) (99)

where u1 is the relative velocity of the particle with respect to the S-frame. The comoving frame of theparticle is called the S-frame.

Figure 5: 3 inertial frame setup for deriving the Lorentz transformations of the momentum 4-vector.

We expect a transformation of the form for the space component:

x = (x ut) = (x u

c(ct))

(100)

(space component) = (

space component uc

(time component))

(101)

So we check from S-frame which is moving at speed u with respect to S-frame,

p1 = mdx1

d(102)

= mdx

d(103)

| recall dx = (u)(dx udt)= m(u)

(dx

d u dt

d

)(104)

| recall time dilation dt = (u1)d= m(u)

(p1

m u(u1)

)(105)

= (u)(p1 u

c(mc(u1))

)(106)

= (u)(p1 u

cp0)

(107)

Indeed p1 transforms like the space component. Now we check the time component which is expected totransform as

t = (t u

c2x)

(108)

ct = (ct u

cx)

(109)

(time component) = (

time component uc

(space component))

(110)

So, p0 ?= (u)(p0 u

cp1)

(111)

The LHS = p0 = mc(u1) (112)

with (u1) =1

1 u21

c2

where u1 is the velocity of the particle (S-frame) with respect to the S-frame.

10

Velocity addition says that u1 =u1 u1 u1uc2

.

The RHS = (u)(p0 u

cp1)

(113)

= (u)

(mc(u1) u

cmdx

d

)(114)

| recall dt = (u1)d and dxdt

= u1

= (u)(mc(u1) u

cmu1(u1)

)(115)

= mc(u)(u1)(

1 uu1c2

)(116)

| recall the identity (u)(u1)(

1 uu1c2

)= (u1)

= mc(u1) (117)

Thus indeed p0 transforms as a time component. We rewrite the contravariant momentum 4-vector as,

p = (p0, p1, p2, p3) =

(E

c, px, py, pz

)= (mc(u1),mu1(u1), 0, 0) (118)

We shall check that E = mc2(u1) is indeed the total energy.

E = mc2(u1) = mc2

(1 u

21

c2

)1/2(119)

| assume the particle moves slowly, so u1

We need to check that they are really 4-vectors by checking their transformations.

x0=

1

c

t=

x

ct

x(129)

=ct

ct

ct+

x

ct

x+ 0 + 0 (130)

| recall Lorentz transformations ct = (ct +

u

cx)

and x = (x + ut)

=

ct+

u

c

x(131)

so if we recall the Lorentz transformation ct = (ct uc x

), we can deduce that ct =

x0 transforms

like the time component of a covariant (lower index) 4-vector. To confirm, we check further,

x1=

x=

ct

x

ct+x

x

x+ 0 + 0 (132)

| recall the Lorentz transformations ct = (ct +

u

cx)

and x = (x + ut)

= u

c

ct+

x(133)

=

(

x+u

c

ct

)(134)

so if we recall the Lorentz transformation x = (x ut), we again realise that the time componentbehaves with a wrong sign which means that it is the time component of a covariant (lower index)4-vector, thus x is a covariant 4-vector.

We can immediately check for x ,

x0= 1

c

t=

((1c

t uc

x

))(135)

x1=

x=

(

x uc

( ct

))(136)

which means x is behaving like a contravariant 4-vector! Hence we denote:

=

x=

(

x0,

x1,

x2,

x3

)=

(1

c

t,

x,

y,

z

)(137)

=

x=

(

x0,

x1,

x2,

x3

)=

(1c

t,

x,

y,

z

)=

( x0

,

x1,

x2,

x3

)(138)

We shall now take the dot product of these 2 vectors. Recall that it is supposed to give us a quantitythat is Lorentz invariant (Lorentz scalar).

= 1c22

t2+

2

x2+

2

y2+

2

z2(139)

= 1c22

t2+ ~2 (140)

where = is called the 4D Laplacian or dAlembertian operator and ~2 is the 3D Laplacianoperator.2

1.4.4 Relativistic Mechanics

Newtons first law is bulit into the first axiom of special relativity in the sense that in inertial frames,there is no net force due to relative motion and so different inertial frames would experience the samephysics.

For Newtons second law, the most intuitive way to fit it into special relativity is to modify themomentum to relativistic momentum.

~F =d~p

dt=

d

dt(m~v(v)) (where v is the speed of the particle) (141)

2In some EM books, the dAlembertian has the symbol 2 = .

12

This relation is for S-frame where the particle is seen to be moving at speed v. How about in S-framewhich is moving at speed u with respect to S-frame along the x axis? We have to transform bothnumerator and denominator! For simplicity, we shall take S-frame to be the (instantaneous) comovingframe of the particle, i.e. v = u. For the y component,

F y =dpydt

=dpy

(u)(dt uc2 dx

) = dpydt(u)

(1 uc2 dxdt

) (142)| where dx

dt= u

=Fy

(u)(1 u2c2

) = (u)Fy (143)Similarly for the z component

F z = (u)Fz (144)

Now for the x component,

F x =dpxdt

=(u)

(dpx uc dp0

)(u)

(dt uc2 dx

) = dpxdt uc dp0dt1 uc2 dxdt

(145)

=Fx uc2 dEdt

1 u2c2(146)

| where dEdt

=d

dtmc2(u) = mc2

2uc2

2(1 u2c2

)3/2 dudt = u ddtmu(u) = uFx= Fx (147)

Thus the ordinary force transforms awkwardly. We try to define another force quantity which is a 4-vector. We could take the clue from the definition of 4-velocity and define the 4-force (or Minkowskiforce) as

K =dp

d(148)

where p is the 4-momentum and is the proper time. The spatial components are

~K =

(dt

d

)d~p

dt= (u)~F (where ~F is the ordinary force) (149)

The time component is

K0 =dp0

d=

1

c

dE

d(150)

where dEd can be interpreted as the rate of energy increase with respect to proper time or proper power.

There is nothing wrong with using ~F = d~pdt where ~p is the relativistic momentum except for awk-wardness in transformation. Therefore we shall use the covariant Minkowski force K for the sake ofconvenience.

2 Electrodynamics recast into a manifestly Lorentz covariantform

2.1 Motivation: Magnetism as a relativistic phenomena

The objective from this point onwards is to make it clear that electrodynamics is meant to be describedin Minkowski space. The ~E and ~B fields are just 2 sides of the same coin. This means that they are 6components of a single object called the field strength tensor F . The ~E and ~B fields shouldnt reallybe seen as 2 different fields. The rewriting of electrodynamics in the language of special relativity doesnot really generalise anything. It shows this unification of ~E and ~B clearly.

13

So in this section, we want to deduce magnetism from the knowledge of electrostatics and specialrelativity to motivate the intimate connection between electrodynamics and special relativity.

Consider, in S-frame, we see a wire and inside the wire, we see a string of positive charges movingto the right with speed v. We assume that a linear charge density can be defined for it. In the samewire, superimposed on the positive string of charges is a negative one, with moving to the left withspeed v. The net current is to the right and given by

I = 2v (151)

For a charge q, distance r outside the wire and moving to the right with speed u(< v), there is noelectrical force since the 2 line charges has no net charge.

We now go into S-frame which is moving to the right with speed u with respect to S-frame. Usingthe velocity addition formula,

speed of positive line charge in S-frame: v+ =v u1 vuc2

(152)

speed of negative line charge in S-frame: v =v + u

1 + vuc2(153)

(154)

The linear charge densities are seen to change by different amounts due to length contraction.

For positive (in S-frame): + = (v+)0 with (v+) =1

1 v2+c2(155)

For negative (in S-frame): = (v)0 with (v) =1

1 v2c2(156)

where 0 is the charge density seen in the comoving frame of the charges,

then, = (v)0 with (v) =1

1 v2c2(157)

and using the velocity addition identities (v+) = (v)(u)(1 uvc2

)and (v) = (v)(u)

(1 + uvc2

), we

can calculate the net charge seen in S-frame

net = + + = 0 ((v+) (v)) (158)= 0(v)(u)

(1 uv

c2 1 uv

c2

)(159)

=2uv

c2

1 u2c2(160)

14

Thus there is an electric force on charge q in S-frame,

F = qE = qnet

2pi0r(161)

= vpi0c2r

qu1 u2c2

(162)

But both S and S-frames must see the same physical outcome so there must also be a force on q inS-frame. We transform the force

F =1

(u)F (refer to the previous section on relativistic mechanics) (163)

= vpi0c2

qu

r(164)

| write 1c2

= 00 and 2v = I

= qu(0I

2pir

)

magnetic field in S-frame

(165)

So this force is the Lorentz force on a charged particle moving in a magnetic field.

2.2 Transformations of ~E and ~B

To begin the rewriting of electrodynamics into Lorentz covariant form 3, first we must know how the ~Eand ~B fields transform and realise that they do not form a 4-vector.

Consider a flat plate capacitor moving with speed u with respect to S-frame. Let the capacitor bein S-frame, i.e. this S-frame is the comoving frame of the capacitor.

Figure 6: Capacitor in S-frame and seen from S-frame. Capacitor plates are parallel to x axis.

In S-frame, the capacitor has length l and width w and surface charge density . Thus the electricfield seen in S-frame:

Ey =

0(166)

In S-frame, the length of the capacitor is measured with length contraction

l =1

(u)l where (u) =

11 u2c2

(167)

Since total charge Q is invariant 4 and width w does not undergo length contraction as it is perpendicularto the direction of motion, the charge density measured in S-frame is = Qlw = (u

) Qlw = (u).

3Lorentz covariant means the expression has the same form in all inertial frames.4I am taking this to be an empirical fact. It can be shown by Noethers theorem which will be discussed at the end of

Part 2.

15

The electric field measured in S-frame is5

Ey =

0= (u)

0= (u)Ey (168)

Now consider the capacitor plates to be perpendicular to the x axis,

Figure 7: Capacitor in S-frame and seen from S-frame. Capacitor plates are perpendicular to x axis.

So the capacitor plate spacing d undergoes length contraction but the electric field (which is now inthe x direction) does not depend on d so,

Ex = Ex (169)

We can now talk about the transformation of magnetic fields because in S-frame, 2 moving chargedplates (we are back to the situation where the capacitor plates are parallel to the x axis) amounts to 2surface currents.

In S-frame, the magnetic field points in the +z direction between the plates using the right handrule. Using Amperes law, the magnetic field between the plates is

In S-frame: Bz = 0|K| = 0u (170)To compare Lorentz transformed fields, we need to bring a 3rd inertial frame and we call it the S-framewhich is moving at speed u with respect to S-frame.

Figure 8: Capacitor in S-frame and seen from S-frame and from S-frame. Capacitor plates are parallelto x axis.

From the addition of velocities, the relative velocity of S-frame with respect to S-frame is u1 =u u1 uuc2

.

In S-frame: Ey =

0= (u1)

0= (u1)E

y (171)

5We may suspect that the electric field may not be perpendicular to the plates but by symmetry considerations, anyparallel component from the + plate will be cancelled by an opposite parallel component from the plate.

16

In S-frame: Bz = 0u1 = 0(u1)u1 (172)

Finally, we need to relate Ey to Ey and Bz and relate Bz to Ey and Bz. Eliminating the double primed

fields will give us the Lorentz transformation of the fields between S-frame and S-frame.

Ey = (u1)Ey (173)

| recall the identity (u1) = (u)(u)(

1 uu

c2

)| and recall (u)Ey = Ey= (u)Ey

(1 u

u

c2

)(174)

= (u)(Ey u

c2uEy

)(175)

= (u)(Ey u

c2u

0

)(176)

| recall that Bz = 0u

= (u)(Ey u

c200Bz

)(177)

| recall 1c2

= 00

= (u) (Ey uBz) (178)Then for Bz

Bz = 0(u1)u1 (179)

| recall identity (u1) = (u)(u)(

1 uu

c2

)and u1 =

u u1 uuc2

and = Ey 0

= 0(u)(u)Ey 0(u

u) (180)| recall (u)Ey = Ey= 00(u

)Ey(u u) (181)= (u) (00Eyu 00Eyu) (182)| recall Ey =

0and Bz = 0u

and 00 =1

c2

= (u)(Bz u

c2Ey

)(183)

We can repeat with the capacitor plates set parallel to the xy plane and get

Ez = (u) (Ez + uBy) (184)

By = (u)(By +

u

c2Ez

)(185)

To get the transformation for Bx, we imagine a solenoid in S-frame, with its axis aligned to the x axis.

Bx = 0nI (186)

| in S-frame, time dilation gives I = 1(u)

I

| in S-frame, length contraction gives n = (u)n= 0nI (187)

= Bx (188)

So for ~E and ~B, the component parallel to motion is unaffected and this is weird because so far we donot have any quantity transforming in this way.

Finally, we collect the whole set of transformations,

Ex = Ex Ey = (u

) (Ey uBz) Ez = (u) (Ez + uBy)Bx = Bx B

y = (u

)(By +

uc2Ez

)Bz = (u

)(Bz uc2Ey

) (189)17

This set of transformation is nowhere similar to a 4-vector transformation. This also gives a strong hintthat ~E and ~B are not different fields as they transform into each other.

2.3 Rewriting electrodynamics or unification of ~E and ~B

2.3.1 Maxwells equations

Transformations of ~E and ~B are not like that of a 4-vector but they are that of an antisymmetric 2-tensor.6 An antisymmetric (4D) 2-tensor has 6 elements and there are exactly 6 field components in~E and ~B. Thus the conclusion is that, there is only one unified field in electrodynamics and it is the2-tensor field strength denoted as F .

We take on a matrix way of writing:

Position 4-vector:

ct = (u)(ct uc x

)x = (u)(x ut)y = yz = z

ct

x

y

z

=

uc 0 0uc 0 00 0 1 00 0 0 1

ctxyz

= x = x (190)Momentum 4-vector:Ec = (u)

(Ec uc px

)px = (u)

(px uc

(Ec

))py = pypz = pz

E/cpxpypz

=

uc 0 0uc 0 00 0 1 00 0 0 1

E/cpxpypz

= p = p(191)

Thus the Lorentz transformation of the field strength 2-tensor F is 7

F = F

(192)

| we write it into matrix multiplication format= F

(193)

= F(T ) (194)

in matrix notation: F = FT where T is the transpose of (195)

It turns out that the correct arrangement of Ex, Ey, Ez, Bx, By and Bz in F so that F = FT gives

the same transformation results with the earlier flying capacitor derivation is

F =

0 Ex/c Ey/c Ez/c

Ex/c 0 Bz ByEy/c Bz 0 BxEz/c By Bx 0

(196)The field strength in S-frame would simply be denoted as

F =

0 Ex/c E

y/c E

z/c

Ex/c 0 Bz ByEy/c Bz 0 BxEz/c By Bx 0

(197)We quickly check 2 examples:

F 10 = 10F

(198)

Ex

c= 10

0F

0 + 110F

1 (199)

= 1001F

01 + 1100F

10 (200)

=(uc)(uc)(Ex

c

)+

(Exc

)(201)

6When I say 4-vector, the 4 indicates dimension and when I say 2-tensor, the 2 indicates number of indices.7At worst, you can take it as the recipe that for every index, you need one Lorentz matrix to transform.

18

Ex = Ex(

1 u2

c2

)2 (202)

Ex = Ex (indeed) (203)

F 20 = 20F

(204)

Ey

c= 22

0F

2 (205)

= 22(00F

20 + 01F21)

(206)

=

(Eyc

) uc(Bz) (207)

Ey = (Ey uBz) (indeed) (208)

You can also directly carry out matrix multiplication F = FT to check every element.Before we see Maxwell equations in covariant form, we first need to realise that there is a current

density 4-vector:Take an infinitesimal volume V which has charge q, the charge density is = qV . Assume that V is

moving along the x axis with speed u, the current density is Jx = u. We define the rest charge densityas 0 =

qV0 . Since only one dimension is contracted, V = 1(u)V0 and so

=q

V0 (u) = 0(u) (209)Jx = u = 0u(u) (210)

We recall the (contravariant) momentum 4-vector p =(Ec , px, py, pz

)= (mc(u1),mu1(u1), 0, 0), we

are inspired to define the (contravariant) current density 4-vector

J = (c, Jx, Jy, Jz) = (J0, J1, J2, J3) = (0c(u), 0u(u), 0, 0) (211)

It must be noted that the charge continuity equation is actually a Lorentz invariant equation.8

t+ ~ ~J = 0 (212)

1

c

J0

t+Jxx

+Jyy

+Jzz

= 0 (213)

J0

x0+J1

x1+J2

x2+J3

x3= 0 (214)

J = 0 (charge continuity equation in Lorentz scalar form) (215)

The Lorentz covariant Maxwell equations are just 2 equations:9

F = 0J

and F + F + F = 0 (216)

We check for example = 0,

F0 = 0J

0 (217)

0F00 + 1F

01 + 2F02 + 3F

03 = 0c (218)

0 +Ex/c

x+Ey/c

y+Ez/c

z= 0c (219)

1

c~ ~E = 0c (220)

recall that1

c2= 00 |~ ~E =

0(221)

8We call such an equation as the 4-divergence of J is zero.9Recall that the charge continuity equation is obtainable from Maxwell equations. In this covariant form, of course the

same can be done. We take 4-divergence of F = 0J to get F = 0J. The LHS is zero because issymmetric and F is antisymmetric, so J = 0.

19

which is Gauss law. The other 3 Maxwell equations shall be checked in the homework.Next is the Lorentz force law. We already have a force 4-vector (or 4-force) called the Minkowski

force K.

K =dp

d(222)

In terms of the fields, or RHS of the law ~F = q( ~E + (~v ~B)) we deduce the relativistic version byreplacing ~v u and ~B F giving the ansatz

K = qFu (223)

We check the spatial part,

K1 = qF 1u (224)

= qF 10u0 + qF11u1 + qF

12u2 + qF13u3 (225)

| note that F 11 = 0| recall u = p

m=

1

m(mc(u),mu(u), 0, 0) where u is in x direction

| now take u to have 3 components, so u = u = (c(u), ux(u), uy(u), uz(u))= q(u)

(c(Exc

)+ uyBz + uz(By)

)(226)

= q(u)(~E + (~u ~B)

)x

(227)

then,

~K =d~p

d= q(u)

(~E + (~u ~B)

)(228)

| recall dt = (u)dd~p

dt= q

(~E + (~u ~B)

)(229)

where ~p is the relativistic momentum. The details of K0 shall be worked out in the homework.Finally, we need to deal with the potentials: and ~A. Turns out that they form the potential 4-vector

(or 4-potential).

A =

(

c,Ax, Ay, Az

)(230)

which means it transforms as follows:

c= (u)

(

c ucAx

)and Ax = (u)

(Ax u

c

(

c

))and Ay = Ay and A

z = Az (231)

and we check that indeed these transformations fulfil the transformation properties of the fields. Recall~E = ~ ~At ,

Ex = x Ax

t(232)

= recall

x= 1 and 1

c

t= 0

= 1+ c0Ax (233)= Ex = 1 + c0Ax (234)

= 1 (( uAx)) + c0((Ax u

c2))

(235)

= (uc0 + 1

)(( uAx)) + c

(0 u

c1

)((Ax u

c2))

(236)

| expand and 4 terms cancel pairwise= 2(0Ax)

(u2

c c) 2(1)

(1 u

2

c2

)(237)

20

| recall 2 = 11 u2c2

= 1+ c0Ax (238)= Ex (indeed) (239)

We can check By, recall ~B = ~ ~A,

By =Axz Az

x= 3Ax 1Az (240)

= By = 3Ax 1Az (241)= 3

((Ax u

c2)) 1Az (242)

= (3Ax u

c23

)(uc0 + 1

)Az (243)

| expand out and use By = 3Ax 1Az and Ez = z Az

t= 3+ c0Az

= (By +

u

c2Ez

)(indeed) (244)

Note the very important patterns

Ex = 1+ c0Ax and By = 3Ax 1AzcF 01 = c0A1 c1A0 F 31 = 3A1 1A3F 01 = 0A1 1A0

In general

F = A A (245)is true.

Recall the Lorenz gauge

~ ~A = 00 t

(246)

1A1 + 2A

2 + 3A3 = 0A0 (247)

A = 0 (248)

Thus the Lorenz gauge is Lorentz invariant. The general gauge transformations can also have a covariantform

~A = ~A+ ~ = t

}Ai = Ai + iA0 = A0 + 0

}A = A + (249)

and the differential equations for the potentials (in Lorenz gauge).

~2 00 2t2 = 0

~2 ~A 00 2 ~At2 = 0 ~J

}(cA0) = J0c0Ai = 0J i

}A0 = 0J0Ai = 0J i

}A = 0J (250)

2.3.2 Conservation Laws

We have already seen the rewriting of the charge continuity equation into a Lorentz invariant (or scalar)form:

Charge continuity equation: J = 0 (251)

We have 2 other continuity equations: energy and momentum and from the definition of 4-momentum,we know that energy and momentum are closely related. Thus we expect the 2 continuity equations tomerge into one and has a similar form to the charge continuity equation, i.e. a 4-divergence form.

We recall from Part 1:

Energy continuity equation: ~E ~J = uEMt

~ ~S (252)

21

Momentum continuity equation: ~E + ~J ~B = ~ T 1c2~S

t(253)

SinceT is already a 3 3 matrix, we can go further and top up

T into a 2-tensor T . We have 4

objects to insert: uEM, Sx, Sy and Sz. SinceT is already symmetric, so T should also be symmetric,

thus we try

Energy-momentum tensor: T =

uEM Sx/c Sy/c Sz/cSx/c

Sy/c T

Sz/c

(254)We recall from Minkowski force: K1 = qF 1u = q(u)

(~E + (~u ~B)

)x

so we try a similar expression

FJ and it should contain ~E + ~J ~B. So finally the ansatz for the combined continuity equation isEnergy-momentum continuity equation: T

= F J (255)We check this ansatz. For = 0,

T0 = F 0J (256)

0T00 + 1T

10 + 2T20 + 3T

30 = (F 01J1 + F 02J2 + F 03J3) (257)1

c

uEMt

+1

c~ ~S = 1

c~E ~J (258)

now for = 1,

T1 = F 1J (259)

0T01 + 1T

11 + 2T21 + 3T

31 = (F 10J0 + F 12J2 + F 13J3) (260)1

c2Sxt

+(~ ( T )

)x

= ((Exc

)(c) +BzJy ByJz

)(261)

= (Ex + ( ~J ~B)x

)(262)

So indeed T is the relativistic generalization of ( T ). Lastly we must ensure that T is really a2-tensor. We take Tij = 0

(EiEj 12ij ~E2

)+ and make a (really intelligent) guess that T should

have a similar form.

T = 10

(FF

+1

4FF

)(263)

We check

T 00 = 10

(0FF

0 +1

400FF

)(264)

| recall that 00 = 1 and FF = 2~E2

c2+ 2 ~B2

= 10

(~E2

c2+

1

2

~E2

c2 1

2~B2

)(265)

=1

20 ~E

2 +1

20~B2 (266)

= uEM (267)

In the homework, you will check FF = 2~E2c2 +2 ~B2, T 0i = Sic and T ij = 0

(EiEj 12ij ~E2

)+ .

At the end of Part 2, I will explicitly take 4-divergence of equation (263) show its conservation law.Finally we just want to quickly mention the angular momentum 3-tensor,

Angular momentum 3-tensor definition: M = Tx Tx (268)Conservation law: M

= 0 (without external sources) (269)

22

2.3.3 Lorentz invariants in electrodynamics

We know that by forming dot products, we can make Lorentz scalars and they are invariant. Theyhave the same numerical value in any inertial frame. Recall

1. dot product of position 4-vector with itself gives the invariant interval

2. dot product of momentum 4-vector with itself gives the Einstein relation

3. dot product of derivative 4-vector with current density 4-vector gives the charge continuityequation

There are 2 invariants in electrodynamics. The first one is the obvious dot product of the field strengthtensor with itself,

FF = 2

(~B2

~E2

c2

)(270)

In the homework, you will check some consequences of it and verify this invariant using the Lorentztransformations of ~E and ~B.

The second invariant is much less obvious. It is deduced from the second Maxwell equation F +F + F = 0 that we can define the so called dual field strength tensor F by

F =1

2F

with being the 4D Levi-Civita symbol (271)

and the Maxwell equation becomes F = 0. We check the 0-component,

F0 =1

20

F (272)

=1

2

(1032

1F 32 + 10231F 23 + 3012

3F 12 + 30213F 21 + 2013

2F 13 + 20312F 31

)(273)

| the even permutations are 3021, 1032 and 2013| the odd permutations are 3012, 1023 and 2031| then use F 12 = F 21, F 23 = F 32 and F 13 = F 31= (1F 23 + 3F 12 + 2F 31) (274)

0 = 1F 23 + 3F 12 + 2F 31 (275)

Other components can be checked similarly. The second invariant is thus the dot product of F withF

FF = FF =

1

2F

F = 4c~E ~B (to be checked in the homework) (276)

In the homework, you will also check some consequences of it and verify this invariant using the Lorentztransformations of ~E and ~B.

2.3.4 Point charge revisited: Lienard-Wiechert Potentials

Although all these rewriting of electrodynamics into Lorentz covariant form does not really contain anynew physics (except maybe relativistic corrections), it does provide a whole new perspective and a newmachinery to uncover more physics.

We shall rederive the Lienard-Wiechert potentials (i.e. retarded potentials for a moving point charge)in relativistic language.

First we need to rewrite the (retarded) Greens function into a Lorentz invariant form.

GR(~r, t;~r, t) = (t t) 14piR

(t t R

c

)(277)

| we need an identity which we will prove now| start with ((x x)2) = ((x0 x0)2 + |~r ~r|2) = (R2 (x0 x0)2)| then = ((R x0 + x0)(R+ x0 x0)) = 1

2R[(R ct+ ct) + (R+ ct ct)]

23

| then = 12Rc

[

(R

c t+ t

)+

(R

c+ t t

)]and since delta function is even

| then = 12Rc

[

(t t R

c

)+

(t t + R

c

)]= (t t) c

2pi((x x)2) (278)

| note that the extra delta function is killed by the step function| note that (t t) has the same effect as (ct ct) = (x0 x0)

GR(x, x) = (x0 x0) c2pi((x x)2) (279)

So the step function must now be explicitly written because we got an extra delta function. The stepfunction is invariant in the sense that if t > t in one inertial frame, then the transformed t >transformed t in all other inertial frames.

Next is the 4-current density. We recall the point charge density and the point charge current density

c(~r, t) = qc3(~r ~r0(t)) and ~J(~r, t) = q~v(t)3(~r ~r0(t)) (280)Note that we take all these coordinates and time to be measured in S-frame. In S-frame, we shallparameterise the charged particles position 4-vector with the proper time ,

(x0) = (ct, ~r0(t)) = (ct(), ~r0(t()) = (ct(), ~r0()) (281)

velocity 4-vector: u =d(x0)

d= (c, ~v) where ~v =

d~r0(t)

dt(282)

To make J = (c, ~J) manifestly covariant, we introduce an extra time delta function,

c(~r, t) = J0 = qc

d(t())3(~r ~r0())(t t()) (283)

| recall that dt = d and put c into (t t())= qc2

d3(~r ~r0())(ct ct()) (284)

| write c = u0

c(x) = J0 = qc

du04(x x0) (285)

similarly, ~J(~r, t) = q

d(t())~v3(~r ~r0())(t t()) (286)

= qc

d~v4(x x0) (287)

so, J(x) = qc

d u4(x x0) (288)

Now, the retarded potentials (recall A =(c ,~A)

),

(~r, t) = 10

GR(~r, t;~r, t)(~r, t)dV dt (289)

| to write into d4x, we convert dt into unit of length(~r, t)

c= 1

0c

((x0 x0) c

2pi((x x)2)

) 1cJ0(x)

1

cd4x (290)

A0(x) =1

2pi0c2

(x0 x0)((x x)2)J0(x)d4x (291)

then, ~A(~r, t) = 0GR(~r, t;~r, t) ~J(~r, t)dV dt (292)

= 0 ((x0 x0) c

2pi((x x)2)

)~J(x)

1

cd4x (293)

| recall that 1c2

= 00

24

=1

2pi0c2

(x0 x0)((x x)2) ~J(x)d4x (294)

So the covariant form is10

A(x) =1

2pi0c2

(x0 x0)((x x)2)J(x)d4x (295)

| insert J(x) = qcdu4(x x0)

| then do the d4x integral using 4(x x0)=

q

2pi0c

d(x0 (x0)0)((x x0)2)u (296)

| the delta imposes the invariant interval (x x0(r))2 = 0| which is called the light cone condition as zero invariant interval is on the light cone!| the step function imposes the retardation condition x0 > (x0)0

| to evaluate further, we use the identity (f()) =i

( i)( dfd )=i

| then dd

(x x0())2 = 2(x x0()) d(x0)

d= 2(x x0())u

| so (f()) = ( r)|2(x x0())u|=r(297)

| Note that ( )u is timelike and so is negative, thus | ( )u| = ( )u

A(x) = q4pi0c

u

(x x0())u=r

(298)

Figure 9: The 4D trajectory intersects the lightcone of the point of observation at 2 points. The retardedpoint is picked out by the step function and is the one of physica interest. Any distance (invariantinterval) on the light cone is zero.

2.3.5 Point charge revisited: Fields of a charged particle in constant velocity

We now revisit the calculation in Part 1 where we calculated the electric and magnetic fields of a chargewith constant velocity. In the context of special relativity, we simply need to have the charge in S-frameand the point of observation to be in S-frame. S-frame is moving with speed v along the x axis relativeto S-frame.

10Strictly speaking, we need to check that the volume element d4x is a Lorentz scalar, i.e. the Jacobian = 1. This willbe checked later when we discusss about the Lagrangian of the EM field.

25

We set the charged particle at the origin of S-frame. The point of observation in S-frame coordinatesare (x, y, z). As the charged particle is not moving with respect to S-frame, there is only electrostaticelectric field and no magnetic field, so Bx, B

y and B

z = 0. The transformation of the fields become:

Ex = Ex Ey = v(Ey vBz) Ez = v(Ez + vBy)

0 = Bx 0 = v(By +

vc2Ez

)0 = v

(Bz vc2Ey

) (299)so, Ey = v

(Ey v

( vc2Ey

))(300)

Ey = vEy (301)

and so, Ez = v(Ez + v

( vc2Ez

))(302)

Ez = vEz (303)

The Coulomb field as seen in S-frame is ~E = q4pi0r2 r. Thus using spherical coordinates in S-frame,

Ex = x ~E = sin cosr q

4pi0r2r (304)

=

x2 + y2

rx

x2 + y2q

4pi0r2(305)

=q

4pi0

x

(x2 + y2 + z2)3/2(306)

| insert x = v(x vt), y = y and z = zEx = E

x =

q

4pi0

v(x vt)(2v(x vt)2 + y2 + z2)3/2

(307)

similarly, Ey = vEy = v y

~E (308)=

q

4pi0

vy

(2v(x vt)2 + y2 + z2)3/2(309)

and, Ez = vEz = v z

~E (310)=

q

4pi0

vz

(2v(x vt)2 + y2 + z2)3/2(311)

So Ex obtained the gamma factor from coordinate transformation while Ey and Ez obtained the gammafactor from field transformation!

To compare with the corresponding example in Part 1, we set the present position of the particle andthe point of observation to be in the xy plane (so z = 0).

Figure 10: Setup where point of observation and present position of particle are in the xy plane ofS-frame. We will shift the origin of S-frame to the present position of the particle.

We shift the origin to the present position of the particle and label the the new x coordinate asxp = x vt. The position vector to the point of observation is ~rp = xpx + yy with xp = rp cos and

26

y = rp sin.

~E = Exx+ Ey y (312)

=q

4pi0v

[xp(

2vx2p + y

2)3/2 x+ y(

2vx2p + y

2)3/2 y

](313)

=q

4pi0

v~rp(2vx

2p + y

2)3/2 (314)

| then, 2vx2p + y2 =x2p

1 v2c2+ y2 =

x2p + y2 v2c2 y2

1 v2c2= 2v

(r2p

v2

c2r2p sin

2

)= r2p

2v

(1 v

2

c2sin2

)=

q

4pi0

v~rp

r3p3v

(1 v2c2 sin2

)3/2 (315)=

q

4pi0

1 v2c2(1 v2c2 sin2

)3/2 rpr2p (which is exactly as before) (316)Thus the reason for the E field to be pointing from the present position instead of from the retardedposition is because both Ex and Ey have the gamma factor and makes it a common factor and allowsthe expression to become v(xpx+ yy) v~rp. The coincidence is that the gamma factors are obtainedin different ways stated earlier.

2.3.6 Point charge revisited: Lienards generalization of Larmors formula

In Part 1, the Larmors formula for point charge radiation is P = 0q2

6pic a2 and it is derived based on the

assumption that it was at instantaneous rest or the particle was moving slowly (but accelerating).We want to generalize it (well, it was generalized by Lienard) by finding out how power P transforms

under Lorentz transfomations and deduce a suitable expression that suits the type of Lorentz object P is.

Our check is that when the speed of the object v is set to zero, we recover Larmors formula P = 0q2

6pic a2.

P =dE

dt=

1

dE

d(317)

| recall that proper power is K0 = 1c

dE

d

=1

cK0 (318)

So if we guess P to be a Lorentz scalar

P =0q

2

6pica2

guess for v 6= 0=

0q2

6pic(aa) where we define a

=du

d(319)

then, K0 =0q

2

6pic2(aa) (320)

| recall u0 = c=

0q2

6pic3(aa)u

0 (321)

which properly makes K0 a 0-component of a 4-vector. Thus power P indeed is a Lorentz scalar so nowwe simply need to work out aa

in terms of ordinary velocity and acceleration.

a =du

d=

dt

d

d

dt(cv, ~vv) (322)

= v

(cdvdt

,d~v

dtv + ~v

dvdt

)(323)

| so dvdt

=d

dt

11 v2c2

= 12(1 v2c2

)3/2 (2~vc2 d~vdt)

= 3v~v ~ac2

=

(4v~v ~ac, 2v~a+ ~v

4v

~v ~ac2

)(324)

27

aa = (a0)2 + ~a2 (325)

= 8v(~v ~a)2c2

+ 4v

(~a+ ~v2v

~v ~ac2

)(~a+ ~v2v

~v ~ac2

)(326)

= 4v

[4v

(~v ~a)2c2

+ a2 + 22v(~v ~a)2c2

+ v24v(~v ~a)2c4

](327)

= 4v

[a2 + (~v ~a)2

( 1c2(1 v2c2

)2 + 2c2 (1 v2c2 ) + v2

c4(1 v2c2

)2)]

(328)

= 4v

[a2 + (~v ~a)2

( c

2

(c2 v2)2 +2

c2 v2 +v2

(c2 v2)2)]

(329)

= 4v

[a2 +

(~v ~a)2c2 v2

](330)

So the generalized power is

P =0q

2

6pic4v

[a2 +

(~v ~a)2c2 v2

](331)

| we need to massage a bit to get to Lienards form=

0q2

6pic6v

[a2(

1 v2

c2

)+

1

c2(~v ~a)2

](332)

=0q

2

6pic6v

[a2 1

c2(v2a2 (~v ~a)2)] (333)

| write ~v ~a = va cos so v2a2 (~v ~a)2 = v2a2(1 cos2 ) = v2a2 sin2 = |~v ~a|2

=0q

2

6pic6v

(a2 |~v ~a|

2

c2

)(334)

This is Lienards formula for the power of a radiating point charge. Obviously when we set ~v = 0, weget Larmors formula.

We shall now deal with the general expression of the power radiated per unit solid angle or dPd . As weare now considering a moving charge, essentially the same geometrical effect in retarded potentials alsooccurs here, i.e. the power radiated by the charge is not the power that is passing through the sphere of

observation. They are related by the same geometrical factor 1 R~vc .

~Spassing through sphere =

(1 R ~v

c

)~Sradiated by particle (335)

For ~v = 0, the Poynting vectors are the same. The expressions in Part 1 are actually ~Sradiated by particle.Now we can continue the calculation for ~v 6= 0 using the equation from Part 1 just before we set ~v = 0,

~Spassing through sphere

=

(1 R ~v

c

)1

0c

(q

4pi0

R

(Rc ~R ~v)3

)2 [~R ((Rc ~v) ~a)

][~R ((Rc ~v) ~a)

]R(336)

| write 1 R ~vc

= 1~R ~vRc

=Rc ~R ~v

Rc

| also write[~R ((Rc ~v) ~a)

][~R ((Rc ~v) ~a)

]= R2

R ((Rc ~v) ~a)2=

1

0c2q2

16pi220

R3R ((Rc ~v) ~a)2

(Rc ~R ~v)5 R (337)

dP

d= ~Spassing through sphere RR2 (338)

=q2

16pi20

R5R ((Rc ~v) ~a)2

(Rc ~R ~v)5 (339)

28

=q2

16pi20

R ((Rc ~v) ~a)2(c R ~v)5 (340)

We shall now consider 2 cases: (i) where ~v and ~a are (instantaneously) colinear and (ii) where ~v and ~aare (instantaneously) perpendicular to each other.11

Case of ~a ~v: 12 When ~a and ~v are parallel, (Rc ~v) ~a = cR ~a so,

dP

d=

q2

16pi20

R (cR ~a)2(c R ~v)5 (341)

| use vector identity ~A ( ~B ~C) = ~B( ~A ~C) ~C( ~A ~B)| so that R (R ~a) = R(R ~a) ~a(R ~R)

=q2c2

16pi20

R(R ~a) ~a2(c R ~v)5 (342)

| andR(R ~a) ~a2 = (R ~a)2 + a2 2(R ~a)2 = a2 (R ~a)2

| let ~v point along the z axis, R ~v = v cos and R ~a = a cos , negative if decelerating

=q2c2

16pi20

a2 sin2

c5(1 vc cos

)5 (343)| then use c20 = 1

0

=0q

2a2

16pi2c

sin2 (1 vc cos

)5 (344)We can integrate this to get the total power but we will just simply apply Lienards formula. Simplywrite ~v ~a = 0 and get

P =0q

2a2

6pic6v (345)

Note that the expressions for dPd and P are applicable for both ~a parallel to ~v (accelerating) and ~aanti-parallel to ~v (decelerating). The lobes (of power) are stretched towards the direction of ~v (here, thez axis) although there is no radiation exactly in the direction of ~v. A typical example is bremsstrahlungwhere an electron hits a metal and decelerates greatly with the emission of radiation.

Case of ~a ~v: 13 We shall set ~v along the z axis, ~a along the x axis so~v = vz,~a = ax and R = sin cosx+ sin siny + cos z (346)

and we shall calculate dPd first.

dP

d=

q2

16pi20

R ((c sin cosx+ c sin siny + (c cos v)z) ax)2(c v cos )5 (347)

11The logic for this entire section may seem messy for a reason. The natural logic should be:

1. include geometrical factor

2. calculate dPd

3. integrate over sphere to get total power P

4. work out 2 special cases of ~a ~v and ~a ~vThe integration over sphere to get P is too difficult to carry out so I changed the logic to:

1. deduce total power P is a Lorentz scalar

2. generalize acceleration to a 4-vector so that P becomes a Lorentz scalar

3. then include geometrical factor to get dPd

4. work out 2 special cases of ~a ~v and ~a ~v12You may take ~a ~v at an instant where time is tr.13You may take ~a ~v at an instant where time is tr.

29

=q2a2

16pi20

R (c sin sinz + (c cos v)y)2c5(1 vc cos

)5 (348)| divide c2 in both numerator and denominator

| the cross product is

x y z

sin cos sin sin cos

0 cos vc sin sin

| which =

(sin2 sin2 + cos

(cos v

c

))x sin2 sin cosy + sin cos

(cos v

c

)z

=q2a2

16pi20c3

(sin2 sin2 + cos

(cos vc

))2+ sin4 sin2 cos2 + sin2 cos2

(cos vc

)2(1 vc cos

)5| terms without v

c= sin4 sin4 + cos4 + 2 sin2 sin2 cos2 + sin4 sin2 cos2 + sin2 cos2 cos2

| |then 1st + 4th = sin4 sin2 (sin2 + cos2 ) = sin4 sin2 | |then 3rd + 5th = sin2 cos2 (2 sin2 + cos2 ) = sin2 cos2(sin2 + 1)| then terms without v

c= sin4 sin2 + cos4 + sin2 cos2 sin2 + sin2 cos2

| |then 1st + 3rd = sin2 sin2 (sin2 + cos2 ) = sin2 sin2 | |then 2nd + 4th = cos2 (cos2 + sin2 ) = cos2 | then terms without v

c= sin2 sin2 + cos2 = sin2 (1 cos2 ) + cos2 = 1 sin2 cos2

| terms with vc

= 2vc

cos3 2vc

sin2 sin2 cos 2vc

sin2 cos cos2

| | = 2vc

(cos3 + sin2 cos (sin2 + cos2 )) = 2vc

cos (cos2 + sin2 ) = 2vc

cos

| terms with v2

c2=v2

c2cos2 +

v2

c2sin2 cos2

| then write 0c2 = 10

=0q

2a2

16pi2c

1 sin2 cos2 2 vc cos + v2

c2 cos2 + v

2

c2 sin2 cos2 (

1 vc cos )5 (349)

| the 3 terms, 1 2vc

cos +v2

c2cos2 =

(1 v

ccos

)2=

0q2a2

16pi2c

(1 vc cos

)2 (1 v2c2 ) sin2 cos2 (1 vc cos

)5 (350)We can integrate this to get the total power but we will just simply apply Lienards formula.

P =0q

2

6pic6v

(a2 |~v ~a|

2

c2

)(351)

| where ~v ~a = vay=

0q2a2

6pic6v

(1 v

2

c2

)(352)

=0q

2a2

6pic4v (353)

We see that when we set = 0, dPd is sharply peaked in the direction of ~v forvc 1. A typical example

is circular motion where the electron emits (so-called) synchrotron radiation tangentially to the circulartrajectory.

30

3 Lagrangian description of a classical relativistic U(1) gaugetheory

3.1 Review of Lagrangian mechanics and Noethers theorem

For a classical system described by n-generalized coordinates: q1(t), . . . , qn(t), the Lagrangian is definedby

L(q1(t), . . . , qn(t); q1(t), . . . , qn(t); t) = T V (354)

where T is the KE and V is the PE. Note that in field theory (where we deal with infinite degrees offreedom), the form L = T V may not really be valid as we will see later.

Hamiltons principle states that the trajectory in configuration space that extremizes (maximum or

minimum) the action S =

tfti

Ldt is the actual (or physical) trajectory of the system.14

Extremum means: S = 0 (355)

= tfti

L(q1(t), . . . , qn(t); q1(t), . . . , qn(t); t)dt = 0 (356)

The requirement of extremum translates into n Euler-Lagrange equations of motion.

L

qi ddt

(L

qi

)= 0 (357)

The next thing we want to mention is Noethers theorem which is the formal way of deriving conservationlaws in a theory. Noethers theorem states: for each symmetry of the theory (i.e. transformations thatdoes not change the Lagrangian), there is a conserved quantity associated to it.

Suppose the Lagrangian is invariant under a change of coordinates with a small parameter ,

1st order change: qi = qi + Ki(q) (358)

where Ki(q) may be a function of all the qis, which we collectively denote as q.We take the Lagrangian to be invariant to first order

L(q, q) = L(q + K, q + K) (359)

L(q, q) +i

(L(q, q)

qiKi +

L(q, q)

qiKi

)(360)

but L(q, q) = 0 (361)i

(L(q, q)

qiKi +

L(q, q)

qiKi

)= 0 (362)

from Euler Lagrange,L

qi=

d

dt

(L

qi

)|

i

(d

dt

(L

qi

)Ki +

L

qi

dKidt

)= 0 (363)

d

dt

(i

L

qiKi

)= 0 (364)

so the conserved quantity isi

L

qiKi.

15

14We assume fixed end points at ti and tf .15The case of cyclic coordinates means its conjugate momenta is conserved, is a special case. Suppose the Lagrangian

does not depend on q3, so q3 = q3 + does not change the Lagrangian. Thus K3 = 1

d

dt

(L

q3K3

)= 0 = dp3

dt= 0 (365)

31

3.2 Relativistic Lagrangian of a charged particle interacting with externalEM field

We start with a free (i.e. non-interacting) relativistic particle for simplicity. However L = T =1

2md~x

dt

d~x

dt=

1

2mv2 is obviously not compatible with special relativity. We have to make an intelligent guess of

the free relativistic particle Lagrangian. To do that, we need to lay out the features that it should have.

Following axiom 1 of Special Relativity, the action S should be a Lorentz scalar so that the actiongives us the same physics in every inertial frame. Then we can write S =

Ldt

d=dt= Ld and

since proper time is invariant, L should be a Lorentz scalar.

In the non-relativistic limit v

= cp2 +m2c2 (376)

=p2c2 +m2c4 (377)

and indeed we get the expected Hamiltonian. We are confident of the free relativistic Lagrangian L =

mc2

1 v2

c2so now we shall include the interaction with an external EM field. We take inspiration

from the electric potential energy q and the magnetic energy~J ~AdV = ~v ~AdV . For a particle

with trajectory ~r(t), = q3(~r ~r(t)) so the magnetic energy is q~v ~A. We let 1 and 2 be possible(dimensionless) constants, so the charged particle interacting with external EM Lagrangian should be

L = mc2

1 v2

c2+ 1q+ 2q~v ~A (378)

We shall determine 1 and 2 by requiring the equations of motion becomed

dtm~v = q ~E + q(~v ~B).

So,

L

~x ddt

L

~v= 0 (379)

1q

~x+ 2q

~v ~A~x

ddt

~v

(mc2

1 v

2

c2

) 2q d

dt

~v~v ~A = 0 (380)

recall that

~v

(mc2

1 v

2

c2

)= m~v |

note that

~x= ~ and

~v~v ~A = ~A |

1q~ 2q d~A

dt ddt

(m~v) + 2q~(~v ~A) = 0 (381)use vector identity ~(~v ~A) = ~v (~ ~A) + ~A (~ ~v) + (~v ~) ~A+ ( ~A ~)~v |

since ~v and ~x are treated as independent, 2nd and 4th terms in the identity are zero |

write ~ ~A = ~B and d~A(~x, t)

dt= ~A

t+ (~v ~) ~A |

1q~ 2q ~A

t 2q(~v ~) ~A d

dt(m~v) + 2q~v ~B + 2q(~v ~) ~A = 0 (382)

for 1~ 2 ~A

t= ~E, we need 1 = 1 and 2 = +1 |

q ~E ddt

(m~v) + q~v ~B = 0 (383)

so, L = mc2

1 v2

c2 q+ q~v ~A (384)

and to get the Hamiltonian, we first need the conjugate momentum

~p =L

~v=

~v

(mc2

1 v

2

c2+ q~v ~A

)= m~v + q ~A (385)

Then perform the Legendre transform

H = ~v ~p L (386)

= ~v (m~v + q ~A

)+mc2

1 v

2

c2+ q q~v ~A (387)

= mv2 +mc2

1 v2

c2+ q (388)

| write m~v = ~p q ~A and replace all v2 = (~p q~A)2c2

(~p q ~A)2 +m2c2

33

=(~p q ~A)2c2 +m2c4 + q (389)

Since we must have L to be a Lorentz scalar, we need to check if the 2 extra terms fulfil it.

L = mc2 q+ q~v ~A (390)= mc2 q(c)

(

c

)+ q~v ~A (391)

| recall 4-velocity u = (c,~v) and 4-potential A =(

c, ~A

)= mc2 + quA (392)

Thus the second term is indeed a Lorentz scalar since (total) charge is Lorentz invariant.The last thing to do in this section is to put the first term into a (manifestly) Lorentz scalar form

and derive the covariant equation of motion K = dp

d = qFu . We look at only the first term (free

particle term) and we also think about Hamiltons principle which is to extremize the trajectory and askWhere is the meaning of trajectory in the action expression S =

(mc2)d ?

To answer that question, we have to go back to the invariant interval which is the trajectory (orworldline) in Minkowski space!17

Recall, ds2 = c2d2 = c2dt2 + dx2 + dy2 + dz2 = dxdx (393)ds =

c2d2 = 1cd (394)

thus, S = mc

11ds (395)

| but ds = dxdx = dxd

dx

dd =

uud (396)

= mc

uu

1 d (397)

= mc uud (398)

Our final invariant action is

S = mc uud + quAd = Ld (399)

We can deduce the (covariant) Euler-Lagrange equations to be

L

x dd

L(dxd

) = 0 (400)

L

x dd

L

u= 0 (401)

dd

u

(mcuu + quA)+ x

(quA) = 0 (402)

dd

mcuuu qdA

d+ qu

Ax

= 0 (403)

note that uu = c2 from beginning of Part 2 |dp

d q dx

d

A

x+ quA = 0 (404)

qu (A A) = dp

d(405)

qu (A A) = dp

d(406)

quF =

dp

d(407)

17Actually we are parameterizing the worldline using the proper time as parameter. This is natural especially forparticles with mass, the notion of proper time is meaningful. In more general terms, we can use some parameter and

write ds =dxdx =

dx

ddx

dd.

34

3.3 Lagrangian of the EM field

The field has infinite degrees of freedom because it has a value (or values) at every spacetime pointand there are infinitely many points. Thus the Lagrangian has to be indicated at every point, i.e. theLagrangian density is the relevant quantity.

Lagrangian: L =

Ld3~x (408)

Action: S =

tfti

Ldt =

tfti

Ld3~xdt (409)

To determine if the Lagrangian density L is a Lorentz scalar or not, we look at how the volume elementd3~xdt transforms under Lorentz transformations. To do that, we look at the Jacobian.

dtdxdydz =

tt

tx

ty

tz

xt

xx

xy

xz

y

ty

xy

yy

zzt

zx

zy

zz

dtdxdydz (410)| with t =

(t u

c2x)

, x = (x ut), y = y and z = z

=

uc2 0 0u 0 00 0 1 00 0 0 1

dtdxdydz (411)=

0 00 1 00 0 1

+ uc2u 0 0

0 1 00 0 1

dtdxdydz (412)

=

(2 2u

2

c2

)dtdxdydz (413)

= dtdxdydz (414)

Thus the volume element is a Lorentz scalar and so for the action to be a Lorentz scalar, the Lagrangiandensity L must be a Lorentz scalar also. The dependences in L shall be deduced as follows:

q(t) A(x)q(t) A(x) (now A(x) depends on 4-position, so all derivatives must be considered)

L(q(t), q(t)) L(A(x), A(x))

To get the Euler-Lagrange equations of motion, we shall carry out the variational calculation,

Variation: A(x) = A(x) + A(x) (415)Fixed endpoints: A(~x, ti) = 0 = A(~x, tf ) (416)

0 = S (417)

=

tfti

d3~xdtL (418)

=

tfti

d3~xdt

(LA

A +L

(A)(A)

)(419)

| the variation and the partial derivative commutes: (A) = (A)| (one dimensional) proof: (xA) = lim

01

(A(x+ )A(x) (A(x+ )A(x)))

| | = lim0

1

(A(x+ )A(x+ ) (A(x)A(x))) = lim

01

(A(x+ ) A(x)) = x(A)

=

tfti

d3~xdt

(LA

A +L

(A)A

)(420)

35

| integrate by parts on the second term

=

tfti

d3~xdt

(LA

A +

(L

(A)A

)

(L

(A)

)A

)(421)

=

tfti

d3~xdt

[LA

(

L(A)

)]A +

d3~x

L(A)

A

tfti

(422)

the last term is zero due to the fixed endpoints condition. Since A is an arbitrary variation, S = 0only if

LA

(

L(A)

)= 0 (423)

which are the desired Euler-Lagrange equations of motion for fields. Since L is a Lorentz scalar, theequations of motion are Lorentz covariant equations (as they should be).

The Euler-Lagrange equations work also for fields interacting with external sources. In the case ofelectrodynamics with external sources, we expect to get F

= 0J. For the free (or non-interacting)

part of the Lagrangian, we only have 2 Lorentz scalars to choose from: FF or FF . We choose

FF . For the interaction part, we choose the Lorentz scalar JA which is the usual vector coupling

term inspired from the previous section.

L = 1FF + 2JA (where 1 and 2 are constants to determine) (424)

LA

(

L(A)

)= 0 (425)

recall that F = A A |(2J

A)

A

[

(A)1(

A A)(A A)]

= 0 (426)

expand and note A(A)

(A)= A

=

A |2J

1(2A 2A 2A + 2A) = 0 (427)2J

41F = 0 (428)(429)

We want F = 0J

so 1 = 14 and 2 = 0.

L = 14FF + 0JA

(430)

Note a bonus identity from the above working: (A)( 14FF) = F .

Next, we shall look at the conservation laws or Noethers theorem in field theory. As usual, we onlylook at 1st order (symmetry) transformations.

Coordinate transformation: x = x + x (431)Corresponding change in field: A(x

) = A(x) + A(x) (432)Define a field variation: A(x) = A

(x)A(x) (433)

which is variation of the field while keeping the coordinates fixed. This variation is useful because itcommutes with the partial derivative (as we have seen earlier). The 2 variations are related by (keepingeverything to 1st order),

A(x) = A(x)A(x) (434)

= A(x)A(x) +A(x)A(x) (435)= (A(x)A(x)) + A(x) (436)= A(x) (A(x+ x)A(x)) (437)= A(x) A

(x)

xx (438)

36

= A(x) (A(x))x (439)| to 1st order, we can write A(x) A(x)= A(x) (A(x))x (440)

We demand that these symmetry transformations leave the action invariant (to 1st order).

0 = S (441)

=

Ld3~xdt (442)

=

(L)d3~xdt+

L(d3~xdt) (443)

| the variation of the volume element is found by calculating the Jacobian

| d3~xdt =

x0x0

x0x1

x1x0

x1x1

......

. . .

d3~xdt =

1 + x0

x0x0

x1 x1

x0 1 +x1

x1 ...

.... . .

d3~xdt| then d3~xdt

(1 +

x

x

)d3~xdt = (1 + x

) d3~xdt so (d3~xdt) = (x)d3~xdt

=

(L+ L(x)) d3~xdt (444)

| recall A(x) = A(x) (A(x))x and infer L = L+ (L)x

=

(L+ L(x) + (L)x

)d3~xdt (445)

=

(L+ (Lx)

)d3~xdt (446)

| so, L = LA

A +L

(A)(A)

| add & subtract a term, L = LA

A (

L(A)

)A +

(L

(A)

)A +

L(A)

A

| then L =[LA

(

L(A)

)]A +

(L

(A)A

)| using Euler-Lagrange equations, the first term is zero=

[L

(A)A + Lx

]d3~xdt (447)

| recall A = A (A)x

0 =

[L

(A)A

(L

(A)(A) L

)x

]d3~xdt (448)

We are thus invited to postulate (not rigorous)18 local conservation laws of the form J = 0. The

so-called Noethers current density is thus defined as19

J =L

(A)A

(L

(A)(A) L

)x (449)

Now with regards to the energy-momentum tensor we saw earlier, we shall apply Noethers theorem tothe special case where the action is invariant under 4D translations,

x = x + = x = (450)and the fields do not change under the 4D translations

A(x) = A(x) = A = 0 (451)

18I have browsed through classical/quantum field theory books and Goldstein, and I am unable to find any satisfactorilyrigorous derivation of this.

19Be careful that Noethers current density and the external current density have the same notation but they may meandifferent quantities.

37

0 = J (452)

0 =

[L

(A)(A) L

] (453)

| since is arbitrary,0 =

[L

(A)(A) L

](454)

| raise the index by multiplying with

0 =

[L

(A)(A) L

](455)

Thus we can define the (canonical) non-symmetric energy-momentum tensor ,20

=L

(A)(A) L (456)

Note that since we are discussing 4D translational invariance, we have to use L = Lfree = 14FF .Interaction with external sources is not translationally invariant.

The quantity that is conserved is revealed when we integrate 0 = J over 3D space.

0 =

J

dV (457)

=

0J

0dV +

~ ~JdV (458)

| use divergence theorem on second term=

d

dx0

J0dV +

~J d ~A (459)

| assume the fields and therefore ~J fall off sufficiently fast=

d

dt

J0dV (460)

ThusJ0dV is the conserved quantity.21 For the (canonical) energy-momentum tensor , where

= 0, the 4 conserved quantities are

0dV . These 4 conserved quantities turn out to be the

energy and the momentum.

1

c

0dV = P =

(E

c, ~P

)(461)

Earlier we have a symmetric energy-momentum tensor T and it is actually related to by a 4-divergence.22

T = + (where we require = ) (462)

This relationship does not affect the conservation law.

T =

+ (463)

| from Noethers theorem, = 0| split = 1

2( + )

=1

2

+1

2

(464)

| in second term, rename indices and =

1

2

+1

2

(465)

20Note that this energy-momentum tensor is actually not gauge invariant!21A quick example is to recall the 4-current J = (c, ~J) where J0 = c and so

J0dV = Q, the total charge, is the

conserved quantity.22If you are interested, you can read pg 112-114 of Electrodynamics and Classical theory of Fields and Particles by

A.O. Barut on how to get T directly from variation.

38

| partial derivatives commute = =

1

2(

+ ) (466)

| recall the requirement = = 0 (467)

which is the conservation law for the non-interacting case. The conserved quantities are also not affected.T 0dV =

(0 +

0)dV (468)

=

0dV +

0

00dV +

i

i0dV (469)

| but 00 = 0 since 00 = 00| use divergence theorem in the 3rd term and assume falls off sufficiently fast=

0dV (470)

Now we shall take the explicit (free field) expressions of and T and work out .

=L

(A)(A) L (471)

| use L = 14FF

and we already knowL

(A)= F

= F(A) + 14FF (472)

T = 10

[FF

+1

4FF

](473)

| replace 14FF =

+ F(A)

= 10

[FF + F(A) +

] (474)

= 10

[FF + F(A) +

] (475)

= 10

[F(A A + A) + ] (476)

= 10

[F(A) + ] (477)

| use product rule on the first term and F = 0 (for free field)= 1

0[(F

A) + ] (478)

| write F = F

= 10

[(FA) + ] (479)

Thus = 10FA which is indeed antisymmetric in the first 2 indices.

As a nice ending to Part 2, we shall take the interacting case and work out explicitly that T =

F J.

T = 1

0

[(

FF) +

1

4(

FF)

](480)

= 10

[(F)F

+ F(F) +

1

4((F)F + F

(F))

](481)

| because we can write F(F) = F(F), the last 2 terms add| the first term F = 0J which are Maxwell equations

39

= 10

[0JF + F(F) + 1

2F(

F)

](482)

| for 1st term: JF = F J, for 2nd term: rename and split it,| for 3rd term: rename = F J 1

20

[F(

F + F + F )]

(483)

| use homogenous Maxwell equation F + F + F = 0= F J 1

20

[F(

F F )] (484)| write F = F

= F J 120

[F(

F + F )]

(485)

| terms in round brackets are symmetric in but F is antisymmetric in = F J (indeed) (486)

40