cooper 3 relativistic electrodynamics

8/8/2019 Cooper 3 Relativistic Electrodynamics

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Department of Physics

University of Cambridge

Part II Experimental and

Theoretical Physics

Relativity and Electrodynamics

Handout 3: Relativistic Electrodynamics

Dr Nigel Cooper

Michaelmas 2001


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RELATIVISTIC ELECTRODYNAMICS i

Contents

6 Relativistic Electrodynamics I 1

6.1 Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.2 Is Charge a Lorentz Invariant? . . . . . . . . . . . . . . . . . . . 36.3 The 4-Gradient . . . . . . . . . . . . . . . . . . . . . . . . . . . 46.4 The 4-Current . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56.5 The 4-Potential and Maxwells Equations . . . . . . . . . . . . . 7

Note on Linear Materials . . . . . . . . . . . . . . . . . . . . . . 8

7 Relativistic Electrodynamics II 9

7.1 Transformation ofE and B . . . . . . . . . . . . . . . . . . . . 97.2 Covariance of Maxwells Equations . . . . . . . . . . . . . . . . 117.3 Magnetism as a Relativistic Effect and the Lorentz Force . . . . 12

7.4 The Field-Strength Tensor . . . . . . . . . . . . . . . . . . . . . 14Transformation ofE and B . . . . . . . . . . . . . . . . . . . . 15Maxwells Equations in Frame-Independent Form and the Lorentz

Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157.5 Electromagnetism as the Simplest Relativistic Theory . . . . . . 17

8 Relativistic Electrodynamics and Radiation 19

8.1 Fields of a Uniformly Moving Charge . . . . . . . . . . . . . . . 198.2 Potentials Due to an Arbitrarily Moving Charge . . . . . . . . . 218.3 Radiation by an Accelerated Charge . . . . . . . . . . . . . . . 23

Approximate Calculation . . . . . . . . . . . . . . . . . . . . . . 248.4 Cyclotron and Synchrotron Radiation . . . . . . . . . . . . . . . 28Cyclotron Radiation . . . . . . . . . . . . . . . . . . . . . . . . 29Synchrotron Radiation . . . . . . . . . . . . . . . . . . . . . . . 29Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

Notes by P. Alexander (2000); modified NRC (2001)


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ii RELATIVITY AND ELECTRODYNAMICS


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RELATIVISTIC ELECTRODYNAMICS 1

6 Relativistic Electrodynamics I

6.1 Background

Let us start our discussion of relativistic electrodynamics by considering wherewe have got to already in our theoretical developments. What follows thissection is going to be an intriguing, and hopefully interesting, discussion ofthe relationship between SR and EM. Indeed it is a very elegant piece oftheory. It is crucial to realise that everything we shall discuss is based veryfirmly on experimental evidence. The theoretical elegance comes from how wecan develop a single physical model which follows using logical deduction fromthe experimental basis

So lets recap on where we have got to:

Experimental observations which contradict the existence of a preferredframe of reference lead directly to the elegant circumvention of the prob-lem by Einsteins proposition of the principle of relativity. This is com-bined with the fact that the speed of light, at least in one frame, isindependent of the motion of the source to give Einsteins law of lightpropagation. From these we deduce that transformation of coordinatesbetween inertial frames must obey Lorentz Transformations:

x

0

x1

x2

x3

=

0 0

0 0

0 0 1 00 0 0 1

x0

x1

x2

x3

.

By introducing the concept of spacetime we gain a powerful mathemat-ical structure. The principle of relativity is embodied simply in the con-cept of a physical 4-vector which has a physical significance independentof the underlying representation in any one inertial frame. The struc-ture of spacetime is embodied in the metric tensor which tells us howto form scalar products; an important property of spacetime and onewhich relates directly to Einsteins law of light propagation. The rep-

resentation of 4-vectors in different frames is just given by the LorentzTransformations.

A B = AA = gAA

g =

1 0 0 00 1 0 00 0 1 00 0 0 1


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2 RELATIVITY AND ELECTRODYNAMICS

x

= x

By requiring all physical quantities to be tensors (invariants) [e.g. scalars,

4-vectors] we build up a theoretical model for the dynamics of particlesin spacetime by calling only upon a few extra ideas such as conservationlaws which we believe must also hold.

Starting from experimental results on currents and charges (e.g. dueto Faraday, Ampere etc.) we follow Maxwell and deduce four equationslinking E and B to their sources (charge and current densities).

E = /0

B = 0

E = BB = 0j + 00E

We then showed that we can re-express the E and B fields in terms of ascalar and vector potential and find that Maxwells equations reduce tofour component equations in terms of these potentials.

B = AE =

t

A

2 1

c2 =

0

2A 1

c2A = 0j

We are able to solve these equations to find the potentials in terms ofthe sources. The result is strongly indicative of the key role of the speedof light; remarkably Einsteins law of light propagation appears to beembodied in these solutions, although we have used nothing yet of thetheory of SR in our development of EM to this point.

(r, t) =1

40

all space

r, t |rr |

c

dV

|r r|

A(r, t) =04

all space

jr, t |rr |

c

dV

|r r|


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Therefore we note that there must be an intimate link between EM and SR aswe have already argued, and these solutions are very suggestive of EM actuallyalready being correct from the point of view of Special Relativity. But, how

do we show this and how do we do calculations? Our success in developingspecial relativity points the way; rewrite our electromagnetic theory explicitlyin 4-vector form then we know that it must be a correctly relativistic theory,and we have a good basis to work with.

So where to next? We can make some guesses at this point which seem totallyreasonable and will turn out to be correct. The fact that we have one scalarand one 3-vector potential is suggestive of them forming a single 4-vector. Afew moments thought about units suggest that an appropriate 4-vector mightbe

A

A =

c

,Aand similarly for current and charge

J J = (c,j) .

But this is not a proof, so we will have to do a little more work.

More importantly what happens if our solutions to Maxwells equations werevalid at low velocity only and that we needed to generalise them in some way?This is equivalent to asking the crucial question, is the charge a LorentzInvariant. We cannot answer this theoretically. We must address this problem

experimentally and therefore this is where we start our detailed discussion.

6.2 Is Charge a Lorentz Invariant?

In this section we will consider the experimental evidence which indicates thatcharge is a Lorentz Invariant to the accuracy of current measurements.

Firstly we note that atoms are very accurately known to be neutral, yet theycontain slow moving nuclei and fast moving electrons and in different atoms theelectrons will move with different velocities, or more strictly have different

momenta which remains a good quantum concept. Further, atoms remainneutral when heated and we are viewing them in a different inertial frame. Wecan quantify this effect with a simple experiment (many such variants can bebased on this idea).

Consider a 10kg block of copper containing about 1026 atoms with a totalcharge in conduction electrons of about 107 Coulombs. Upon heating, the nu-clear motion is increased more significantly than the electron motion (phonons


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are induced). If the charge on the protons changed due to this motion thenthe copper would charge up. If the copper is in the form of a sphere and wecould detect a change in potential of say 1mV (easily done) then we would be

sensitive to a net charge of about 1014

Coulomb (the sphere is about 65 mmradius). No such effect is observed hence the charges cannot differ by morethan 1 part in 1021!

Within particle accelerators and mass spectrometers the dynamics of electronsdepend on e/m and their dynamics are correctly predicted by using the analysiswe have already developed and taking e as a constant. This has been verifiedto the highest energies attainable with modern accelerators.

We therefore know to very high accuracy that charge is a LI.

6.3 The 4-Gradient

We will start our detailed theoretical analysis by considering a 4-vector oper-ator, the 4-gradient:

=

1

c

t,

.

(The symbols and represent the 4-gradient operator in symbolic andcomponent notations, respectively.) The possible importance of this operatoris suggested by the form of the wave equations satisfied by the vector andscalar potentials. We have already seen that differentiation of a 4-vector by a

coordinate does not result in a 4-vector and we had to consider differentiationby LI or scalars. Clearly this operator is not a scalar, therefore is it itself a4-vector? We shall prove that it is by showing directly that it transforms aswe would expect it to as a 4-vector.

Consider the time-like component in a frame S and relate this to the frame Susing the chain rule

1

c

t=

1

c

t

t

t+

x

t

x+

y

t

y+

z

t

z

.

We can now use the Lorentz transformations to evaluate each of the partialdifferentials between the coordinates

t

t= ,

x

t= V ,

y

t= 0 and

z

t= 0

hence1

c

t=

1

c

t+ V

x

=

1

c

t

x

.


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Similarly we can show that

x

=

x

1

c

t and

y

=

y

,

z

=

z

.

Therefore the 4-gradient transforms like a physical 4-vector and hence mustbe a 4-vector.

Consider the scalar product of the 4-gradient with some 4-vector X.

gX = X

=1

c

X0

t+

X1

x+

X2

y+

X3

z=

X

x.

In our earlier language is a contravariant operator. Its covariant form, ,

is given by: =

x= g

=

1

c

t,

.

We can therefore see immediately that 2 = must be a scalarand Lorentz Invariant

2 =

1

c22

t22 = 1

c22

t2

2

x2

2

y2

2

z2

This operator is known as the dAlembertian.1 It is the spacetime equivalent

of the 3-space Laplacian.

6.4 The 4-Current

Our preliminary discussions suggested that a possible form for the a current 4-vector (which we shall call the 4-current or more strictly the 4-current-density)might be:

J = (c,j).

This we shall show is indeed the correct form and is a physical 4-vector. Let

us consider the scalar product of the gradient 4-vector with J

J = J = t

+ j1Be careful, some texts use the symbol to represent the dAlembertian itself (what

we call 2). In practice, there should be no ambiguity, as the dAlembertian is a scalar,while is a 4-vector. The meaning of the symbol in any equation can therefore bededuced by reading off the number of components it is required to have.


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Now we know that in the rest frame of the charge density that t

+ j = 0: this is just the equation of continuity from electromagnetism whichexpresses the conservation of charge. If we take the conservation of charge to

be something which should indeed hold in all frames then we can identify Jas a physical 4-vector since we know that the 4-gradient is a 4-vector. This isnot, however, something we should take so lightly, so let us consider this in alittle more detail.

Imagine a macroscopic charge, Q, which occupies some volume V0 in its instan-taneous rest frame S. In S the charge density is 0 = Q/V0. Now in a generalframe, S, in which the charge is moving at speed u the volume occupied by thecharge Q (which is unchanged since it is a Lorentz Invariant) will be given by

V = V0/

x

0

z

S uQ

V

z

x

SQ

V

and the charge density in this frame is therefore

= 0

Now consider two nearby points on the world line of Q separated by the 4-vector (cdt, dr). The current in S is zero, but in S

j = u = 0dr

dt

Therefore in S

J = (c,j) =

c0, 0dr

dt

= 0

d

dt (ct, r) = 0d

d(ct, r)

where we have used d = dt/, with the proper time. Since 0 is a LorentzInvariant (it is defined in the rest frame of the charge) and the proper time isa Lorentz Invariant, then J must be a 4-vector since (ct, r) is a 4-vector.

Therefore we have proven that J, the 4-current, is a physical 4-vector, andhence that the conservation of charge must hold in all frames.


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J = t

+ j = 0

The scalar product of the 4-current density with itself gives a Lorentz Invariant

J J = c22 j2 = c220

where again 0 is the charge density in the rest frame of the charges.

Applying the Lorentz transformations to the current 4-vector we can easilyfind the transformation properties for the current and charge density to be

c = (c jx)jx = (jx c)jy = jy

jz = jz .

Alternatively we can write these same equation in a vector form noting thecomponent ofj parallel to the direction of motion to give

c = (c j) = (c j)j

= (j c) + j .

6.5 The 4-Potential and Maxwells Equations

Let us return and consider Maxwells equations in terms of the scalar andvector potentials

2 1

c2 =

0

2

A1

c2A = 0j.

We will now show that these equations can be written in terms of 4-vectorquantities. We will introduce (again as we have already postulated) the 4-potential:

A A =

c,A

,


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and rewrite our Maxwells equations in terms of the potential using thedAlembertian

1

c 1

c2

2

= 1

c2

2

t2

2c

=

c0 = 0c

1

c2A2A =

1

c22

t22

A = 0j

These equations can clearly be combined by introducing the 4-current and4-potential to give

2A =

1

c22

t22

A = 0J

or simply

2A = 0J

Since 2 is a Lorentz invariant and J is a 4-vector this proves that A is a 4-vector. We have now succeeded in writing Maxwells equations in a way whichclearly demonstrates that they represent a set of Lorentz Invariant equations.We say that we have written Maxwells equations in a manifestly covariantform. We must also have that the scalar product of and A is an invariant

A = A = 1c

t

c

( A) = invariant = 0this is just the Lorenz Gauge where we choose the invariant to be zero in allframes, and as before this reduces to the Coulomb gauge in the rest frame ofthe charges.

Note on Linear Materials

We have considered throughout this analysis the effects of charges and currentsin vacuo. We can easily generalise to linear materials by making the followingsubstitutions

= =

00

t,

A = A = (

00 ,A)

J = J =

00,j

.


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7 Relativistic Electrodynamics II

7.1 Transformation ofE and B

In the discussion of section 6 mention of the electric and magnetic field wereconspicuous because of their absence. To some extent this is because our 4-potential and 4-current provide the natural physical quantities with which wecan talk about relativistic electrodynamics, however if we wish to considerforces on charged particles it is essential that we now consider the electric andmagnetic field explicitly.

We will start by deriving expressions for E and B in two frames S and S andhence derive their transformation properties. To do this we will start withthe 4-potential and 4-gradient transformation laws. We have already had the

results necessary for the 4-gradient and the transformation of the 4-potentialis simply derived by applying the Lorentz transformations to A =

c

,A

= ( V Ax), Ax =

Ax V c2

,Ay = Ay, A

z = Az

and for the 4-gradient

1

c

t=

1

c

t+ V

x

,

x=

x+

V

c2

t

y

= y

, z

= z

We find B and E in the usual way

B = A, E = A

t

Consider first the components ofB.

Bx =Azy

Ay

z =

Azy

Ayz

= Bx

and

By =Axz

Az

x=

Axz

Vc2

z

Azx

+V

c2Azt

=

Axz

Azx

+

V

c2

z Az

t

=

By +

V

c2Ez


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Similarly

Bz =

Bz V

c2Ey

Now consider the components ofE

Ex =

x A

x

t

=

x+

V

c2

t

( V Ax)

t+ V

x

Ax V

c2

= 2

x V Ax

x+

V

c2

t V

2

c2Ax

t+

Axt

Vc2

t+ V

Axx

V2

c2

x

=

21V2

c2

x+

Ax

t

= Ex

and for Ey

Ey =

y A

y

t

= y

( V Ax)

t+ V

x

Ay

=

2y

VAx

y+

Ay

t+ V

Ax

x

= (Ey V Bz)

Similarly

Ez = (Ez + V By)

Collecting these results

Ex = Ex Bx = Bx

Ey = (Ey V Bz) B

y =

By +

Vc2 Ez

Ez = (Ez + V By) B

z =

Bz Vc2 Ey

We can also express these results in terms of the components of E and Bparallel to and perpendicular to the velocity V indicating the transformationbetween the two frames:


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E = E E = (E+ VB)

B = B B =

B 1

c2(VE)

The transformation between B and E is very revealing. It suggests that themagnetic and electric fields are two aspects of the same phenomenon whichjust depend upon the frame in which we view the process. This is perhapsmost apparent when we note the similarity between the transformations invector form and the familiar form of the Lorentz force on a particle moving inE and B fields. We will return to this shortly.

7.2 Covariance of Maxwells Equations

We have in fact already shown that Maxwells equations are covariant (preservetheir form) under relativistic transformations by showing that we could writethem in terms of 4-vector quantities alone. Here we will show the same physicalresult explicitly by transforming the field between frames as an example ofusing the transformation laws we have just derived.

Let us consider Maxwells 4th equation

B 00Et

= 0j

We want to prove that we have exactly the same form of the equation insome other frame S. Since the algebra is very tedious we will do this forjust one vector component of just this Maxwell equation; lets consider the x

component in the frame S.

B 00E

t

x

=B zy

By

z 1

c2Ext

=

y

Bz V

c2Ey

z

By +

V

c2Ez

c2

t

+ V

xEx

=

Bzy

Byz

Vc2

Eyy

+Ezz

+Exx

1

c2Ex

t

=

B 1

c2E

t

x

Vc2 E

=

0jx V

c2

0


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= 0(jx V )= 0j

x

Where we have used the transformation laws for the fields, currents and thegradient operator. Of course we can show that this holds for all the other com-ponents as well and hence we conclude that Maxwells equations are covariant.It is nice to actually see that this works, but it also emphasises the powerof using the arguments concerning physical 4-vectors which demonstrated thesame physics certainly more elegantly and (perhaps) more clearly.

7.3 Magnetism as a Relativistic Effect and the Lorentz

Force

In this section we will investigate in more detail what was hinted at by thetransformation laws we have derived for the field components, namely that theelectric and magnetic fields are different manifestations of the same underlyingphysics. Consider a current carrying wire and a charge q at a location P.

At P we have a magnetic field found using Amperes theorem

B =0I

2r.

To detect this magnetic field the charge must be moving; let us take the particlevelocity to be u.

In Lab, S: = 0 jx = I/a

In rest frame of charge, S:

= (0 ujx/c2) jx = (jx u 0)= uI

c2a(a is LI)


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Therefore in the rest frame we see a radial electric field which we can find fromGausss theorem:

E

y =

1

2rluI

0c2a al = 0uI

2r

and the force on the charge is

dpydt

= qu 0I2r

(inwards)

But py = py andd

dt= d

dthence the force equation gives

dpydt

= quB

which is just one component of the Lorentz force, f = q(uB), in S. Whatappears as a force due to the magnetic field in one frame appears as an electricforce in the rest frame.

By considering variations on the geometry we could easily demonstrate theother components of the Lorentz force. Further if there were an electric fieldin S we can now state that the correct form for the force on a charge q in the

presence of electric and magnetic fields is:

f= q (E+ uB)

Consider now a stationary charge adjacent to a line charge and no current

In a frame moving with velocity V in the x direction


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Consider in S the ratio of the magnetic to electric forces

|magnetic force on q||electrostatic force on q| = V Bx

Ey= V

2

001

= V2

c2 1023

where the numerical estimate comes from taking a typical carrier drift velocityin a wire to be 103ms1. How is it that we ever detect the effects of a magneticfield? The answer is that conductors are neutral to an accuracy much betterthan one part in 1023.

Magnetic fields are a manifestation of a relativisticeffect with 1011.

7.4 The Field-Strength Tensor

In the analysis of the preceding three sections we have lost much of the eleganceand simplicity of earlier analysis based on the 4-potential and 4-current-density.More importantly since we are no longer working explicitly with 4-vectors thecovariant or frame independent nature of the equations in not immediatelyobvious, although as we showed in Section 7.2 it is present. However we cannotavoid introducing the electric and magnetic fields since they are what lead tomeasurable effects on charges. Lets re-examine the definition ofEand B and

relate it directly to the 4-gradient,

, and 4-potential A

.Recall

A =

c,A

and =

1

c

t,

x,

y,

z

Then we can write for the x-component of the electric field

Ex = Axt

x

= c 0A1 1A0


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and for the x-component of the magnetic field

cBx = c

Azy

Ayz

= c

2A3 3A2

.

We can define an anti-symmetric second-rank tensor which therefore containsall of the E and B components as an outer product

F = c

A Awhich in matrix form has components which we can easily identify as

F =

0 Ex Ey EzEx 0 cBz cByEy cBz 0 cBxEz cBy cBx 0

Transformation ofE and B

Since F was formed from 4-vector components then it must be a properphysical 4-tensor. This means that it must transform according to the LorentzTransformations. We therefore have

F

=

F

and the transformation of the field-strength tensor immediately gives the trans-formations of E and B. Lets see how this works by considering the F 02

componentF

02

= 02

F = 0

2

2F2 = 0

02

2F02 + 0

12

2F12

i.e. Ey = (Ey) + (cBz)or Ey = (Ey V Bz)and of course the other components follow similarly.

Maxwells Equations in Frame-Independent Form and the Lorentz

Force

Let us take the scalar product of the field-strength tensor with the 4-gradient

F = c (

A A) = cA = c2A = c0J

[NB A = 0 because we use the Lorenz Gauge.]

Hence we can recover Maxwells equations in terms of the field-strength tensor,and therefore in terms of the fields from:


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F = c0J

Writing out the components we easily see that these four equations are, interms of the 3-vector fields just

E = /0B 00E = 0j

i.e. two of Maxwells equations. What has happened to the other two? Werecover them by noting that

F + F + F = 0

which follows from the definition of the field strength tensor in terms of the4-potential. Expanding this in components shows that it is equivalent to

B = 0E = B

Hence we have been able to write Maxwells original equations for the fieldsexplicitly in a frame independent or covariant form. We have come full cir-cle, but en route shown the complete unity between Special Relativity andElectromagnetism.

For completeness we can now write the Lorentz force in a frame independentway

dP

d=

q

cFU

It is a useful exercise to check that these four equations do reproduce theappropriate dynamics for a (relativistic) particle in electric and magnetic fields.


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7.5 Electromagnetism as the Simplest Relativistic The-

ory

The analysis we have seen strongly hints that Electromagnetism may be specialfrom the point of view of SR. It is certainly a relativistically correct theory.Is it an unavoidable consequence of SR? In fact, it is nota direct consequenceof SR. However, as we now show, the whole of EM can be derived based ononly a few, simplifying, assumptions on the form a relativistic theory can take.

We start by recalling the scalar product of the 4-force, F,2 with the 4-velocity,U, is identically zero

F U = 0.We can then ask the question: What is the simplest form that F can take tosatisfy this equation? Clearly, since this must hold for all U, F can neither

be arbitrary nor be independent of U. The simplest form is to consider Fto be linearly dependent on U. Since U U is not zero, it cannot be simplyproportional to U, hence the simplest form is:

F = GU

where is just some scalar constant we introduce for later notational conve-nience. For F U = 0 to hold we have GUU = 0 and hence we require:

G = G .From G we can construct a new 4-vector, K

K G (1)which has zero divergence

K = G

= 0

due to the antisymmetry ofG. K is therefore a 4-vector which has the prop-erties of a conserved 4-current. What is more, by (1) it acts as the appropriatesource in what looks like Maxwells equations in terms of F.

However, we have not quite finished. The rest of Maxwells equations followfrom one last assumption: that the tensor G can simply be formed from thegradient of a 4-Vector B. Antisymmetry demands

G = B BIf you now make the identifications: = q, B = A, G = F/c, K = 0J,you will find that we have constructed all of electromagnetism. In this sense,electromagnetism is the simplest relativistic theory one can imagine.

2Be sure not to confuse the 4-vector F = F with the field strength tensor F . If indoubt, count the indices.


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8 Relativistic Electrodynamics and Radiation

8.1 Fields of a Uniformly Moving Charge

We shall start be calculating the fields of a uniformly moving charge

r

r

(x,y)S

(x,y)

x

yS

x

y V

V

We shall call the frame in which the charge is moving at a speed V the labframe, S; the rest frame of the charge is S . We consider the instant t = t = 0when the charge is at the origin of both frames.

In the (x, y)-plane of S we have for the fields (since the charge is at rest it isjust the static fields)

E =q

40 x

r3,

y

r3, 0 and B

= 0

Applying the Lorentz Transformations to these (Section 7.1) gives in S

Ex = Ex =

qx

40r3

; Ey = Ey =

qy

40r3

; Ez = 0

with x = x and y = y. The system has polar symmetry and therefore wecan take the x-axis as the polar axis and write x = r cos , y = r sin giving

r2

= x2

+ y2

= 2r2(1 sin2 ) + r2 sin2

= 2

r2

1V2

c2 sin2

[where 1 12

= V2

c2]

Therefore

E =q cos

40 2r2

1 V2c2

sin2 3/2 E = q sin

40 2r2

1 V2c2

sin2 3/2


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Br = B = 0, but B =V

c2Ey =

V

c2E

What do we note?

There is now a magnetic field which is purely azimuthal which arisesfrom the current due to the moving charge.

The electric field has the usual 1/r2 fall off. The electric field is purely radial, although its magnitude has an angular

dependence which depends on the Lorentz factor .

Lets think about this last point a little. We see from the above that

EE

= sin cos

,

i.e. the field is directed away from the origin which is the present position ofthe charge.

This is perhaps a little surprising since one might expect it to depend onwhere the charge was some time ago. However, we are dealing with a steady-state situation in which the charge is in a uniform state of motion. There istherefore no characteristic time in the problem (set for example by the startof the particles motion) and therefore the only defining point at any time is

the particles current position. This argument makes it less surprising thatthe electric field radiates from the current particle position and that the fieldsare carried along by the particle much like fields fixed in a solid body. Belowwe shall consider the case when the electron accelerates and then we shall seethat things are very different.

yS Direction of EField

lV/c

lx

present positionposition at retarded time

Lets examine the magnitude of the field as a function of the Lorentz factor.


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ct

light cone

Ox

velocity,

(particle trajectory)(t)w

(ct , )R Rw

V

In Section 4.8, we determined the potentials in terms of integrals over theretarded charge and current densities. The fields one observes at O arise fromthe charge distribution at each position r at the earlier retarded time t =|r|/c. In the present case of a single point-particle, with position w(t),simplifications arise because only a single space-time point on the particlesworld-line contributes to the potential. The retarded time and position aredetermined by the condition

tR = |w(tR)|/c

i.e. the point where the trajectory cuts the (backward pointing) light-coneof the observation point O (see diagram). We denote this space-time point(ctR,wR), and the instantaneous velocity of the particle at this retarded timetR by

V w(tR).We choose axes such that this velocity is along the x-axis of the frame S,V = Vx, and calculate the potentials in the inertial frame S moving withrespect to S at velocity V in the standard configuration (the space-timeorigin of S coincides with that of S),4

A

=

c ,A

= q

40rc, 0, 0, 0

.4It may help to note that for a point-particle only a single instant in the world-line

contributes to the potential, so we may equally well consider the potentials to be generatedby a particle that is in uniform motion at the retarded velocity V, passing through thepoint (ctR,wR). Be careful: this is correct for the potentials; it does notapply for the fields,which are gradients of the potential with respect to the observation point. The variation ofthe retardation time tR as a function of observation point causes the fields to depend on thecurvature of the trajectory i.e. the particles acceleration.


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In the lab frame, S, this transforms to

A = q

40r

c

,q

40r

c

, 0, 0 .The length r is the spatial distance between the charge and the point ofobservation in S. We wish to determine this length for the present problemin terms of quantities measured in S. Following convention, we choose r todenote the vector from the charge to the point of observation. In the case ofinterest, r = wR in S at a time tR = |r|/c, and r may be determined bythe Lorentz transformation of the event (c(|r|/c) ,r) LT (c(|r|/c),r).Applying the LT to the time coordinate we have

|r| = (|r|+ x)= (r r)

where we have introduced the vector = V/c. The length r in this expressionis of course the distance between the point at which the potential is determinedand the point where the charge was allowing for the light-travel time, r =|wR|. In other words we have to evaluate this expression at the retarded time.Inserting the expression for r we now have our results for the potential of anarbitrarily moving charge:

=q

40[r r] A =0q[V]

4[r r]

These potentials are called the Lienard-Wiechert potentials, where the brackets[ ] indicate that the quantities are to be evaluated at the retarded time asusual. As we have seen in the previous section if the particle is in a uniformstate of motion then we get fields which just move along with the charge. Thesituation is very much more interesting if the charge accelerates.

8.3 Radiation by an Accelerated Charge

The following diagram shows how the field lines develop for a charge initiallyin steady motion with a speed V = 0.8c to the right is decelerated and comesto rest.


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(a) shows the field for a uniformly moving charge. Only the fields in the upperhalf plane are shown in all of the figures.

(b) The charge is brought to rest. The field now develops a kink which propa-

gates out at the speed of light. To a distant observer the field emanates fromthe location the charge would have moved to, P1.

(c) The fields a short time later.

Approximate Calculation

We will now do an approximate calculation ofE and B in the kink. Thiswas first done by J.J. Thomson in 1903, 2 years before Einstein published hisSpecial Theory of Relativity.

Imagine a charge that has been travelling at a constant velocity, v, such thatx = vt, until the moment t = 0 at which point it decelerates at a rate |a| = v/to come to rest at a time t = (and position x = v /2). It then remains atrest for t > .

We shall assume that v c, which allows several simplifications. In particular,the final position of the particle x = v /2 is small compared to other relevantlengthscales, and may be considered to coincide with the origin. Also, we


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are permitted to neglect the relativistic compression effects discussed in 8.1( 1).Consider the configuration of the electric field at a time t = T

. Observers

further from the origin than R = cT cannot know that the particle has decel-erated. The field in this region is that of a charge moving (and appearing tocontinue to move) at a constant speed v. The field appears to emanate froma point charge at position x = vT.

An observer whose distance from the origin is less than c(T ) cannot knowthat the particle had previously been moving. In this region the field is thatof a charge at rest close to the origin (actually at x = v /2).

The field is clearly kinked between these two regions, in the manner illustratedin the above figure. We now calculate the electric field in the shell of width

c, with reference to the following diagram.

x

vT

R=cT

c

sin

A

B

C

D

Fields of Fields of themoving chargethe stationary charge

x=vT

The lines AB and CD represent electric field lines in the inner and outerregions, which both make the same angle to the x-axis about the points fromwhich the fields appear to emanate (x = v /2 0 and x = vT respectively).Consider the flux through the bases of the cones formed by rotating the linesAB and CD around the x-axis. The field patterns in the two regions are justthe uniform radial fields from the charge (since we are neglecting relativisticcompression), so the flux enclosed by these two cones is identical. Hence, since E= 0 in free space, for all the flux within AB in the inner region to emergewithin CD in the outer region, AB and CD must be part of the same field line.


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This tells us the direction of the field in the thin shell is parallel to BC (thesegment BC may be considered straight, for v c).Geometry then gives the ratio of the transverse and radial components.

ErE

=c

vT sin =

c

(|a|)(R/c)sin =c2

|a|R sin Now Er must take the same value as it does in the inner region close to B(integrating E = 0 over a small shell containing B), such that Er =Q/40R

2 from Gausss theorem. This then gives

E |a|R sin c2

Q

40R2 0Q|a| sin

4R.

Note that the field falls of like 1/R as is appropriate for a radiation field.

We can now estimate the B-field again in the kink. Maxwells 4th equation inintegral form gives in free space:

B dl = 00 t

E dS= 00

tE

Let us apply this to the path formed by rotating the point C around the x-

axis. This is a circle around the direction of the particle velocity and in a planeperpendicular to it. The area will be the spherical surface bounded on the leftby C. The electric flux, just before the arrival of the kink, is given by:

E =Q

40R24R2(1 cos )

2,

where we have used the expression for the field due to the charge at x = vT.After a time the kink has swept past the point C and the flux through thesurface will be smaller value

E =

Q

40R24R2(1

cos )

2

originating from the static charge close to the origin. However, from geometrywe have

R( ) vT sin = vT

Rsin = v

csin


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Thus for small v/c, ( ) is small and we may writedE

dt

E E

Q

20sin ( ) = v

c

Q

20sin2 = |a|

c

Q

20sin2

ButB dl =2R sin B, hence we find

B 12R sin

00dE

dt= 0Q|a| sin

4Rc= E

c.

Therefore both E and B vary like radiation fields: i.e. decay as 1/R, aremutually perpendicular, and are in phase with relative amplitude c. The ac-celerated charge is radiating and the power radiated is given by (being carefulto choose the outward-going Poynting vector)

N =BE

0=

0Q2|a|2 sin2

162

R2

c

.

The total power radiated is

P =

N2R2 sin d =

0Q2|a|2

6c

This is known as the Larmor Formula.

This result is exact in the limit as v tends to zero, i.e. for an accelerating

charge in its instantaneous rest frame. We can find the result in a generalframe by noting that (i) the total power radiated is a Lorentz invariant, 5 and(ii) in the IRF a2 = aa where a

is here the acceleration 4-vector. But wehave (problem 8) a general expression for this scalar product, hence

P =0Q

2(A A)6c

=0Q

2

6c(6a2|| +

4a2)

This result is the relativistic Larmor formula, or Lienard formula. Here parallel

and perpendicular refer to the direction of the velocity.5That the power radiated is a Lorentz invariant can be seen by considering a small

amount of energy dE emitted in a time dt in the IRF of the charge (t is therefore theproper time). The power loss in the IRF is then P dE/dt. In the IRF, the radiation isemitted with a sin2 distribution and carries no net linear momentum; the four-momentumlost is dP = (dE/c,0, 0, 0). Transforming to a frame S, we therefore find dE= dE, anddt = dt (as usual). Hence the power loss in S is P dE/dt = (dE)/(dt) = P. Thepower loss is the same in all inertial frames.


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8.4 Cyclotron and Synchrotron Radiation

Consider a charged particle, of mass m and charge q, moving in a uniform

magnetic field B = Bez. The equation of motion isd(mu)

dt= quB

mu + mu = qBu ezTake the dot product of this equation with the velocity u

mu2 + mu u = 0By direct differentiation, = 3uu/c2 u u = (c2/3), so inserting in theabove equation we find

mu2

+ m(c2

/3

) = 0mc2(2 + 1/2) = 0

mc2 = 0

Thus, the particle has constant energy = 0, and

mu = qBu ezThis describes circular motion in the xy-plane6 at the angular frequency

=eB

m

The radius of the orbit is R = u/, where u = |u|.Substituting the acceleration a =

quBm

into the relativistic Larmor formulagives

P =0q

42B2u2

6cm2

We can write this in an interesting way by re-introducing the Thomson crosssection

T =20q4

6m2,

and the energy density in the magnetic field, UB = B2

/20. Then we have forthe power radiatedP = 2cT

22UB

Two limits are important: when the particle velocity is very much less than,or close to the speed of light

6The particle can also have a constant velocity along ez. Here, we consider motions onlyin the xy-plane.


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Cyclotron Radiation

In this case 1. As the particle moves on its orbit the direction of theacceleration continuously changes as seen by a distant observer. The chargeradiates as if it is two dipoles in quadrature with a frequency = qB/m(this quantity is sometimes called the Larmor frequency or gyrofrequency,g qB/m). [See problem 21 for a discussion of the radiation fields.]

Synchrotron Radiation

In this case 1. The emission is then far from isotropic in the lab frame.In the IRF the power-pattern from the accelerated charge is

[i.e. with sin2 distribution with respect to the axis of the acceleration]. Inthe lab frame this is significantly aberrated as we have already seen

Therefore as the charge orbits we see a succession of pulses. We can use ourearlier results on aberration to show that the angular width of the pulse isapproximately 2/. To calculate what an observer will actually see, we mustallow for light travel time effects.


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The leading edge of the observed pulse originates from radiation that is emittedwhen the particle is at X, and the radiation forming the trailing edge of thepulse is emitted when the particle is at Y.

1/

X Yobserver

R

The start of the pulse is detected by an observer a distance r away at a time

tX +r + R/

c

where tX is the time when the particle is at X. The end of pulse is detected atthe time

tX + 2Ru

+ r R/c

The duration of the pulse is therefore

t =1

c

r R

r R

+

2R

u=

2R

1

u 1

c

=2

(1 u

c) 1

3

where we have used (1

u/c)

1/22. The observed spectrum contains

significant contributions at a characteristic frequency given by:

s 1t

= 3

The spectrum of synchrotron radiation is shown in the figure, with y = /swhere s (3/2)3.


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1 2 3 4 5y

0.1

0.2

0.3

0.4

0.5

Intensity

Examples

Significant energy loss in electron synchrotrons. Radiation losses limitthe energy that can be reached in circular accelerators.

Synchrotrons now used as a source of highly collimated X-rays for spec-troscopy and X-ray diffraction studies; possible to have powerful lab-scaleequipment.

Natural sources of synchrotron radiation are very important astronomi-cally, such as supernovae and radio galaxies.

cooper 3 relativistic electrodynamics

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