(9.234)
2
jj'
j'
t c
V
j
j' 1
S
where dAj = njdA = njd 2 r is the j th component of the elementary area vector dA = n dA = n d 2r that is normal to the volume’s surface, and directed out of the volume – see Fig. 17.76
dF
dA n dA
n
volume V
occupied by the field dAj
dA
Fig. 9.17. The force dF exerted on a boundary
r
j
element dA of the volume V occupied by the field.
surface S
Since, according to Eq. (5.10), the vector f in Eq. (234) is nothing other than the density of volume-distributed Lorentz forces exerted by the field on the charged particles, we can use the 2nd Newton law, in its relativistic form (144), to rewrite Eq. (234), for a stationary volume V, as Field
d S
momentum’s
3
d r p
,
(9.235)
2
part F
dynamics
dt c
V
where ppart is the total mechanical (relativistic) momentum of all particles in the volume V, and the vector F is defined by its Cartesian components:
Force via
3
the Maxwell
(M)
F dA .
(9.236)
j
tensor
jj'
j'
j' 1
S
Relations (235)-(236) are our main new results. The first of them shows that the vector
75 Just like the Poynting theorem (233), Eq. (234) may be obtained directly from the Maxwell equations, without resorting to the 4-vector formalism – see, e.g., Sec. 8.2.2 in D. Griffiths, Introduction to Electrodynamics, 3rd ed., Prentice-Hall, 1999. However, the derivation discussed above is superior because it shows the wonderful unity between the laws of conservation of energy and momentum.
76 The same notions are used in the mechanical stress theory – see, e.g., CM Sec. 7.2.
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S
g
,
(9.237)
2
c
already discussed in Sec. 6.8 without derivation, may be indeed interpreted as the density of momentum of the electromagnetic field (per unit volume). This classical relation is consistent with the quantum-mechanical picture of photons as ultra-relativistic particles, with a momentum of magnitude E/ c, because then the flux of the momentum carried by photons through a unit normal area per unit time may be represented either as Sn/ c or as gnc. It also allows us to revisit the Poynting vector paradox that was discussed in Sec. 6.8 – see Fig. 611 and its discussion. As was emphasized in that discussion, in this case, the vector S = EH does not correspond to any measurable energy flow. However, the corresponding momentum of the field, equal to the integral of the density (237) over a volume of interest,77 is not only real but may be measured by the recoil impulse it gives to the field sources – say, to a magnetic coil inducing the field H, or to the capacitor plates creating the field E.
Now let us turn to our second result, Eq. (236). It tells us that the 33-element Maxwell stress tensor complies with the general definition of the stress tensor78 characterizing the forces exerted on the boundaries of a volume, in our current case the volume occupied by the electromagnetic field (Fig. 17).
Let us use this important result to analyze two simple examples of static fields.
(i) Electrostatic field’s effect on a perfect conductor. Since Eq. (235) has been derived for a free space region, we have to select volume V outside the conductor, but we may align one of its faces with the conductor’s surface (Fig. 18).
V
z
E
Fig. 9.18. The electrostatic field
near a conductor’s surface.
From Chapter 2, we know that the electrostatic field just outside the conductor’s surface has to be normal to it. Selecting the z-axis in this direction, we have Ex = Ey =0, Ez = E, so that only diagonal elements of the tensor (228) are not equal to zero:
(M)
(M)
0
2
(M)
0
2
E ,
E ,
(9.238)
xx
yy
2
zz
2
Since the elementary surface area vector has just one non-zero component, dAz, according to Eq. (236), only the last component (that is positive regardless of the sign of E) gives a contribution to the surface force F. We see that the force exerted by the conductor (and eventually by the external forces that hold the conductor in its equilibrium position) on the field is normal to the conductor and directed out of the field volume: dFz 0. Hence, by the 3rd Newton law, the force exerted by the field on the conductor’s surface is directed toward the field-filled space:
77 It is sometimes called hidden momentum.
78 See, e.g., CM Sec. 7.2.
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Electric
dF
dF
0
E 2 dA .
(9.239)
field’s
surface
z
2
pull
This important result could be obtained by simpler means as well. (Actually, this was the task of one of the exercise problems assigned in Chapter 2.) For example, one could argue, quite convincingly, that the local relation between the force and the field should not depend on the global configuration creating the field, and thus consider the simplest configuration, a planar capacitor (see, e.g. Fig. 2.3) with surfaces of both plates charged by equal and opposite charges of density = 0 E. According to the Coulomb law, the charges should attract each other, pulling each plate toward the field region, so that the Maxwell-tensor result gives the correct direction of the force. Now the force’s magnitude given by Eq. (239) may be verified either by the direct integration of the Coulomb law or by the following simple reasoning. In the plane capacitor, the inner field Ez = /0 is equally contributed by two surface charges; hence the field created by the negative charge of the counterpart plate (not shown in Fig. 18) is E- = –/20, and the force it exerts of the elementary surface charge dQ = dA of the positively charged plate is dF surface = EdQ = –2 dA/20 = 0 E 2 dA/2, in accordance with Eq. (239).79
Quantitatively, even for such a high electric field as E = 105 V/m (close to the electric breakdown’s threshold in the air at a frequency of 10 GHz80), the “negative pressure” ( dF/ dA) given by Eq. (239) is of the order of 0.05 Pa (N/m2), i.e. many orders below the ambient atmospheric pressure of 1 bar 105 Pa. Still, this negative pressure may be substantial (well above 1 bar) in some cases, for example in good dielectrics (such as the high-quality SiO2 grown at high temperature, which is broadly used in integrated circuits), which can withstand electric fields up to ~109 V/m.
(ii) Static magnetic field’s effect on its source81 – say a solenoid’s wall or a superconductor’s surface (Fig. 19). With the Cartesian coordinates’ choice shown in that figure, we have Bx = B, By = Bz =
0, so that the Maxwell stress tensor (228) is diagonal again:
1
1
(M)
2
(M)
(M)
2
B ,
B .
(9.240)
xx
2
yy
zz
2
0
0
However, since for this geometry, only dAz differs from 0 in Eq. (236), the sign of the resulting force is opposite to that in electrostatics: dFz 0, and the force exerted by the magnetic field upon the conductor’s surface,
Magnetic
1
field’s
dF
dF
B 2 dA ,
(9.241)
surface
z
push
20
79 By the way, repeating these arguments for a plane capacitor filled with a linear dielectric, we may readily see that Eq. (239) may be generalized for this case by replacing 0 with . A similar replacement (0 ) is valid for Eq. (241) in a linear magnetic medium.
80 Note that the breakdown field E t in is a strong function of frequency. In the ambient air, it drops from its dc value of ~3106 V/m to ~1.5105 V/m at microwave frequencies and then rises to as much as ~6109 V/m at optical frequencies. The reason of the rise is that at very high frequencies, the amplitude of the field-induced oscillations of the rare free electrons becomes much smaller than their mean free path, inhibiting the bulk impact-ionization of neutral atoms. (Because of this reason, E t also depends on the air’s pressure.) 81 The causal relation is not important here. Especially in the case of a superconductor, the magnetic field may be induced by another source, with the surface supercurrent j just shielding the superconductor’s bulk from its penetration – see Sec. 6.
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corresponds to positive pressure. For good laboratory magnets ( B ~ 10 T), this pressure is of the order of 4107 Pa 400 bars, i.e. is very substantial, so the magnets require solid mechanical design.
B
x
j
z
Fig. 9.19. The magnetostatic field near
V
a current-carrying surface.
The direction of the force (241) could be also readily predicted using elementary magnetostatics arguments. Indeed, we can imagine the magnetic field volume limited by another, parallel wall with the opposite direction of surface current. According to the starting point of magnetostatics, Eq. (5.1), such surface currents of opposite directions have to repulse each other – doing that via the magnetic field.
Another explanation of the fundamental sign difference between the electric and magnetic field pressures may be provided using the electric circuit language. As we know from Chapter 2, the potential energy of the electric field stored in a capacitor may be represented in two equivalent forms, CV 2
Q 2
U
.
(9.242)
e
2
C
2
Similarly, the magnetic field energy of an inductive coil is
LI 2
2
U
.
(9.243)
m
2
2 L
If we do not want to consider the work of external sources at a virtual change of the system dimensions, we should use the last forms of these relations, i.e. consider a galvanically detached capacitor ( Q =
const) and an externally-shorted inductance ( = const).82 Now if we let the electric field forces (239) drag the capacitor’s plates in the direction they “want”, i.e. toward each other, this would lead to a
reduction of the capacitor thickness, and hence to an increase of its capacitance C, and hence to a decrease of U e. Similarly, for a solenoid, allowing the positive pressure (241) to move its walls from each other would lead to an increase of the solenoid’s volume, and hence of its inductance L, so that the potential energy U m would be also reduced – as it should be. It is remarkable (actually, beautiful!) how the local field formulas (239) and (241) “know” about these global circumstances.
Finally, let us see whether the major results (237) and (241) obtained in this section, match each other. For that, let us return to the normal incidence of a plane, monochromatic wave from the free space upon the plane surface of a perfect conductor (see, e.g., Fig. 7.8 and its discussion), and use those results to calculate the time average of the pressure dF surface/ dA imposed by the wave on the surface. At elastic reflection from the conductor’s surface, the electromagnetic field’s momentum retains its amplitude but reverses its sign, so that the average momentum transferred to a unit area of the surface in a unit time (i.e. the average pressure) is
82 Of course, this condition may hold “forever” only for solenoids with superconducting wiring, but even in normal-metal solenoids with practicable inductances, the flux relaxation constants L/ R may be rather large (practically, up to a few minutes), quite sufficient to carry out the force measurement.
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dF
S
EH
surface
incident
*
2 cg
2 c
2 c
E
,
(9.244)
incident
2
2
H
dA
c
c
where E and H are complex amplitudes of the incident wave. Using the relation (7.7) between these amplitudes (for = 0 and = 0 giving E = cB), we get
2
*
dF
1
B
B
surface
cB
.
(9.245)
dA
c
0
0
On the other hand, as was discussed in Sec. 7.3, at the surface of a perfect mirror the electric field vanishes while the magnetic field doubles, so that we can use Eq. (241) with B B( t) =
2Re[ Bexp{- i t}]. Averaging the pressure given by Eq. (241) over time, we get 2
dF
1
2
surface
2Re i
B
t
B e
,
(9.246)
dA
2
0
0
i.e. the same result as Eq. (245).
For physics intuition development, it is useful to evaluate the electromagnetic radiation pressure.
Even for a relatively high wave intensity Sn of 1 kW/m2 (close to that of the direct sunlight at the Earth’s surface), the pressure 2 cgn = 2 Sn/ c is somewhat below 10-5 Pa ~ 10-10 bar. Still, this extremely small effect was experimentally observed (by P. Lebedev) as early as 1899, giving one more confirmation of Maxwell’s theory. Currently, there are ongoing attempts to use the pressure of the Sun’s light for propelling small spacecraft, e.g., the LightSail 2 satellite with a 32-m2 sail, launched in 2019.
9.9. Exercise problems
9.1. Use the pre-relativistic picture of the Doppler effect, in which light propagates with velocity c in a Sun-bound aether, to derive Eq. (4).
9.2. Show that two successive Lorentz space/time transforms in the same direction, with velocities u’ and v, are equivalent to a single transform with the velocity u given by Eq. (25).
9.3. N + 1 reference frames, numbered by index n (taking values 0, 1, …, N), move in the same direction as a particle. Express the particle’s velocity in the frame number 0 via its velocity uN in the frame number N and the set of velocities vn of the frame number n relative to the frame number ( n – 1).
9.4. A spaceship moving with a constant velocity v directly from the Earth, sends back brief flashes of light with a period ts – as measured by the spaceship's clock. Calculate the period with that Earth-based observers may receive these signals – as measured by their clock.
9.5. From the point of view of observers in a “moving” reference y
y'
frame 0 ' , a straight thin rod, parallel to the x'- axis, is moving without rotation u '
with a constant velocity u ' directed along the y' -axis. The reference frame 0 ' is itself moving relative to another ("lab") reference frame 0 with a constant v
velocity v along the x- axis, also without rotation – see the figure on the right.
0
x
'
0
x'
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Calculate:
(i) the direction of the rod's velocity, and
(ii) the orientation of the rod on the [ x, y] plane,
both as observed from the lab reference frame. Is the velocity, in this frame, perpendicular to the rod?
9.6. Starting from the rest at t = 0, a spaceship moves directly from the Earth, with a constant acceleration as measured in its instantaneous rest frame. Find its displacement x( t) from the Earth, as measured from Earth’s reference frame, and interpret the result.
Hint: The instantaneous rest frame of a moving particle is the inertial reference frame that, at the considered moment of time, has the same velocity as the particle.
9.7. Analyze the twin paradox for the simplest case of 1D travel with a piecewise-constant acceleration.
Hint: You may use an intermediate result of the solution of the previous problem.
9.8. Suggest a natural definition of the 4-vector of linear acceleration (commonly called the 4-acceleration) of a point, and use it to calculate the acceleration of a relativistic point moving with velocity u = u( t).
9.9. Calculate the first relativistic correction to the frequency of a harmonic oscillator as a function of its amplitude.
9.10. An atom, with an initial rest mass m, has been excited to an internal state with an additional energy E, still being at rest. Next, it returns to its initial state, emitting a photon. Calculate the photon’s frequency, taking into account the relativistic recoil of the atom.
Hint: In this problem, and also in Problems 11-13, treat photons as classical ultra-relativistic point particles with zero rest mass, energy E = , and momentum p = k.
9.11. A particle of mass m, initially at rest, decays into two particles with rest masses m 1 and m 2.
Calculate the total energy of the first decay product, in the reference frame moving with that particle.
9.12. A relativistic particle with a rest mass m, moving with velocity u, decays into two particles with zero rest mass.
(i) Calculate the smallest possible angle between the decay product velocities (in the lab frame, in that the velocity u is measured).
(ii) What is the largest possible energy of one product particle?
9.13. A relativistic particle flying in free space with velocity u, decays into two photons.83
Calculate the angular dependence of the probability of photon detection, as measured in the lab frame.
83 Such a decay may happen, for example, with a neutral pion.
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p
9.14. A photon with wavelength is scattered by an
m e
electron, initially at rest. Calculate the wavelength ’ of the
scattered photon as a function of the scattering angle –
see the figure on the right.84
'
9.15. Calculate the threshold energy of a -photon for the reaction
0
γ p p π ,
if the proton was initially at rest.
Hint: For protons, m p c 2 938 MeV, while for neutral pions, m c 2 135 MeV.
9.16. Calculate the largest possible velocity of the electrons provided at the so-called - decays, n p e ,
e
of neutrons at rest.
Hint: Electron neutrinos and antineutrinos are virtually massless (on the energy scale of this problem); the rest energies E mc 2 of the other involved particles are: 939.565 MeV for the neutron, 938.272 MeV for the proton, 0.511 MeV for the electron.
9.17. A relativistic particle with a rest mass m and energy E, collides with a similar particle, initially at rest in the laboratory reference frame. Calculate:
(i) the final velocity of the center of mass of the system, in the lab frame,
(ii) the total energy of the system, in the center-of-mass frame, and
(iii) the final velocities of both particles (in the lab frame), if they move along the same direction.
9.18. A “primed” reference frame moves, relative to the “lab” frame, with a reduced velocity
v/ c = n x. Use Eq. (109) to express the elements T’ 00 and T’ 0 j (with j = 1, 2, 3) of an arbitrary contravariant 4-tensor T via its elements in the lab frame.
9.19. Prove that quantities E 2 – c 2 B 2 and EB are Lorentz-invariant.
9.20. Static fields E and B are uniform but arbitrary (both in magnitude and in direction). What should be the velocity of an inertial reference frame to have the vectors E ’ and B ’, observed from that frame, parallel? Is this solution unique?
q
u
9.21. Two charged particles, moving with equal constant velocities u, are
1
offset by distance R = { a, b} (see the figure on the right), as measured in the lab b
frame. Calculate the forces between the particles – also in the lab frame.
u
q 2
a
84 This is the famous Compton scattering effect, whose discovery in 1923 was one of the major motivations for the development of quantum mechanics – see, e.g., QM Sec. 1.1.
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9.22. Each of two thin, long, parallel particle beams of the same velocity u, separated by distance d, carries electric charge with a constant density per unit length, as measured in the reference frame moving with the particles.
(i) Calculate the distribution of the electric and magnetic fields in the system (outside the beams), as measured in the lab reference frame.
(ii) Calculate the interaction force between the beams (per particle) and the resulting acceleration, both in the lab reference frame and in the frame moving with the particles. Compare the results and give a brief discussion of their relation.
9.23. Spell out the Lorentz transform of the scalar-potential and vector-potential components, and use the result to calculate the potentials of a point charge q moving with a constant velocity u, as measured in the lab reference frame.
9.24. Calculate the scalar and vector potentials created by a time-independent electric dipole p, as measured in a reference frame that moves relative to the dipole with a constant velocity v, with the shortest distance (“impact parameter”) equal to b.
9.25. Solve the previous problem, in the limit v << c, for a time-independent magnetic dipole m.
9.26. Assuming that the magnetic monopole does exist and has a magnetic charge g, calculate the change of the magnetic flux in a superconductor ring due to the passage of a single monopole through it. Evaluate for the monopole charge conjectured by P. Dirac, g = ng 0 n(2/ e), where n is an integer; compare the result with the magnetic flux quantum 0 (6.62) with q = e, and discuss their relationship.
Hint: For simplicity, you may consider the monopole’s passage along the symmetry axis of a thin, round superconducting ring, in otherwise free space.
9.27. Re-derive Eq. (9.161) of the lecture notes for the simplest case p(0) = 0, using the 4-vector form (9.145) of the equation of motion and the notion of rapidity tanh-1 that was briefly discussed in Sec. 9.2.
9.28.* Calculate the trajectory of a relativistic particle in a uniform electrostatic field E, for an arbitrary direction of its initial velocity u(0), using two different approaches – at least one of them different from the approach used in Sec. 6 for the case u(0) E.
9.29. A charged relativistic particle with velocity u performs planar cyclotron rotation in a uniform external magnetic field of magnitude B. How much would the velocity and the orbit’s radius change at a slow change of the field to a new magnitude B' ?
9.30.* Analyze the motion of a relativistic particle in uniform, mutually perpendicular fields E
and B, for the particular case when E is exactly equal to cB.
9.31.* Find the law of motion of a relativistic particle in uniform, parallel, static fields E and B.
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9.32. An external Lorentz force F is exerted on a relativistic particle with an electric charge q and a rest mass m, moving with velocity u relatively some “lab” frame. Calculate its acceleration as observed from that frame.
9.33. Neglecting relativistic kinetic effects, calculate the smallest voltage V that has to be applied between the anode and cathode of a magnetron (see Fig. 13 and its discussion) to enable electrons to reach the anode, at negligible electron-electron interactions (including the space-charge effects) and collisions with the residual gas molecules. You may:
(i) model the cathode and anode as two coaxial round cylinders, of radii R 1 and R 2, respectively; (ii) assume that the magnetic field B is uniform and directed along their common axis; and (iii) neglect the initial velocity of the electrons emitted by the cathode.
(After the solution, estimate the validity of the last assumption and of the non-relativistic approximation, for reasonable values of parameters.)
9.34. A charged relativistic particle has been injected into a uniform electric field whose magnitude oscillates in time with frequency . Calculate the time dependence of the particle’s velocity, as observed from the lab reference frame.
9.35.* A plane, linearly-polarized electromagnetic wave of frequency is incident on an otherwise free relativistic particle with electric charge q. Analyze the dynamics of the particle’s momentum and compare the result with those of the previous problem and Problem 7.5.
9.36. Analyze the motion of a non-relativistic particle in a region where the electric and magnetic fields are both uniform and constant in time, but not necessarily parallel or perpendicular to each other.
9.37. A static distribution of electric charge in otherwise free space has created a time-independent distribution E(r) of the electric field. Use two different approaches to express the field energy density u’ and the Poynting vector S ’, as observed from a reference frame moving with constant velocity v, via the components of the vector E. In particular, is S ’ equal to (–v u’)?
9.38. A plane wave of frequency and intensity S, is normally incident on a perfect mirror moving with velocity v in the same direction as the wave.
(i) Calculate the reflected wave’s frequency, and
(ii) use the Lorentz transform of the fields to calculate the reflected wave’s intensity
– both as observed from the lab reference frame.
9.39. Perform the second task of the previous problem by using general relations between the wave’s energy, power, and momentum.
Hint: As a byproduct, this approach should also give you the pressure exerted by the wave on the moving mirror.
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9.40. Consider the simple model of a capacitor’s charging by a lumped
current source, shown in the figure on the right, and prove that the momentum
given by the constant, uniform external magnetic field B to the current-
B
carrying conductor is equal and opposite to the momentum of the
I ( t)
electromagnetic field that the current I( t) builds up in the capacitor. (You may assume that the capacitor is plane and very broad, and neglect the fringe field
effects.)
9.41. Consider an electromagnetic plane wave packet propagating in free space, with its electric field represented as the Fourier integral
E(r, t) Re E ei k dk,
with
kz t
and
,
c k
.
k
k
k
k
Express the full linear momentum (per unit area of wave’s front) of the packet via the complex amplitudes E k of its Fourier components. Does the momentum depend on time? (In contrast with Problem 7.8, in this case the wave packet is not necessarily narrow.)
9.42. Calculate the forces exerted on well-conducting walls of a waveguide with a rectangular ( a b) cross-section, by a wave propagating along it in the fundamental ( H 10) mode. Give an interpretation of the results.
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Chapter 10. Radiation by Relativistic Charges
The discussion of special relativity in the previous chapter enables us to revisit the analysis of electromagnetic radiation by charged particles, now for arbitrary velocities. For a single point particle, it turns out to be possible to calculate the radiated wave fields in an explicit form and analyze the results for such important particular cases as synchrotron radiation and the “Bremsstrahlung” (brake radiation). After that, we will discuss the apparently unrelated effect of the so-called Coulomb losses of energy by a particle moving in condensed matter, because this discussion will naturally lead us to such important phenomena as the Cherenkov radiation and the transitional radiation. At the end of the chapter, I will briefly review the effects of the back action of the emitted radiation on the emitting particle, whose analysis reveals some limitations of classical electrodynamics.
10.1. Liénard-Wiechert potentials
A convenient starting point for the discussion of radiation by relativistic charges is provided by Eqs. (8.17) for the retarded potentials. In free space, these formulas with the integration variable notation changed from r ’ to r ” for the clarity of what follows, are reduced to 1
( "
r , t R / c)
3
0
j( "
r , t R / c)
r
( , t)
d r" , A r
( , t)
d 3 r" ,
R
with r "
r
. (10.1a)
4
R
4
R
0
As a reminder, Eqs. (1a) were derived from the Maxwell equations without any restrictions, and are very natural for situations with continuous distributions of the electric charge and/or current. However, for a single charged particle, whose charge and current distributions may be described as
r, t q r ' r,
jr, t qu r ' r,
u
with '
r ,
(10.1b)
where r ’ = r ’( t) is the instantaneous position of the charge, it is more convenient to recast Eqs. (1a) into an explicit form that would not require integration in each particular case. Indeed, as Eqs. (1) show, the potentials at a given observation point {r, t} are contributed by only one specific point {r ’( t ret), t ret} of the particle’s 4D trajectory (called its world line), which satisfies the following condition: R
ret
t t
,
(10.2)
ret
c
where t ret is called the retarded time, and R ret is the length of the following distance vector R
r t '
r t
(10.3)
ret
ret
– physically, the distance covered by the electromagnetic wave from its emission to observation.
The reduction of Eqs. (1a) to such a simpler form, however, requires some care. Their naïve integration over r ” would yield the following apparent but wrong results:
q
r t
qc
qu
r, t
1
,
0
,
i.e.
;
Ar, t
0
ret
, (WRONG!)
(10.4)
4 R
c
4 R
4 R
0
ret
ret
ret
© K. Lik
harev
EM: Classical Electrodynamics
where uret is the particle’s velocity at the retarded point r ’( t ret). Eqs. (4) is a good example of how the relativity theory (even the special one :-) cannot be taken too lightly. Indeed, the strings (9.84)-(9.85), formed from the apparent potentials (4), would not obey the Lorentz transform rule (9.91), because according to Eqs. (2)-(3), the distance R ret also depends on the reference frame it is measured in.
In order to correct the error, we need, first of all, to discuss the conditions (2)-(3). Combining them by eliminating R ret, we get the following equation for t ret:
c( t t ) r( t) '
r ( t ) .
(10.5)
Retarded
ret
ret
time
Figure 1 depicts the graphical solution of this self-consistency equation as the only1 point of intersection of the light cone of the observation point (see Fig. 9.9 and its discussion) and the particle’s world line.
{r ' t
t
adv
particle' s
,
}
adv
time
word line
{r ' t' , t' }
{r, t}
r
{r ' t
t
ret ,
}
r
j '
ret
Fig. 10.1. Graphical
j
R
solution of Eq. (5).
ret
In Eq. (5), just as in Eqs. (1)-(3), all variables have to be measured in the inertial (“lab”) reference frame in which the observation point r rests. Now let us write Eqs. (1) for a point charge in another inertial frame the frame 0 ’ whose velocity (as measured in the lab frame) coincides, at the moment t’ = t ret, with the velocity uret of the charge.2 In that frame, the charge rests, so that, as we know from the electro- and magnetostatics,
q
'
,
A ' 0 .
(10.6a)
4 R'
0
(Remember that this R’ may not be equal to R ret, because the latter distance is measured in the “lab”
reference frame.) Let us use the identity 1/0 0 c 2 again to rewrite Eqs. (6a) in the form of components of a 4-vector similar in structure to the last two of Eqs. (4):
'
qc
0
,
'
A 0 .
(10.6b)
c
4 R'
Now it is easy to guess the correct answer for the 4-potential for an arbitrary reference frame: 1 As Fig. 1 shows, there is always another, “advanced” point {r ’( t adv), t adv} of the particle’s world line, with t adv > t, which is also a solution of Eq. (5), but it does not fit Eqs. (1), because the observation, at the point {r, t < t adv}, of the field induced at the advanced point, would violate the causality principle.
2 This is just a particular case of the instantaneous reference frame –the notion that was encountered in several exercise problems of the previous chapter, and indeed was implied (though admittedly not sufficiently advertised) as the derivation of the key Eq. (9.60).
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0 qcu
A
,
(10.7)
4 u R
where (as a reminder) A {/ c, A}, u { c, u}, and R is the 4-vector of the inter-event distance, formed similarly to that of a single event – cf. Eq. (9.48):
R { c( t t' ), '
R } { c( t t' ), r '
r } .
(10.8)
Indeed, we needed the 4-vector A that would:
(i) obey the Lorentz transform,
(ii) have its spatial components Aj scaling, at low velocity, as uj, and (iii) be reduced to the correct result (6) in the instantaneous reference frame of the charge.
Eq. (7) evidently satisfies all these requirements, because the scalar product in its denominator is just u R
c,
u c( t t' ), R [ 2
c ( t t' ) u R] c
R β R cR1 β n, (10.9)
where n R/ R is a unit vector in the observer’s direction, u/ c is the normalized velocity of the particle, and 1/(1- u 2/ c 2)1/2. In the instantaneous reference frame of the charge (in which = 0 and
= 1), the expression (9) is reduced to cR, so that Eq. (7) is correctly reduced to Eq. (6b). Now let us spell out the components of Eq. (7) for the lab frame (in which t’ = t ret and R = R ret):
q
r, t
1
1
1
q
,
(10.10a)
Liénard-
4 ( R β R)
4
R 1
( β n)
0
ret
0
ret
Wiechert
potentials
u
β
u
Ar, t
0
0
q
qc
r, t ret . (10.10b)
2
4 R β R
4
R 1
( β n)
c
ret
ret
These formulas are called the Liénard-Wiechert potentials.3 In the non-relativistic limit, they coincide with the naïve guess (4), but in the general case include the additional factor 1/(1 – n)ret. Its physical origin may be illuminated by one more formal calculation – whose result we will need anyway.
Let us differentiate the geometric relation (5), rewritten as
R c( t t ) ,
(10.11)
ret
ret
over t ret and then, independently, over t, assuming that r is fixed. For that, let us first differentiate, over t
2
ret, both sides of the identity R ret = RretRret:
R
R
ret
ret
2 R
2R
.
(10.12)
ret
ret
t
t
ret
ret
If r is fixed, then Rret/ t ret (r – r ’)/ t ret = –r ’/ t ret –uret, and Eq. (12) yields
R
R
R
ret
ret
ret
n u .
(10.13)
ret
t
R
t
ret
ret
ret
Now let us differentiate the same R ret over t. On one hand, Eq. (11) yields 3 They were derived in 1898 by Alfred-Marie Liénard and (independently) in 1900 by Emil Wiechert.
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R
t
ret c c ret .
(10.14)
t
t
On the other hand, according to Eq. (5), at the partial differentiation over time, i.e. if r is fixed, t ret is a function of t alone, so that (using Eq. (13) at the second step), we may write
R
R
t
t
ret
ret
ret n u
ret .
(10.15)
t
t
t
ret
t
ret
ret
Now requiring Eqs. (14) and (15) to give the same result, we get:4
t
c
1
ret
.
(10.16) t ret/ t
t
c n u 1
β n
ret
ret
This important relation may be readily re-derived (and more clearly understood) for the particular case when the charge’s velocity is directed straight toward the observation point. In this case, its vector u resides in the same space-time plane as the observation point’s world line r = const – say, the plane [ x, t] shown in Fig. 2.
dt
t
dt dt dx / c
ret
ret
dt ret
dx / c
ret
Fig. 10.2. Deriving Eq. (16)
for the case n = .
x
dx u dt
ret
ret
ret
Let us consider an elementary time interval dt ret dt’, during which the particle would travel the space interval dx ret = u ret dt ret. In Fig. 2, the corresponding segment of its world line is shown with a solid vector. The dotted vectors in this figure show the world lines of the radiation emitted by the particle in the beginning and at the end of this interval, and propagating with the speed of light c. As it follows from the drawing, the time interval dt between the instants of the arrival of the radiation from these two points to any time-independent spatial point of observation is
dx
u
dt
1
1
ret
ret
ret
dt dt
dt
dt ,
that
so
.
(10.17)
ret
ret
ret
c
c
dt
1 u / c
1
ret
ret
This expression coincides with Eq. (16) for our particular case when the directions of the vectors u/ c and n R/ R (both taken at time t ret) coincide, and hence (n)ret = ret. The difference between Eqs.
(16) and (17) may be interpreted by saying that the particle’s velocity in the transverse directions (normal to the vector n) is not important for this kinematic effect5 – the fact almost evident from Fig. 1.
4 This relation may be used for an alternative derivation of Eqs. (10) directly from Eqs (1) – the exercise highly recommended to the reader.
5 Note that this effect (linear in ) has nothing to do with the Lorentz time dilation (9.21), which is quadratic in .
(Indeed, all our arguments above referred to the same, lab frame.) Rather, it is close in nature to the Doppler effect.
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So, the additional factor in the Liénard-Wiechert potentials is just the derivative t ret/ t. The reason for its appearance in Eqs. (10) is usually interpreted along the following lines. Let the charge q be spread along the direction of the vector Rret (in Fig. 2, along the x-axis) by an infinitesimal speed-independent interval x ret, so that the linear density of its charge is proportional to 1/ x ret. Then the time rate of charge’s arrival at some spatial point is u ret = dx ret/ dt ret, i.e. scales as 1/ dt ret. However, the rate of radiation’s arrival at the observation point scales as 1/ dt, so that due to the non-zero velocity uret of the particle, this rate differs from the charge arrival rate by the factor of dt ret/ dt, given by Eq. (16). (If the particle moves toward the observation point, (n)ret > 0, as shown in Fig. 2, this factor is larger than 1.) This radiation compression effect leads to the field change (at (n)ret > 0, its enhancement) by the same factor (16) – as described by Eqs. (10).
So, the 4-vector formalism was very instrumental for the calculation of field potentials. It may be also used to calculate the fields E and B – by plugging Eq. (7) into Eq. (9.124) to calculate the field strength tensor. This calculation yields
q
0
1
d
R u
R u
F
.
(10.18)
4
u R d
u R
Now using Eq. (9.125) to identify the elements of this tensor with the field components, we may bring the result to the following vector form: 6
q
n β
n (n β)
β
E
,
(10.19)
2
3
2
3
Relativistic
4
1
( β n) R
1
( β n) cR
0
ret
particle’s
fields
n E
n E
ret
ret
B
,
i.e. H
.
(10.20)
c
Z 0
Thus the magnetic and electric fields of a relativistic particle are always proportional and perpendicular to each other, and related just as in a plane wave – cf. Eq. (7.6), with the difference that now the vector nret may be a function of time. Superficially, this result contradicts the electro- and magnetostatics, because for a particle at rest, B should vanish while E stays finite. However, note that according to the Coulomb law for a point charge, in this case E = Enret, so that B nretE nretnret =
0. (Actually, in these relations, the subscript “ret” is unnecessary.)
As a sanity check, let us use Eq. (19) as an alternative way to find the electric field of a charge moving without acceleration, i.e. uniformly, along a straight line – see Fig. 9.11a reproduced, with minor changes, in Fig. 3. (This calculation will also illustrate the technical challenges of practical applications of the Liénard-Wiechert formulas for even simple cases.) In this case, the vector does not change in time, so that the second term in Eq. (19) vanishes, and all we need to do is to spell out the Cartesian components of the first term.
6 An alternative way of deriving these formulas (highly recommended to the reader as an exercise) is to plug Eqs.
(10) into the general relations (9.121), and carry out the required temporal and spatial differentiations directly, using Eq. (16) and its spatial counterpart (which may be derived absolutely similarly):
n
t
.
ret
c1βnret
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y r
R c( t t )
ret
ret
b
n
ret
'
r ( t )
β
'
r ( t)
ret
Fig. 10.3. The linearly moving charge
0
ut
x
problem.
ut
u( t t )
ret
ret
Let us select the coordinate axes and the time origin as shown in Fig. 3, and make a clear distinction between the actual position, r ’ ( t) = { ut, 0, 0} of the charged particle at the instant t we are considering, and its position r ’( t ret) at the retarded instant defined by Eq. (5), i.e. the moment when the particle’s field had to be radiated to reach the observation point r at the given time t, propagating with the speed of light. In these coordinates
β , ,
0
0 , r ,
0 ,
b
0 , r ' ( t ) ut
n
(10.21)
ret
, ,0
ret
0 ,
ret
cos , sin ,
0 ,
with cos = – ut ret/ R ret, so that [(n – ) x]ret = – ut ret /R ret – , and Eq. (19) yields, in particular: q
ut / R β
q
ut βR
ret
ret
E
.
(10.22)
x
2
4
1
( β n) R
4
1
( β n) R
0
3
2
ret
ret
2
ret
0
3
3 ret
But according to Eq. (5), the product R ret may be represented as c( t – t ret) u( t – t ret). Plugging this expression into Eq. (22), we may eliminate the explicit dependence of Ex on time t ret: q
ut
E
(10.23)
x
4
2
0
1( βn R .
) 3ret
The only non-zero transverse component of the field also has a similar form:
q
sin
q
b
E
(10.24)
y
4
2
1( β n)3 2
R
4
2
0
β n R
ret
0
1(
,
) 3ret
while Ez = 0. From Fig. 3, – nret = cos = – ut ret/ R ret, so that (1 – n) R ret R ret + ut ret , and we may again use Eq. (5) to get (1 – n) R ret = c( t – tret) + ut ret ct – ct ret/2. What remains is to calculate t ret from the self-consistency equation (5), whose square in our current case (Fig. 3) takes the form 2
2
2
2
2
R b ( ut ) c ( t t ) .
(10.25)
ret
ret
ret
This is a simple quadratic equation for t ret, which (with the appropriate negative sign before the square root, to get t ret < t) yields:
1/ 2
2
t t t t b / c
t
u t b
,
(10.26)
ret
2
2
2 2
2
2
2
2 2 2 21/2
c
so that the only retarded-function combination that participates in Eqs. (23)-(24) is
c
1
( β n) R
u t b
(10.27)
ret
2 2 2 21/2,
2
and, finally, the electric field components are
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q
ut
q
b
E
, E
,
E 0 .
(10.28)
x
4
y
4
z
0 2
2
2 2
b u t 3/2
0 2
2
2 2
b u t 3/2
But these are exactly Eqs. (9.139),7 which had been obtained in Sec. 9.5 by much simpler means, without the necessity to solve the self-consistency equation (5). However, that alternative approach was essentially based on the inertial motion of the particle, and cannot be used in problems in which it moves with acceleration. In such problems, the second term in Eq. (19), dropping with distance more slowly, as 1/ R ret, and hence describing wave radiation, is frequently the most important one.
10.2. Radiation power
Let us calculate the angular distribution of the particle’s radiation. For that, we need to return to Eqs. (19)-(20) to find the Poynting vector S = EH, and in particular, its radial component Sn = Snret, at large distances R from the particle. Following tradition,8 let us express the result as the energy radiated into unit solid angle per unit time interval dt rad of the radiation, rather than that ( dt) of its measurement.
(We will need to return to the measurement time t in the next section to calculate the observed radiation spectrum.) Using Eq. (16), we get
d P
d E
t
2
R S
.
(10.29)
n
E H 2
R n 1 β n
ret
ret
Ω
d
Ω
d dt
t
ret
ret
At sufficiently large distances from the particle, i.e. in the limit R ret (in the radiation zone), the contribution of the first (essentially, the Coulomb-field) term in the square brackets of Eq. (19) vanishes as 1/ R 2, and the substitution of the remaining term into Eqs. (20) and then Eq. (29) yields the following formula, which is valid for an arbitrary law of the particle’s motion:9
Radiation
2
n
power
d P
Z q
n ββ 2
0
.
(10.30)
density
d Ω
4 2
1 nβ5
Now, let us apply this important result to some simple cases. First of all, Eq. (30) says that a charge moving with a constant velocity does not radiate at all. This might be expected from our analysis of this case in Sec. 9.5 because in the reference frame moving with the charge it produces only the Coulomb electrostatic field, i.e. no radiation.
Next, let us consider a linear motion of a point charge with a non-zero acceleration directed along the straight line of the motion. In this case, with the coordinate axes selected as shown in Fig. 4a, each of the vectors involved in Eq. (30) has at most two non-zero Cartesian components:
7 A similar calculation of magnetic field components from Eq. (20) gives results identical to Eqs. (9.140).
8 This tradition may be reasonably justified. Indeed, we may say that the radiation field “detaches” from the particle at times close to t ret, while the observation time t depends on the detector’s position, and hence is less relevant for the radiation process as such.
9 If the direction of radiation, n, does not change in time, this formula does not depend on the observer’s position R. Hence, from this point on, the index “ret” may be safely dropped for brevity, though we should always remember that in Eq. (30) is the reduced velocity of the particle at the instant of the radiation’s emission, not of its observation.
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n sin , ,
0 cos,
β ,
0 ,
0 ,
β ,
0 ,
0 ,
(10.31)
where is the angle between the directions of the particle’s motion and of the radiation’s propagation.
Plugging these expressions into Eq. (30) and performing the vector multiplications, we readily get 2
2
d P
Z q
sin
0
2
.
(10.32)
Ω
d
4 2
1 cos 5
Figure 4b shows the angular distribution of such radiation, for three values of the particle’s speed u.
(a)
(b)
x
n
Fig. 10.4. Particle’s radiation
0
z
at its linear acceleration: (a)
β
β
1
the problem’s geometry, and
.
0 3
(b) the last fraction of Eq. (32)
as a function of the angle .
0.5
If the speed is relatively low ( u << c, i.e. << 1), the denominator in Eq. (32) is very close to 1
for all observation angles , so that the angular distribution of the radiation power is close to sin2 –
just as it follows from the general non-relativistic Larmor formula (8.26), for our current case with =
. However, as the velocity is increased, the denominator becomes less than 1 for < /2, i.e. for the forward-looking directions, and larger than 1 for back directions. As a result, the radiation in the direction of the particle’s motion is increased (somewhat counter-intuitively, regardless of the acceleration’s sign!), while that in the back direction is suppressed. For ultra-relativistic particles (
1), this trend is strongly exacerbated, and radiation to very small forward angles dominates. To describe this main part of the angular distribution, we may expand the trigonometric functions of participating in Eq. (32) in the Taylor series in small , and keep only their leading terms: sin , cos 1 – 2/2, so that (1 – cos) (1 + 2 2)/22. The resulting expression,
d P
2
2
Z q
0
2
8
2
,
(10.33)
Ω
2
d
1 2
, for
1
5
2
describes a narrow “hollow cone” distribution of radiation, with its maximum at the angle 1
1.
(10.34)
0
2
Another important aspect of Eq. (33) is how extremely fast (as 8) the radiation density grows with the Lorentz factor , i.e. with the particle’s energy E = mc 2.
Still, the total radiated power P (into all observation angles) at linear acceleration is not too high for any practicable values of parameters. To show this, let us first calculate P for an arbitrary motion of the particle. To start, let me demonstrate how P may be found (or rather guessed) from the general relativistic arguments. In Sec. 8.2, we have derived Eq. (8.27) for the power of the electric dipole radiation for a non-relativistic particle motion. That result is valid, in particular, for one charged particle, Chapter 10
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whose electric dipole moment’s derivative over time may be expressed as d( qr)/ dt = ( q/ m)p, where p is the particle’s linear mechanical momentum ( not its electric dipole moment). As the result, the Larmor formula (8.27) in free space, i.e. with v = c (but u << c) reduces to 2
2
Z
q dp
Z q
dp d
0
0
P
p
,
for u c.
(10.35)
6 2
c
m dt
6
2
2
m
c dt dt
This is evidently not a Lorentz-invariant result, but it gives a clear hint of how such an invariant, that would be reduced to Eq. (35) in the non-relativistic limit, may be formed:
2
Z q
dp
dp
2
Z q
2
2
dp
1 d
0
0
E
P
.
(10.36)
2
2
6 m c
d
d
2
2
6 m c
2
d
c
d
Using the relativistic expressions p = mc, E = mc 2, and d = dt/, the last formula may be recast into the so-called Liénard extension of the Larmor formula:10
2
2
Total
Z q
Z q
radiation
0
P
β2
6
β β2
0
β2
4
β β2
2
.
(10.37)
power via
6
6
It may be also obtained by direct integration of Eq. (30) over the full solid angle, thus confirming our guess.
However, for some applications, it is beneficial to express P via the time evolution of the particle’s momentum alone. For that, we may differentiate the fundamental relativistic relation (9.78), E 2 = ( mc 2)2 + ( pc)2, over the proper time to get
d E
dp
d E
c 2
2
p dp
dp
E
2
c
2 p
,
i.e.
u
,
(10.38)
d
d
d
E
d
d
where the last step used the relativistic relation c 2p/E = u mentioned in Sec. 9.3. Plugging Eq. (38) into Eq. (36), we may rewrite it as
Total
2
Z q
2
2
dp
dp
0
2
radiation
P
.
(10.39)
2
2
power via p
6 m c
d
d
Please note the difference between the squared derivatives in this expression: in the first of them we have to differentiate the momentum’s vector p first, and only then form a scalar by squaring the resulting vector derivative, while in the second case, only the magnitude of the vector has to be differentiated. For example, for circular motion with a constant speed (to be analyzed in detail in the next section), the second term vanishes, while the first one does not.
However, if we return to the simplest case of linear acceleration (Fig. 4), then ( dp/ d)2 =
( dp/ d)2, and Eq. (39) is reduced to
10 The second form of Eq. (10.37), which is frequently more convenient for applications, may be readily obtained from the first one by applying MA Eq. (7.7a) to the vector product.
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2
2
2
2
2
Z q
dp
Z q
dp 1
Z q
dp
0
P
2
1
2
0
0
,
(10.40)
2
2
2
2
2
2
2
6 m
c d
6 m
c d
6
m
c dt ret
i.e. formally coincides with the non-relativistic relation (35). To get a better feeling of the magnitude of this radiation, we may combine Eq. (9.144) with B = 0, and Eq. (9.148) with E u to get dp/ dt ret =
d E/ dz’, where z’ is the particle’s coordinate at the moment t ret. The last relation allows us to rewrite Eq.
(40) in the following form:
2
2
2
2
Z q
d E
Z q
d E d E dt
Z q
d E d
0
0
ret
0
E
P
.
(10.41)
2
2
2
2
2
2
6 m
c dz
6 m
c dz' dt
dz'
6 m
c u dz' dt
ret
ret
For the most important case of ultra-relativistic motion ( u c), this result reduces to P
2 d(E /
2
mc )
,
(10.42)
d E / dt
3 d( z' / r )
ret
c
where r c is the classical radius of the particle, defined by Eq. (8.41). This formula shows that the radiated power, i.e. the change of the particle’s energy due to radiation, is much smaller than that due to the accelerating field unless energy as large as ~ mc 2 is gained on the classical radius of the particle. For example, for an electron, with r c 310-15 m and mc 2 = m e c 2 0.5 MeV, such an acceleration would require the accelerating electric field of the order of (0.5 MV)/(310-15 m) ~ 1014 MV/m, while practicable accelerating fields are below 102 MV/m – limited by the electric breakdown effects. (As described by the factor m 2 in the denominator of Eq. (41), for heavier particles such as protons, the relative losses are even lower.) Such negligible radiative losses of energy are actually a large advantage of linear accelerators – such as the famous two-mile-long SLAC,11 which can accelerate electrons or positrons to energies up to 50 GeV, i.e. to 105. If obtaining radiation from the accelerated particles is the goal, it may be readily achieved by bending their trajectories using additional magnetic fields – see the next section.
10.3. Synchrotron radiation
Now let us consider a charged particle being accelerated in the direction perpendicular to its velocity u (for example by the magnetic component of the Lorentz force), so that its speed u, and hence the magnitude p of its momentum, do not change. In this case, the second term in the square brackets of Eq. (39) vanishes, and it yields
2
2
2
2
Z q
dp
Z q
dp
0
0
2
P
.
(10.43)
2
2
2
2
6 m c
d
6
m c dt ret
Comparing this expression with Eq. (40), we see that for the same acceleration magnitude, the electromagnetic radiation is a factor of 2 larger. For modern accelerators, with ~ 104-105, such a factor creates an enormous difference. For example, if a particle is on a cyclotron orbit in a constant magnetic field (as was analyzed in Sec. 9.6), both u and p = mu obey Eq. (9.150), so that 11 See, e.g., https://www6.slac.stanford.edu/.
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dp
u
2
mc 2
p
p
,
(10.44)
dt
c
R
R
ret
(where R is the orbit’s radius), so that for the power of this synchrotron radiation, Eq. (43) yields Synchrotron
2
2
4
2
Z q
c
1 2 q B
radiation:
0
4
4
2
2
P
.
(10.45)
total power
2
2
6
R
4 3 m c
0
Note that for ultrarelativistic particles ( 1), the power grows as 2, i.e. as the square of the particle’s energy E . For example, for typical parameters of the first electron cyclotrons (such as the General Electric’s machine in which the synchrotron radiation was first noticed in 1947), R ~ 1 m, E ~
0.3 GeV ( ~ 600), Eq. (45) gives a very modest electron energy loss per one revolution: P T P(2 R/ u)
2P R/ c ~ 1 keV. However, already by the mid-1970s, electron accelerators, with R ~ 100 m, could give each particle energy E ~10 GeV, and the energy loss per revolution grew to ~ 10 MeV, becoming the major energy loss mechanism. For proton accelerators, such energy loss is much less of a problem, because the of an ultra-relativistic particle (at fixed E) is proportional to 1/ m, so that the estimates, at the same R, should be scaled back by ( m p/ m e)4 ~ 1013. Nevertheless, in the giant modern accelerators such as the LHC (with R 4.3 km and E up to 7 TeV), the synchrotron radiation loss per revolution is rather noticeable (P T ~ 6 keV), leading not as much to particle deceleration as to a substantial photoelectron emission from the beam tube’s walls, creating harmful defocusing effects.
However, what is bad for particle accelerators and storage rings is good for the so-called synchrotron light sources – the electron accelerators designed specially for the generation of intensive synchrotron radiation – with the spectrum extending well beyond the visible light range. Let us analyze the angular and spectral distributions of such radiation. To calculate the angular distribution, let us select the coordinate axes as shown in Fig. 5, with the origin at the current location of the orbiting particle, the z-axis directed along its instant velocity (i.e. the vector ), and the x-axis, toward the orbit’s center.
y
n
P
x
z
β
Fig. 10.5. The synchrotron
β
radiation problem’s geometry.
0
In the general case, when the unit vector n toward the radiation’s observer is not within any of the coordinate planes, it has to be described by two angles – the polar angle , and the azimuthal angle
between the x-axis and the projection 0P of the vector n onto the [ x, y]-plane. Since the length of the segment 0P is sin, the Cartesian components of the relevant vectors are as follows:
n sin cos, sin sin, cos,
β ,
0 ,
0 ,
and β , ,
0
0 .
(10.46)
Plugging these expressions into the general Eq. (30), we get
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Essential Graduate Physics
EM: Classical Electrodynamics
d P
2
2
Z q
2
0
6
β f ,
,
where
2
Synchrotron
d
Ω
(10.47) radiation:
angular
f
2
2
,
1
sin
cos
1
,
distribution
8 6
1
( cos )3
2
1
( cos )2
According to this result, just as at the linear acceleration, in the ultra-relativistic limit, most radiation goes into a narrow cone (of a width ~ -1 << 1) around the vector , i.e. around the instant direction of the particle’s propagation. For such small angles, and >> 1,
1
2
2
2
4 cos
f ,
1
.
(10.48)
2
2 3
2
2 2
1
( )
1
( )
The left panel of Fig. 6 shows a color-coded contour map of this angular distribution f(, ), as observed on a distant plane normal to the particle’s instant velocity (in Fig. 5, parallel to the [ x, y]-plane), while its right panel shows the factor f as a function of in two perpendicular directions: within the particle’s rotation plane (in the direction parallel to the x-axis, i.e. at = 0) and perpendicular to this plane (along the y-axis, i.e. at = /2). The result shows, first of all, that, in contrast to the case of linear acceleration, the narrow radiation cone is now not hollow: the intensity maximum is reached at = 0, i.e. exactly in the direction of the particle’s motion direction. Second, the radiation cone is not axially symmetric: within the particle rotation plane, the intensity drops faster (and even has nodes at = 1/).
0.8
sin
0.6
f ( ,)
0.4
off-plane
( = /2)
0.2 in-plane
( = 0)
0
0
0.5
1
1.5
2
cos
Fig. 10.6. The angular distribution of the synchrotron radiation at >> 1.
The angular distribution (47) of the synchrotron radiation was calculated for the (inertial) reference frame whose origin coincides with the particle’s position at this particular instant, i.e. its radiation pattern is time-independent in the frame moving with the particle. This pattern enables a semi-quantitative description of the radiation by an ultra-relativistic particle from the point of view of a stationary observer: if the observation point is on (or very close to) the rotation plane,12 it is being 12 It is easy (and hence is left for the reader’s exercise) to show that if the observation point is much off-plane (say, is located on the particle orbit’s axis), the radiation is virtually monochromatic, with frequency c. (As we know from Sec. 8.2, in the non-relativistic limit u << c, this is true for any observation point.) Chapter 10
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EM: Classical Electrodynamics
“struck” by the narrow radiation cone once each rotation period T 2 R/ c, each “strike” giving a field pulse of a short duration t ret << 1/c – see Fig. 7.13
(a)
(b)
0
T 2 /c
β( t )
2
n
t
'
r ( t ) r ' ( t )
1
2
β( t )
3
1
t
~ T /
Fig. 10.7. (a) The synchrotron radiation cones (at >> 1) for two close values of t ret, and (b) the in-plane component of the electric field observed in the rotation plane, as a function of time t – schematically.
The evaluation of the time duration t of each pulse requires some care: its estimate t ret ~ 1/c is correct for the duration of the retarded time interval during which its cone is aimed at the observer.
However, due to the time compression effect discussed in detail in Sec. 1 and described by Eq. (16), the pulse duration as seen by the observer is a factor of 1/(1 – ) shorter, so that
1
1
t 1
( ) t ~
~
~ 3
T ,
for 1.
(10.49)
ret
3
c
c
From the Fourier theorem, we can expect the frequency spectrum of such radiation to consist of numerous ( N ~ 3 >> 1) harmonics of the particle rotation frequency c, with comparable amplitudes.
However, if the orbital frequency fluctuates even slightly (c/c > 1/ N ~ 1/3), as it happens in most practical systems, the radiation pulses are not coherent, so that the average radiation power spectrum may be calculated as that of one pulse, multiplied by the number of pulses per second. In this case, the spectrum is continuous, extending from low frequencies all the way to approximately
~ 1/ t
~ 3
.
(10.50)
max
c
In order to verify and quantify this result, let us calculate the spectrum of radiation due to a single pulse. For that, we should first make the general notion of the radiation spectrum quantitative. Let us represent an arbitrary electric field (say that of the synchrotron radiation we are studying now) observed at a fixed point r, as a function of the observation time t, as a Fourier integral:14
E t
E e i t dt
.
(10.51)
13 The fact that the in-plane component of each electric field’s pulse E( t) is antisymmetric with respect to its central point, and hence vanishes at that point (as Fig. 7b shows), readily follows from Eq. (19).
14 In contrast to the single-frequency case (i.e. a monochromatic wave), we may avoid taking the real part of the complex function (E e-i t) by requiring that in Eq. (51), E - = E*. However, it is important to remember the factor ½ required for the transition to a monochromatic wave of frequency 0 and with real amplitude E0: E = E0
[( – 0) + ( + 0)]/2.
Chapter 10
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EM: Classical Electrodynamics
This expression may be plugged into the formula for the total energy of the radiation pulse (i.e. of the loss of particle’s energy E) per unit solid angle:15
2
d E
2
R
S t() R dt
2
E( )
.
(10.52)
n
t dt
d Ω
Z
0
This substitution, followed by a natural change of the integration order, yields
2
