.
(10.66)
q
q
j
j
But the last derivative is nothing else than the generalized momentum pj, so that S
p .
(10.67)
j
q
j
(As a reminder, both parts of this relation refer to the final moment t of the trajectory.) As a result, the full derivative of the action S[ t, qj( t)] over time takes the form dS
S
S
S
q
p q .
(10.68)
j
dt
t
j q
j
j
t
j
j
Now, by the definition of S, the full derivative dS/ dt is nothing more than the Lagrangian function L, so that Eq. (67) yields
S
L p q .
(10.69)
j
j
t
j
However, according to the definition (2) of the Hamiltonian function H, the right-hand side of Eq. (69) is just (– H), and we get an extremely simply-looking Hamilton-Jacobi equation Hamilton-S
Jacobi
H.
(10.70)
equation
t
This simplicity is, however, rather deceiving, because to use this equation for the calculation of the function S( t, qj) for any particular problem, the Hamiltonian function has to be first expressed as a function of time t, generalized coordinates qj, and the generalized momenta pj (which may be, according Chapter 10
Page 13 of 16
CM: Classical Mechanics
to Eq. (67), represented just as the derivatives S/ qj). Let us see how this procedure works for the simplest case of a 1D system with the Hamiltonian function given by Eq. (10). In this case, the only generalized momentum is p = S/ q, so that
2
2
p
1 S
H
U ( q, t)
U ( q, t),
(10.71)
2
ef
m
2
ef
m q
ef
ef
and Eq. (70) is reduced to the following partial differential equation,
2
S
1 S
U ( q, t) 0 .
(10.72)
t
2
ef
m
q
ef
Its solution may be readily found in the easiest case of time-independent potential energy U ef =
U ef ( q). In this case, Eq. (72) is evidently satisfied by the following variable-separated solution: S t
( , q) S ( q) const t .
(10.73)
0
Plugging this solution into Eq. (72), we see that since the sum of the two last terms on the left-hand side of that equation represents the full mechanical energy E, the constant in Eq. (73) is nothing but (– E).
Thus for the function S 0( q) we get an ordinary differential equation
2
1 dS
0
E
U ( q) .
0
(10.74)
2
ef
m
dq
ef
Integrating it, we get
S
m E U
q
dq
(10.75)
0
2 ef ( )
ef
1/ 2
const,
so that, finally, the action is equal to
S 2 m E U q
dq Et
(10.76)
ef
( )
ef
1/ 2
const.
For the case of 1D motion of a single 1D particle, i.e. for q = x, m ef = m, U ef( q) = U( x), this solution is just the 1D case of the more general Eqs. (59)-(60), which were obtained above in a much more simple way. (In particular, S 0 is just the abbreviated action.)
This particular example illustrates that the Hamilton-Jacobi equation is not the most efficient way for the solution of most practical problems of classical mechanics. However, it may be rather useful for studies of certain mathematical aspects of dynamics.22 Moreover, in the early 1950s this approach was extended to a completely different field – the optimal control theory, in which the role of the action S is played by the so-called cost function – a certain functional of a system (understood in a very general sense of this term), that should be minimized by an optimal choice of a control signal – a function of time that affects the system’s evolution in time. From the point of view of this theory, Eq. (70) is a particular case of a more general Hamilton-Jacobi-Bellman equation.23
22 See, e.g., Chapters 6-9 in I. C. Percival and D. Richards, Introduction to Dynamics, Cambridge U. Press, 1983.
23 See, e.g., T. Bertsekas, Dynamic Programming and Optimal Control, vols. 1 and 2, Aetna Scientific, 2005 and 2007. The reader should not be intimidated by the very unnatural term “dynamic programming”, which was invented by the founding father of this field, Richard Bellman, to lure government bureaucrats into funding his research, deemed too theoretical at that time. (Presently, it has a broad range of important applications.) Chapter 10
Page 14 of 16
Essential Graduate Physics
CM: Classical Mechanics
10.5. Exercise problems
In each of Problems 10.1-10.3, for the given system:
(i) derive the Hamilton equations of motion, and
(ii) check whether these equations are equivalent to those derived from the Lagrangian
formalism.
10.1. Our “testbed” system: a bead on a ring, being rotated with a fixed angular
velocity about its vertical diameter – see Fig. 2.1, reproduced on the right.
R
mg
x 0( t)
10.2. The system considered in Problem 2.3: a pendulum hanging from a horizontal
support whose motion law x 0( t) is fixed – see the figure on the right. (No vertical-plane l
constraint.)
g
m
10.3. The system considered in Problem 2.8: a block of mass m that
m
can slide, without friction, along the inclined surface of a heavy wedge of
m'
mass m’. The wedge is free to move, also without friction, along a horizontal
g
surface – see the figure on the right. (Both motions are within the vertical plane containing the steepest slope line.)
10.4. Derive and solve the equations of motion of a particle with the following Hamiltonian function:
1
H
p ar2 ,
2 m
where a is a constant scalar.
10.5. Let L be the Lagrangian function, and H the Hamiltonian function, of the same system.
What three of the following four statements,
dL
L
dH
H
i
(
)
,
0
(ii)
,
0
(iii)
,
0
(iv)
,
0
dt
t
dt
t
are equivalent? Give an example when those three equalities hold, but the fourth one does not.
10.6. Calculate the Poisson brackets of a Cartesian component of the angular momentum L of a particle moving in a central force field and its Hamiltonian function H, and discuss the most evident implication of the result.
Chapter 10
Page 15 of 16
Essential Graduate Physics
CM: Classical Mechanics
10.7. After small oscillations had been initiated in the point pendulum shown in
Fig. on the right, the supporting string is being pulled up slowly, so that the pendulum’s length l is being reduced. Neglecting dissipation,
(i) prove by a direct calculation that the oscillation energy is indeed changing l( t)
proportionately to the oscillation frequency, as it follows from the constancy of the corresponding adiabatic invariant (40); and
m
g
(ii) find the l-dependence of the amplitudes of the angular and linear deviations from the equilibrium.
10.8. The mass m of a small body that performs 1D oscillations in the potential well U( x) = ax 2 n, with n > 0, is being changed slowly. Calculate the oscillation energy E as a function of m.
10.9. A stiff ball is bouncing vertically from the floor of an elevator whose upward acceleration changes very slowly. Neglecting the energy dissipation, calculate how much the bounce height h changes during the acceleration’s increase from 0 to g. Is your result valid for an equal but abrupt increase of the elevator’s acceleration?
10.10.* A 1D particle of a constant mass m moves in a time-dependent potential U( q, t) =
m2( t) q 2/2, where ( t) is a slow function of time, with 2
. Develop the approximate method for
the solution of the corresponding equation of motion, similar to the WKB approximation used in quantum mechanics.24 Use the approximation to confirm the conservation of the action variable (40) for this system.
Hint: You may like to look for the solution to the equation of motion in the form q t
exp t i t,
where and are some real functions of time, and then make proper approximations in the resulting equations for these functions.
24 See, e.g., QM Sec. 2.4.
Chapter 10
Page 16 of 16
Konstantin K. Likharev
Essential Graduate Physics
Lecture Notes and Problems
Beta version
Open online access at
http://commons.library.stonybrook.edu/egp/
and
https://sites.google.com/site/likharevegp/
Part EM:
Classical Electrodynamics
Last corrections: September 30, 2022
A version of this material was published in 2018 under the title
Classical Electrodynamics: Lecture notes
IOPP, Essential Advanced Physics – Volume 3, ISBN 978-0-7503-1405-3,
with the model solutions of the exercise problems published under the title
Classical Electrodynamics: Problems with solutions
IOPP, Essential Advanced Physics – Volume 4, ISBN 978-0-7503-1408-4
However, by now this online version of the lecture notes
and the problem solutions available from the author, have been better corrected
About the author:
https://you.stonybrook.edu/likharev/
© K. Likharev
EM: Classical Electrodynamics
Table of Contents
Chapter 1. Electric Charge Interaction (20 pp.)
1.1. The Coulomb law
1.2. The Gauss law
1.3. Scalar potential and electric field energy
1.4. Exercise problems (20)
Chapter 2. Charges and Conductors (68 pp.)
2.1. Polarization and screening
2.2. Capacitance
2.3. The simplest boundary problems
2.4. Using other orthogonal coordinates
2.5. Variable separation – Cartesian coordinates
2.6. Variable separation – polar coordinates
2.7. Variable separation – cylindrical coordinates
2.8. Variable separation – spherical coordinates
2.9. Charge images
2.10. Green’s functions
2.11. Numerical approach
2.12. Exercise problems (47)
Chapter 3. Dipoles and Dielectrics (28 pp.)
3.1. Electric dipole
3.2. Dipole media
3.3. Polarization of dielectrics
3.4. Electrostatics of linear dielectrics
3.5. Electric field energy in a dielectric
3.6. Exercise problems (30)
Chapter 4. DC Currents (16 pp.)
4.1. Continuity equation and the Kirchhoff laws
4.2. The Ohm law
4.3. Boundary problems
4.4. Energy dissipation
4.5. Exercise problems (15)
Chapter 5. Magnetism (42 pp.)
5.1. Magnetic interaction of currents
5.2. Vector potential and the Ampère law
5.3. Magnetic flux, energy, and inductance
5.4. Magnetic dipole moment, and magnetic dipole media
5.5. Magnetic materials
5.6. Systems with magnetic materials
5.7. Exercise problems (29)
Table of Contents
Page 2 of 4
EM: Classical Electrodynamics
Chapter 6. Electromagnetism (38 pp.)
6.1. Electromagnetic induction
6.2. Magnetic energy revisited
6.3. Quasistatic approximation, and the skin effect
6.4. Electrodynamics of superconductivity, and the gauge invariance
6.5. Electrodynamics of macroscopic quantum phenomena
6.6. Inductors, transformers, and ac Kirchhoff laws
6.7. Displacement currents
6.8. Finally, the full Maxwell equation system
6.9. Exercise problems (30)
Chapter 7. Electromagnetic Wave Propagation (68 pp.)
7.1. Plane waves
7.2. Attenuation and dispersion
7.3. Reflection
7.4. Refraction
7.5. Transmission lines: TEM waves
7.6. Waveguides: H and E waves
7.7. Dielectric waveguides, optical fibers, and paraxial beams
7.8. Resonance cavities
7.9. Energy loss effects
7.10. Exercise problems (39)
Chapter 8. Radiation, Scattering, Interference, and Diffraction (38 pp.)
8.1. Retarded potentials
8.2. Electric dipole radiation
8.3. Wave scattering
8.4. Interference and diffraction
8.5. The Huygens principle
8.6. Fresnel and Fraunhofer diffraction patterns
8.7. Geometrical optics placeholder
8.8. Fraunhofer diffraction from more complex scatterers
8.9. Magnetic dipole and electric quadrupole radiation
8.10. Exercise problems (28)
Chapter 9. Special Relativity (56 pp.)
9.1. Einstein postulates and the Lorentz transform
9.2. Relativistic kinematic effects
9.3. 4-vectors, momentum, mass, and energy
9.4. More on 4-vectors and 4-tensors
9.5. The Maxwell equations in the 4-form
9.6. Relativistic particles in electric and magnetic fields
9.7. Analytical mechanics of charged particles
9.8. Analytical mechanics of electromagnetic field
9.9. Exercise problems (42)
Table of Contents
Page 3 of 4
EM: Classical Electrodynamics
Chapter 10. Radiation by Relativistic Charges (40 pp.)
10.1. Liénard-Wiechert potentials
10.2. Radiation power
10.3. Synchrotron radiation
10.4. Bremsstrahlung
10.5. Coulomb losses
10.6. Density effects and the Cherenkov radiation
10.7. Radiation’s back-action
10.8. Exercise problems (15)
* * *
Additional file (available from the author upon request):
Exercise and Test Problems with Model Solutions (295 + 54 = 349 problems; 464 pp.) Table of Contents
Page 4 of 4
Essential Graduate Physics
EM: Classical Electrodynamics
Chapter 1. Electric Charge Interaction
This chapter reviews the basics of electrostatics – the description of interactions between stationary (or relatively slowly moving) electric charges. Much of this material should be known to the reader from their undergraduate studies; 1 because of that, the explanations are very brief.
1.1. The Coulomb law
A quantitative discussion of classical electrodynamics, starting from the electrostatics, requires common agreement on the meaning of the following notions:2
- electric charges qk, as revealed, most explicitly, by observation of electrostatic interaction between the charged particles;
- point charges – the charged particles so small that their position in space, for the given problem, may be completely described (in the given reference frame) by their radius-vectors r k; and
- electric charge conservation – the fact that the algebraic sum of all charges qk inside any closed volume is conserved unless the charged particles cross the volume’s border.
I will assume that these notions are well known to the reader. Using them, the Coulomb law 3 for the interaction of two stationary point charges may be formulated as follows:
r r
q q
k
k'
k
k'
F q q
n ,
kk'
k
k'
3
2
kk'
r r
R
Coulomb
k
k '
kk'
(1.1)
law
R
with R
r r ,
kk '
n
, R R
,
kk'
k
k'
kk'
kk '
kk '
Rkk'
where F kk’ denotes the electrostatic ( Coulomb) force exerted on the charge number k by the charge number k’, separated from it by distance Rkk’ – see Fig. 1.
qk
F
'
k'k
R
r r
r
kk'
k
k '
n kk'
q
k '
k
F kk'
r k
0
Fig. 1.1. Coulomb force directions (for the case qkqk’ > 0).
1 For remedial reading, I can recommend, for example, D. Griffiths, Introduction to Electrodynamics, 4th ed., Pearson, 2015.
2 On top of the more general notions of the classical Newtonian space, point particles and forces, as used in classical mechanics – see, e.g., CM Sec. 1.1.
3 Formulated in 1785 by Charles-Augustin de Coulomb, on basis of his earlier experiments, in turn rooting in prior studies of electrostatic phenomena, with notable contributions by William Gilbert, Otto von Guericke, Charles François de Cisternay Du Fay, Benjamin Franklin, and Henry Cavendish.
© K. Likharev
EM: Classical Electrodynamics
I am confident that this law is very familiar to the reader, but a few comments may still be due: (i) Flipping the indices k and k’, we see that Eq. (1) complies with the 3rd Newton law: the reciprocal force is equal in magnitude but opposite in direction: F k’k = –F kk’.
(ii) Since the vector R kk’ r k – r k’, by its definition, is directed from point r k’ toward point r k (Fig. 1), Eq. (1) correctly describes the experimental fact that charges of the same sign (i.e. with qkqk’ > 0) repulse, while those with opposite signs ( qkqk’ < 0) attract each other.
(iii) In some textbooks, the Coulomb law (1) is given with the qualifier “in free space” or “in vacuum”. However, actually, Eq. (1) remains valid even in the presence of any other charges – for example, of internal charges in a quasi-continuous medium that may surround the two charges (number k and k’) under consideration. The confusion stems from the fact, to be discussed in detail in Chapter 3
below, that in some cases it is convenient to formally represent the effect of the other charges as an effective (rather than actual!) modification of the Coulomb law.
(iv) The constant in Eq. (1) depends on the system of units we use. In the Gaussian units, is set to 1, for the price of introducing a special unit of charge (the statcoulomb) that would make experimental data compatible with Eq. (1) if the force F kk’ is measured in the Gaussian units ( dynes). On the other hand, in the International System (“SI”) of units, the charge’s unit is one coulomb (abbreviated C), and is different from 1:
1
,
(1.2)
in
SI
4
SI units
0
where 0 8.85410-12 is called the electric constant.4
Unfortunately, the continuing struggle between zealous proponents of these two systems of units bears all not-so-nice features of a religious war, with a similarly slim chance for any side to win it in any foreseeable future. In my humble view, each of these systems has its advantages and handicaps (to be noted on several occasions below), and every educated physicist should have no problem with using any of them. Following insisting recommendations of international scientific unions, I am using the SI units throughout my series. However, for the readers’ convenience, in this course (where the difference between the Gaussian and SI systems is especially significant) I will write the most important formulas with the constant (2) clearly displayed – for example, Eq. (1) as
1
r r
k
k'
F
q q
,
(1.3)
kk'
k
k'
3
40
r r
k
k '
so that the transfer to the Gaussian units may be performed just by the formal replacement 40 1. (In the rare cases when the transfer is not obvious, I will duplicate formulas in the Gaussian units.) Besides Eq. (3), another key experimental law of electrostatics is the linear superposition principle: the electrostatic forces exerted on some point charge (say, qk) by other charges add up as vectors, forming the net force
4 Since 2018, one coulomb is defined, in the “legal” metrology, as a certain, exactly fixed number of the fundamental electric charges e, and the “legal” SI value of 0 is not more exactly equal to 107/4 c 2 (where c is the speed of light) as it was before that, but remains extremely close to that fraction, with the relative difference of the order of 10-10 – see appendix CA: Selected Physical Constants. In this series, this minute difference is ignored.
Chapter 1
Page 2 of 20
EM: Classical Electrodynamics
F
F ,
(1.4)
k
kk'
k ' k
where the summation is extended over all charges but qk, and the partial force F kk’ is described by Eq.
(3). The fact that the sum is restricted to k’ k means that a point charge, in statics, does not interact with itself. This fact may look obvious from Eq. (3), whose right-hand side diverges at r k r k’, but becomes less evident (though still true) in quantum mechanics – where the charge of even an elementary particle is effectively spread around some volume, together with the particle’s wavefunction.5
Now we may combine Eqs. (3) and (4) to get the following expression for the net force F acting on a probe charge q located at point r:
1
r r
Fr q
k'
q
.
(1.5)
k'
3
40 r r
r r
k '
k'
This equality implies that it makes sense to introduce the notion of the electric field (as an entity independent of q), whose distribution in space is characterized by the following vector: Electric
F r
field:
Er
,
(1.6)
definition
q
formally called the electric field strength – but much more frequently, just the “electric field”. In these terms, Eq. (5) becomes
1
r r
Electric field of
E r
( )
point charges
k'
q
.
(1.7)
k '
3
40 r r
r r
k '
k'
Just convenient is electrostatics, the notion of the field becomes virtually unavoidable for the description of time-dependent phenomena (such as electromagnetic waves, see Chapter 7 and on), where the electromagnetic field shows up as a specific form of matter, different from the usual “material” particles
– even though quantum electrodynamics (to be reviewed in QM Chapter 9) offers their joint description.
Many real-world problems involve multiple point charges located so closely that it is possible to approximate them with a continuous charge distribution. Indeed, let us consider a group of many ( dN >> 1) close charges, located at points r k’, all within an elementary volume d 3 r’. For relatively distant field observation points, with r – r k’ >> dr’, the geometrical factor in the corresponding terms of Eq. (7) is essentially the same. As a result, these charges may be treated as a single elementary charge dQ(r ’).
Since at dN >> 1, this elementary charge is proportional to the elementary volume d 3 r’, we can define the local 3D charge density (r ’) by the following relation:
( '
r ) d 3 r' dQ(r') q ,
(1.8)
k '
r d 3 r'
k '
and rewrite Eq. (7) as an integral (over the volume containing all essential charges):
Electric field of
1
r '
r
continuous
E(r)
( '
r )
3
d r' .
(1.9)
charge
4
3
0
r '
r
5 Note that some widely used approximations, e.g., the density functional theory (DFT) of multiparticle systems, essentially violate this law, thus limiting their accuracy and applicability – see, e.g., QM Sec. 8.4.
Chapter 1
Page 3 of 20
Essential Graduate Physics
EM: Classical Electrodynamics
Note that for a continuous, smooth charge density (r ’), the integral in Eq. (9) does not diverge at R
r – r ’ 0, because in this limit, the fraction under the integral increases as R-2, i.e. slower than the decrease of the elementary volume d 3 r’, proportional to R 3.
Let me emphasize the dual use of Eq. (9). In the case when (r) is a continuous function representing the average charge defined by Eq. (8), Eq. (9) is not valid at distances r – r k’ of the order of the distance between the adjacent point charges, i.e. does not describe rapid variations of the electric field at these distances. Such approximate, smoothly changing field E(r), is called macroscopic; we will repeatedly return to this notion in the following chapters. On the other hand, Eq. (9) may be also used for the description of the exact (frequently called microscopic) field of discrete point charges, by employing the notion of Dirac’s delta function, which is the mathematical description of a very sharp function equal to zero everywhere but one point, and still having a finite integral (equal to 1).6 Indeed, in this formalism, a set of point charges qk’ located in points r k’ may be represented by the pseudo-continuous density
( '
r ) q ( ' r r .)
(1.10)
k '
k '
k '
Plugging this expression into Eq. (9), we return to its exact, discrete version (7). In this sense, Eq. (9) is exact, and we may use it as the general expression for the electric field.
1.2. The Gauss law
Due to the extension of Eq. (9) to point (“discrete”) charges, it may seem that we do not need anything besides it for solving any problem of electrostatics. In practice, however, this is not quite true –
first of all, because the direct use of Eq. (9) frequently leads to complex calculations. Indeed, let us try to solve a problem that is conceptually very simple: find the electric field induced by a spherically-symmetric charge distribution with density ( r’) – see Fig. 2.
( r' d 3
) r'
r'
R
dE E
d cos
'
h
z
0
r
Fig. 1.2. One of the simplest problems of
a
r a
E
d
electrostatics: the electric field produced by
a spherically-symmetric charge distribution.
We may immediately use the problem’s symmetry to argue that the electric field should be also spherically-symmetric, with only one component in the spherical coordinates: E(r)= E( r)n r, where n r
r/ r is the unit vector in the direction of the field observation point r. Taking this direction for the polar 6 See, e.g., MA Sec. 14. The 2D ( areal) charge density and the 1D ( linear) density may be defined absolutely similarly to the 3D ( volumic) density : dQ = d 2 r, dQ = dr. Note that the approximations in that either 0
or 0 imply that is formally infinite at the charge location; for example, the model in that a plane z = 0 is charged with areal density 0, means that = ( z), where ( z) is Dirac’s delta function.
Chapter 1
Page 4 of 20
Essential Graduate Physics
EM: Classical Electrodynamics
axis of a spherical coordinate system, we can use the evident axial symmetry of the system to reduce Eq.
(9) to
1
r'
2
( )
E
2 sin 'd ' r' dr'
cos ,
(1.11)
4
2
R
0
0
0
where , ’, and R are the geometrical parameters marked in Fig. 2. Since and R may be readily expressed via r’ and ’, using the auxiliary parameters a and h,
r a
cos
,
2
2
R h ( r r' cos )2 ,
where a r' cos '
, h r' sin '
,
(1.12)
R
Eq. (11) may be eventually reduced to an explicit integral over r’ and ’, and worked out analytically, but that would require some effort.
For other problems, the integral (9) may be much more complicated, defying an analytical solution. One could argue that with the present-day abundance of computers and numerical algorithm libraries, one can always resort to numerical integration. This argument may be enhanced by the fact that numerical integration is based on the replacement of the required integral by a discrete sum, and the summation is much more robust to the (unavoidable) rounding errors than the finite-difference schemes typical for the numerical solution of differential equations. These arguments, however, are only partly justified, since in many cases the numerical approach runs into a problem sometimes called the curse of dimensionality – the exponential dependence of the number of needed calculations on the number of independent parameters of the problem.7 Thus, despite the proliferation of numerical methods in physics, analytical results have an everlasting value, and we should try to get them whenever we can.
For our current problem of finding the electric field generated by a fixed set of electric charges, large help may come from the so-called Gauss law.
To derive it, let us consider a single point charge q inside a smooth closed surface S (Fig. 3), and calculate the product End 2 r, where d 2 r is an elementary area of the surface (which may be well approximated with a plane fragment of that area), and En En is the component of the electric field at that point, normal to the plane.
(a)
(b)
En
E
Eout
d 2 r
2
S
r
d r'
S
d 0
2
d r cos
d
q
Ein
d 0
q
Fig. 1.3. Deriving the Gauss law: a point charge q (a) inside the volume V, and (b) outside of that volume.
7 For a more detailed discussion of this problem, see, e.g., CM Sec. 5.8.
Chapter 1
Page 5 of 20
EM: Classical Electrodynamics
This component may be calculated as E cos, where is the angle between the vector E and the unit vector n normal to the surface. Now let us notice that the product cos d 2 r is nothing more than the area d 2 r’ of the projection of d 2 r onto the plane normal to the vector r connecting the charge q with the considered point of the surface (Fig. 3), because the angle between the elementary areas d 2 r’ and d 2 r is also equal to . Using the Coulomb law for E, we get
2
2
1
q
E d r E cos
2
d r
d r' .
(1.13)
n
4
2
r
0
But the ratio d 2 r’/ r 2 is nothing more than the elementary solid angle d under which the areas d 2 r’ and d 2 r are seen from the charge point, so that End 2 r may be represented just as a product of d by a constant ( q/40). Summing these products over the whole surface, we get
q
q
2
E d r
d Ω
,
(1.14)
n
4
S
0 S
0
since the full solid angle equals 4. (The integral on the left-hand side of this relation is called the flux of electric field through the surface S.)
Relation (14) expresses the Gauss law for one point charge. However, it is only valid if the charge is located inside the volume V limited by the surface S. To find the flux created by a charge located outside of this volume, we still can use Eq. (13), but have to be careful with the signs of the elementary contributions EndA. Let us use the common convention to direct the unit vector n out of the closed volume we are considering (the so-called outer normal), so that the elementary product End 2 r =
(En) d 2 r and hence d = End 2 r’/ r 2 is positive if the vector E is pointing out of the volume (like in the example shown in Fig. 3a and at the upper-right area in Fig. 3b), and negative in the opposite case (for example, at the lower-left area in Fig. 3b). As the latter panel shows, if the charge is located outside of the volume, for each positive contribution d there is always an equal and opposite contribution to the integral. As a result, at the integration over the solid angle, the positive and negative contributions cancel exactly, so that
2
E d r .
0
(1.15)
n
S
The real power of the Gauss law is revealed by its generalization to the case of several, especially many charges. Since the calculation of flux is a linear operation, the linear superposition principle (4) means that the flux created by several charges is equal to the (algebraic) sum of individual fluxes from each charge, for which either Eq. (14) or Eq. (15) are valid, depending on whether the charge is in or out of the volume. As the result, for the total flux we get:
1
1
2
Q
E d r V
q
( '
r ) d 3 r' ,
(1.16) Gauss
n
j
law
S
0
0 r V
j
0 V
where QV is the net charge inside volume V. This is the full version of the Gauss law.8
In order to appreciate the problem-solving power of the law, let us revisit the problem shown in Fig. 2, i.e. the field of a spherical charge distribution. Due to its symmetry, which had already been 8 The law is named after the famed Carl Gauss (1777-1855), even though it was first formulated earlier (in 1773) by Joseph-Louis Lagrange who was also the father-founder of analytical mechanics – see, e.g., CM Chapter 2.
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discussed above, if we apply Eq. (16) to a sphere of a certain radius r, the electric field has to be normal to the sphere at each its point (i.e., En = E), and its magnitude has to be the same at all points: En = E( r).
As a result, the flux calculation is elementary:
2
E d r 4
2
r E( r)
.
(1.17)
n
Now applying the Gauss law (16), we get:
r
2
1
3
4
4 r E( r)
( r' )
2
d r'
r' ( r' ) dr' ,
(1.18)
0 r' r
0 0
so that, finally,
1 r
Q
2
1
E( r)
r' ( r' ) dr'
r
,
(1.19)
2
r
4
2
r
0 0
0
where Qr is the full charge inside the sphere of radius r:
r
Q
r' d r'
r' r' dr'
(1.20)
r
3
4
( ) 2
.
r' r
0
In particular, this formula shows that the field outside of a sphere of a finite radius R is exactly the same as if all its charge Q = Q( R) is concentrated in the sphere’s center. (Note that this important result is only valid for a spherically-symmetric charge distribution.) For the field inside the sphere, finding the electric field still requires the explicit integration (20), but this 1D integral is much simpler than the 2D integral (11), and in some important cases may be readily worked out analytically. For example, if the charge Q is uniformly distributed inside a sphere of radius R, Q
Q
( r' )
,
(1.21)
V
(4 / )
3 3
R
then the integration is elementary:
r
r
1 Qr
2
E( r)
r' dr'
.
(1.22)
2
3
r
3
4 R
0 0
0
0
We see that in this case, the field is growing linearly from the center to the sphere’s surface, and only at r > R starts to decrease in agreement with Eq. (19) with constant Q( r) = Q. Note also that the electric field is continuous for all r (including r = R) – as for all systems with finite volumic density, In order to underline the importance of the last condition, let us consider one more elementary but very important example of Gauss law’s application. Let a thin plane sheet (Fig. 4) be charged uniformly, with a finite areal density = const. In this case, it is fruitful to use the Gauss volume in the form of a planar “pillbox” of thickness 2 z (where z is the Cartesian coordinate perpendicular to the plane) and certain area A – see the dashed lines in Fig. 4. Due to the symmetry of the problem, it is evident that the electric field should be: (i) directed along the z-axis, (ii) constant on each of the upper and bottom sides of the pillbox, (iii) equal and opposite on these sides, and (iv) parallel to the side surfaces of the box. As a result, the full electric field flux through the pillbox’s surface is just 2 AE( z), so the Gauss law (16) yields 2 AE( z) = QA/0 A/0, and we get a very simple but important formula
E( z)
.
const
(1.23)
2 0
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E
z
z
Fig. 1.4. The electric field of
A
a charged plane.
E
Notice that, somewhat counter-intuitively, the field magnitude does not depend on the distance from the charged plane. From the point of view of the Coulomb law (5), this result may be explained as follows: the farther the observation point from the plane, the weaker the effect of each elementary charge, dQ = d 2 r, but the more such elementary charges give contributions to the z-component of vector E, because they are “seen” from the observation point at relatively small angles to the z-axis.
Note also that though the magnitude E E of this electric field is constant, its component En normal to the plane (for our coordinate choice, Ez) changes its sign at the plane, experiencing a discontinuity (jump) equal to
E E
.
(1.24)
z
z z
0
Ez z 0
0
This jump disappears if the surface is not charged. Returning for a split second to our charged sphere problem (Fig. 2), solving it we have considered the volumic charge density to be finite everywhere, including the sphere’s surface, so that on it = 0, and the electric field should be continuous – as it is.
Admittedly, the integral form (16) of the Gauss law is immediately useful only for highly symmetrical geometries, such as in the two problems discussed above. However, it may be recast into an alternative, differential form whose field of useful applications is much wider. This form may be obtained from Eq. (16) using the divergence theorem of the vector algebra, which is valid for any space-differentiable vector, in particular E, and for the volume V limited by any closed surface S:9
E d 2 r ( E) d 3 r ,
(1.25)
n
S
V
where is the del (or “nabla”) operator of spatial differentiation.10 Combining Eq. (25) with the Gauss law (16), we get
E
d 3 r
.
0
(1.26)
V
0
For a given spatial distribution of electric charge (and hence of its electric field), this equation should be valid for any choice of the volume V. This can hold only if the function under the integral vanishes at each point, i.e. if11
9 See, e.g., MA Eq. (12.2). Note also that the scalar product under the volumic integral in Eq. (25) is nothing else than the divergence of the vector E – see, e.g., MA Eq. (8.4), hence the theorem’s name.
10 See, e.g., MA Secs. 8-10.
11 In the Gaussian units, just as in the initial Eq. (6), 0 has to be replaced with 1/4, so that the Maxwell equation (27) looks like E = 4, while Eq. (28) stays the same.
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Inhomo-
geneous
Maxwell
E
.
(1.27)
equation
0
for E
Note that in sharp contrast with the integral form (16), Eq. (27) is local: it relates the electric field’s divergence to the charge density at the same point. This equation, being the differential form of the Gauss law, is frequently called one of the famed Maxwell equations 12 – to be discussed again and again later in this course.
In the mathematical terminology, Eq. (27) is inhomogeneous, because it has a right-hand side independent (at least explicitly) of the field E that it describes. Another, homogeneous Maxwell equation’s “embryo” (this one valid for the stationary case only!) may be obtained by noticing that the curl of the point charge’s field, and hence that of any system of charges, equals zero:13
Homo-
geneous
Maxwell
E 0 .
(1.28)
equation
for E (We will arrive at two other Maxwell equations, for the magnetic field, in Chapter 5, and then generalize all the equations to their full, time-dependent form at the end of Chapter 6. However, Eq. (27) will stay the same.)
Just to get a better gut feeling of Eq. (27), let us apply it to the same example of a uniformly charged sphere (Fig. 2). Vector algebra tells us that the divergence of a spherically symmetric vector function E(r) = E( r)n r may be simply expressed in spherical coordinates:14 E = [ d( r 2 E)/ dr]/ r 2. As a result, Eq. (27) yields a linear ordinary differential equation for the scalar function E( r): 1 d
r
R
2
/ , for ,
( r E)
(1.29)
2
0
r dr
0,
for
r R,
which may be readily integrated on each of these segments:
1 1
2
r dr
3
r / 3 c , for r R,
E( r)
(1.30)
2
1
r
0
c ,
for r .
R
2
To determine the integration constant c 1, we can use the following boundary condition: E(0) = 0. (It follows from the problem’s spherical symmetry: in the center of the sphere, the electric field has to vanish, because otherwise, where would it be directed?) This requirement gives c 1 = 0. The second constant, c 2, may be found from the continuity condition E( R – 0) = E( R + 0), which has already been discussed above, giving c 2 = R 3/3 Q/4. As a result, we arrive at our previous results (19) and (22).
We can see that in this particular, highly symmetric case, using the differential form of the Gauss law is a bit more complex than its integral form. (For our second example, shown in Fig. 4, it would be even less natural.) However, Eq. (27) and its generalizations are more convenient for asymmetric charge 12 Named after the genius of classical electrodynamics and statistical physics, James Clerk Maxwell (1831-1879).
13 This follows, for example, from the direct application of MA Eq. (10.11) to any spherically-symmetric vector function of type f(r) = f( r)n r (in particular, to the electric field of a point charge placed at the origin), giving f = f
= 0 and fr/ = fr/ = 0 so that all components of the vector f vanish. Since nothing prevents us from placing the reference frame’s origin at the point charge’s location, this result remains valid for any position of the charge.
14 See, e.g., MA Eq. (10.10) for the particular case / = / = 0.
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distributions, and are invaluable in the cases where the distribution (r) is not known a priori and has to be found in a self-consistent way. (We will start discussing such cases in the next chapter.) 1.3. Scalar potential and electric field energy
One more help for solving problems of electrostatics (and electrodynamics as a whole) may be obtained from the notion of the electrostatic potential, which is just the electrostatic potential energy U
of a probe point charge q placed into the field in question, normalized by its charge: U
Electro-
.
(1.31) static
q
potential
As we know from classical mechanics,15 the notion of U (and hence ) makes the most sense for the case of potential forces – for example, those depending just on the particle’s position. Eqs. (6) and (9) show that stationary electric fields fall into this category. For such a field, the potential energy may be defined as a scalar function U(r) that allows the force to be calculated as its gradient (with the opposite sign):
F
U .
(1.32)
Dividing both sides of this equation by the probe charge, and using Eqs. (6) and (31), we get16
Electrostatic
E
.
(1.33) field as a
gradient
To calculate the scalar potential, let us start from the simplest case of a single point charge q placed at the origin. For it, Eq. (7) takes the simple form
1
r
1
n
E
q
q r .
(1.34)
3
2
4
r
4
r
0
0
It is straightforward to verify that the last fraction in the last form of Eq. (34) is equal to –(1/ r).17
Hence, according to the definition (33), for this particular case
1 q
.
(1.35) Potential of a
4 r
point charge
0
(In the Gaussian units, this result is spectacularly simple: = q/ r.) Note that we could add an arbitrary constant to this potential (and indeed to any other distribution of discussed below) without changing the field, but it is convenient to define the potential energy so it would approach zero at infinity.
In order to justify the introduction and the forthcoming exploration of U and , let me demonstrate (I hope, unnecessarily :-) how useful the notions are, on a very simple example. Let two similar charges q be launched from afar, with the same initial speed v 0 << c each, straight toward each other (i.e. with the zero impact parameter) – see Fig. 5. Since, according to the Coulomb law, the 15 See, e.g., CM Sec. 1.4.
16 Eq. (28) could be also derived from this relation because according to vector algebra, any gradient field has no curl – see, e.g., MA Eq. (11.1).
17 This may be done either by Cartesian components or using the well-known expression f = ( df/ dr)n r valid for any spherically-symmetric scalar function f( r) – see, e.g., MA Eq. (10.8) for the particular case / = / = 0.
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Essential Graduate Physics
EM: Classical Electrodynamics
charges repel each other with increasing force, they will stop at some minimum distance r min from each other, and then fly back. We could of course find r min directly from the Coulomb law. However, for that, we would need to write the 2nd Newton law for each particle (actually, due to the problem symmetry, they would be similar), then integrate them over time to find the particle velocity v as a function of distance, and only then recover r min from the requirement v = 0.
v
v
0
0
m, q
r
?
m, q
min
Fig. 1.5. A simple problem of charged particle motion.
The notion of potential allows this problem to be solved in one line. Indeed, in the field of potential forces, the system’s total energy E = T + U T + q is conserved. In our non-relativistic case v
<< c, the kinetic energy T is just mv 2/2. Hence, equating the total energy of two particles at the points r
= and r = r min, and using Eq. (35) for , we get
2
2
mv
1
q
0
2
0 0
,
(1.36)
2
4 r
0
min
immediately giving us the final answer: r
2
min = q 2/40 mv 0 . So, the notion of scalar potential is indeed very useful.
With this motivation, let us calculate for an arbitrary configuration of charges. For a single charge in an arbitrary position (say, at point r k’), r r in Eq. (35) should be evidently replaced with
r – r k’. Now, the linear superposition principle (3) allows for an easy generalization of this formula to the case of an arbitrary set of discrete charges,
1
q
r
( )
k '
.
(1.37)
40 r r r r
k '
k '
Finally, using the same arguments as in Sec. 1, we can use this result to argue that in the case of an arbitrary continuous charge distribution
Potential
1
( '
r )
of a charge .
(r)
d 3 r'
.
(1.38)
distribution
4
r '
r
0
Again, Dirac’s delta function allows using the last equation to recover Eq. (37) for discrete charges as well, so that Eq. (38) may be considered as the general expression for the electrostatic potential.
For most practical calculations, using this expression and then applying Eq. (33) to the result, is preferable to using Eq. (9), because is a scalar, while E is a 3D vector, mathematically equivalent to three scalars. Still, this approach may lead to technical problems similar to those discussed in Sec. 2. For example, applying it to the spherically-symmetric distribution of charge (Fig. 2), we get the integral
1
2
( r' )
2
sin 'd ' r' dr'
cos ,
(1.39)
4
R
0
0
0
which is not much simpler than Eq. (11).
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EM: Classical Electrodynamics
The situation may be much improved by recasting Eq. (38) into a differential form. For that, it is sufficient to plug the definition of , Eq. (33), into Eq. (27):
()
.
(1.40)
0
The left-hand side of this equation is nothing else than the Laplace operator of (with the minus sign), so that we get the famous Poisson equation 18 for the electrostatic potential:
Poisson
2
.
(1.41) equation
for
0
(In the Gaussian units, the Poisson equation is 2 = –4.) This differential equation is so convenient for applications that even its particular case for = 0,
Laplace
2
0,
(1.42) equation
for
has earned a special name – the Laplace equation.19
In order to get a gut feeling of the Poisson equation’s value as a problem-solving tool, let us return to the spherically-symmetric charge distribution (Fig. 2) with a constant charge density .
Exploiting this symmetry, we can represent the potential as ( r), and hence use the following simple expression for its Laplace operator:20
2
1 d 2 d
r
,
(1.43)
r 2 dr
dr
so that for the points inside the charged sphere ( r R) the Poisson equation yields 1 d
d
d
d
2
2
2
r
,
i.e.
r
r .
(1.44)
2
r dr
dr
dr
dr
0
0
Integrating the last form of the equation over r once, with the natural boundary condition d/ dr r = 0 = 0
(because of the condition E(0) =0, which has been discussed above), we get
d
r
r
1 Qr
2
( r)
r' dr'
.
(1.45)
2
3
dr
r
3
4 R
0 0
0
0
Since this derivative is nothing more than – E( r), in this formula we can readily recognize our previous result (22). Now we may like to carry out the second integration to calculate the potential itself: r
2
Q
Qr
( r)
r'dr' c
c .
(1.46)
4
3
1
R
8
3
1
R
0
0
0
18 Named after Siméon Denis Poisson (1781-1840), also famous for the Poisson distribution – one of the central results of the probability theory – see, e.g., SM Sec. 5.2.
19 Named after the famous mathematician (and astronomer) Pierre-Simon Laplace (1749-1827) who, together with Alexis Clairault, is credited for the development of the very concept of potential.
20 See, e.g., MA Eq. (10.8) for / = / = 0.
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Before making any judgment on the integration constant c 1, let us solve the Poisson equation (in this case, just the Laplace equation) for the range outside the sphere ( r > R): 1 d
2 d
r
0 .
(1.47)
2
r dr
dr
Its first integral,
d
c
2
( r)
,
(1.48)
2
dr
r
also gives the electric field (with the minus sign). Now using Eq. (45) and requiring the field to be continuous at r = R, we get
c
Q
d
Q
2
,
i.e.
( r)
,
(1.49)
2
R
4
2
R
dr
4
2
r
0
0
in an evident agreement with Eq. (19). Integrating this result again,
Q
dr
Q
( r)
c ,
for r R,
(1.50)
4
2
r
4
3
r
0
0
we can select c 3 = 0, so that () = 0, in accordance with the usual (though not compulsory) convention.
Now we can finally determine the constant c 1 in Eq. (46) by requiring that this equation and Eq. (50) give the same value of at the boundary r = R. (According to Eq. (33), if the potential had a jump, the electric field at that point would be infinite.) The final answer may be represented as
2
2
Q
R r
( r)
1 ,
for r .
R
(1.51)
4 R 2 2
R
0
This calculation shows that using the Poisson equation to find the electrostatic potential distribution for highly symmetric problems may be a bit more cumbersome than directly finding the electric field – say, from the Gauss law. However, we will repeatedly see below that if the electric charge distribution is not fixed in advance, using Eq. (41) may be the only practicable way to proceed.
Returning now to the general theory of electrostatic phenomena, let us calculate the potential energy U of an arbitrary system of point electric charges qk. Despite the apparently simple relation (31) between U and , the result is not that straightforward. Indeed, let us assume that the charge distribution has a finite spatial extent, so that at large distances from it (formally, at r = ) the electric field tends to zero, so that the electrostatic potential tends to a constant. Selecting this constant, for convenience, to equal zero, we may calculate U as a sum of the energy increments Uk created by bringing the charges, one by one, from infinity to their final positions r k – see Fig. 6.21 According to the integral form of Eq.
(32), such a contribution is
r
r
k
k
U
F r
( ) dr
q
E r
( ) dr q r ,
(1.52)
k
k
k
k
21 Indeed, by the very definition of the potential energy of a system, it should not depend on the way we are arriving at its final configuration.
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Essential Graduate Physics
EM: Classical Electrodynamics
where E(r) is the total electric field, and (r) is the total electrostatic potential during this process, besides the field created by the very charge qk that is being moved.
from
q , r
1 1
q , r
k
k
q , r
2
2
external
charges
Eext r
q , r
q , r ,
k'
k '
Fig. 1.6. Deriving Eqs. (55) and
3
3
(60) for potential energies of a
k
with ' k
system of several point charges.
the system of charges under analysis
This expression shows that the increment Uk, and hence the total potential energy U, depend on the source of the electric field E. If the field is dominated by an external field Eext, induced by some external charges, not being a part of the charge configuration under our analysis (whose energy we are calculating, see Fig. 6), then the spatial distribution (r) is determined by this field, i.e. does not depend on how many charges we have already brought in, so that Eq. (52) is reduced to
r
U q
E
' d '
(1.53)
k
k ext r k ,
where
r
( )
ext
r
( ) r
.
ext
Summing up these contributions, we get what is called the charge system’s energy in the external field:22
U
.
(1.54)
ext
U q
r
k
k ext k
k
k
Now repeating the argumentation that has led us to Eq. (9), we see that for a continuously distributed charge, this sum turns into an integral:
Energy:
U
(r) (r)
.
(1.55) external
ext
d 3 r
ext
field
(As was discussed above, using the delta-functional representation of point charges, we may always return from here to Eq. (54), so that Eq. (55) may be considered as a final, universal result.) The result is different in the opposite limit when the electric field E(r) is created only by the very charges whose energy we are calculating. In this case, (r k) in Eq. (52) is the potential created only by the charges with numbers k’ = 1, 2, …, ( k – 1) that are already in place when the k th charge is moved in (in Fig. 6, the charges inside the dashed boundary), and we may use the linear superposition principle to write
U q
r
U
U
q r .
(1.56)
k
k
(
,
)
that
so
k'
k
k
( )
k k '
k
k' k
k
k , k '
( k ' k )
This result is so important that it is worthy of rewriting in several other forms. First, we may use Eq.
(35) to represent Eq. (56) in a more symmetric form:
22 An alternative, perhaps more accurate term for U ext is the energy of the system’s interaction with the external field.
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EM: Classical Electrodynamics
1
q q
U
k k' .
(1.57)
4
r
r
0
k , k '
k
k '
( k ' k )
The expression under this sum is evidently symmetric with respect to the index swap, so that it may be extended into a different form,
1 1
q q
U
k k' ,
(1.58)
4 2
r
r
0
k ,' k
k
k '
( k ' k )
where the interaction between each couple of charges is described by two equal terms under the sum, and the front coefficient ½ is used to compensate for this double-counting. The convenience of the last form is that it may be readily generalized to the continuous case:
1 1
3
3
r
( ) ( '
r )
U
d r d r'
.
(1.59)
4 2
r '
r
0
(As before, in this case the restriction expressed in the discrete charge case as k k’ is not important, because if the charge density is a continuous function, the integral (59) does not diverge at point r = r ’.) To represent this result in one more form, let us notice that according to Eq. (38), the inner integral over r’ in Eq. (59), divided by 4 0, is just the full electrostatic potential at point r, and hence Energy:
1
charge
U (r)(r d 3
) r .
(1.60)
interaction
2
For the discrete charge case, this result is
1
U q (r ) ,
(1.61)
k
k
2 k
but here it is important to remember that here the “full” potential’s value (r k) should exclude the (infinite) contribution from the point charge k itself. Comparing the last two formulas with Eqs. (54) and (55), we see that the electrostatic energy of charge interaction within the system, as expressed via the charge-by-potential product, is twice less than that of the energy of charge interaction with a fixed (“external”) field. This is the result of the fact that in the case of mutual interaction of the charges, the electric field E in the basic Eq. (52) is proportional to the charge’s magnitude, rather than constant.23
Now we are ready to address an important conceptual question: can we locate this interaction energy in space? This task may seem trivial: Eqs. (58)-(61) seem to imply that non-zero contributions to U come only from the regions where the electric charges are located. However, one of the most beautiful features of physics is that sometimes completely different interpretations of the same mathematical result are possible. To get an alternative view of our current result, let us write Eq. (60) for a volume V
so large that the electric field on the limiting surface S is negligible, and plug into it the charge density expressed from the Poisson equation (41):
23 The nature of this additional factor ½ is absolutely the same as in the well-known formula U = (½) x 2 for the potential energy of an elastic spring providing the returning force F = – x, proportional to its displacement x from the equilibrium position.
Chapter 1
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EM: Classical Electrodynamics
U 0 2 d 3 r .
(1.62)
2 V
This expression may be integrated by parts as24
U 0 d 2 r
2
d 3 r .
(1.63)
n
2 S
V
According to our condition of negligible field E = – at the surface, the first integral vanishes, and we get a very important formula
U 0
2 d 3 r 0 E 2 d 3 r .
(1.64)
2
2
This result, represented in the following equivalent form:25
Energy:
U u(r) 3
d r, with u(r)
0
2
E (r),
(1.65) electric
2
field
certainly invites an interpretation very much different than Eq. (60): it is natural to consider u(r) as the spatial density of the electric field energy, which is continuously distributed over all the space where the field exists – rather than just its part where the charges are located.
Let us have a look at how these two alternative pictures work for our testbed problem, a uniformly charged sphere. If we start with Eq. (60), we may limit the integration by the sphere volume (0 r R) where 0. Using Eq. (51), and the spherical symmetry of the problem (giving d 3 r =
4 r 2 dr), we get
R
1
2
1
Q
R
R 2 r 2
2
6 1 Q 2
U 4 r dr 4
1 r dr
.
(1.66)
2
2
4 R 2 R 2
5 4
2 R
0
0
0
0
On the other hand, if we use Eq. (65), we need to integrate the energy density everywhere, i.e. both inside and outside of the sphere:
