Y
J
( )
,
(2.149)
sin
49 See, e.g., MA Eq. (6.7a). Note that ( s) at s 0, –1, –2,…
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called the Bessel functions of the second kind, or more often the Weber functions,50 and then to follow the limit n. At this, both the numerator and denominator of the right-hand side of Eq. (149) tend to zero, but their ratio tends to a finite value called Yn( x). It may be shown that the resulting functions are still the solutions of the Bessel equation and are linearly independent of Jn( x), though are related just as those functions if the sign of n changes:
Y ( ) ( )
1 nY ( ) .
(2.150)
n
n
Figure 19 shows a few Weber functions of the lowest integer orders.
1
n
1
0
3
2
0.5
Y ( )
n
0
0.5
Fig. 2.19. A few Bessel
functions of the second kind
(a.k.a. the Weber functions,
1
a.k.a. the Neumann functions).
0
5
10
15
20
The plots show that their asymptotic behavior is very much similar to that of the functions Jn( ):
2 1/2
(
n
Y
)
sin
(2.151)
n
,
for ,
4
2
but with the phase shift necessary to make these Bessel functions orthogonal to those of the fist order –
cf. Eq. (135). However, for small values of argument , the Bessel functions of the second kind behave completely differently from those of the first kind:
2
ln ,
for
n ,
0
2
Y ( )
(2.152)
n
n
( n )!
1
,
for
n
0,
2
where is the so-called Euler constant, defined as follows:
50 Sometimes, they are called the Neumann functions and denoted as N().
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1 1
1
lim
1
...
ln n
0.577157
(2.153)
n
2 3
n
As Eqs. (152) and Fig. 19 show, the functions Yn( ) diverge at 0 and hence cannot describe the behavior of any physical variable, in particular the electrostatic potential.
One may wonder: if this is true, when do we need these functions in physics? Figure 20 shows an example of a simple boundary problem of electrostatics, whose solution by the variable separation method involves both functions Jn( ) and Yn( ).
(a)
(b)
0
R
R
1
2
Fig. 2.20. A simple boundary
1
2
problem that cannot be solved
using just one kind of Bessel
0
R
R
functions.
1
2
Here two round, conducting coaxial cylindrical tubes are kept at the same (say, zero) potential, but at least one of two lids has a different potential. The problem is almost completely similar to that discussed above (Fig. 17), but now we need to find the potential distribution in the free space between the tubes, i.e. for R 1 < < R 2. If we use the same variable separation as in the simpler counterpart problem, we need the radial functions R() to satisfy two zero boundary conditions: at = R 1 and =
R 2. With the Bessel functions of just the first kind, Jn(), it is impossible to do, because the two boundaries would impose two independent (and generally incompatible) conditions, Jn( R 1) = 0, and Jn( R 2) = 0, on one “stretching parameter” . The existence of the Bessel functions of the second kind immediately saves the day, because if the radial function solution is represented as a linear combination, R c J () c Y (),
(2.154)
J
n
Y n
two zero boundary conditions give two equations for and the ratio c cY/ cJ.51 (Due to the oscillating character of both Bessel functions, these conditions would be typically satisfied by an infinite set of discrete pairs {, c}.) Note, however, that generally none of these pairs would correspond to zeros of either Jn or Yn, so that having an analog of Table 1 for the latter function would not help much. Hence, even the simplest problems of this kind (like the one shown in Fig. 20) typically require the numerical solution of transcendental algebraic equations.
51 A pair of independent linear functions, used for the representation of the general solution of the Bessel equation, may be also chosen differently, using the so-called Hankel functions
,
1
( 2)
H
( ) J ( ) iY ( ) .
n
n
n
For representing the general solution of Eq. (130), this alternative is completely similar, for example, to using the pair of complex functions exp{ i x} cos x i sin x instead of the pair of real functions {cos x, sin x} for the representation of the general solution of Eq. (89) for X( x).
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In order to complete the discussion of variable separation in the cylindrical coordinates, one more issue to address is the so-called modified Bessel functions: of the first kind, I(), and of the second kind, K(). They are two linearly-independent solutions of the modified Bessel equation, 2
d R
1
2
dR
Modified
1
R 0 ,
(2.155)
2
2
Bessel
d
d
equation
which differs from Eq. (130) “only” by the sign of one of its terms. Figure 21 shows a simple problem that leads (among many others) to this equation: a round thin conducting cylindrical pipe is sliced, perpendicular to its axis, to rings of equal height h, which are kept at equal but sign-alternating potentials.
z
V
/ 2
h
V
/ 2
t
Fig. 2.21. A typical boundary problem whose
V
/ 2
solution may be conveniently described in
terms of the modified Bessel functions.
If the system is very long (formally, infinite) in the z-direction, we may use the variable separation method for the solution of this problem, but now evidently need periodic (rather than exponential) solutions along the z-axis, i.e. linear combinations of sin kz and cos kz with various real values of the constant k. Separating the variables, we arrive at a differential equation similar to Eq.
(129), but with the negative sign before the separation constant:
2
d R
1
2
d
R ( 2
k
) R 0 .
(2.156)
2
2
d
d
The same radial coordinate’s normalization, k, immediately leads us to Eq. (155), and hence (for
= n) to the modified Bessel functions In() and Kn().
Figure 22 shows the behavior of such functions, of a few lowest orders. One can see that at
0 the behavior is virtually similar to that of the “usual” Bessel functions – cf. Eqs. (132) and (152), with Kn() multiplied (by purely historical reasons) by an additional coefficient, /2:
ln , for n ,
0
1
n
2
I ( )
,
K ( )
(2.157)
n
n
n! 2
n
( n )!
1
, for n ,
0
2
2
However, the asymptotic behavior of the modified functions is very much different, with In( x) exponentially growing, and Kn() exponentially dropping at :
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1 1/2
1/2
I
( )
e ,
K
( )
.
(2.158)
n
e
n
2
2
3
3
2
2
I ()
n 0
1
2
K ( )
n
n
1
1
Fig. 2.22. The modified Bessel
functions of the first kind (left
panel) and the second kind
0
0
(right panel).
0
1
2
3
0
1
2
3
This behavior is completely natural in the context of the problem shown in Fig. 21, in which the electrostatic potential may be represented as a sum of terms proportional to In() inside the thin pipe, and of terms proportional to Kn() outside it.
To complete our brief survey of the Bessel functions, let me note that all of them discussed so far may be considered as particular cases of Bessel functions of the complex argument, say Jn(z) and Yn(z), or, alternatively, H (1,2)
n
(z) Jn(z) iYn(z).52 At that, the “usual” Bessel functions Jn() and Yn() may be considered as the sets of values of these generalized functions on the real axis (z = ), while the modified functions as their particular case on the imaginary axis, i.e. at z = i, also with real :
I ( ) i J ( i ),
K ( )
1
)
1
(
i H ( i )
.
(2.159)
2
Moreover, this generalization of the Bessel functions to the whole complex plane z enables the use of their values along other directions on that plane, for example under angles /4 /2. As a result, one arrives at the so-called Kelvin functions:
ber i bei J ( i / 4
e
),
(2.160)
ker i kei
)
1
(
i H (
i
3 / 4
e
),
2
which are also useful for some important problems in physics and engineering. Unfortunately, I do not have time/space to discuss these problems in this course.53
52 These complex functions still obey the general relations (143) and (146), with replaced with z.
53 In the QM part of in this series we will run into the so-called spherical Bessel functions jn() and yn(), which may be expressed via the Bessel functions of semi-integer orders. Surprisingly enough, these functions turn out to be simpler than Jn() and Yn().
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2. 8. Variable separation – spherical coordinates
The spherical coordinates are very important in physics, because of the (at least approximate) spherical symmetry of many physical objects – from nuclei and atoms, to water drops in clouds, to planets and stars. Let us again require each component k of Eq. (84) to satisfy the Laplace equation.
Using the full expression for the Laplace operator in spherical coordinates,54 we get
1
k
k
2
1
1
2
r
sin
k 0 .
(2.161)
2
2
r r
r r sin
2
r sin2
2
Let us look for a solution of this equation in the following variable-separated form:
R ( r)
,
(2.162)
k
P (cos ) F ()
r
Separating the variables one by one, starting from , just like this has been done in cylindrical coordinates, we get the following equations for the partial functions participating in this solution: 2
d R l( l )
1
R 0 ,
(2.163)
2
2
dr
r
2
d
dP
1
(
2
)
l( l )
1
P 0
,
(2.164)
d
d
1
2
2
d
F
2
F 0,
(2.165)
2
d
where cos is a new variable used in lieu of (so that –1 +1), while 2 and l( l+1) are the separation constants. (The reason for selection of the latter one in this form will be clear in a minute.) One can see that Eq. (165) is very simple, and is absolutely similar to the Eq. (107) we have got for the polar and cylindrical coordinates. Moreover, the equation for the radial functions is simpler than in the cylindrical coordinates. Indeed, let us look for its partial solution in the form cr – just as we have done with Eq. (106). Plugging this solution into Eq. (163), we immediately get the following condition on the parameter :
1 l( l )
1 .
(2.166)
This quadratic equation has two roots, = l + 1 and = – l, so that the general solution of Eq. (163) is
b
l 1
l
R a r
.
(2.167)
l
l
r
However, the general solution of Eq. (164) (called either the general or associated Legendre equation) cannot be expressed via what is usually called elementary functions.55 Let us start its discussion from the axially-symmetric case when / =0. This means F() = const, and thus = 0, so that Eq. (164) is reduced to the so-called Legendre differential equation:
54 See, e.g., MA Eq. (10.9).
55 Again, there is no generally accepted line between the “elementary” and “special” functions.
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d
d
Legendre
P
1
(
2
)
l( l )
1 P 0
equation
.
(2.168)
d
d
One can readily verify that the solutions of this equation for integer values of l are specific ( Legendre) polynomials56 that may be described by the following Rodrigues’ formula:
Legendre
1 d l
polynomials
P ( )
( 2
)
1 l , with l ,
0 ,
1 ,...
2 .
(2.169)
l
2 l !
l d l
According to this formula, the first few Legendre polynomials are pretty simple:
P ( ) ,
1
0
P ( ) ,
1
1
P ( ) 3 2
(2.170)
2
1,
2
1
P ( )
3
5 3 3 ,
2
1
P ()
4
35 4 30 2 3,..,
8
though such explicit expressions become more and more bulky as l is increased. As Fig. 23 shows, all these polynomials, which are defined on the [-1, +1] segment, end at the same point: Pl(+1) = + 1, while starting either at the same point or at the opposite point: Pl(-1) = (-1) l. Between these two endpoints, the l th Legendre polynomial has l zeros. It is straightforward to use Eq. (169) for proving that these polynomials form a full, orthogonal set of functions, with the following normalization rule: 1
2
P ( ) P ( ) d
,
(2.171)
l
l '
ll '
2 l 1
1
so that any function f() defined on the segment [-1, +1] may be represented as a unique series over the polynomials.57
Thus, taking into account the additional division by r in Eq. (162), the general solution of any axially-symmetric Laplace problem may be represented as
Variable
separation
in spherical
b
l
l
coordinates
( r, )
a r
P
.
(2.172)
l
(cos )
l1
l
(for axial
l 0
r
symmetry) Note a strong similarity between this solution and Eq. (112) for the 2D Laplace problem in the polar coordinates. However, besides the difference in the angular functions, there is also a difference (by one) in the power of the second radial function, and this difference immediately shows up in problem solutions.
56 Just for the reference: if l is not an integer, the general solution of Eq. (2.168) may be represented as a linear combination of the so-called Legendre functions (not polynomials!) of the first and second kind, Pl() and Ql().
57 This is why, at least for the purposes of this course, there is no good reason for pursuing (more complicated) solutions to Eq. (168) for non-integer values of l, mentioned in the previous footnote.
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1
l 0
l 3
P ( )
l
l 4
0
l 2
Fig. 2.23. A few lowest Legendre
l 1
polynomials Pl().
1 1
0
1
Indeed, let us solve a problem similar to that shown in Fig. 15: find the electric field around a conducting sphere of radius R, placed into an initially uniform external field E0 (whose direction I will now take for the z-axis) – see Fig. 24a.
(a)
(b)
E
z
0
R
0
Fig. 2.24. Conducting sphere in a uniform electric field: (a) the problem’s geometry, and (b) the equipotential surface pattern given by Eq. (176). The pattern is qualitatively similar but quantitatively different from that for the conducting cylinder in a perpendicular field – cf. Fig. 15.
If we select the arbitrary constant in the electrostatic potential so that z=0 = 0, then in Eq. (172) we should take a 0 = b 0 = 0. Now, just as has been argued for the cylindrical case, at r >> R the potential should approach that of the uniform field:
E z E r cos ,
(2.173)
0
0
so that in Eq. (172), only one of the coefficients al survives: al = – E 0 l, 1. As a result, from the boundary condition on the surface, ( R,) = 0, we get the following equation for the coefficients bl:
b
b
1
E R
cos l P (cos) 0.
(2.174)
0
2
l 1
R
l
l2 R
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Now repeating the argumentation that led to Eq. (117), we may conclude that Eq. (174) is satisfied if 3
b E R ,
(2.175)
l
0
l 1
,
so that, finally, Eq. (172) is reduced to
3
R
E r
cos .
(2.176)
0
2
r
This distribution, shown in Fig. 24b, is very much similar to Eq. (117) for the cylindrical case (cf. Fig.
15b, with the account for a different plot orientation), but with a different power of the radius in the second term. This difference leads to a quantitatively different distribution of the surface electric field:
E
3 E cos ,
(2.177)
n
r R
0
r
so that its maximal value is a factor of 3 (rather than 2) larger than the external field.
Now let me briefly (mostly just for the reader’s reference) mention the Laplace equation solutions in the general case – with no axial symmetry. If the conductor-free space surrounds the origin from all sides, the solutions to Eq. (165) have to be 2-periodic, and hence = n = 0, 1, 2,…
Mathematics says that Eq. (164) with integer = n and a fixed integer l has a solution only for a limited range of n:58
l n l .
(2.178)
These solutions are called a ssociated Legendre functions (generally, they are not polynomials). For n
0, these functions may be defined via the Legendre polynomials, using the following formula:59
n
d
n
P
.
(2.179)
l
( )
1 n 1
(
2
) n/2
P ( )
n l
d
On the segment [-1, +1], each set of the associated Legendre functions with a fixed index n and non-negative values of l form a full, orthogonal set, with the normalization relation,
1
2
l
( n)!
n
n
P
( ) P
( ) d
,
(2.180)
l
l '
ll'
l
2 1 l
( n)!
1
that is evidently a generalization of Eq. (171).
Since these relations may seem a bit intimidating, let me write down explicit expressions for a few P nl (cos) with the three lowest values of l and n 0, which are most important for applications.
l 0 :
0
P
;
(2.181)
0 cos
1
58 In quantum mechanics, the letter n is typically reserved for the “principal quantum number”, while the azimuthal functions are numbered by index m. However, here I will keep using n as their index because for this course’s purposes, this seems more logical, in view of the similarity of the spherical and cylindrical functions.
59 Note that some texts use different choices for the front factor (called the Condon-Shortley phase) in the functions P m
m
l , which do not affect the final results for the spherical harmonics Yl .
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0
P
1 cos
cos
,
l 1:
(2.182)
1
P
1 cos
sin
;
0
P (cos )
2
3cos2 1/
,
2
l 2 : 1
P (cos ) 2sin cos
,
(2.183)
2
2
2
P (cos ) 3cos
.
2
The reader should agree there is not much to fear in these functions – they are just certain sums of products of cos and sin (1 – 2)1/2. Fig. 25 shows the plots of a few lowest functions P nl ().
n
Pl
1
l 0
l 1
l 2
l 3
l 4
Fig. 2.25. A few lowest
associated Legendre functions.
(Adapted from an original by
n 0
Geek3,
available
at
n 1
https://en.wikipedia.org/wiki/Ass
n 2
ociated_Legendre_polynomials,
n 3
under
the
GNU
Free
n 4
Documentation License.)
1
Using the associated Legendre functions, the general solution (162) to the Laplace equation in the spherical coordinates may be expressed as
Variable
separation
l
in spherical
l
b
( r, ,) a r
l
n
.
(2.184) coordinates
P (cos) F ( ,)
F
( ) c cos n s sin n
l
l 1
l
n
(general
l0
r
n
n
n
n 0
case)
Since the difference between the angles and is somewhat artificial, physicists prefer to think not in terms of the functions P and F in separation, but directly about their products that participate in this solution.60
60 In quantum mechanics, it is more convenient to use a slightly different alternative set of basic functions of the same problem, namely the following complex functions called the spherical harmonics: 1/ 2
n
l
2
1 l
(
n)!
n
in
Y
( ,)
P (cos ) e
,
l
4
l
( n
l
)!
which are defined for both positive and negative n (within the limits – l n + l) – see, e.g., QM Secs. 3.6 and 5.6.
(Note again that in that field, our index n is traditionally denoted as m, and called the magnetic quantum number.) Chapter 2
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As a rare exception for my courses, to save time I will skip giving an example of using the associated Legendre functions in electrostatics, because quite a few examples of these functions’
applications will be given in the quantum mechanics part of this series.
2.9. Charge images
So far, we have discussed various methods of solution of the Laplace boundary problem (35).
Let us now move on to the discussion of its generalization, the Poisson equation (1.41). We need it when besides conductors, we also have stand-alone charges with a known spatial distribution (r). (Its discussion will also allow us, better equipped, to revisit the Laplace problem in the next section.) Let us start from a somewhat limited, but very useful charge image (or “image charge”) method.
Consider a very simple problem: a single point charge near a conducting half-space – see Fig. 26.
z
q
r 1
r
'
r
d
0
r 2
0
Fig. 2.26. The simplest problem readily solvable by the
d
charge image method. The points’ colors are used, as
"
r
before, to denote the charges of the original (red) and
q
opposite (blue) sign.
Let us prove that its solution, above the conductor’s surface ( z 0), may be represented as: 1 q
q
q
1
1
r
( )
,
(2.185)
4
0 r
r
1
2
4 0 r ' r
r "
r
or in a more explicit form, using the cylindrical coordinates shown in Fig. 26:
q
1
1
(r)
,
(2.186)
4
1/ 2
1/ 2
2
0
( z 2
d)
2 ( z 2
d)
where is the distance of the field observation point r from the “vertical” line on which the charge is located. Indeed, this solution satisfies both the boundary condition = 0 at the surface of the conductor ( z = 0), and the Poisson equation (1.41), with the single -functional source at point r ’ = {0, 0, + d} on its right-hand side, because the second singularity of the solution, at point r ” = {0, 0, – d}, is outside the region of the solution’s validity ( z 0). Physically, this solution may be interpreted as the sum of the fields of the actual charge (+ q) at point r ’, and an equal but opposite charge (– q) at the “mirror image”
point r ” (Fig. 26). This is the basic idea of the charge image method.
Before moving on to more complex problems, let us discuss the situation shown in Fig. 26 in a little bit more detail, due to its fundamental importance. First, we can use Eqs. (3) and (186) to calculate the surface charge density:
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q
1
1
q
2 d
. (2.187)
0
z0
z
4 z
2
2
( z d) 1/2 2
2
( z d) 1/2
4
d
z0
2 23/2
From this, the total surface charge is
q
2 d
Q 2
d r 2 ()
d
d .
(2.188)
3 / 2
2
2
2
S
0
0
d
This integral may be easily worked out using the substitution 2/ d 2 (giving d = 2 d/ d 2): q
d
Q
q
(2.189)
2
3 / 2
0
.
1
This result is very natural: the conductor brings as much surface charge from its interior to the surface as necessary to fully compensate for the initial charge (+ q) and hence kill the electric field at large distances as efficiently as possible, hence reducing the total electrostatic energy (1.65) to the lowest possible value.
For a better feeling of this polarization charge of the surface, let us take our calculations to the extreme – to the q equal to one elementary change e, and place a particle with this charge (for example, a proton) at a macroscopic distance – say 1 m – from the conductor’s surface. Then, according to Eq.
(189), the total polarization charge of the surface equals that of an electron, and according to Eq. (187), its spatial extent is of the order of d 2 = 1 m2. This means that if we consider a much smaller part of the surface, A << d 2, its polarization charge magnitude Q = A is much less than one electron! For example, Eq. (187) shows that the polarization charge of quite a macroscopic area A = 1 cm2 right under the initial charge ( = 0) is e A/2 d 2 1.610-5 e. Can this be true, or our theory is somehow limited to the charges q much larger than e? (After all, the theory is substantially based on the approximate macroscopic model (1); maybe it is the culprit?)
Surprisingly enough, the answer to this question has become clear (at least to some physicists :-) only as late as in the mid-1980s when several experiments demonstrated, and theorists accepted (some of them rather grudgingly) that the usual polarization charge formulas are valid for elementary charges as well, i.e., such the polarization charge Q of a macroscopic surface area may differ from a multiple of e. The underlying reason for this paradox is the physical nature of the polarization charge of a conductor’s surface: as was discussed in Sec. 1, it is due not to new charged particles brought into the conductor (such charge would be in fact a multiple of e), but to a small shift of the free charges of a conductor by a very small distance from their equilibrium positions that they had in the absence of the external field induced by charge q. This shift is not quantized, at least on the scale relevant to our problem, and hence neither is Q.
This understanding has paved the way toward the invention and experimental demonstration of several new devices including so-called single-electron transistors,61 which may be used, in particular, for ultrasensitive measurement of polarization charges as small as ~10-6 e. Another important class of single-electron devices is the dc and ac current standards based on the fundamental relation I = – ef, 61 Actually, this term (for which the author of these notes may be blamed :-) is misleading: the operation of the
“single-electron transistor” is based on the interplay of discrete charges (multiples of e) transferred between conductors, and sub-single-electron polarization charges – see, e.g., K. Likharev, Proc. IEEE 87, 606 (1999).
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where I is the dc current carried by electrons transferred with the frequency f. The experimentally achieved62 relative accuracy of such standards is of the order of 10-7, and is not too far from that provided by the competing approach based on a combination of the Josephson effect and the quantum Hall effect.63
Second, let us find the potential energy U of the charge-to-surface interaction. For that, we may use the value of the electrostatic potential (185) at the point of the charge itself (r = r ’), of course ignoring the infinite potential created by the real charge, so that the remaining potential is that of the image charge
1
q
( '
r )
.
(2.190)
image
4 2 d
0
Looking at the electrostatic potential’s definition given by Eq. (1.31), it may be tempting to immediately write U = qimage = – (1/40)( q 2/2 d) [WRONG!], but this would be incorrect. The reason is that the potential image is not independent of q, but is actually induced by this charge. This is why the correct approach is to calculate U from Eq. (1.61), with just one term:
1
1 q 2
U q
,
(2.191)
2
image
4 4 d
0
giving twice lower energy than the wrong result cited above. To double-check Eq. (191), and also get a better feeling of the factor ½ that distinguishes it from the wrong guess, we can calculate U as the integral of the force exerted on the charge by the conductor’s surface charge (i.e., in our formalism, by the image charge):
d
d
1
q 2
1 q 2
U F( z) dz
dz
.
(2.192)
4
(2 z)2
4 4 d
0
0
This calculation clearly accounts for the gradual build-up of the force F, as the real charge is being brought from afar (where we have opted for U =0) toward the surface.
This result has several important applications. For example, let us plot the electrostatic energy U
of an electron, i.e. a particle with charge q = – e, near a metallic surface, as a function of d. For that, we may use Eq. (191) until our macroscopic model (1) becomes invalid, and U transitions to some negative constant value (-) inside the conductor – see Fig. 27a. Since our calculation was for an electron with zero potential energy at infinity, at relatively low temperatures, k B T << , electrons in metals may occupy only the states with energies below – (the so-called Fermi level 64). The positive constant is called the workfunction because it describes the smallest work needed to remove the electron from a metal. As was discussed in Sec. 1, in good metals the electric field screening takes place at interatomic distances a 0 ~ 10-10 m. Plugging d = 110-10 m and q = – e –1.610-19 C into Eq. (191), we get
610–19 J 3.5 eV. This crude estimate is in surprisingly good agreement with the experimental values of the workfunction, ranging between 4 and 5 eV for most metals.65
62 See, e.g., M. Keller et al., Appl. Phys. Lett. 69, 1804 (1996) ; F. Stein et al., Metrologia 54, 1 (2017).
63 J. Brun-Pickard et al., Phys. Rev. X 6, 041051 (2016).
64 More discussion of these states may be found in SM Secs. 3.3 and 6.3.
65 More discussion of the workfunction, and its effect on the electrons’ kinetics, is given in SM Sec. 6.3.
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U
(a)
(b)
U
a 0
0
0
d
d
eE d
0
Fig. 2.27. (a) The origin
of the workfunction, and
1
(b) the field emission of
U
electrons (schematically).
d
Next, let us consider the effect of an additional uniform external electric field E0 applied normally to a metallic surface, on this potential profile. For that, we may the potential energy that the field gives to the electron at distance d from the surface, U ext = – eE 0 d, to that created by the image charge. (As we know from Eq. (1.53), since the field E0 is independent of the electron’s position, its recalculation into the potential energy does not require the coefficient ½.) As the result, the potential energy of an electron near the surface becomes
1
e 2
U ( d ) eE d
,
for d >> a
0
4 4 d
0,
(2.193)
0
with a similar crossover to U = – inside the conductor – see Fig. 27b. One can see that at the appropriate sign, and a sufficient magnitude of the applied field, it lowers the potential barrier that prevents electrons from leaving the conductor. At E 0 ~ / a 0 (for metals, ~1010 V/m), this suppression becomes so strong that electrons with energies at, and just below the Fermi level start quantum-mechanical tunneling through the remaining thin barrier. This is the field electron emission (or just
“field emission”) effect, which is used in vacuum electronics to provide efficient cathodes that do not require heating to high temperatures.66
Returning to the basic electrostatics, let us find some other conductor geometries where the method of charge images may be effectively applied. First, let us consider a right-angle corner (Fig.
28a). Reflecting the initial charge in the vertical plane, we get the image shown in the top left corner of that panel. This image makes the boundary condition = const satisfied on the vertical surface of the corner. However, for the same to be true on the horizontal surface, we have to reflect both the initial charge and the image charge in the horizontal plane, flipping their signs. The final configuration of four charges, shown in Fig. 28a, satisfies all boundary conditions. The resulting potential distribution may be readily written as an evident generalization of Eq. (185). From it, the electric field and electric charge distributions, and the potential energy and forces acting on the charge may be calculated exactly as above – an easy exercise left for the reader.
Next, consider a corner with the angle /4 (Fig. 28b). Here we need to repeat the reflection operation not two but four times before we arrive at the final pattern of eight positive and negative charges. (Any attempt to continue this process would lead to overlap with the already existing charges.) 66 The practical use of such “cold” cathodes is affected by the fact that, as it follows from our discussion in Sec. 4, any nanoscale irregularity of a conducting surface (a protrusion, an atomic cluster, or even a single “adatom”
stuck to it) may cause a strong increase of the local field well above the applied uniform field E 0, making the electron emission reproducibility and stability in time significant challenges. In addition, the impact-ionization effects may lead to avalanche-type electric breakdown at dc fields as low as ~3106 V/m.
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This reasoning may be readily extended to corners of angles = / n, with any integer n, which require 2 n charges (including the initial one) to satisfy all the boundary conditions.
(a)
(b)
(c)
d
q
2 a d
q
q
q
q
q
x
q
q
q
q
q
0
a
2 a
(d)
(e)
Fig. 2.28. The charge images for (a, b) the corners with angles /2 and /4, (c) a plane capacitor, and (d) a rectangular box; (e) typical equipotential surfaces for the last system.
Some configurations require an infinite number of images but are still tractable. The most important of them is a system of two parallel conducting surfaces, i.e. an unbiased plane capacitor of infinite area (Fig. 28c). Here the repeated reflection leads to an infinite system of charges q at points x 2 aj d ,
(2.194)
j
where d (with 0 < d < a) is the position of the initial charge, and j is an arbitrary integer. The resulting infinite sum for the potential of the real charge q, created by the field of its images, 1
q
q
q 1
2
d
1
( d)
(2.195)
4 2 d
j
d x
d
a j j j d a
0
0
4
2
3
2
2
j
0
1
( / ,
)
is converging (in its last form) very fast. For example, the exact value, ( a/2) = –2ln2( q/40 a), differs by less than 5% from the approximation using just the first term of the sum.
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The same method may be applied to 2D (cylindrical) and 3D rectangular conducting boxes that require, respectively, 2D or 3D infinite rectangular lattices of images; for example in a 3D box with sides a, b, and c, charges q are located at points (Fig. 28d) r 2 ja 2 kb lc
2 '
r ,
(2.196)
jkl
where r ’ is the location of the initial (real) charge, and j, k, and l are arbitrary integers. Figure 28e shows a typical result of the summation of the potentials of such charge set, including the real one, in a 2D box (within the plane of the real charge). One can see that the equipotential surfaces, concentric near the charge, are naturally leaning along the conducting walls of the box, which have to be equipotential.
Even more surprisingly, the image charge method works very efficiently not only for rectilinear geometries but also for spherical ones. Indeed, let us consider a point charge q at distance d from the center of a conducting, grounded sphere of radius R (Fig. 29a), and try to satisfy the boundary condition
= 0 for the electrostatic potential on the sphere’s surface using an imaginary charge q’ located at some point beyond the surface, i.e. inside the sphere.
(a)
(b)
q, d
r
q' , d'
R
Fig. 2.29. Method of charge images for
0
a conducting sphere: (a) the idea, and
(b) the resulting potential distribution
in the central plane containing the
charge, for the particular case d = 2 R.
From the problem’s symmetry, it is clear that the point should be at the line passing through the real charge and the sphere’s center, at some distance d’ from the center. Then the total potential created by the two charges at an arbitrary point of free space, i.e. at r R (Fig. 29a) is
1
q
q'
( r, )
.
(2.197)
4
1/ 2
1/ 2
2
0
r 2
d 2 rd cos
2 r 2
d' 2 rd' cos
This expression shows that we can make the two involved fractions equal and opposite at all points on the sphere’s surface (i.e. for any at r = R) if we take67
R 2
R
d'
,
q' q .
(2.198)
d
d
Since the solution of any Poisson boundary problem is unique, Eqs. (197) and (198) give us the final solution for this problem. Fig. 29b shows a typical equipotential pattern following from this solution. It may be surprising how formulas that simple may describe such an elaborate field distribution.
67 In geometry, such points with dd’ = R 2, are referred to as the result of mutual inversion in a sphere of radius R.
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Now we can calculate the total charge Q on the grounded sphere’s surface, induced by the external charge q. We could do this, as we have done for the conducting plane, by the brute-force integration of the surface charge density = –0/ r r = R. It is more elegant, however, to use the following Gauss law argument. Equality (197) is valid (at r R) regardless of whether we are dealing with our real problem (charge q and the conducting sphere) or with the equivalent charge configuration
– with the point charges q and q’, but no sphere at all. Hence, according to Eq. (1.16), the Gaussian integral over a surface with radius r = R + 0, and the total charge inside the sphere should be also the same. Hence we immediately get
R
Q q'
q .
(2.199)
d
A similar argumentation may be used to calculate the charge-to-sphere interaction force: 2
2
q'
q
R
1
q
Rd
F qE
( d) q
.
(2.200)
image
2
2
2
2
2 2
4 ( d d' )
4 d ( d R / d)
4 ( d R )
0
0
0
(Note that this expression is legitimate only at d > R.) At large distances, d >> R, this attractive force decreases as 1/ d 3. This unusual dependence arises because, as Eq. (199) specifies, the induced charge of the sphere, responsible for the force, is not constant but decreases as 1/ d. In the next chapter, we will see that such force is also typical for the interaction between a point charge and a dipole.
All previous formulas were for a sphere that is grounded to keep its potential equal to zero. But what if we keep the sphere galvanically insulated, so that its net charge is fixed, for example, equals zero? Instead of solving this problem from scratch, let us use (again!) the almighty linear superposition principle. For that, we may add to the previous problem an additional charge, equal to – Q = – q’, to the sphere, and argue that this addition gives, at all points, an additional, spherically-symmetric potential that does not depend on the potential induced by the external charge q, and was calculated in Sec. 1.2 –
see Eq. (1.19). For the interaction force, such addition yields
2
qq'
qq'
q
Rd
R
F
.
(2.201)
2
2
2
2 2
3
4 ( d d')
4 πε d
4
( d
R )
d
0
0
0
At large distances, the two terms proportional to 1/ d 3 cancel each other, giving F 1/ d 5, so that the potential energy of such interaction behaves as U 1/ d 4. Such a rapid force decay is due to the fact that the field of the uncharged sphere is equivalent to that of two (equal and opposite) induced charges + q’
and – q’, and the distance between them ( d – d’ = d – R 2/ d) tends to zero at d .
2.10. Green’s functions
I have spent so much time/space discussing potential distributions created by a single point charge in various conductor geometries because for any of the geometries, the generalization of these results to the arbitrary distribution (r) of free charges is straightforward. Namely, if a single charge q, located at some point r ’, creates at point r the electrostatic potential 1
(r)
qG(r, '
r ) ,
(2.202)
40
then, due to the linear superposition principle, an arbitrary charge distribution creates the potential Chapter 2
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1
1
Spatial
(r)
q G(r, r )
3
(r ) (r, r )
.
(2.203)
j
j
' G ' d r'
Green’s
4
4
0
j
0
function
The function G(r, r ’) is called the (spatial) Green’s function – the notion very fruitful and hence popular in all fields of physics.68 Evidently, as Eq. (1.35) shows, in the unlimited free space
G r
( , '
r
1
)
,
(2.204)
r '
r
i.e. the Green’s function depends only on one scalar argument – the distance between the field-observation point r and the field-source (charge) point r ’. However, as soon as there are conductors around, the situation changes. In this course, I will only discuss the Green’s functions defined to vanish as soon as the radius-vector r points to the surface ( S) of any conductor:69
G(r, '
r )
.
0
r S
(2.205)
With this definition, it is straightforward to deduce the Green’s functions for the solutions of the last section’s problems in which conductors were grounded, i.e. had potential = 0. For example, for a semi-space z 0 limited by a grounded conducting plane z = 0 (Fig. 26), Eq. (185) yields 1
1
G
,
with "
ρ '
ρ
and
z" z' ,
(2.206)
r '
r
r "
r
where is the 2D radius vector. We see that in the presence of conductors (and, as we will see later, any other polarizable media), the Green’s function may depend not only on the difference r – r ’, but on each of these two arguments in a specific way.
So far, this is just re-naming our old results. The really non-trivial result of the Green’s function formalism in electrostatics is that, somewhat counter-intuitively, the knowledge of this function for a system with grounded conductors (Fig. 30a) enables the calculation of the field created by voltage-biased conductors (Fig. 30b), with the same geometry. To show this, let us use the so-called Green’s theorem of the vector calculus.70 The theorem states that for any two scalar, differentiable functions f(r) and g(r), and any volume V,
f 2 g
g 2 f d 3 r f g g f
d 2 r ,
(2.207)
n
V
S
where S is the surface limiting the volume. Applying the theorem to the electrostatic potential (r) and the Green’s function G (also considered as a function of r), let us use the Poisson equation (1.41) to replace 2 with (-/0), and notice that G, considered as a function of r, obeys the Poisson equation with the -functional source:
2
G(r, '
r ) 4 (r '
r ) .
(2.208)
68 See, e.g., CM Sec. 5.1, QM Secs. 2.2 and 7.4, and SM Sec. 5.5.
69 G so defined is sometimes called the Dirichlet function.
70 See, e.g., MA Eq. (12.3). Actually, this theorem is a ready corollary of the better-known divergence (“Gauss”) theorem, MA Eq. (12.2).
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(Indeed, according to its definition (202), this function may be formally considered as the field of a point charge q = 40.) Now swapping the notation of the radius-vectors, r r ’, and using the Green’s function symmetry, G(r, r ’) = G(r ’, r),71 we get
( ' r)
3
G(r, '
r )
( '
r )
4(r)
G(r, '
r ) d r'
2
( ' r)
G(r, '
r )
d r' .
(2.209)
n'
n'
V
0
S
(a)
(b)
0
1
0
q
q
2
2
2
q
q
q
j
q
j
1
1
Fig. 2.30. Green’s function method allows the solution of a simpler boundary problem (a) to be used for the solution of a more complex problem (b), for the same conductor geometry.
Let us apply this relation to the volume V of free space between the conductors, and the boundary S drawn immediately outside of their surfaces. In this case, by its definition, Green’s function G(r, r ’) vanishes at the conductor surface, i.e. at r S – see Eq. (205). Now changing the sign of ∂ n’ (so that it would be the outer normal for conductors, rather than free space volume V), dividing all terms by 4, and partitioning the total surface S into the parts (numbered by index j) corresponding to different conductors (possibly, kept at different potentials k), we finally arrive at the famous result:72
1
Potential
3
1
G
(r, '
r )
(r)
( '
r G
) (r, '
r ) d r'
via Green’s
d 2 'r .
(2.210)
k
4
4
function
0
k
n'
V
Sk
While the first term on the right-hand side of this relation is a direct and evident expression of the superposition principle, given by Eq. (203), the second term is highly non-trivial: it describes the effect of conductors with arbitrary potentials k (Fig. 30b), using the Green’s function calculated for the similar system with grounded conductors, i.e. with all k = 0 (Fig. 30a). Let me emphasize that since our volume V excludes conductors, the first term on the right-hand side of Eq. (210) includes only the stand-alone charges in the system (in Fig. 30, marked q 1, q 2, etc.), but not the surface charges of the conductors
– which are taken into account, indirectly, by the second term.
In order to illustrate what a powerful tool Eq. (210) is, let us use to calculate the electrostatic field in two systems. In the first of them, a plane, circular, conducting disk of radius R, separated with a very thin cut from the remaining conducting plane, is biased with potential = V, while the rest of the plane is grounded – see Fig. 31.
71 This symmetry, evident for the particular cases (204) and (206), may be readily proved for the general case by applying Eq. (207) to the functions f (r) G(r, r ’) and g(r) G(r, r ” ). With this substitution, the left-hand side of that equality becomes equal to –4[ G(r ” , r ’) – G(r ’, r ” )], while the right-hand side is zero, due to Eq. (205).
72 In some textbooks, the sign before the surface integral is negative, because their authors use the outer normal to the free-space region V rather than those occupied by conductors – as I do.
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z
'
r
r
0
V
0
0
R
"
r
Fig. 2.31. A voltage-biased conducting disk separated from the rest of a conducting plane.
If the width of the gap between the disk and the rest of the plane is negligible, we may apply Eq.
(210) without stand-alone charges, (r ’) = 0, and the Green’s function for the uncut plane – see Eq.
(206).73 In the cylindrical coordinates, with the origin at the disk’s center (Fig. 31), the function is 1
G(r, '
r )
2
2
2
'
2 '
cos( '
) ( z z' ) 1/2
1
.
(2.211)
2
2
2
'
2 ' cos( '
) ( z z' ) 1/2
(The sum of the first three terms under each square root in Eq. (211) is just the squared distance between the horizontal projections and ’ of the vectors r and r ’ (or r ” ) correspondingly, while the last terms are the squares of their vertical displacements.)
Now we can readily calculate the derivative participating in Eq. (210), for z 0: G
G
2 z
.
(2.212)
S
z' 0
n'
z'
2
2
2
'
2 '
cos( '
) z 3/2
Due to the axial symmetry of the system, we may take for zero. With this, Eqs. (210) and (212) yield V
2
R
G(r, r')
2
Vz
'd '
d r'
d '
.
(2.213)
4
n
2
'
2
' 2
2
'
cos '
z 2 3/2
S
0
0
This integral is not overly pleasing, but may be readily worked out at least for points on the symmetry axis ( = 0, z 0):74
2
R
R / 2
'd '
V
z
d
z
Vz
V
(2.214)
3 / 2
2
2
' z
R z
0
2
3 / 2
0
1
1
2 2 ,
1 / 2
This result shows that if z 0, the potential tends to V (as it should), while at z >> R, 2
R
V
.
(2.215)
2
2 z
73 Indeed, if all parts of the cut plane are grounded, a narrow cut does not change the field distribution, and hence the Green’s function, significantly.
74 There is no need to repeat the calculation for z 0: from the symmetry of the problem, (– z) = ( z).
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Now let us use the same Eq. (210) to solve the (in :-)famous problem of the cut sphere (Fig. 32).
Again, if the gap between the two conducting semi-spheres is very thin ( t << R), we may use Green’s function for the grounded (and uncut) sphere. For a particular case r ’ = dn z, this function follows from Eqs. (197)-(198); generalizing the former relation for an arbitrary direction of vector r ’, we get 1
R / r'
G
,
for , , (2.216)
r 2 r' 2 2 rr' cos 1/2 r 2 ( R 2 / r' )2 2 r( R 2 / r' )cos
r r'
R
1/ 2
where is the angle between the vectors r and r ’, and hence r ” – see Fig. 32.
z
R
r
V
/ 2
'
r
"
r
t R
Fig. 2.32. A system of two separated, oppositely
biased semi-spheres.
V
/ 2
Now, calculating the Green’s function’s derivative,
2
2
G
( r R )
r'
,
