Applications of conformal transformations

APPLICATIONS OF COMPLEX VARIABLES
A conducting circular cylinder of radius a is placed with its centre line passing through
the origin and perpendicular to a uniform electric field E in the x-direction. Find the charge
per unit length induced on the half of the cylinder that lies in the region x < 0.
As mentioned immediately following the previous example, the appropriate complex
potential for this problem is f(z) = −E(z − a2 /z). Writing z = r exp iθ this becomes
a2
f(z) = −E r exp iθ −
exp(−iθ)
r
a2
a2
cos θ − iE r +
sin θ,
= −E r −
r
r
so that on r = a the imaginary part of f is given by
ψ = −2Ea sin θ.
Therefore the induced charge q per unit length on the left half of the cylinder, between
θ = π/2 and θ = 3π/2, is given by
q = 20 Ea[sin(3π/2) − sin(π/2)] = −40 Ea. 25.2 Applications of conformal transformations
In section 24.7 of the previous chapter it was shown that, under a conformal
transformation w = g(z) from z = x + iy to a new variable w = r + is, if a solution
of Laplace’s equation in some region R of the xy-plane can be found as the real
or imaginary part of an analytic function§ of z, then the same expression put in
terms of r and s will be a solution of Laplace’s equation in the corresponding
region R of the w-plane, and vice versa. In addition, if the solution is constant
over the boundary C of the region R in the xy-plane, then the solution in the
w-plane will take the same constant value over the corresponding curve C that
bounds R .
Thus, from any two-dimensional solution of Laplace’s equation for a particular
geometry, typified by those discussed in the previous section, further solutions for
other geometries can be obtained by making conformal transformations. From
the physical point of view the given geometry is usually complicated, and so
the solution is sought by transforming to a simpler one. However, working from
simpler to more complicated situations can provide useful experience and make
it more likely that the reverse procedure can be tackled successfully.
§
In fact, the original solution in the xy-plane need not be given explicitly as the real or imaginary
part of an analytic function. Any solution of ∇2 φ = 0 in the xy-plane is carried over into another
solution of ∇2 φ = 0 in the new variables by a conformal transformation, and vice versa.
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25.2 APPLICATIONS OF CONFORMAL TRANSFORMATIONS
s
y
s
r
x
(a) z-plane
(b) w-plane
r
(c) w-plane
Figure 25.3 The equipotential lines (broken) and field lines (solid) (a) for an
infinite charged conducting plane at y = 0, where z = x + iy, and after the
transformations (b) w = z 2 and (c) w = z 1/2 of the situation shown in (a).
Find the complex electrostatic potential associated with an infinite charged conducting
plate y = 0, and thus obtain those associated with
(i) a semi-infinite charged conducting plate (r > 0, s = 0);
(ii) the inside of a right-angled charged conducting wedge (r > 0, s = 0 and
r = 0, s > 0).
Figure 25.3(a) shows the equipotentials (broken lines) and field lines (solid lines) for the
infinite charged conducting plane y = 0. Suppose that we elect to make the real part of
the complex potential coincide with the conventional electrostatic potential. If the plate is
charged to a potential V then clearly
φ(x, y) = V − ky,
(25.6)
where k is related to the charge density σ by k = σ/0 , since physically the electric field E
has components (0, σ/0 ) and E = −∇φ.
Thus what is needed is an analytic function of z, of which the real part is V − ky. This
can be obtained by inspection, but we may proceed formally and use the Cauchy–Riemann
relations to obtain the imaginary part ψ(x, y) as follows:
∂ψ
∂φ
=
=0
∂y
∂x
and
∂ψ
∂φ
=−
= k.
∂x
∂y
Hence ψ = kx + c and, absorbing c into V , the required complex potential is
f(z) = V − ky + ikx = V + ikz.
(25.7)
(i) Now consider the transformation
w = g(z) = z 2 .
(25.8)
This satisfies the criteria for a conformal mapping (except at z = 0) and carries the upper
half of the z-plane into the entire w-plane; the equipotential plane y = 0 goes into the
half-plane r > 0, s = 0.
By the general results proved, f(z), when expressed in terms of r and s, will give a
complex potential whose real part will be constant on the half-plane in question; we
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APPLICATIONS OF COMPLEX VARIABLES
deduce that
F(w) = f(z) = V + ikz = V + ikw 1/2
2
(25.9)
2 1/2
is the required potential. Expressed in terms of r, s and ρ = (r + s )
1/2
1/2 ρ+r
ρ−r
+i
w 1/2 = ρ1/2
,
2ρ
2ρ
and, in particular, the electrostatic potential is given by
1/2
k .
Φ(r, s) = Re F(w) = V − √ (r2 + s2 )1/2 − r
2
,w
1/2
is given by
(25.10)
(25.11)
The corresponding equipotentials and field lines are shown in figure 25.3(b). Using results
(25.3)–(25.5), the magnitude of the electric field is
|E| = |F (w)| = | 12 ikw −1/2 | = 12 k(r2 + s2 )−1/4 .
(ii) A transformation ‘converse’ to that used in (i),
w = g(z) = z 1/2 ,
has the effect of mapping the upper half of the z-plane into the first quadrant of the
w-plane and the conducting plane y = 0 into the wedge r > 0, s = 0 and r = 0, s > 0.
The complex potential now becomes
F(w) = V + ikw 2
= V + ik[(r 2 − s2 ) + 2irs],
(25.12)
showing that the electrostatic potential is V −2krs and that the electric field has components
E = (2ks, 2kr).
(25.13)
Figure 25.3(c) indicates the approximate equipotentials and field lines. (Note that, in both
transformations, g (z) is either 0 or ∞ at the origin, and so neither transformation is
conformal there. Consequently there is no violation of result (ii), given at the start of
section 24.7, concerning the angles between intersecting lines.) The method of images, discussed in section 21.5, can be used in conjunction with
conformal transformations to solve some problems involving Laplace’s equation
in two dimensions.
A wedge of angle π/α with its vertex at z = 0 is formed by two semi-infinite conducting
plates, as shown in figure 25.4(a). A line charge of strength q per unit length is positioned
at z = z0 , perpendicular to the z-plane. By considering the transformation w = z α , find the
complex electrostatic potential for this situation.
Let us consider the action of the transformation w = z α on the lines defining the positions
of the conducting plates. The plate that lies along the positive x-axis is mapped onto the
positive r-axis in the w-plane, whereas the plate that lies along the direction exp(iπ/α) is
mapped into the negative r-axis, as shown in figure 25.4(b). Similarly the line charge at z0
is mapped onto the point w0 = z0α .
From figure 25.4(b), we see that in the w-plane the problem can be solved by introducing
a second line charge of opposite sign at the point w0∗ , so that the potential Φ = 0 along
the r-axis. The complex potential for such an arrangement is simply
q
q
ln(w − w0 ) +
ln(w − w0∗ ).
F(w) = −
2π0
2π0
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