Static Potentials in Vacuum – Laplace’s Equation
5.1 Laplace’s Equation
5.2.5 Schwarz-Christoffel Transformations
terms of x andy. Instead, we separate the equation forz into real and imaginary components to get
x= a
2π (1 +u+eucosv) (5–57)
and y= a
2π(v+eusinv) (5–58)
which we can evaluate parametrically by varyingu to obtain the constantv (con- sequently also the constant V =V0v/2π) curves. Similarly, the electric field lines are obtained in thex-y plane by varyingv. The field and potential lines are plotted in Figure 5.12.
Figure 5.12:The equipotential and field lines near the edge of a semi-infinite plate capacitor.
Verifying that a given mapping does indeed map a particular surface onto an- other is fairly straightforward. It is not, however, obvious how mappings for given boundaries are to be obtained, other than perhaps by trial and error. For polygonal boundaries, a general method of constructing the mapping is offered by Schwarz- Christoffel transformations. For nonpolygonal boundaries, one must rely on “dic- tionaries” of mappings.15
To begin, we note that a polygon is a figure made up of line segments, each of which has a constant slope, terminated by a vertex where the slope suddenly changes to a new value. We therefore consider a differential equation (an equation for the slope) relatingf andz. In particular, consider the differential equation
dz df =A
1n i=1
(f−ui)βi (5–59)
where the ui are points on the real axis in thef plane. We see immediately that conformality fails at every point f = ui. The real axis of the f plane will be mapped in continuous segments between these points, to distinct line segments in thez plane. Recall that a complex number may be represented in polar form,Reiθ. Theargumentof the number,θ, gives the inclination of a line joining the origin to z. The angle of the segments in z plane may be determined from the argument of the derivative (5–59) above. Using arg(f−ui)βi = arg(Reiθ)βi = arg(Rβieiθβi) = βiarg(eiθ) =βiarg(f −ui), we may write
arg dz
df
= argA+β1arg(f−u1) +β2arg(f−u2) +· · ·+βnarg(f−un) (5–60) Since we were interested in mappings of the real axis, we restrictf to the real axis.
Whenu < u1,f−ui<0, implying each of the arguments equalsπ. Asu increases to greater thanu1, the argument of (f−u1) vanishes. Further increases will make more and more of the terms vanish asu passes each of the rootsui. Withf on the real axisdf=duso that
arg dz
df
= arg
dx+idy du
= arg (dx+idy) = tan−1 dy
dx
(5–61) In other words, the argument of the derivative is just the slope of the line segment atx. We abbreviate this slope in the interval u∈(ui, ui+1) asθi.
Let us evaluate this angle whenulies betweenuiandui+1. uis larger than every point to the left, meaning the argument is zero and it is smaller than every point on the right, meaning the argument for all factors of the product with subscript> i isπ. Therefore
θi= argA+π(βi+1+βi+2+· · ·+βn) (5–62) We conclude that the image of the segment lying between ui and ui+1 in the f plane is mapped to a segment in the z plane inclined at angle θi to thex axis. If we subtract θi from θi+1 to find the difference between two successive segments’
slopes, we obtain the bend at the point where the two segments meet,
θi+1−θi=−πβi+1 (5–63)
We can relate this change in slope to the interior angle,αi+1of the polygon at the point mapped fromui+1 as indicated in Figure 5.13.
αi+1=π+πβi+1 (5–64)
Figure 5.13:The angleαi+1is complementary to−πβi+1.
which we solve forβi+1to give
βi+1=αi+1
π −1 (5–65)
In terms of the interior angles, (5–59) becomes dz
df =A 1n i=1
(f −ui)αi/π−1 (5–66)
The argument of the (complex) factor A will rotate the whole figure and the magnitude will scale it. The equation (5–66) is useful provided it can be integrated in terms of elementary functions which is frequently not possible. We will illustrate the procedure with several examples.
Example 5.11: Use the Schwarz-Christoffel transformation to produce the 90◦ bent plane mapping of page 117.
Solution: We wish to produce the mapping that maps the real axis of thef plane to +y axis for u <0 and the +x axis foru >0. This means we want to produce a bend ofπ/2 atu= 0. The mapping is according to (5–66)
dz
df =A(f −0)1/2−1=Af−1/2 (Ex 5.11.1) which is readily integrated to give
z=Af1/2+k (Ex 5.11.2)
Choosingk= 0 and A real gives the required mapping.
Example 5.12: Find the mapping that maps the real axis onto the real axis and a line parallel to the real axis a distancea above it.
Solution: We choose as before, the point zero on the u-axis to produce the singular point, and this time as we want parallel lines we chooseα= 0. (5–66) then becomes
dz
df =A(f −0)0−1=Af−1 (Ex 5.12.1)
which we integrate to obtain
z=Alnf+k (Ex 5.12.2)
Again we can eliminatek without loss (it merely identifiesz=−∞withf = 0). For positiveu, we havez=Aln|u| whereas for negativeu, we havez=A(ln|u|+iπ).
Thus the negativeu axis maps to a line Aπabove the x axis. ChoosingA =a/π gives the required mapping.
Example 5.13: Find the mapping that carries the real axis onto the real axis with a gap of width 2a straddling the origin. (This would be a slotted plane in three dimensions.)
Figure 5.14: The plane with a slot can be formed by cutting the infinite plane at the center and folding the edges back at±b.
Solution: It is help to imagine how the real axis might be bent and cut to achieve this mapping. Figure 15.14 illustrates the operation. The real axis is cut atu= 0.
At u = ±b, the line is folded back, meaning the image of the segment of the u axis straddling zero has a ±π ‘bend’ while the image of ±b has a 2π bend. The prescription for the mapping then gives us
dz
df =A(f+b)(f−b)(f−0)−2=Af2−b2
f2 (Ex 5.13.1)
Integrating (Ex 13.1) gives
z=A
f+b2 f
+B (Ex 5.13.2)
Whenf =±bwe wishz=±awhich givesB= 0 andA=a/2bso that the required mapping is
z= a 2
f b + b
f
(Ex 5.13.3) As a final example we will obtain the mapping of the top half of the semi-finite parallel plate capacitor.
Example 5.14: Obtain a transformation to map the real axis onto the real axis and z=idforx <0.
Solution: For this mapping we need two vertices (the vertices at±∞are irrelevant), one to split the line and one to fold the upper ‘electrode’ back onto itself in order to have it terminate. Let the image of the vertex producing the 2πfold lie ataand the image of displacement vertex with 0 vertex angle atb on the real axis in the f plane. The mapping then obeys
dz
df =A(f−a)(f−b)−1 (Ex 5.14.1) Although we have three constants to determine, there is only one significant con- stant in the geometry, namelyd. Therefore any two of the constants may be given convenient values with the third left to fit the correct boundary. For convenience then, we chooseb= 0 and a=−1, resulting in
dz df =A
f+ 1 f
(Ex 5.14.2) which, when integrated gives
z=A(f+ lnf) +B (Ex 5.14.3)
For large positive real f, z is real so that we conclude B is real. For negative real f, z=A(u+iπ+ ln|u|) +B so that we conclude Aπ =d. Letting our fold point u=−1 correspond tox= 0 we obtainB=d/π. The required mapping is then
z= d
π(1 +f+ lnf) (Ex 5.14.4)
Before leaving the example above entirely, we note that the mapping used earlier mapped the semi-infinite plate not from the real axis but from a line above the real axis. Substituting for f the results from example 5.13 will complete the transfor- mation. In particular, letting lnf =wwith w=u+iv we can recast (Ex 5.14.4) as
z= d
π(1 +w+ew)
which mapsw=u+iπonto the top ‘plate’ andw=uonto the x axis. Cascading maps in this fashion is frequently useful.