Siméon Denis Poisson, 1781 to 1840 |
Pierre-Simon Laplace, 1749 to 1827 |
In this chapter we'll be working in cylindrical coordinates. We'll look for solutions to Laplace's equation. Later we'll apply boundary conditions to find specific solutions. Traditionally, ρ is used for the radius variable in cylindrical coordinates, but in electrodynamics we use ρ for the charge density, so we'll use s for the radius. φ will be the angular dimension, and z the third dimension.
s lies in the x-y plane, an angle φ up from the x axis. The relevant equations are:
x = s cos φ
y = s sin φ z = z | s = √ (x^{2} + y^{2})
φ = arctan(y/x) z = z |
Laplace's equation for the potential in the absence of sources is:
Ñ^{2}Φ = 0
Ñ^{2} in cylindrical coordinates is:
1 | ∂ | ∂ | 1 | ∂^{2} | ∂^{2} | |||||
Ñ^{2} | = | s | + | + | ||||||
s | ∂s | ∂s | s^{2} | ∂φ^{2} | ∂z^{2} |
First we'll solve a pseudo two dimensional problem. We'll presume the given geometry extends infinitely in the z direction, so that the solution does not depend on z. We'll also presume that our solution is separable, that is Φ( s, φ, z ) = S(s)Q(φ). We can check the separability assumption later. Laplace's equation is a second order differential equation, so if we find two unique solutions we've done it. Generally speaking, if the boundary conditions are separable, there's a good chance the solution is separable. If the boundary conditions are not separable, most likely we're hosed. Now,
Ñ^{2}Φ(s,φ,z) = Ñ^{2}S(s)Q(φ).
Q is independent of s, so it can be pulled through the s partial derivatives. Similarly S is independent of φ, so it can be pulled through the φ partial derivatives. Φ is independent of z, so the z partial derivatives are 0.
Q | ∂ | ∂ | S | ∂^{2} | ||||||
Ñ^{2}SQ | = | s | S + | Q | = 0 | |||||
s | ∂s | ∂s | s^{2} | ∂φ^{2} |
Now multiply the entire equation by s^{2} / SQ:
s | ∂ | ∂ | 1 | ∂^{2} | ||||||
Ñ^{2}SQ | = | s | S + | Q | = 0 | |||||
S | ∂s | ∂s | Q | ∂φ^{2} |
Since there's an expression that depends only on s and another that depends only on φ, and since the sum is zero for all s and φ, it must be that the individual expressions are constants. We'll start by setting them to 0.
s | ∂ | ∂ | ∂ | ∂ | ∂S | ||||||||
s | S | = 0 = | s | S | Þ | = | |||||||
S | ∂s | ∂s | ∂s | ∂s | ∂s | s |
1 | ∂^{2} | ∂^{2}Q | ||
Q | = 0 = | |||
Q | ∂φ^{2} | ∂φ^{2} |
We can solve these immediately. S = a_{0} + b_{0} ln s. Q = A_{0} + B_{0} φ.
Another possibility is that the expression for S is equal to some constant, and the expression for Q is equal to the negative of that same constant. Since these are second order equations, we'll anticipate and set the constant to η^{2}. Now,
s | ∂ | ∂ | ∂ | ∂ | ||||||
s | S | = η^{2} Þ s | s | S | = η^{2} S | |||||
S | ∂s | ∂s | ∂s | ∂s |
∂^{2} | ||
Q | = -η^{2}Q | |
∂φ^{2} |
We can solve the second equation immediately. Q = A_{η} cos ηφ + B_{η} sin ηφ.
For the equation in S, we'll try a polynomial s^{α}. Immediately we see that the solutions to this equation are S = a_{η}s^{η} + b _{η}s^{-η}.
We've arbitrarily put the plus sign on the S expression, and the minus sign on the Q expression. Let's see what happens if we reverse this.
s | ∂ | ∂ | ∂ | ∂ | ||||||
s | S | = -η^{2} Þ s | s | S | = -η^{2} S | |||||
S | ∂s | ∂s | ∂s | ∂s |
∂^{2} | ||
Q | = η^{2}Q | |
∂φ^{2} |
Again, we can solve the Q equation immediately - this is just e^{ηφ} and e ^{-ηφ}. We'll choose to represent the solutions as Q = A_{η } cosh ηφ + B_{η} sinh ηφ.
The S equation is a bit trickier. Again, we'll try a polynomial s^{α}. Immediately we see that the solutions to this equation are S = a_{η}s^{iη } + b_{η}s^{-iη}. This is not very pleasant, so we'll work on these solutions a bit. s^{iη} = e ^{iη ln s}. This suggests, perhaps "obviously" in retrospect, that we should have tried cos( η ln s ). So, our solutions to the equation will be
S = a_{η} cos( η ln s ) + b_{η} sin( η ln s ).
These solutions oscillate - unlike the solutions for S above, which are strictly non-negative, these solutions are both positive and negative. Both the Q and the S solutions are either symmetric or anti-symmetric, so we can assign these solutions to negative values of η and any minus sign will be absorbed into the coefficients. Similarly, the other solutions we got were either symmetric or anti-symmetric, so we can assign those solutions to the positive values of η.
Now, our complete solution to the 2d + z Laplace equation is:
-1 | ∞ | |||||||
Φ | = | Σ | (a_{η} cos( η ln s ) + b_{η} sin( η ln s )) | (A_{η} cosh ηφ + B_{η} sinh ηφ ) + | Σ | ( a_{η}s^{η} + b_{η}s^{-η} ) | ( A_{η} cos ηφ + B_{η} sin ηφ ) + | (a_{0} + b_{0} ln s) (A_{0} + B_{0}φ) |
-∞ | 1 |
Suppose we have conductor filling all of space except for a wedge cut out of it. OK, this is a pretty artificial problem. Unfortunately, artificial problems are pretty much what we can actually solve. Importantly, if you had a piece of metal with a wedge cut out of it, this solution would be quite accurate as long as you were reasonable far from the edges.
Figure 5.1: A boundary value problem
Suppose the metal is at a potential V. Now, we have three boundary conditions:
Condition 1 tells us:
a_{η} = b_{η} = 0 for η < 0, as the sine and cosine functions are not well behaved as their arguments go to minus infinity.
b_{η} = 0 for η > 0, as otherwise we have a rather embarrassing divide by zero.
b_{0} = 0 or we have another embarrassing infinity
B_{0} = because the constant term must be independent of φ.
a_{0} A_{0} = V
Now our remaining solution is:
∞ | |||||
Φ | = | Σ | ( a_{η}s^{η} ) | ( A_{η} cos ηφ + B_{η} sin ηφ ) + | V |
1 |
Condition 2 tells us:
A_{η} = 0, otherwise the voltage on the lower face will be a function of s. We could have set a_{η} to 0 instead, but this takes all the fun out of the solution.
Condition 3 tells us:
ηβ = nπ, n integer.
Now, our complete solution is:
∞ | ||||
Φ | = V + | Σ | a_{n}s^{nπ/β} | sin( nπφ/β ) |
1 |
Now let's have a closer look at this solution. If s is very small, then we can ignore all the terms with n > 1. The electric field is E = -ÑΦ, which is
E = -a_{1} s^{(π/β-1)} ( sin πφ/β s + cos πφ/β φ )
As φ goes to 0, E goes to -a_{1} s^{(<π/β-1)} = 4πσ.
Immediately we see that if β < π, E -> 0; if β > pi;, E -> ∞. This is the well known result that the electric field and charge density go to zero at an inside corner, and they go to infinity at a point. This is why lightning rods work: the lightning is attracted to the very high field at a point on an object. We should not be too concerned by the infinity - after all, there's no such thing as an infinitely sharp point.