Solving an Epic Boundary Value Problem from J.D. Jackson’s "Classical Electrodynamics" – Laplace’s Equation in Polar Coordinates!
2025-07-20
Introduction
Today I present the solution to one of my favorite problems from J.D. Jackson’s "Classical Electrodynamics." The problem reads as follows
Two conducting planes at zero potential meet along the z axis, making an angle \beta between them, as in Fig. 2.14. A unit line charge parallel to the z axis is located between the planes at position (\rho',\phi').
Show that (4\pi\epsilon_0) times the potential in the space between the planes, that is, the Dirichlet Green function G(\rho,\phi;\rho',\phi'), is given by the infinite series
G(\rho,\phi;\rho',\phi')=4\sum_{m=1}^{\infty}\frac{1}{m}\rho_{<}^{m\pi/\beta}\rho_{>}^{-m\pi/\beta}\sin(m\pi\phi/\beta)\sin(m\pi\phi'/\beta)
By means of complex-variable techniques or other means, show that the series can be summed to give a closed form
G(\rho,\phi;\rho',\phi')=\ln(\frac{(\rho)^{2\pi/\beta}+(\rho')^{2\pi/\beta}-2(\rho\rho')^{\pi/\beta}\cos\left[\pi(\phi+\phi')/\beta\right]}{(\rho)^{2\pi/\beta}+(\rho')^{2\pi/\beta}-2(\rho\rho')^{\pi/\beta}\cos\left[\pi(\phi-\phi')/\beta\right]})
While this problem seems intimidating, it can be very elegantly reduced to basic mathematical operations. The solution employs many of my favorite tricks, so I was very pleased to encounter this problem. I hope the reader will find the solution as entertaining. If you want to attempt the problem on your own, please do so now – the next section will contain spoilers :)
Laplace’s Equation in Polar Coordinates
Given the problem description, it is obvious we are solving Laplace’s Equation. The differential form of Gauss’ Law tells us \nabla \cdot \mathbf{E}=\frac{\rho}{\epsilon_0}
But we also know that \mathbf{E}=-\nabla \Phi. Therefore, \nabla \cdot \mathbf{E}=-\nabla \cdot (\nabla \Phi)=-\nabla^2{\Phi}
In regions free of charge (such as the region described in this problem), \rho=0 so Gauss’ law tells us \nabla \cdot \mathbf{E}=0
Therefore,
\nabla^2{\Phi}=0
This is the famed Laplace’s Equation that Jackson readers must worship. In two dimensions, the Laplacian takes the simple form \nabla^2=\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}
The geometry of this problem makes it convenient to use polar coordinates. Through a lengthy derivation, it can be found that in polar coordinates, Laplace’s equation is
\nabla^2{\Phi}=\frac{1}{\rho}\frac{\partial }{\partial \rho}\left(\rho\frac{\partial \Phi}{\partial \rho}\right)+\frac{1}{\rho^2}\frac{\partial^2\Phi}{\partial\phi^2}
The proof is left as an exercise for the morbidly curious reader. If you do it during English class and send proof that you got in trouble for it, I’ll nominate you for the Nobel Prize if I ever get on the committee.
We can use separation of variables to easily solve this differential equation. Assume the solution is of the form \Phi(\rho,\phi)=R(\rho)Q(\phi)
Then
\nabla^2{\Phi}=\frac{Q(\phi)}{\rho}\frac{\partial }{\partial \rho}\left(\rho\frac{dR}{d\rho}\right)+\frac{R(\rho)}{\rho^2}\frac{d^2Q}{d\phi^2}=0
Note that, since \nabla^2{\Phi}=0, we also know that
\nabla^2{\Phi}\left(\frac{\rho^2}{\Phi}\right)=0
We can thus multiply by \frac{\rho^2}{\Phi} on both sides to obtain
\nabla^2{\Phi}=\frac{\rho}{R(\rho)}\frac{\partial }{\partial \rho}\left(\rho\frac{dR}{d\rho}\right)+\frac{1}{Q(\phi)}\frac{d^2Q}{d\phi^2}=0
This has separated the radial and angular dependence of the equation. We can set both equal to a constant and solve the differential equations separately now. Let
\frac{1}{Q(\phi)}\frac{d^2Q}{d\phi^2}=-v^2
Start with the case where v>0
There are two linearly independent solutions to this differential equation, so by superposition, the general solution must be
Q(\phi)=A_ve^{iv\phi}+B_ve^{-iv\phi}
The radial equation is a bit more complicated, but it has the familiar form of an Euler-Cauchy differential equation once we expand the partial derivative.
\frac{\rho}{R(\rho)}\frac{\partial }{\partial \rho}\left(\rho\frac{dR}{d\rho}\right)=\frac{\rho}{R(\rho)}\left(\frac{dR}{d\rho}+\rho\frac{d^2R}{d\rho^2}\right)
We set this equal to the constant v^2 to obtain
\frac{\rho}{R(\rho)}\left(\frac{dR}{d\rho}+\rho\frac{d^2R}{d\rho^2}\right)=v^2
\rho^2\frac{d^2R}{d\rho^2}+\rho\frac{dR}{d\rho}=v^2R(\rho)
Now use the ansatz R(\rho)=C\rho^n
This yields
C\left(\rho^2n(n-1)\rho^{n-2}+\rho n\rho^{n-1}\right)=v^2C\rho^n
n(n-1)\rho^n+n\rho^n=v^2\rho^n
(n(n-1)+n)\rho^n=v^2\rho^n n^2=v^2 n=\pm v
Thus, we find
R(\rho)=a_v\rho^v+b_v\rho^{-v}
For the case where v=0, we find
\frac{d^2Q}{d\rho^2}=0
which means
Q_{v=0}(\phi)=A_0+B_0\phi
And we then find that the radial equation becomes
\rho\frac{d^2R}{d\rho^2}+\frac{dR}{d\rho}=0
The first solution to this is the trivial R_1(\rho)=a_0
If we let u=\frac{dR}{d\rho} and assume u=Cx^n, we can find the second solution
\frac{nu}{x}=-\frac{u}{x} n=-1 \frac{dR}{d\rho}=\frac{C}{x} R_2(\rho)=b_0\ln(\rho)
The complete solution is the superposition of these two,
R_{v=0}(\rho)=a_0+b_0\ln(\rho)
Now remember our solution is of the form \Phi(\rho,\phi)=R(\rho)Q(\phi). We have found infinitely many solutions corresponding to different values of v, so the complete solution must be their linear superposition
\Phi(\rho,\phi)=\sum_{v=0}^{\infty}R_v(\rho)Q_v(\phi) =(a_0+b_0\ln(\rho))(A_0+B_0\phi)+\sum_{v>0}(a_v\rho^v+b_v\rho^{-v})(A_ve^{iv\phi}+B_v^{-iv\phi})
Now we can successively apply boundary conditions to find the coefficients.
Applying boundary conditions to determine the coefficients
The potential must equal zero on the surface of both planes, so we can start with the bottom plane (\phi=0) and find
\Phi(\rho,\phi=0)=0
A_0(a_0+b_0\ln(\rho))+\sum_{v>0}(a_v\rho^v+b_v\rho^{-v})(A_v+B_v)=0
For the first term to equal zero regardless of \rho, it is necessary that A_0=0
For the terms with v>0 to equal zero, we need
A_v+B_v=0
A_v=-B_v
We can now absorb the A_v into the a_v and b_v since (a_v\rho^v+b_v\rho^{-v})(A_ve^{iv\phi}-A_ve^{-iv\phi})=(A_va_v\rho^v+A_vb_v\rho^{-v})(e^{iv\phi}-e^{-iv\phi})
A_va_v and A_vb_v is just another series of constants dependent only on v, so we can make the change a_v\rightarrow A_va_v and b_v\rightarrow A_vb_v
Also note that e^{iv\phi}-e^{-iv\phi}=C\sin(v\phi) – so we can rewrite that part of the equation and ignore the C by again absorbing it into the a_v and b_v constants.
Thus, from using \Phi(\rho,\phi=0)=0, we have found
\Phi(\rho,\phi)=(a_0+b_0\ln(\rho))(B_0\phi)+\sum_{v>0}(a_v\rho^v+b_v\rho^{-v})\sin(v\phi)
Again note that B_0 can be absorbed into a_0 and b_0 to give the slightly simpler form
\Phi(\rho,\phi)=(a_0+b_0\ln(\rho))\phi+\sum_{v>0}(a_v\rho^v+b_v\rho^{-v})\sin(v\phi)
Now we apply the next boundary condition, that \Phi=0 on the top plane. The top plane is defined by \phi=\beta, so we have \Phi(\rho,\phi=\beta)=0
(a_0+b_0\ln(\rho))\beta+\sum_{v>0}(a_v\rho^v+b_v\rho^{-v})\sin(v\beta)=0
For a_0+b_0\ln(\rho)=0 to be true regardless of \rho, we need a_0=b_0=0 since the constants can’t vary with \rho
Similarly, we need \sin(v\beta)=0 for v>0
For this to hold true, it is necessary that
v\beta=m\pi v=\frac{m\pi}{\beta} \text{ for } m=1,2,\ldots
Now we find
\Phi(\rho,\phi)=\sum_{m>0}\left(a_m\rho^{\frac{m\pi}{\beta}}+b_m\rho^{-\frac{m\pi}{\beta}}\right)\sin(\frac{m\pi\phi}{\beta})
Now we employ some mathematically shady trickery. We can arbitrarily divide the region into two subregions with different functions for \Phi. Let us define \Phi_1 for \rho<\rho' and \Phi_2 for \rho>\rho'
When \rho<\rho', we are close to the origin – to have a well-defined solution near \rho=0, we must avoid the undefined behavior of \rho^{-\frac{m\pi}{\beta}}. Therefore, we set b_m=0 for all m>0 in \Phi_1
Similarly, \Phi_2 is defined for arbitrarily large \rho>\rho'. \rho^{\frac{m\pi}{\beta}} will go to infinity as \rho grows, so to avoid undefined behavior, we must have a_m=0 for m>0 in \Phi_2
Thus, we find
\Phi_1(\rho,\phi)=\sum_{m>0}a_m\rho^{\frac{m\pi}{\beta}}\sin(\frac{m\pi\phi}{\beta})
\Phi_2(\rho,\phi)=\sum_{m>0}b_m\rho^{-\frac{m\pi}{\beta}}\sin(\frac{m\pi\phi}{\beta})
For the next boundary condition, we use the formula for the discontinuity of the electric field’s normal component across a boundary
(\mathbf{E}_2-\mathbf{E}_1)\cdot\mathbf{\hat{n}}=\frac{\sigma}{\epsilon_0}
where \sigma is the charge density.
For the uninitiated, this formula can easily be derived by applying Gauss’ Law to an infinitesimally small Gaussian pillbox straddling a boundary.
Note that \mathbf{E}=-\nabla \Phi so
\mathbf{E}\cdot\mathbf{\hat{n}}=-(\nabla \Phi)\cdot\mathbf{\hat{n}}=-\frac{\partial \Phi}{\partial \mathbf{\hat{n}}}
At the boundary defined by \rho=\rho', the unit outward normal is the unit vector in the radial direction, so
\frac{\partial \Phi}{\partial \mathbf{\hat{n}}}\Big|_{\rho=\rho'}=\frac{\partial \Phi}{\partial \rho}\Big|_{\rho=\rho'}
Therefore, we find that
\left(-\frac{\partial \Phi_2}{\partial \rho}+\frac{\partial \Phi_1}{\partial \rho}\right)\Big|_{\rho=\rho'}=\frac{\sigma}{\epsilon_0}
The charge density at \rho=\rho' is
\sigma=\frac{\lambda\delta(\phi-\phi')}{\rho'}
where \lambda is the total charge contained in the line charge. Note that the factor of \frac{1}{\rho'} appears because
\delta(\mathbf{x}-\mathbf{x}')=\frac{1}{\rho'}\delta(\rho-\rho')\delta(\phi-\phi')
This formula arises from the necessity that \int_{\mathbb{R}^2}\delta(\mathbf{x}-\mathbf{x'})\,d\mathbf{x}=1
We can combine all these formulas and evaluate the partial derivatives to find
\sum_{m>0}\left(\frac{a_mm\pi}{\beta}(\rho')^{\frac{m\pi}{\beta}-1}+\frac{b_mm\pi}{\beta}(\rho')^{-\frac{m\pi}{\beta}-1}\right)\sin(\frac{m\pi\phi}{\beta})=\frac{\lambda\delta(\phi-\phi')}{\rho'\epsilon_0}
We can multiply by \frac{\rho'\beta}{\pi} on both sides to get
\sum_{m>0}\left(a_mm(\rho')^{\frac{m\pi}{\beta}}+b_mm(\rho')^{-\frac{m\pi}{\beta}}\right)\sin(\frac{m\pi\phi}{\beta})=\frac{\beta\lambda\delta(\phi-\phi')}{\pi\epsilon_0}
The presence of \delta(\phi-\phi') on the right hand side suggests multiplying by \sin(\frac{n\pi\phi}{\beta}) on both sides and integrating over \phi.
How do we know to do this? Frankly speaking, because I’ve seen this trick used several times in other boundary value problems. It isn’t immediately obvious to a newbie, but if you’ve done enough problems, the idea just naturally pops into your head when it is applicable. This is also one of my favorite tricks in boundary value problems, so that’s probably another reason why it comes to me so quickly.
The integral on the right side is
\frac{\beta\lambda}{\pi\epsilon_0}\int_{0}^{\beta}\sin(\frac{n\pi\phi}{\lambda})\delta(\phi-\phi')\,d\phi=\frac{\beta\lambda}{\pi\epsilon_0}\sin(\frac{n\pi\phi'}{\lambda})
The integral on the left side is
\sum_{m>0}m\left(a_m(\rho')^{\frac{m\pi}{\beta}}+b_m(\rho')^{-\frac{m\pi}{\beta}}\right)\int_{0}^{\beta}\sin(\frac{m\pi\phi}{\beta})\sin(\frac{n\pi\phi}{\beta})\,d\phi
The observant reader will immediately see this integral can be solved by recalling the orthogonality of sine functions. For readers who don’t know what that means, try solving the integral on your own and isolating the cases where n=m and n\ne m – the orthogonality is easily derived.
I omit the calculation because it’s a standard result. We find that the result is
\sum_{m>0}\frac{m\beta}{2}\left(a_m(\rho')^{\frac{m\pi}{\beta}}+b_m(\rho')^{-\frac{m\pi}{\beta}}\right)\delta_{mn}=\frac{n\beta}{2}\left(a_n(\rho')^{\frac{n\pi}{\beta}}+b_n(\rho')^{-\frac{n\pi}{\beta}}\right)
Thus, equating the left and right sides again,
\left(a_n(\rho')^{\frac{n\pi}{\beta}}+b_n(\rho')^{-\frac{n\pi}{\beta}}\right)=\frac{2\lambda}{n\pi\epsilon_0}\sin(\frac{n\pi\phi'}{\beta})
We can change n to m to make this result mesh better with the rest of our solution.
\left(a_m(\rho')^{\frac{m\pi}{\beta}}+b_m(\rho')^{-\frac{m\pi}{\beta}}\right)=\frac{2\lambda}{m\pi\epsilon_0}\sin(\frac{m\pi\phi'}{\beta})
This result seems useless, but it becomes incredibly powerful when combined with the result of applying another boundary condition. We next consider the continuity of the tangential component of the electric field across a boundary, i.e.
\mathbf{E}_2\times\mathbf{\hat{n}}=\mathbf{E}_2\times\mathbf{\hat{n}}
Again, this is a standard result of electrodynamics. For the uninitiated, this result can be derived by considering the integral form of Faraday’s Law applied to an infinitesimally small rectangular loop around a boundary.
Using properties of the cross product or a simple geometric argument (the details are omitted because it can be done mentally), it can be seen that the above formula is equivalent to
\frac{\partial \Phi_1}{\partial \phi}\Big|_{\rho=\rho'}=\frac{\partial \Phi_2}{\partial \phi}\Big|_{\rho=\rho'}
Evaluating these partial derivatives yields
\sum_{m>0}\frac{a_mm\pi}{\beta}(\rho')^{\frac{m\pi}{\beta}}\cos(\frac{m\pi\phi}{\beta})=\sum_{m>0}\frac{b_mm\pi}{\beta}(\rho')^{-\frac{m\pi}{\beta}}\cos(\frac{m\pi\phi}{\beta})
For these series to equal each other independent of \phi, we need
a_m(\rho')^{\frac{m\pi}{\beta}}=b_m(\rho')^{-\frac{m\pi}{\beta}}
b_m=a_m(\rho')^{\frac{2m\pi}{\beta}}
Incredible progress! Now go back to our result from the earlier boundary condition and plug this new result in:
\left(a_m(\rho')^{\frac{m\pi}{\beta}}+b_m(\rho')^{-\frac{m\pi}{\beta}}\right)=\frac{2\lambda}{m\pi\epsilon_0}\sin(\frac{m\pi\phi'}{\beta})
2a_m(\rho')^{\frac{m\pi}{\beta}}=\frac{2\lambda}{m\pi\epsilon_0}\sin(\frac{m\pi\phi'}{\beta})
a_m=(\rho')^{-\frac{m\pi}{\beta}}\frac{\lambda}{m\pi\epsilon_0}\sin(\frac{m\pi\phi'}{\beta})
This gives us formulas for a_m and b_m! Now we can plug them into our formula for \Phi and find \begin{aligned} & \Phi(\rho,\phi;\rho',\phi') =\\ &\frac{\lambda}{\pi\epsilon_0}\sum_{m>0}\frac{1}{m}\sin(\frac{m\pi\phi'}{\beta})\sin(\frac{m\pi\phi}{\beta})\cdot\begin{cases} (\rho')^{-\frac{m\pi}{\beta}}\rho^{\frac{m\pi}{\beta}} & \text{ for } \rho<\rho' \\ (\rho')^{\frac{m\pi}{\beta}}\rho^{-\frac{m\pi}{\beta}} & \text{ for } \rho>\rho' \end{cases} \end{aligned}
Now if we let \rho_{<}=\text{min}(\rho,\rho') and \rho_{>}=\text{max}(\rho,\rho'), we have
\Phi(\rho,\phi;\rho',\phi') = \frac{\lambda}{\pi\epsilon_0}\sum_{m>0}\frac{1}{m} \rho_{<}^{\frac{m\pi}{\beta}}\rho_{>}^{-\frac{m\pi}{\beta}}\sin(\frac{m\pi\phi'}{\beta})\sin(\frac{m\pi\phi}{\beta})
Now if we let \lambda=1 (the total charge in the line charge) and multiply by 4\pi\epsilon_0, we get the Green function for this problem
G(\rho,\phi;\rho',\phi')=4\sum_{m=1}^{\infty}\frac{1}{m}\rho_{<}^{m\pi/\beta}\rho_{>}^{-m\pi/\beta}\sin(m\pi\phi/\beta)\sin(m\pi\phi'/\beta)
Voila, part a of the problem is solved! I found this solution incredible, and I hope you do too. We used a staggering amount of techniques to reduce a complicated problem to an elegant series solution – that, to my mind, is the beauty of boundary value problems.
Now we solve part b of the problem and find a closed form through even more elegant manipulations!
Finding a closed form – wizardry with complex numbers
Let’s start by rewriting the sine product as a sum of cosines, i.e.
\sin(\frac{m\pi\phi'}{\beta})\sin(\frac{m\pi\phi}{\beta})=\frac{1}{2}\left(\cos(\frac{m\pi}{\beta}(\phi-\phi'))-\cos(\frac{m\pi}{\beta}(\phi+\phi'))\right)
=\frac{1}{2}\left(\Re(e^{\frac{im\pi}{\beta}(\phi-\phi')})-\Re(e^{\frac{im\pi}{\beta}(\phi+\phi')})\right)
G=2\Re\left(\sum_{m=1}^{\infty}\frac{1}{m}\left(\frac{\rho_{<}}{\rho_{>}}\right)^{\frac{m\pi}{\beta}}e^{\frac{im\pi}{\beta}(\phi-\phi')}-\sum_{m=1}^{\infty}\frac{1}{m}\left(\frac{\rho_{<}}{\rho_{>}}\right)^{\frac{m\pi}{\beta}}e^{\frac{im\pi}{\beta}(\phi+\phi')}\right)
Now define
z_1=\left(e^{i(\phi-\phi')}\frac{\rho_{<}}{\rho_{>}}\right)^{\frac{\pi}{\beta}}
z_2=\left(e^{i(\phi+\phi')}\frac{\rho_{<}}{\rho_{>}}\right)^{\frac{\pi}{\beta}}
This allows us to rewrite the series in the cleaner form
G=2\Re\left(\sum_{m=1}^{\infty}\frac{z_1^m}{m}-\sum_{m=1}^{\infty}\frac{z_2^m}{m}\right)
This is another point where a good memory will help solve the problem. Recall the Taylor series expansion for \ln(1+x)
\ln(1+x)=\sum_{m>0}(-1)^{m+1}\frac{x^m}{m}=x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots
Plugging in -x gives
\ln(1-x)=-\sum_{m>0}\frac{x^m}{m}=-x-\frac{x^2}{2}-\frac{x^3}{3}-\cdots
Now it becomes clear that
-\ln(1-x)=\sum_{m>0}\frac{x^m}{m}
Thus,
G=2\Re\left(-\ln(1-z_1)+\ln(1-z_2)\right)=2\Re\left(\ln(\frac{1-z_2}{1-z_1})\right)
Excellent! We are close to the structure desired by the problem’s description. We have the logarithm of a quotient, but the tricky part is taking the real part.
Consider the real part of such a logarithm:
\Re(\ln(1-z))=\Re\biggl[\ln(|1-z|)+i\mathop{\mathrm{Arg}}(1-z)\biggr] =\ln(|1-z|)=\ln(\sqrt{(\Re(1-z))^2+(\Im(1-z))^2}) =\frac{1}{2}\ln\left[\left(1-|z|\cos(\mathop{\mathrm{Arg}}z)\right)^2+\left(|z|\sin(\mathop{\mathrm{Arg}}z)\right)^2\right] =\frac{1}{2}\ln\left[1+|z|^2-2\Re(z)\right]
Great, this is a clean way of expanding the real part of our logarithm!
Note that
|z_1|=\left(\frac{\rho_{<}}{\rho_{>}}\right)^{\frac{\pi}{\beta}}
|z_2|=\left(\frac{\rho_{<}}{\rho_{>}}\right)^{\frac{\pi}{\beta}}
\Re(z_1)=\left(\frac{\rho_{<}}{\rho_{>}}\right)^{\frac{\pi}{\beta}}\cos(\frac{\pi}{\beta}(\phi-\phi'))
\Re(z_2)=\left(\frac{\rho_{<}}{\rho_{>}}\right)^{\frac{\pi}{\beta}}\cos(\frac{\pi}{\beta}(\phi+\phi'))
Then we have
G=\ln\left(\frac{1+|z_2|^2-2\Re(z_2)}{1+|z_1|^2-2\Re(z_1)}\right) =\ln\left(\frac{1+\left(\frac{\rho_{<}}{\rho_{>}}\right)^{2\pi/\beta}-2\left(\frac{\rho_{<}}{\rho_{>}}\right)^{\pi/\beta}\cos\left(\frac{\pi}{\beta}(\phi+\phi')\right)}{ 1+\left(\frac{\rho_{<}}{\rho_{>}}\right)^{2\pi/\beta}-2\left(\frac{\rho_{<}}{\rho_{>}}\right)^{\pi/\beta}\cos\left(\frac{\pi}{\beta}(\phi-\phi')\right) }\right) =\ln\left(\frac{(\rho_{>})^{2\pi/\beta}+\left(\rho_{<}\right)^{2\pi/\beta}-2\left(\rho_{<}\rho_{>}\right)^{\pi/\beta}\cos\left(\frac{\pi}{\beta}(\phi+\phi')\right)}{ (\rho_{>})^{2\pi/\beta}+\left(\rho_{<}\right)^{2\pi/\beta}-2\left(\rho_{<}\rho_{>}\right)^{\pi/\beta}\cos\left(\frac{\pi}{\beta}(\phi-\phi')\right) }\right)
Now note that, since \rho_{>} and \rho_{<} have the same exponent everywhere, we can just replace them with \rho and \rho'.
G(\rho,\phi;\rho',\phi')=\ln\left(\frac{(\rho)^{2\pi/\beta}+\left(\rho'\right)^{2\pi/\beta}-2\left(\rho\rho'\right)^{\pi/\beta}\cos\left(\frac{\pi}{\beta}(\phi+\phi')\right)}{ (\rho)^{2\pi/\beta}+\left(\rho'\right)^{2\pi/\beta}-2\left(\rho\rho'\right)^{\pi/\beta}\cos\left(\frac{\pi}{\beta}(\phi-\phi')\right) }\right)
Behold! The epic closed form solution to the problem. My mind was blown away by how such a complicated series could be reduced to a closed form. The final solution looks cool – which, let’s be honest, is all we care about – but is also surprisingly well-behaved. There’s a singularity at (\rho',\phi') but, other than that, the function is everywhere finite and continuous for 0\le\phi\le\beta and 0\le\rho<\infty
What is especially surprising is continuity, since our series solution was a piecewise function. It’s incredible that such a gnarly series produced such an elegant closed form.
Conclusion
I hope the reader has found this problem as entertaining as I did. As always, if you have any questions or thoughts about this post, feel free to contact me on Discord or via email. I have a few other Jackson-related posts I plan to write soon, but it takes a long time to write LaTeX, so the post schedule is chaotic.
I tried to be thorough with my explanations at the start of this post, but after a few hours of writing, my patience likely wore out. Apologies if some parts of this post are less understandable than others :) Please message me to ask about anything that’s unclear or confusing.
Also, if there are any other people reading Jackson or people who would want to talk about the book with me – please contact me!! I’m tired of getting stuck on one equation for an hour, only to realize there’s a typo or something similarly stupid. It’s easier to read this book in a group than to read it alone, and I’d love to work with another devotee to the holy church of Jackson. If I was doing this in college, I’d have Jackson buddies, but instead I’m hoping someone on the internet reads this haha.
Thanks for reading, please contact me if this is interesting (I’m lonely) !!!