OPhO 2023, Теория — Задача 2. T2: The Complex Potential

• A Conformal Transformed Electrostatic System, (d) $\frac{df}{dz} = -E_x + iE_y$

• A Conformal Transformed Electrostatic System, (g) $\phi_c(x,y) = \frac{\phi_0}{d}y$

• A Conformal Transformed Electrostatic System, (i) $\phi_I(r) = -\frac{\lambda}{2\pi\epsilon_0}\ln(r/R)$

• A Conformal Transformed Electrostatic System, (j) $\phi(x,y) = \frac{\lambda}{2\pi\epsilon_0}\ln\sqrt{\frac{1+\frac{2}{h}y+\frac{x^2+y^2}{h^2}}{1-\frac{2}{h}y+\frac{x^2+y^2}{h^2}}}$

• A Conformal Transformed Electrostatic System, (k) $|f'(\gamma(t_0))|$

• On the Electrostatic Interaction between a Line Charge and a Conducting Wedge, (n) $\vec F = -k\frac{Q^2}{L}\left[\hat r + \frac{\pi}{\theta}\cot\left(\frac{\pi}{\theta}\alpha\right)\hat\varphi\right],\quad |\vec F| = \frac{\sqrt{13}}{2}\frac{kQ^2}{L} \approx 1.8028\,\frac{kQ^2}{L}$

  Magnitude for θ = 120°, α = 30°.

Solution 2:

Part A

Has hinted in the problem, we define differentiability of complex functions just as we would in single variable calculus on the reals. A function $f : \mathbb{C} \to \mathbb{C}$, is complex-differentiable or holomorphic at a point $z_0 \in \mathbb{C}$ if the following limit converges to a number $f'(z_0) \in \mathbb{C}$:

We call the number $f'(z_0)$ the derivative of $f$ at $z_0$. Note that the limit $h \to 0$ is taken by any sequence of points in the complex plane that converges to the origin. As in multivariable calculus on $\mathbb{R}^2$, $h$ can approach the origin in two different ways: along the real axis and along the imaginary axis – both limits should yield the same result if the limit in (1) indeed converges. If $h$ moves along the real axis, $h = \delta x \in \mathbb{R}$. The limit in (1) becomes:

where $z_0 = x_0 + iy_0$. If we write $f(x + iy) = w(x, y) + iu(x, y)$, the limit then becomes:

but we see that these are just partial derivatives evaluated at $z_0$:

Now, if we move along the imaginary axis, $h = i\delta y \in i\mathbb{R}$. The limit in (1) becomes:

Following a similar procedure from before:

Since $\frac{1}{i} = -i$, the limit can be written as:

As discussed, if $f$ is complex-differentiable at $z_0$, these limits should be equal. Equating real and imaginary parts, we find:

Part B

From the conditions in (2), we observe that:

It can similarly be shown that $\nabla^2 u = 0$. We therefore find that a holomorphic function has real and imaginary parts that satisfy the 2D Laplace equation.

Part C

The ‘tangent vectors’ along $\gamma_1$ at $t_1$ and $\gamma_2$ at $t_2$ are represented by the complex numbers $\gamma_1'(t_1)$ and $\gamma_2'(t_2)$, respectively. Let us denote the ‘dot product’ of two complex numbers $z$ and $w$ as: $\langle z, w\rangle \equiv \mathrm{Re}[z]\mathrm{Re}[w] + \mathrm{Im}[z]\mathrm{Im}[w]$. The cosine of the angle $\alpha$ is therefore:

The transformed curves $f(\gamma_1(t))$ and $f(\gamma_2(t))$, at the intersection $f(p)$, will have tangent vectors $w_1 = \frac{d}{dt}f(\gamma_1(t))\big|_{t_1}$ and $w_2 = \frac{d}{dt}f(\gamma_2(t))\big|_{t_2}$, respectively. Let us write $\gamma_i(t) = x_i(t) + y_i(t)$ for $i = 1, 2$ and $\tilde p = (x_1(t_1), y_1(t_1)) = (x_2(t_2), y_2(t_2)) \in \mathbb{R}^2$ . Using the chain rule,

And similarly for $w_2$:

Now, let us consider the cosine of the angle formed between $w_1$ and $w_2$:

Denoting the $t$ derivatives with dots and omitting the input points/parameter values,

Hence,

Note that since $f$ is complex-differentiable, we require $\left\|\frac{\partial f}{\partial x}\right\|^2 = \left\|\frac{\partial f}{\partial y}\right\|^2 = \|f'(p)\|^2$. Applying the condition found in (a) to the second term, we find that it vanishes. Hence,

Now, observe that $w_1 = f'(p)\gamma_1'(t_1)$ and $w_2 = f'(p)\gamma_2'(t_2)$

As was to be shown.

Part D

Note that for a holomorphic complex potential $f = \phi + i\psi$, taking the derivative in the real direction:

$\psi$ was chosen such that $\partial_x\psi = -\partial_y\phi = E_y$ so

Part E

There was a domain issue in the original problem statement – $\pi$ should be added/subtracted depending on the domain, but this shouldn’t matter in verifying the 2D Laplace equation since it involves derivatives. A better statement would be $\phi(r, \theta) = \frac{\phi_0}{\pi}\theta$ for $0 \le \theta \le \pi$, which clearly satisfies the given boundary conditions $\phi(x > 0, 0) = 0$, $\phi(x < 0, 0) = \phi_0$. In spherical coordinates, the Laplacian operator is

Hence,

By the uniqueness theorem, the true, unique electric field respecting the boundary conditions is modeled by $-\nabla\phi$.

Part F

A radial line on the upper half plane at angle $\theta \in (0, \pi)$ is the set of points $L_\theta = \{re^{i\theta} \in \mathcal{H}, r \in (0, \infty)\}$. Under the logarithm map, this set is mapped to $\log(L_\theta) = \{\log|r| + i\theta, r \in (0, \infty)\}$ which is clearly the horizontal line at $\mathrm{Im}[z] = \theta$ extending $-\infty < \mathrm{Re}[z] < \infty$. Since the upper half plane can be thought as the collection of the sets $L_\theta$ for $\theta \in (0, \pi)$, $\log\mathcal{H}$ is clearly the strip bounded by $\mathrm{Im}[z] = 0$ and $\mathrm{Im}[z] = \pi$. Let us call this strip $\mathcal{S}$. We are given that $\log(z)$ is holomorphic on $\mathcal{H}$ and that $e^z : \mathcal{S} \to \mathcal{H}$

Part G

Note that the boundary condition in (e), under the logarithm map, becomes the desired boundary conditions of two infinite capacitor plates. Let us crudely denote the ‘potential space’ as $\Phi$ – the space to which complex potentials map. For example, if $f$ is an appropriate complex potential to (e), we may write $f : \mathcal{H} \to \Phi$ with the boundary condition $\phi(x > 0, 0) = 0$ and $\phi(x < 0, 0) = \phi_0$ defined on the upper half plane $\mathcal{H}$. We observe that taking a map $f \circ e^z : \mathcal{S} \to \mathcal{H} \to \Phi$ defines a complex potential with the boundary conditions $\phi(x, 0) = 0$ and $\phi(x, \pi) = \phi_0$ defined on $\mathcal{S}$. For an arbitrary separation $d$, we can consider $f(e^{\frac{\pi}{d}z})$. Since $e^{\frac{\pi}{d}z}$ and $f$ are holomorphic, so is $f(e^{\frac{\pi}{d}z}) = \phi(e^{\frac{\pi}{d}z}) + i\psi(e^{\frac{\pi}{d}z})$. As verified in (b), this means that the real part, $\phi(e^{\frac{\pi}{d}z})$, satisfies the 2D Laplace equation, as well as the boundary conditions $\phi(x, 0) = 0$ and $\phi(x, d) = \phi_0$, hence a valid potential for this problem. We have:

for $y \in (0, d)$. This is indeed the expected result from basic electrostatics.

Part H

This problem really isn’t approachable without rigorous mathematical reasoning. The unit disc is the set defined by $\{z \in \mathbb{C}, |z| < 1\}$. The upper half plane is defined by the set $\{z \in \mathbb{C}, \mathrm{Im}[z] \ge 0\}$. We first show that $f(\mathcal{D}) \subseteq \mathcal{H}$. For any $\xi \in \mathcal{D}$, we observe that

Clearly, if $|\xi| < 1$, we have

implying $f(\xi) \in \mathcal{H}$. Since for an arbitrary $f(\xi) \in f(\mathcal{D})$ we have $f(\xi) \in \mathcal{H}$, $f(\mathcal{D}) \subseteq \mathcal{H}$. Now we’d like to show that $\mathcal{H} \subseteq f(\mathcal{D})$. Consider an arbitrary $\omega \in \mathcal{H}$. We’d like to show that there exists a number $\xi \in \mathcal{D}$ such that $f(\xi) = \omega$, which would imply $\omega \in f(\mathcal{D})$. Observe that:

Now, note that $\mathrm{Re}[i - \omega] = -\mathrm{Re}[i + \omega] = \mathrm{Re}[\omega]$, so when comparing $|i - \omega|$ and $|i + \omega|$, we need only compare the imaginary parts. Denoting $\omega_i \equiv \mathrm{Im}[\omega] > 0$,

since $\omega_i > 0$, we can get rid of the absolute value symbol – $|\omega_i| = \omega_i$ – and make the above a strict inequality. This suggests that $|\mathrm{Im}[i-\omega]| < |\mathrm{Im}[i+\omega]|$, which implies $|i-\omega| < |i+\omega|$, and we conclude:

so for all $\omega \in \mathcal{H}$, there exists a $\xi \in \mathcal{D}$ with $f(\xi) = \omega$. We’ve shown that $f(\mathcal{D}) \subseteq \mathcal{H}$ and that $f(\mathcal{H}) \subseteq \mathcal{D}$, implying $f(\mathcal{D}) = \mathcal{H}$. The way we found $\xi$ above hints that the $\omega \in \mathcal{H}$, $\xi \in \mathcal{D}$ pair is defined uniquely – that there is a one-to-one correspondence such that $f$ is bijective. To show this, we can first suppose that $f$ has an inverse $h$ defined by:

$h$ is clearly single-valued, and it’s easy to verify that $h(f(z)) = z$ and $f(h(w)) = w$. This would be a direct proof that $f$ is a bijection between $\mathcal{D}$ and $\mathcal{H}$. Now we observe the mapping of the circle $C = \{|z| = 1\}$. Note that there should be problems with $f$ around $z = -1 \in C$ since $f$ is not holomorphic at $z = -1$. For an arbitrary point $e^{i\theta} \in C$ with $\theta \in [0, 2\pi)$, it’s easy to simplify expressions down to:

Graphing this on desmos suggests that the arc $(0, \pi)$ maps to the positive real line and that $(\pi, 2\pi)$ maps to the negative real line. Notice there’s a jump from $\infty$ to $-\infty$ as we move from the top arc to the bottom arc at $\pi$ – this is where $f$ “breaks”.

Part I

This is a standard electrostatics problem. The potential $\phi_I$ in the region between the rod and the cylindrical shell can be found by integrating the appropriate electric field $\boldsymbol{E}$.

where $R$ is the radius of the concentric cylindrical shell.

Part J

From our results in (h), let us define a map $g : \mathcal{D}_R \to \mathcal{H}$, where $\mathcal{D}_R$ is the disc of radius $R$ centered at the origin, as:

Under this map, the set up in (i) is mapped to the boundary conditions on $\mathcal{H}$: $\phi = 0$ on the real line, infinite rod placed at $z = ih$. These are the exact boundary conditions that we require for this problem. The inverse map of $g$, denoted $H$, is given by:

$H$ is holomorphic everywhere on the complex plane except at $-ih$, which isn’t included in $\mathcal{H}$, so $H$ holomorphic on the upper half plane. Since the boundary condition satisfied by $\phi_I$ has the geometry of $\mathcal{D}_R$, we can consider a complex potential $f = \phi_I + i\psi_I$ that maps $f : \mathcal{D}_R \to \Phi$. As argued in (g), $f \circ H : \mathcal{H} \to \mathcal{D}_R \to \Phi$ is a holomorphic function on $\mathcal{H}$ that respects all necessary boundary conditions. Since $\phi_I$ is only dependent on $r$, we’d like to consider $|H(w)|$:

If we identify $\mathrm{Im}[w] = y$ and $\|w\|^2 = x^2 + y^2$, the desired potential function is:

A method of images results in the same expression.

Part K

If we parameterize the curve $C$ as $\gamma(t) = x(t) + iy(t)$, the length element $dl$ at $\gamma(t_0)$ can be written as:

with all derivatives evaluated at $t_0$. We can use the calculations in (c) to write $dl'$ in terms of $dt$:

Hence, we find the scaling factor is $|f'(\gamma(t_0))|$.

Part L

If $\hat n$ is orthogonal to $dl$, the surface charge at $dl$ is proportional to $\nabla\phi \cdot \hat n$. Since the surface is a conductor, $\nabla\phi$ is parallel to $\hat n$ near the surface. From previous problems, we know that $\phi' = \phi \circ f^{-1}$ is the appropriate electrostatic potential in $f(C)$, and $\nabla\phi'$ is parallel to the mapping of $\hat n$, which remains orthogonal to the curve, as shown in (c). There’s a theorem that states that if $f$ is a bijective holomorphic map, $f'$ never vanishes, $f^{-1}$ is holomorphic, and the derivative of $f^{-1}$ is given by $\frac{1}{f'}$. However, since we deal with the individual real and imaginary components of complex functions in this question, we may crudely rely on calculus on $\mathbb{R}^2$. The surface charge at $dl'$ is proportional to $\nabla\phi' \cdot \hat n' = |\nabla\phi'|$. Carefully applying the chain rule results in:

Part M

If we take the charge at points to remain the same even after conformal transformations (think about stretching charged surfaces with the charges glued in place) we see that the scale factor found in (j) results in the scaling of the surface charge found in (l). Hence, given that functions behave nicely at boundaries, we can determine the surface charge of other surfaces simply by looking at the form of $f$ and its derivative.

Part N

The electrostatic system of interested is effectively two-dimensional. For simplicity, set $L = 1$ and $Q = +1$. We can recover the general answer with dimensional analysis later on.

Consider a configuration in $z$-space in which the grounded conductor fills the area $\mathrm{Re}(z) < 0$ and a charge is placed at position $e^{+i\pi\alpha/\theta}$. This configuration can be mapped to our puzzle in $w$-space by the conformal transformation $w(z) = z^{\theta/\pi}$ as shown in the figure below.

In $z$-space, the electric potential in open space is as if created by the original charge and the image charge $-1$ located at position $e^{-i\pi\alpha/\theta}$. Thus, the complex potential is given by:

After the transformation $z \to w(z)$, in $w$-space it becomes $\phi(z) \to \phi'(w)$:

Choose a polar coordinate $w = (r, \varphi)$, then we can obtain the real potential as:

Subtract this by the electric potential created by the original charge in the $w$-space, we have a regularized complex potential which is not singular at the location of that charge:

We can then calculate the electrical field created at the location of the original charge by the induced charge distributed on the conducting wedge:

The total electrostatic force acting on the original charge is therefore $\vec F = \vec E$, and after putting back $Q$ and $L$, we arrive at the expression:

For $\theta = 120^o$ and $\alpha = 30^o$, the magnitude of this force in the unit of $kQ^2/L$ is $\sqrt{13}/2 \approx 1.8028$.

* This part was created with helps from Quy C. Tran and Nam H. Nguyen.

Grading Scheme

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