Problems for Chapter 4

Problem IV-2. Verify the following identities.

(a) $\nabla(fg) = f \nabla g + g \nabla f$

$$ \begin{align*} \nabla(fg) &= \frac{\partial(fg)}{\partial x} \mathbf{i} + \frac{\partial(fg)}{\partial y} \mathbf{j} + \frac{\partial(fg)}{\partial z} \mathbf{k} \\[1em] &= g \frac{\partial f}{\partial x} \mathbf{i} + f \frac{\partial g}{\partial x} \mathbf{i} + g \frac{\partial f}{\partial y} \mathbf{j} + f \frac{\partial g}{\partial y} \mathbf{j} + g \frac{\partial f}{\partial z} \mathbf{k} + f \frac{\partial g}{\partial z} \mathbf{k} \\[1em] &= f \left( \frac{\partial g}{\partial x} \mathbf{i} + \frac{\partial g}{\partial y} \mathbf{j} + \frac{\partial g}{\partial z} \mathbf{k} \right) + g \left( \frac{\partial f}{\partial x} \mathbf{i} + \frac{\partial f}{\partial y} \mathbf{j} + \frac{\partial f}{\partial z} \mathbf{k} \right) \\[1em] &= f \nabla g + g \nabla f \end{align*} $$

(b) $\nabla(\mathbf{F} \cdot \mathbf{G}) = (\mathbf{G} \cdot \nabla) \mathbf{F} + (\mathbf{F} \cdot \nabla) \mathbf{G} + \mathbf{F} \times (\nabla \times \mathbf{G}) + \mathbf{G} \times (\nabla \times \mathbf{F})$

Solution. Click to Expand

We have:

$$ \begin{align*} \left( (\mathbf{G} \cdot \nabla) \mathbf{F} \right)_x &= G_x \frac{\partial F_x}{\partial x} + G_y \frac{\partial F_x}{\partial y} + G_z \frac{\partial F_x}{\partial z} \\[1em] \left( (\mathbf{F} \cdot \nabla) \mathbf{G} \right)_x &= F_x \frac{\partial G_x}{\partial x} + F_y \frac{\partial G_x}{\partial y} + F_z \frac{\partial G_x}{\partial z} \\[1em] \left(\mathbf{F} \times (\nabla \times \mathbf{G})\right)_x &= F_y \left( \frac{\partial G_y}{\partial x} - \frac{\partial G_x}{\partial y} \right) - F_z \left( \frac{\partial G_x}{\partial z} - \frac{\partial G_z}{\partial x} \right) \\[1em] \left(\mathbf{G} \times (\nabla \times \mathbf{F})\right)_x &= G_y \left( \frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} \right) - G_z \left( \frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x} \right) \end{align*} $$

Adding these together, we get:

$$ G_x \frac{\partial F_x}{\partial x} + F_x \frac{\partial G_x}{\partial x} + G_y \frac{\partial F_x}{\partial y} + F_y \frac{\partial G_x}{\partial y} + G_z \frac{\partial F_x}{\partial z} + F_z \frac{\partial G_x}{\partial z} + $$

which is equal to:

$$ \frac{\partial}{\partial x} (F_x G_x + F_y G_y + F_z G_z) $$

which is the $x$-component of $\nabla(\mathbf{F} \cdot \mathbf{G})$.

Similarly, we can prove the equality for the $y$ and $z$ components.

(c) $\nabla \cdot (f \mathbf{F}) = f \nabla \cdot \mathbf{F} + \mathbf{F} \cdot \nabla f$

Solution. Click to Expand

$$ \begin{align*} \nabla \cdot (f \mathbf{F}) &= \frac{\partial (f F_x)}{\partial x} + \frac{\partial (f F_y)}{\partial y} + \frac{\partial (f F_z)}{\partial z} \\[1em] &= f \frac{\partial F_x}{\partial x} + F_x \frac{\partial f}{\partial x} + f \frac{\partial F_y}{\partial y} + F_y \frac{\partial f}{\partial y} + f \frac{\partial F_z}{\partial z} + F_z \frac{\partial f}{\partial z} \\[1em] &= f \nabla \cdot \mathbf{F} + \mathbf{F} \cdot \nabla f \end{align*} $$

(d) $\nabla \cdot (\mathbf{F} \times \mathbf{G}) = \mathbf{G} \cdot (\nabla \times \mathbf{F}) - \mathbf{F} \cdot (\nabla \times \mathbf{G})$

Solution. Click to Expand

We have:

$$ \begin{align*} \mathbf{G} \cdot (\nabla \times \mathbf{F}) &= G_x \left( \frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z} \right) + G_y \left( \frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x} \right) + G_z \left( \frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} \right) \\[1em] \mathbf{F} \cdot (\nabla \times \mathbf{G}) &= F_x \left( \frac{\partial G_z}{\partial y} - \frac{\partial G_y}{\partial z} \right) + F_y \left( \frac{\partial G_x}{\partial z} - \frac{\partial G_z}{\partial x} \right) + F_z \left( \frac{\partial G_y}{\partial x} - \frac{\partial G_x}{\partial y} \right) \end{align*} $$

Then:

$$ \begin{align*} \mathbf{G} \cdot (\nabla \times \mathbf{F}) &- \mathbf{F} \cdot (\nabla \times \mathbf{G}) = \\[1em] & \left(G_z \frac{\partial F_y}{\partial x} + F_y \frac{\partial G_z}{\partial x}\right) \- \left(G_y \frac{\partial F_z}{\partial x} + F_z \frac{\partial G_y}{\partial x}\right) \\[1em] \+ &\left(G_x \frac{\partial F_z}{\partial y} + F_z \frac{\partial G_x}{\partial y}\right) \- \left(G_z \frac{\partial F_x}{\partial y} + F_x \frac{\partial G_z}{\partial y}\right) \\[1em] \+ &\left(G_y \frac{\partial F_x}{\partial z} + F_x \frac{\partial G_y}{\partial z}\right) \- \left(G_x \frac{\partial F_y}{\partial z} + F_y \frac{\partial G_x}{\partial z}\right) \end{align*} $$

Then:

$$ \begin{align*} \mathbf{G} \cdot &(\nabla \times \mathbf{F}) - \mathbf{F} \cdot (\nabla \times \mathbf{G}) \\[1em] &= \frac{\partial}{\partial x} (F_y G_z - F_z G_y) + \frac{\partial}{\partial y} (F_z G_x - F_x G_z) + \frac{\partial}{\partial z} (F_x G_y - F_y G_x) \\[1em] &= \nabla \cdot (\mathbf{F} \times \mathbf{G}) \end{align*} $$

(e) $\nabla \times (f \mathbf{F}) = f \nabla \times \mathbf{F} + (\nabla f) \times \mathbf{F}$

Solution. Click to Expand

$$ \begin{align*} \nabla \times (f \mathbf{F}) &= \mathbf{i} \left( \frac{\partial (f F_z)}{\partial y} - \frac{\partial (f F_y)}{\partial z} \right) \\[1em] &\quad + \mathbf{j} \left( \frac{\partial (f F_x)}{\partial z} - \frac{\partial (f F_z)}{\partial x} \right)\\[1em] &\quad + \mathbf{k} \left( \frac{\partial (f F_y)}{\partial x} - \frac{\partial (f F_x)}{\partial y} \right) \\[1em] &= \mathbf{i} \left( f \frac{\partial F_z}{\partial y} + F_z \frac{\partial f}{\partial y} - f \frac{\partial F_y}{\partial z} - F_y \frac{\partial f}{\partial z} \right) \\[1em] &\quad + \mathbf{j} \left( f \frac{\partial F_x}{\partial z} + F_x \frac{\partial f}{\partial z} - f \frac{\partial F_z}{\partial x} - F_z \frac{\partial f}{\partial x} \right) \\[1em] &\quad + \mathbf{k} \left( f \frac{\partial F_y}{\partial x} + F_y \frac{\partial f}{\partial x} - f \frac{\partial F_x}{\partial y} - F_x \frac{\partial f}{\partial y} \right) \\[1em] &= f \left( \mathbf{i} \left( \frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z} \right) + \mathbf{j} \left( \frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x} \right) + \mathbf{k} \left( \frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} \right) \right) \\[1em] &\quad + \left( \frac{\partial f}{\partial y} F_z - \frac{\partial f}{\partial z} F_y \right) \mathbf{i} \+ \left( \frac{\partial f}{\partial z} F_x - \frac{\partial f}{\partial x} F_z \right) \mathbf{j} + \left( \frac{\partial f}{\partial x} F_y - \frac{\partial f}{\partial y} F_x \right) \mathbf{k} \\[1em] &= f \nabla \times \mathbf{F} + (\nabla f) \times \mathbf{F} \end{align*} $$

Problem IV-3. Show that $\nabla \times \nabla f = 0$, where $f(x, y, z)$ is an arbitrary differentiable scalar function. Assume that mixed second-order partial derivatives are independent of the order of differentiation.

Solution. Click to Expand

$$ \begin{align*} \nabla \times \nabla f &= \mathbf{i} \left( \frac{\partial^2 f}{\partial y \partial z} - \frac{\partial^2 f}{\partial z \partial y} \right) + \mathbf{j} \left( \frac{\partial^2 f}{\partial z \partial x} - \frac{\partial^2 f}{\partial x \partial z} \right) + \mathbf{k} \left( \frac{\partial^2 f}{\partial x \partial y} - \frac{\partial^2 f}{\partial y \partial x} \right)\\[1em] &= \mathbf{0} \end{align*} $$

Problem IV-6.

(a) An electric dipole of strength $p$ situated at the origin and oriented in the positive $z$-direction gives rise to the electric field

$$ \mathbf{E(r, \theta, \phi)} = \frac{1}{4 \pi \epsilon_0} \frac{p}{r^3} \left( 2 \hat{\mathbf{e}}_r \cos \phi + \hat{\mathbf{e}}_\phi \sin \phi \right) $$

Show that the dipole potential is given by

$$ \Phi(r, \theta, \phi) = \frac{1}{4 \pi \epsilon_0} \frac{p \cos \phi}{r^2} $$

In spherical coordinates,

$$ \hat{\mathbf{t}} = \hat{\mathbf{e}}_r \frac{dr}{ds} + \hat{\mathbf{e}}_\phi r \frac{d\phi}{ds} + \hat{\mathbf{e}}_\theta r \sin \phi \frac{d\theta}{ds} $$

Solution. Click to Expand

We have:

$$ \mathbf{E} \cdot \hat{\mathbf{t}} = \frac{1}{4 \pi \epsilon_0} \frac{p}{r^3} \left( 2 \cos \phi \frac{dr}{ds} + r \sin \phi \frac{d\phi}{ds} \right) $$

Since path integral of $\mathbf{E}$ is path independent, then:

$$ \begin{align*} \Phi &= - \int_{(r_0, \theta_0, \phi_0)}^{(r, \theta, \phi)} \mathbf{E} \cdot \hat{\mathbf{t}} ds \\[1em] &= - \left( \int_{(r_0, \theta_0, \phi_0)}^{(r, \theta_0, \phi_0)} \mathbf{E} \cdot \hat{\mathbf{t}} ds + \int_{(r, \theta_0, \phi_0)}^{(r, \theta, \phi_0)} \mathbf{E} \cdot \hat{\mathbf{t}} ds + \int_{(r, \theta, \phi_0)}^{(r, \theta, \phi)} \mathbf{E} \cdot \hat{\mathbf{t}} ds \right) \\[1em] &= - \frac{p}{4 \pi \epsilon_0} \left( \int_{r_0}^r \frac{2 \cos \phi_0}{r^3} dr + 0 + \int_{\phi_0}^{\phi} \frac{\sin \phi}{r^2} d\phi \right) \\[1em] &= - \frac{p}{4 \pi \epsilon_0} \left( - \frac{\cos \phi_0}{r^2} +\frac{\cos \phi_0}{r_0^2} - \frac{\cos \phi}{r^2} + \frac{cos \phi_0}{r^2} \right) \\[1em] &= \frac{1}{4 \pi \epsilon_0} \frac{p \cos \phi}{r^2} + \text{constant} \end{align*} $$

(b) Calculate the flux of the dipole field through a sphere of radius $R$ centered at the origin.

Solution. Click to Expand

In a spherical surface, $\hat{\mathbf{n}} = \hat{\mathbf{e}}_r$. Then:

$$ \begin{align*} \Phi_{\mathbf{E}} &= \iint_S \mathbf{E} \cdot \hat{\mathbf{n}} dS \\[1em] &= \iint_S \frac{1}{4 \pi \epsilon_0} \frac{2p}{R^3} \cos \phi dS \\[1em] &= \frac{1}{4 \pi \epsilon_0} \frac{2p}{R^3} \iint_S \cos \phi dS \\[1em] \end{align*} $$

We have $dS = R^2 \sin \phi \\, d\phi \\, d \theta$. Then:

$$ \begin{align*} \Phi_{\mathbf{E}} &= \frac{1}{4 \pi \epsilon_0} \frac{2p}{R^3} \int_{0}^{\pi} \int_{0}^{2\pi} R^2 \cos \phi \sin \phi \, d\theta \, d\phi \\[1em] &=\frac{1}{\epsilon_0} \frac{p}{R} \int_{0}^{\pi} \cos \phi \sin \phi \, d\phi \\[1em] &= 0 \end{align*} $$

(c) What is the flux of the dipole field over any closed surface that doesn’t pass through the origin?

Solution. Click to Expand

Divergence in spherical coordinates is given by:

$$ \nabla \cdot \mathbf{F} = \frac{1}{r^2} \frac{\partial}{\partial r} \left( r^2 F_r \right) + \frac{1}{r \sin \phi} \frac{\partial}{\partial \phi} \left( F_\phi \sin \phi \right) + \frac{1}{r \sin{\phi}} \frac{\partial F_\theta}{\partial \theta} $$

Then:

$$ \begin{align*} \nabla \cdot \mathbf{E} &= \frac{p}{4 \pi \epsilon_0} \left( \frac{1}{r^2} \frac{\partial}{\partial r} \left( r^2 \frac{2}{r^3} \cos \phi \right) + \frac{1}{r \sin \phi} \frac{\partial}{\partial \phi} \left( \frac{\sin^2 \phi}{r^3} \right) + 0 \right) \\[1em] &= \frac{p}{4 \pi \epsilon_0} \left( \frac{1}{r^2} \left(-\frac{2}{r^2} \cos \phi \right) + \frac{1}{r \sin \phi} \left( \frac{2 \sin \phi \cos \phi}{r^3} \right) \right) \\[1em] &= \frac{p}{4 \pi \epsilon_0} \left( -\frac{2 \cos \phi}{r^4} + \frac{2 \cos \phi}{r^4} \right) \\[1em] &= 0 \end{align*} $$

Now, given any closed surface $S$ that doesn’t pass through the origin, there exists a sphere with $R > 0$ inside $S$ centered at the origin. Using part (b), flux of $\mathbf{E}$ over the sphere is 0. Using problem II-27, since $\nabla \cdot \mathbf{E} = 0$ then flux of $\mathbf{E}$ over $S$ is equal to the flux of $\mathbf{E}$ over the sphere, which is 0.

Problem IV-8. Fick’s law states that in certain diffusion processes the current density $\mathbf{J}$ is proportional to the negative of the gradient of the density $\rho$; that is, $\mathbf{J} = -k \nabla \rho$, where $k$ is a positive constant. If a substance of density $\rho(x, y, z, t)$ and velocity $\mathbf{v}(x, y, z, t)$ diffuses according to Fick’s law, show that the flow is irrotational. That is, $\nabla \times \mathbf{v} = 0$.

Solution. Click to Expand

Since $\mathbf{J} = \rho \mathbf{v}$ (using the definition at page 52 of the book) and $\mathbf{J} = -k \nabla \rho$ (using the equation at the problem statement), then:

$$ \mathbf{v} = -k \frac{\nabla \rho}{\rho} $$

Then using the identity IV-2e we have:

$$ \nabla \times \mathbf{v} = -k \left(\underbrace{\frac{1}{\rho} \nabla \times \nabla \rho}_{A} + \underbrace{\nabla\left(\frac{1}{\rho}\right) \times \nabla \rho}_{B}\right) $$

Curl of a gradient is zero, so $A = 0$. For $B$, we have:

$$ \begin{align*} \nabla\left(\frac{1}{\rho}\right) &= \mathbf{i} \left( -\frac{1}{\rho^2} \frac{\partial \rho}{\partial x} \right) + \mathbf{j} \left( -\frac{1}{\rho^2} \frac{\partial \rho}{\partial y} \right) + \mathbf{k} \left( -\frac{1}{\rho^2} \frac{\partial \rho}{\partial z} \right) \\[1em] &= -\frac{1}{\rho^2} \nabla \rho \end{align*} $$

Since this is parallel to $\nabla \rho$, then $B$ is zero as well. Therefore:

$$ \nabla \times \mathbf{v} = 0 $$

Problem IV-9. (a) A substance diffuses according to Fick’s law. Assuming the diffusing matter is conserved, derive the diffusion equation

$$ \frac{\partial \rho}{\partial t} = k \nabla^2 \rho $$

Solution. Click to Expand

From page 52 of the book, if matter is conserved, then:

$$ \frac{\partial \rho}{\partial t} + \nabla \cdot \mathbf{J} = 0 $$

Using Fick’s law, $\mathbf{J} = -k \nabla \rho$, we have:

$$ \frac{\partial \rho}{\partial t} + \nabla \cdot (-k \nabla \rho) = 0 $$

which is equivalent to:

$$ \frac{\partial \rho}{\partial t} = k \nabla^2 \rho $$