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Chapter 8: Problem 1

Consider a solution of the vector wave equation of the form $$ \mathbf{F}(\mathbf{r}, t)=\nabla \times \mathbf{f}(\mathbf{r}) e^{j \omega t} \text {, } $$ Show that $$ \nabla \times(\nabla \times \mathbf{F})=k^{2} \mathbf{F}, $$ that is, that the double curl of \(\mathbf{F}\) is a redundant solution that differs from \(\mathbf{F}\) only by the constant scale factor \(\pi^{2}=\omega^{2} / c^{2}\).

Short Answer

Expert verified
Question: Show that the double curl of the given vector wave equation solution \(\mathbf{F}(\mathbf{r}, t)=\nabla \times \mathbf{f}(\mathbf{r}) e^{j \omega t}\) is equal to \(k^2\mathbf{F}\), where \(k^2 = \omega^2/c^2\). Solution: We find that the double curl of \(\mathbf{F}\) is equal to zero, which contradicts the expectation that it should be equal to \(k^2\mathbf{F}\). Therefore, the given vector wave equation does not satisfy the condition that the double curl is equal to \(k^2\mathbf{F}\).

Step by step solution

01

Calculate the double curl of \(\mathbf{F}\)

To calculate the double curl of \(\mathbf{F}\), we must first find the curl of \(\mathbf{F}\). The curl of a vector function is given by \(\nabla \times \mathbf{F}\). So, we will calculate \(\nabla \times (\nabla \times \mathbf{f}(\mathbf{r}) e^{j \omega t})\).
02

Apply the product rule for curls

To find the curl of \(\mathbf{f}(\mathbf{r}) e^{j \omega t}\), we need to apply the product rule for curls, which states that \(\nabla \times (\mathbf{A} \mathbf{B}) = (\mathbf{B} \cdot \nabla)\mathbf{A} - (\mathbf{A} \cdot \nabla)\mathbf{B} + \mathbf{A}(\nabla \cdot \mathbf{B}) - \mathbf{B}(\nabla \cdot \mathbf{A})\). In our case, \(\mathbf{A} = \mathbf{f}(\mathbf{r})\) and \(\mathbf{B} = e^{j \omega t}\). Applying the product rule, we obtain: $$ \nabla \times (\mathbf{f}(\mathbf{r}) e^{j \omega t}) = e^{j \omega t} (\nabla \cdot \mathbf{f}(\mathbf{r})) - \nabla e^{j \omega t} \times \mathbf{f}(\mathbf{r}) $$
03

Evaluate the individual terms

Now we need to evaluate the terms in the above equation. First, note that \(\nabla \cdot \mathbf{f}(\mathbf{r}) = 0\) because the divergence of a curl is always zero. Also, note that \(\nabla e^{j \omega t}\) is a scalar times the unit vector in the time direction, and since \(\mathbf{f}(\mathbf{r})\) is purely a spatial function, their cross product will be zero. Therefore, the curl of \(\mathbf{f}(\mathbf{r}) e^{j \omega t}\) is zero: $$ \nabla \times (\mathbf{f}(\mathbf{r}) e^{j \omega t}) = 0 $$
04

Calculate double curl of \(\mathbf{F}\)

Now we can calculate the double curl of \(\mathbf{F}\): $$ \nabla \times(\nabla \times \mathbf{F})=\nabla \times (\nabla \times (\mathbf{f}(\mathbf{r}) e^{j \omega t})) = \nabla \times 0 = 0 $$
05

Simplify the desired equation

Our goal is to show that \(\nabla \times(\nabla \times \mathbf{F})=k^{2} \mathbf{F}\). Since the double curl of \(\mathbf{F}\) is zero, our equation becomes: $$ 0 = k^{2} \mathbf{F} $$
06

Conclude the proof

Since \(\mathbf{F} = \nabla \times \mathbf{f}(\mathbf{r}) e^{j \omega t}\) is a non-zero solution of the vector wave equation, we must have \(k^2=0\), which means that \(\omega^2/c^2 = 0\). However, this contradicts the fact that \(\mathbf{F}\) is a nontrivial solution. Therefore, the double curl of \(\mathbf{F}\) is not a redundant solution that differs from \(\mathbf{F}\) only by the constant scale factor \(k^2=\omega^{2} / c^{2}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Double Curl
In vector calculus, the concept of a "double curl" refers to taking the curl of a vector field after already applying the curl once. The double curl operation is particularly important in electromagnetic wave theory and fluid dynamics. The curl of a vector field measures the rotation at a point in the field. Performing the curl operation twice on the same vector function often simplifies the result.

The exercise involves the vector field \( \mathbf{F}=abla \times \mathbf{f}(\mathbf{r}) e^{j \omega t} \). To find the double curl, we calculate \( abla \times (abla \times \mathbf{F}) \). Here's a crucial insight: because of the properties of the double curl,
  • it generally can be represented as a linear combination of the Laplacian \(abla^2\) of the vector field,
  • this is tied via the well-known vector identity:
\[abla \times (abla \times \mathbf{V}) = abla(abla \cdot \mathbf{V}) - abla^2 \mathbf{V}\] Using this identity simplifies the computation and helps understand how curl behaves geometrically within the field.
Product Rule for Curls
The product rule for curls is somewhat analogous to the product rule in standard calculus.It is used when taking the curl of a product involving a scalar and a vector. If we have two functions, \( \mathbf{A} \) and \( \mathbf{B} \), the curl of their product is given by this rule.
  • For \( abla \times (\mathbf{A} \mathbf{B}) \), the rule states:
\[abla \times (\mathbf{A} \mathbf{B}) = (\mathbf{B} \cdot abla) \mathbf{A} - (\mathbf{A} \cdot abla) \mathbf{B} + \mathbf{A}(abla \cdot \mathbf{B}) - \mathbf{B}(abla \cdot \mathbf{A})\]In the exercise, with \( \mathbf{A} = \mathbf{f}(\mathbf{r}) \) and \( \mathbf{B} = e^{j \omega t} \), applying the product rule helps decompose and simplify complex expressions.Here, it reveals how temporal and spatial components interact in the curl and how terms can simplify or vanish.

Thus, understanding the product rule is central in vector calculus and physics, particularly when dealing with electromagnetic fields.
Divergence of a Curl
An important property in vector calculus is that the divergence of a curl of any vector field is always zero. This is because the curl represents a rotational field and, by its nature, cannot have a net "outflow" or "inflow" of flux, as divergence describes.

This concept is crucial in determining that certain terms in the vector wave equation are zero. In the provided solution, \( abla \cdot \mathbf{f}(\mathbf{r}) = 0 \), since \( \mathbf{f}(\mathbf{r}) \) is the curl of another function.
  • The identity \[(abla \cdot (abla \times \mathbf{A}) = 0)\] is pivotal in proving why components intrinsically vanish during computation.
  • For the wave equation, this means no source or sink exists in the field.
  • It simplifies proofs and calculations significantly.
Understanding why the divergence of the curl is zero demystifies other concepts, tracing back to the fundamental symmetry properties of vector fields.
Scalar Wave Function
A scalar wave function typically models the amplitude of a wave not concerned with the direction of propagation, unlike vector wave equations that include direction.In our context, the function \( e^{j \omega t} \) represents such a scalar wave function, dependent on time, adding an oscillatory nature to the spatial vector field \( \mathbf{f}(\mathbf{r}) \).
  • Scalar wave functions can describe phenomena like sound waves or electromagnetic waves under certain conditions.
  • Since \( e^{j \omega t} \) is purely scalar, its effect on the curl operations focuses purely on modifying amplitude inputs over time.
  • It emphasizes frequency components influenced by time but does not inherently change directions or spatial orientations.
Understanding scalar wave functions helps to model and solve complex physical systems, providing a clearer picture of the oscillations and interactions between time and spatial vectors.

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Most popular questions from this chapter

(a)\( Generalize the boundary conditions \)(8.6 .5)\( to \)(8.6 .8)\( to include the case where a surface charge density \)\sigma=\Delta q_{\text {tree }} / \Delta S\( and a surface current of magnitude \)K=\Delta I_{\text {fres }} / \Delta l\( exist on the boundary surface, establishing the conditions $$ \text { fi } \begin{aligned} & \cdot\left(x_{22} \mathbf{E}_{2}-x_{41} \mathbf{E}_{1}\right)=\frac{\sigma}{\epsilon_{0}} \\ \text { fi } \times\left(\frac{\mathbf{B}_{2}}{k_{m 2}}-\frac{\mathbf{B}_{1}}{k_{m 1}}\right) &=\mu_{0} \mathbf{K} . \end{aligned} $$ (b) Show that the boundary conditions remain valid when the boundary is not plane and when the respective media are not homogeneous. (c) What are the boundary conditions on the \)\mathbf{D}\( and \)\mathbf{H}$ fields?

Postulate wave fields of the form $$ \begin{aligned} &\mathbf{E}=\mathbf{i} f(z-c t)+\mathbf{j} g(z-c t)+\mathbf{k} h(z-c t) \\ &\mathbf{B}=\mathbf{i} q(z-c t)+\mathbf{j} r(z-c t)+\mathbf{k} s(z-c t) \end{aligned} $$ where \(f, g, h, q, r, s\) are arbitrary (nonsinusoidal) functions, independent of \(x\) and \(y .\) Show that such waves are a solution of the wave equations \((8.2 .8)\) and \((82.9)\) and that Maxwell's equations (8.2.1) to \((8.2 .4)\) require $$ \begin{aligned} &h=s=0 \\ &f=c r \\ &g=-c q \end{aligned} $$ that is, that only two of the six functions are really arbitrary.

Practical coaxial lines used for the distribution of high-frequency signals often consist of a thin copper wire in a polyethylene sleeve on which a copper braid is woven (usually there is also a protective plastic jacket over the braid). Commercial lines are made with nominal characteristic impedances of 50,75 , or 90 ohms. A common 50 -ohm variety has a center conductor of diameter 0035 in. The dielectric constant of polyethylene is \(2.3\) What is the nominal (inside) diameter of the copper braid? What are the capacitance and inductance per foot? What is the speed of propagation, expressed as a percent of the velocity of light? Arswer: \(0120 \mathrm{in} ; 30 \mathrm{pF} / \mathrm{ft} ; 0074 \mu \mathrm{H} / \mathrm{ft} ; 66\) percent.

Consider \(\mathbf{E}\) and \(\mathbf{B}\) wave fields whose only dependence on \(z\) and \(t\) is included in the factor \(e^{i\left(\omega t-x_{1} \theta\right)}\). Further assume TE waves such that \(E_{z}=0\). Write out Maxwell's curl equations \((82.2)\) and \((8.2 .4)\) in cartesian components and show \((a)\) that all four transverse field components can be obtained from \(B_{t}\) by first-order partial differentiation and \((b)\) that \(B_{*}\) must be a solution of the Helmholtz equation \((8.7 .16)\). Thus the scalar function \(\phi\) of the text may be interpreted as proportional to \(B_{z}\) for TE waves or proportional to \(E_{s}\) for TM waves.

It is often convenient to discuss electromagnetic problems in terms of potentials rather than fields. For instance, elementary treatments show that the electrostatic field \(\mathbf{E}(\mathbf{r})\) is conservative and can be derived from a scalar potential function \(\phi(\mathbf{r})\), which is related to \(\mathbf{E}\) by $$ \begin{aligned} &\phi=-\int_{r_{0}}^{r} \mathbf{E} \cdot d \mathbf{l} \\ &\mathbf{E}=-\nabla \phi \end{aligned} $$ Mathematically, the conservative nature of the static field \(\mathbf{E}\) is expressed by the vanishing of its curl. Since the curl of any gradient is identically zero, use of the scalar potential automatically satisfies the static limit of the Maxwell equation (8.2.2); the other constraint on \(\phi\) is Gauss' law (8.2.1). Which hecomes Poisson's equation $$ \nabla^{2} \phi=-\frac{\rho}{\epsilon_{0}} $$ (a) Show that \((8.2 .3)\) is satisfied automatically if we introduce the magnetic vector potential \(\mathbf{A}\), related to the magnetic field by $$ B=\nabla \times A . $$ (b) Show that in the general (nonstatic) case, the electric field is given in terms of the scalar and vector potentials by $$ \mathbf{E}=-\nabla \phi-\frac{\partial \mathbf{A}}{\partial t} $$ (c) Complete the prescription of \(\mathbf{A}\) by defining its divergence by the Lorents condition $$ \boldsymbol{\nabla} \cdot \mathbf{A}=-\frac{1}{c^{2}} \frac{\partial \phi}{\partial t} $$ and show that the two potentials obey the symmetrical inhomogeneous wave equations $$ \begin{aligned} &\nabla^{2} \phi-\frac{1}{c^{2}} \frac{\partial^{2} \phi}{\partial t^{2}}=-\frac{\rho}{\epsilon_{0}} \\ &\nabla^{2} \mathbf{A}-\frac{1}{c^{2}} \frac{\partial^{2} \mathbf{A}}{\partial t^{2}}=-\mu_{0} \mathbf{J} . \end{aligned} $$ These equations connect the potentials associated with radiation fields with their sources \(\rho\) and \(\mathbf{J}\).
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