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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Most popular questions from this chapter

Show that the right-hand side of \((84.16)\) can be written in the form $$ \int_{V} \nabla \cdot \mathbf{T} d v $$ where \(\mathrm{T}\) is the Maxwell stress tensor $$ \mathbf{T}={ }_{0} \mathbf{E} \mathbf{E} \mathbf{E}+\frac{\mathbf{1}}{\mu_{0}} \mathbf{B B}-\mathbf{1}\left(\frac{1}{2} \epsilon_{0} E^{2}+\frac{B^{2}}{2 \mu_{0}}\right) $$ Then apply Gauss' theorem to convert the integral to the surface-integral form $$ \int_{S} \mathbf{T} \cdot d \mathbf{S} . $$

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