Show that $$ \mathbf{E}=\nabla \times(\mathbf{r} \psi)=-\mathbf{r} \times \nabla \psi $$ is a solenoidal solution of the vector wave equation (8.7.1) such that \(\mathbf{E}\) is everywhere tangential to a spherical boundary. Show that $$ \mathbf{E}^{\prime}=\nabla \times\left(\nabla \times \mathbf{r} \psi^{\prime}\right) \quad \text { or } \quad \mathbf{B}^{\prime}=\nabla \times \mathbf{r} \psi^{\prime \prime} $$ is also a solution, with tangential B. Show that in either case the \(\mathbf{E}\) and \(\mathbf{B}\) fields are orthogonal. (This form of solution is the most useful general solution of the spherical vector wave problem. \(\dagger\) )

Given that the electric field \(\mathbf{E} = \nabla \times (\mathbf{r}\psi)\), we need to show that this satisfies the vector wave equation: $$ \nabla^2 \mathbf{E} - \frac{1}{c^2}\frac{\partial^2 \mathbf{E}}{\partial t^2} = 0 $$ Where \(\nabla^2\) is the Laplacian operator and \(c\) is the speed of light. First, let's find the curl of \(\mathbf{E}\): $$ \nabla\times\mathbf{E} = \nabla\times\left(\nabla\times (\mathbf{r}\psi)\right) $$ Now, using the Curl-Curl identity, we have: $$ \nabla\times\mathbf{E} = \nabla\left(\nabla\cdot(\mathbf{r}\psi)\right) - \nabla^2(\mathbf{r}\psi) $$ Next, let's evaluate the time derivative in the wave equation: $$ \frac{\partial \mathbf{E}}{\partial t} = \frac{\partial}{\partial t}(\nabla\times(\mathbf{r}\psi)) $$ Since \(\mathbf{r}\) is a position vector and does not depend on time: $$ \frac{\partial \mathbf{E}}{\partial t} = \nabla\times(\mathbf{r}\frac{\partial \psi}{\partial t}) $$ Taking the second-order time derivative: $$ \frac{\partial^2 \mathbf{E}}{\partial t^2} = \nabla\times(\mathbf{r}\frac{\partial^2 \psi}{\partial t^2}) $$ Now substitute the expressions for the Laplacian of \(\mathbf{E}\) and second-order time derivative of \(\mathbf{E}\) in the vector wave equation.

Substituting the expressions found in Step 1 into the vector wave equation: $$ \nabla^2(\mathbf{r}\psi) - \frac{1}{c^2}(\nabla\times(\mathbf{r}\frac{\partial^2 \psi}{\partial t^2})) = 0 $$ Now, apply the curl on both sides: $$ \Rightarrow \nabla\times\Big(\nabla^2(\mathbf{r}\psi) - \frac{1}{c^2}(\nabla\times(\mathbf{r}\frac{\partial^2 \psi}{\partial t^2}))\Big) = \mathbf{0} $$ By the linearity of the curl operator, we have: $$ \Rightarrow \nabla\times\nabla^2(\mathbf{r}\psi) - \frac{1}{c^2}\nabla\times(\nabla\times(\mathbf{r}\frac{\partial^2 \psi}{\partial t^2})) = \mathbf{0} $$ Now, utilize the Curl-Curl identity in reverse and rearrange the terms: $$ \Rightarrow \nabla\times\left(\nabla\left(\nabla\cdot(\mathbf{r}\psi)\right) - \nabla^2(\mathbf{r}\psi)\right) = \frac{1}{c^2}\nabla\times(\nabla\times(\mathbf{r}\frac{\partial^2 \psi}{\partial t^2})) $$ Looking at the equivalence: $$ \mathbf{E} = \nabla\times(\mathbf{r}\psi) $$ We can say that the given \(\mathbf{E}\) field satisfies the vector wave equation, and thus is a solenoidal solution.

The expressions for \(\mathbf{E'}\) and \(\mathbf{B'}\) are given as: $$ \mathbf{E'}=\nabla \times\left(\nabla \times \mathbf{r} \psi^{\prime}\right) \quad \text { and } \quad \mathbf{B'}=\nabla \times \mathbf{r} \psi^{\prime \prime} $$ We need to show that these are solenoidal solutions with tangential \(\mathbf{B}\) field. We will use the divergence-curl identity to show that both these fields are solenoidal: $$ \nabla\cdot\mathbf{E'} = \nabla\cdot\left(\nabla \times\left(\nabla \times \mathbf{r} \psi^{\prime}\right)\right) = 0 $$ $$ \nabla\cdot\mathbf{B'} = \nabla\cdot\left(\nabla \times \mathbf{r} \psi^{\prime \prime}\right) = 0 $$ Therefore, both \(\mathbf{E'}\) and \(\mathbf{B'}\) are solenoidal. Next, we show that the \(\mathbf{B}\) field is tangential by analyzing its dot product with the \(\mathbf{E}\) field: $$ \mathbf{E}\cdot\mathbf{B'} = (\nabla\times(\mathbf{r}\psi))\cdot(\nabla\times(\mathbf{r}\psi^{\prime\prime})) $$ Use vector triple product identity: \(\mathbf{A}\cdot(\mathbf{B}\times\mathbf{C}) = \mathbf{B}\cdot(\mathbf{C}\times\mathbf{A})\). Therefore, $$ \mathbf{E}\cdot\mathbf{B'} = (\mathbf{r}\psi^{\prime\prime})\cdot(\nabla\times(\mathbf{r}\psi)) = \mathbf{r}\cdot(\nabla\times(\mathbf{r}\psi))\psi^{\prime\prime} = 0 $$ Since the dot product is zero, the \(\mathbf{E}\) and \(\mathbf{B}\) fields are orthogonal. Thus, the given \(\mathbf{E}\) and \(\mathbf{B'}\) are solenoidal solutions with orthogonal fields.