Retarded Potentials and Far-Field Radiation

Source: Cambridge Electromagnetism, Chapter 7 exercises, Question 4; accompanies David Tong’s Electromagnetism.

Let $\phi$ be the retarded scalar potential given by \[\phi(t,\mathbf{x}) = \frac{1}{4\pi\varepsilon_0}\int d^3y\, \frac{\rho(t_{\mathrm{ret}},\mathbf{y})}{R},\tag{1.1}\] where $R = |\mathbf{x}-\mathbf{y}|$, $t_{\mathrm{ret}} = t - R/c$, and $\hat{\mathbf{R}} = (\mathbf{x}-\mathbf{y})/R$. Show that \[\frac{\partial}{\partial t}\phi(t,\mathbf{x}) = \frac{1}{4\pi\varepsilon_0}\int d^3y\, \frac{\dot{\rho}(t_{\mathrm{ret}},\mathbf{y})}{R},\tag{1.2}\] where $\dot{\rho}(t_{\mathrm{ret}},\mathbf{y})$ is $\partial\rho(t,\mathbf{y})/\partial t$ evaluated at $t_{\mathrm{ret}}$. Show further that \[\nabla\phi(t,\mathbf{x}) = -\frac{1}{4\pi\varepsilon_0}\int d^3y\,\hat{\mathbf{R}} \left(\frac{1}{R^2}\rho(t_{\mathrm{ret}},\mathbf{y}) + \frac{1}{cR}\dot{\rho}(t_{\mathrm{ret}},\mathbf{y})\right).\tag{1.3}\] Hence verify, using $\nabla^2(1/R) = -4\pi\delta^{(3)}(\mathbf{x}-\mathbf{y})$, that $\phi$ satisfies the wave equation: \[\Box\,\phi(t,\mathbf{x}) = -\frac{1}{\varepsilon_0}\rho(t,\mathbf{x}).\tag{1.4}\]

Write down a similar retarded solution for the vector potential $\mathbf{A}$ in terms of the current density $\mathbf{J}$.

Now assume that $\rho$ and $\mathbf{J}$ are non-zero only in a finite region. Setting $\hat{\mathbf{x}} = \mathbf{x}/|\mathbf{x}|$, show that the leading terms in the far-field expansion are \[\begin{aligned}\mathbf{E}(t,\mathbf{x}) &\approx \frac{\mu_0}{4\pi|\mathbf{x}|}\int d^3y \left(\hat{\mathbf{x}}c\dot{\rho}(t_{\mathrm{ret}},\mathbf{y}) - \dot{\mathbf{J}}(t_{\mathrm{ret}},\mathbf{y})\right) \notag\\ &= \frac{\mu_0}{4\pi|\mathbf{x}|}\,\hat{\mathbf{x}}\times \left(\hat{\mathbf{x}}\times\int d^3y\, \dot{\mathbf{J}}(t_{\mathrm{ret}},\mathbf{y})\right),\end{aligned}\tag{1.5}\] where conservation of current, integration by parts, and discarding of a surface integral have all been used, and \[\mathbf{B}(t,\mathbf{x}) \approx -\frac{\mu_0}{4\pi c|\mathbf{x}|}\, \hat{\mathbf{x}}\times\int d^3y\,\dot{\mathbf{J}}(t_{\mathrm{ret}},\mathbf{y}) = \frac{1}{c}\hat{\mathbf{x}}\times\mathbf{E}(t,\mathbf{x}).\tag{1.6}\] (Note that these results do not assume the dipole approximation.) Determine the Poynting vector.

Hint: when using current conservation, and integration by parts, be careful with the $\mathbf{y}$-dependence of $t_{\mathrm{ret}}$.

Following from the retarded potential, we have, trivially, \[\frac{\partial}{\partial t}\phi(t,\mathbf{x}) = \frac{1}{4\pi\varepsilon_0}\int d^3y\, \frac{\dot{\rho}(t_{\mathrm{ret}},\mathbf{y})}{R}.\tag{1.7}\] Then, note that \[\nabla\left(\frac{1}{R}\right) = -\frac{\hat{\mathbf{R}}}{R^2} \qquad \text{and} \qquad \nabla \rho(t_\text{ret},\mathbf{y}) = \frac{\partial\rho(t_\text{ret},\mathbf{y})}{\partial R} = -\frac{1}{c}\dot{\rho}(t_{\mathrm{ret}},\mathbf{y})\hat{\mathbf{R}}.\tag{1.8}\] Thus, we have \[\nabla\phi(t,\mathbf{x}) = -\frac{1}{4\pi\varepsilon_0}\int d^3y\,\hat{\mathbf{R}} \left(\frac{1}{R^2}\rho(t_{\mathrm{ret}},\mathbf{y}) + \frac{1}{cR}\dot{\rho}(t_{\mathrm{ret}},\mathbf{y})\right).\tag{1.9}\].

Now, first consider the second-order time derivative, which is simply given by \[\frac{\partial^2}{\partial t^2}\phi(t,\mathbf{x}) = \frac{1}{4\pi\varepsilon_0}\int d^3y\, \frac{\ddot{\rho}(t_{\mathrm{ret}},\mathbf{y})}{R}.\tag{1.10}\] For the Laplacian, it is slightly trickier; we find \[\nabla^2\phi(t,\mathbf{x}) = -\frac{1}{\varepsilon_0}\rho(t,\mathbf{x}) +\frac{1}{4\pi\varepsilon_0c^2}\int d^3y\, \frac{\ddot{\rho}(t_{\mathrm{ret}},\mathbf{y})}{R}.\tag{1.11}\] Putting them together, we obtain \[\square\phi(t,\mathbf{x}) = -\frac{1}{\varepsilon_0}\rho(t,\mathbf{x}).\tag{1.12}\]

By simple analogy, we have \[\square\mathbf{A}(t,\mathbf{x}) = -\mu_0\mathbf{J}(t,\mathbf{x}),\tag{1.13}\] \[\frac{\partial}{\partial t}\mathbf{A}(t,\mathbf{x}) = \frac{\mu_0}{4\pi}\int d^3y\, \frac{\dot{\mathbf{J}}(t_{\mathrm{ret}},\mathbf{y})}{R},\tag{1.14}\] and \[\nabla A_i(t,\mathbf{x}) = -\frac{\mu_0}{4\pi}\int d^3y\,\hat{\mathbf{R}} \left(\frac{1}{R^2}J_i(t_{\mathrm{ret}},\mathbf{y}) + \frac{1}{cR}\dot{J_i}(t_{\mathrm{ret}},\mathbf{y})\right)\tag{1.15}\]

In the far-field, we have $R\gg 1$ and $|\mathbf{x}|\approx R$, which gives \[\begin{aligned}\mathbf{E}(t,\mathbf{x}) &= -\nabla\phi(t,\mathbf{x})-\frac{\partial\mathbf{A}(t,\mathbf{x})}{\partial t}\\ &\approx\frac{1}{4\pi\varepsilon_0}\int d^3y\,\hat{\mathbf{x}} \frac{1}{c|\mathbf{x}|}\dot{\rho}(t_{\mathrm{ret}},\mathbf{y}) - \frac{\mu_0}{4\pi}\int d^3y\,\hat{\mathbf{x}} \frac{1}{|\mathbf{x}|}\dot{J_i}(t_{\mathrm{ret}},\mathbf{y})\\ &=\frac{\mu_0}{4\pi|\mathbf{x}|}\int d^3y \left(\hat{\mathbf{x}}c\dot{\rho}(t_{\mathrm{ret}},\mathbf{y}) - \dot{\mathbf{J}}(t_{\mathrm{ret}},\mathbf{y})\right).\end{aligned}\tag{1.16}\]

To simplify the first term, consider the continuity equation, \[\dot{\rho}(t_\textbf{ret},\mathbf{y}) + \nabla_y\cdot \mathbf{J}(t,\mathbf{y})\bigg|_{t=t_\text{ret}}=0.\tag{1.17}\] However, to perform an integration by parts, we need \[\nabla_y\cdot\mathbf{J}(t_\text{ret},\mathbf{y}) = \nabla_y\cdot \mathbf{J}(t,\mathbf{y})\bigg|_{t=t_\text{ret}} + \dot{\mathbf{J}}(t_\text{ret},\mathbf{y})\cdot\nabla t_\text{ret} = \nabla_y\cdot \mathbf{J}(t,\mathbf{y})\bigg|_{t=t_\text{ret}} + \dot{\mathbf{J}}(t_\text{ret},\mathbf{y})\cdot\frac{\hat{\mathbf{R}}}{c}.\tag{1.18}\] Thus, the first term integral becomes \[\frac{\mu_0}{4\pi|\mathbf{x}|}\int d^3y\, \hat{\mathbf{x}}\dot{\mathbf{J}}(t_\text{ret},\mathbf{y})\cdot\hat{\mathbf{R}} - \frac{\mu_0}{4\pi|\mathbf{x}|}\oint d^2y\,\mathbf{J}(t_\text{ret},\mathbf{y})\cdot \hat{\mathbf{n}}.\tag{1.19}\] Since $\mathbf{J}$ is bounded, the surface integral vanishes. Therefore, we can write \[E(t, \mathbf{x}) \approx \frac{\mu_0}{4\pi|\mathbf{x}|}\int d^3y \left(\hat{\mathbf{x}}(\dot{\mathbf{J}}(t_\text{ret},\mathbf{y})\cdot\hat{\mathbf{R}}) - \dot{\mathbf{J}}(t_\text{ret},\mathbf{y})\right),\tag{1.20}\] Again, for far-field, we have $\hat{\mathbf{R}} \approx \hat{\mathbf{x}}$. Finally, using the vector identity, we find \[\mathbf{E}(t,\mathbf{x}) \approx \frac{\mu_0}{4\pi|\mathbf{x}|}\,\hat{\mathbf{x}}\times \left(\hat{\mathbf{x}}\times\int d^3y\, \dot{\mathbf{J}}(t_{\mathrm{ret}},\mathbf{y})\right).\tag{1.21}\]

For the magnetic field, we have \[B_i(t,\mathbf{x}) = \epsilon_{ijk}\partial_jA_k \approx -\frac{\mu_0}{4\pi c|\mathbf{x}|}\epsilon_{ijk}\int d^3y\,\hat{\mathbf{x}}_j\dot{J}_k(t,\mathbf{x}),\tag{1.22}\] where we have made a far-field approximation again. Therefore, we find \[\mathbf{B}(t,\mathbf{x}) \approx -\frac{\mu_0}{4\pi c|\mathbf{x}|}\, \hat{\mathbf{x}}\times\int d^3y\,\dot{\mathbf{J}}(t_{\mathrm{ret}},\mathbf{y}) = \frac{1}{c}\hat{\mathbf{x}}\times\mathbf{E}(t,\mathbf{x}).\tag{1.23}\]

The Poynting vector of the radiation is, therefore, given by \[\mathbf{S(t,\mathbf{x})} = \frac{1}{\mu_0 c}\mathbf{E}(t,\mathbf{x})\times\left(\hat{\mathbf{x}}\times\mathbf{E}(t,\mathbf{x})\right) = \frac{1}{\mu_0c}[\hat{\mathbf{x}}|\mathbf{E}(t,\mathbf{x})|^2 - \mathbf{E}(t,\mathbf{x})(\hat{\mathbf{x}}\cdot\mathbf{E}(t,\mathbf{x}))].\tag{1.24}\] Note that $\hat{\mathbf{x}}\cdot\mathbf{E}=0$, and hence we have \[\mathbf{S(t,\mathbf{x})}=\frac{\mu_0}{16\pi^2c|\hat{\mathbf{x}}|^2}\left|\int d^3y\dot{\mathbf{J}}(t_{\mathrm{ret}},\mathbf{y})\right|^2\sin^2\theta \,\hat{\mathbf{x}},\tag{1.25}\] where $\theta$ is the angle between the integral and $\hat{\mathbf{x}}$.