Let's carefully **prove that the given fields** satisfy the **homogeneous Maxwell equations in vacuum**. Given: - \(\mathbf{E}(\mathbf{x}, t) = G(\mathbf{k} \cdot \mathbf{x} - ckt)\) - \(\mathbf{B}(\mathbf{x}, t) = c^{-1} \mathbf{e}_k \times \mathbf{E}(\mathbf{x}, t)\) - where \(\mathbf{e}_k = \frac{\mathbf{k}}{k}\), \(k = |\mathbf{k}|\) - \(\nabla \cdot \mathbf{k} = 0\), so \(\mathbf{k}\) is constant. The **homogeneous Maxwell equations** in vacuum are: \[ \nabla \cdot \mathbf{B} = 0 \] \[ \nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t} \] --- ### **Step 1: Compute \(\nabla \cdot \mathbf{B}\)** Given: \[ \mathbf{B} = c^{-1} \mathbf{e}_k \times \mathbf{E} \] \[ \nabla \cdot \mathbf{B} = \nabla \cdot [c^{-1} \mathbf{e}_k \times \mathbf{E}] \] But \(\mathbf{e}_k\) is a constant vector (since \(\mathbf{k}\) is constant), and \(c\) is a constant: \[ \nabla \cdot (\mathbf{a} \times \mathbf{F}) = \mathbf{a} \cdot (\nabla \times \mathbf{F}) - \mathbf{F} \cdot (\nabla \times \mathbf{a}) = \mathbf{a} \cdot (\nabla \times \mathbf{F}) \] since \(\nabla \times \mathbf{a} = 0\). Therefore, \[ \nabla \cdot \mathbf{B} = c^{-1} \mathbf{e}_k \cdot (\nabla \times \mathbf{E}) \] But let's see what \(\nabla \times \mathbf{E}\) is. --- ### **Step 2: Compute \(\nabla \times \mathbf{E}\)** \[ \mathbf{E}(\mathbf{x}, t) = G(\mathbf{k} \cdot \mathbf{x} - ckt) \] So, \(\mathbf{E}\) is a function of \(\phi = \mathbf{k} \cdot \mathbf{x} - ckt\). The gradient operator acts as: \[ \nabla G(\phi) = (\nabla \phi) G'(\phi) = \mathbf{k} G'(\phi) \] But we need the curl: \[ \nabla \times \mathbf{E} \] If \(\mathbf{E}\) is **parallel to some direction** (say, polarization vector \(\mathbf{E}_0\)), then: \[ \mathbf{E}(\mathbf{x}, t) = \mathbf{E}_0 G(\phi) \] where \(\mathbf{E}_0\) is a **constant vector** perpendicular to \(\mathbf{k}\). So, \[ \nabla \times \mathbf{E} = \nabla \times [\mathbf{E}_0 G(\phi)] = \mathbf{E}_0 \times \nabla G(\phi) = \mathbf{E}_0 \times \mathbf{k} G'(\phi) \] --- ### **Step 3: Compute \(\frac{\partial \mathbf{B}}{\partial t}\)** Recall: \[ \mathbf{B} = c^{-1} \mathbf{e}_k \times \mathbf{E} \] So, \[ \frac{\partial \mathbf{B}}{\partial t} = c^{-1} \mathbf{e}_k \times \frac{\partial \mathbf{E}}{\partial t} \] But, \[ \frac{\partial \mathbf{E}}{\partial t} = \mathbf{E}_0 \frac{\partial}{\partial t} G(\phi) = \mathbf{E}_0 G'(\phi) \frac{\partial \phi}{\partial t} = \mathbf{E}_0 G'(\phi) (-ck) = -ck \mathbf{E}_0 G'(\phi) \] Thus, \[ \frac{\partial \mathbf{B}}{\partial t} = c^{-1} \mathbf{e}_k \times [-ck \mathbf{E}_0 G'(\phi)] = -k \mathbf{e}_k \times \mathbf{E}_0 G'(\phi) \] --- ### **Step 4: Compare \(\nabla \times \mathbf{E}\) and \(-\frac{\partial \mathbf{B}}{\partial t}\)** We have: \[ \nabla \times \mathbf{E} = \mathbf{E}_0 \times \mathbf{k} G'(\phi) = -\mathbf{k} \times \mathbf{E}_0 G'(\phi) \] and \[ -\frac{\partial \mathbf{B}}{\partial t} = k \mathbf{e}_k \times \mathbf{E}_0 G'(\phi) \] But \(\mathbf{e}_k = \mathbf{k}/k\), so \[ k \mathbf{e}_k \times \mathbf{E}_0 = \mathbf{k} \times \mathbf{E}_0 \] Therefore, \[ \nabla \times \mathbf{E} = -\mathbf{k} \times \mathbf{E}_0 G'(\phi) = -\frac{\partial \mathbf{B}}{\partial t} \] **First homogeneous Maxwell equation is satisfied.** --- ### **Step 5: Compute \(\nabla \cdot \mathbf{B}\) again** Recall: \[ \nabla \cdot \mathbf{B} = c^{-1} \mathbf{e}_k \cdot (\nabla \times \mathbf{E}) \] But as above, \[ \nabla \times \mathbf{E} = -\mathbf{k} \times \mathbf{E}_0 G'(\phi) \] So, \[ \mathbf{e}_k \cdot (\mathbf{k} \times \mathbf{E}_0) = (\mathbf{e}_k \times \mathbf{k}) \cdot \mathbf{E}_0 = 0 \] because \(\mathbf{e}_k\) is parallel to \(\mathbf{k}\); their cross product is zero. Thus, \[ \nabla \cdot \mathbf{B} = 0 \] **Second homogeneous Maxwell equation is satisfied.** --- ## **Final Answer** **The given fields:** \[ \mathbf{E}(\mathbf{x}, t) = G(\mathbf{k} \cdot \mathbf{x} - ckt) \] \[ \mathbf{B}(\mathbf{x}, t) = c^{-1} \mathbf{e}_k \times \mathbf{E}(\mathbf{x}, t) \] **with \(\mathbf{e}_k = \mathbf{k}/k\), \(k = |\mathbf{k}|\), and \(\mathbf{E}_0 \cdot \mathbf{k} = 0\),** **satisfy the homogeneous Maxwell equations in vacuum:** \[ \nabla \cdot \mathbf{B} = 0, \quad \nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t} \] **as shown above.**