← 2025 Paper 1

UPSC Maths 2025 Paper 1 Q6c-ii — Solution

10 marks · Section B

Question

If $\nabla \cdot \vec{E} = 0$, $\nabla \cdot \vec{H} = 0$, $\nabla \times \vec{E} = -\dfrac{\partial \vec{H}}{\partial t}$ and $\nabla \times \vec{H} = \dfrac{\partial \vec{E}}{\partial t}$, then show that $\nabla^2 \vec{H} = \dfrac{\partial^2 \vec{H}}{\partial t^2}$ and $\nabla^2 \vec{E} = \dfrac{\partial^2 \vec{E}}{\partial t^2}$.

Technique

Apply the vector identity $\nabla\times(\nabla\times\vec A) = \nabla(\nabla\cdot\vec A) - \nabla^2\vec A$ to the source-free Maxwell equations, then eliminate the other field using the curl equations. These are the wave equations (in units where $c=1$).

Solution

Use the standard identity, valid for any sufficiently smooth vector field $\vec A$: \nabla\times(\nabla\times\vec A) = \nabla(\nabla\cdot\vec A) - \nabla^2\vec A. \tag{$\ast$}

Wave equation for $\vec E$

Take the curl of $\nabla\times\vec E = -\dfrac{\partial\vec H}{\partial t}$: $\nabla\times(\nabla\times\vec E) = -\,\nabla\times\frac{\partial\vec H}{\partial t} = -\frac{\partial}{\partial t}\bigl(\nabla\times\vec H\bigr).$ (Space and time derivatives commute.) Substitute $\nabla\times\vec H = \dfrac{\partial\vec E}{\partial t}$: $\nabla\times(\nabla\times\vec E) = -\frac{\partial}{\partial t}\!\left(\frac{\partial\vec E}{\partial t}\right) = -\frac{\partial^2\vec E}{\partial t^2}.$ Now apply $(\ast)$ on the left, using $\nabla\cdot\vec E = 0$: $\nabla(\underbrace{\nabla\cdot\vec E}_{=0}) - \nabla^2\vec E = -\frac{\partial^2\vec E}{\partial t^2} \;\Longrightarrow\; -\nabla^2\vec E = -\frac{\partial^2\vec E}{\partial t^2}.$ Hence $\nabla^2\vec E = \frac{\partial^2\vec E}{\partial t^2}.$

Wave equation for $\vec H$

Take the curl of $\nabla\times\vec H = \dfrac{\partial\vec E}{\partial t}$: $\nabla\times(\nabla\times\vec H) = \frac{\partial}{\partial t}\bigl(\nabla\times\vec E\bigr) = \frac{\partial}{\partial t}\!\left(-\frac{\partial\vec H}{\partial t}\right) = -\frac{\partial^2\vec H}{\partial t^2},$ using $\nabla\times\vec E = -\dfrac{\partial\vec H}{\partial t}$. Apply $(\ast)$ with $\nabla\cdot\vec H = 0$: $\nabla(\underbrace{\nabla\cdot\vec H}_{=0}) - \nabla^2\vec H = -\frac{\partial^2\vec H}{\partial t^2} \;\Longrightarrow\; \nabla^2\vec H = \frac{\partial^2\vec H}{\partial t^2}.$

Answer

$\boxed{\;\nabla^2\vec E = \frac{\partial^2\vec E}{\partial t^2},\qquad \nabla^2\vec H = \frac{\partial^2\vec H}{\partial t^2}.\;}$