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UPSC 2025 Maths Optional Paper 1 Q6c-ii — Step-by-Step Solution 10 marks · Section B
Vector identities (curl of grad, div of curl, product rules) · Vector Analysis · asked 4× in 14 yrs · Read the full method →
Question
If ∇ ⋅ E ⃗ = 0 \nabla \cdot \vec{E} = 0 ∇ ⋅ E = 0 ∇ ⋅ H ⃗ = 0 \nabla \cdot \vec{H} = 0 ∇ ⋅ H = 0 ∇ × E ⃗ = − ∂ H ⃗ ∂ t \nabla \times \vec{E} = -\dfrac{\partial \vec{H}}{\partial t} ∇ × E = − ∂ t ∂ H ∇ × H ⃗ = ∂ E ⃗ ∂ t \nabla \times \vec{H} = \dfrac{\partial \vec{E}}{\partial t} ∇ × H = ∂ t ∂ E ∇ 2 H ⃗ = ∂ 2 H ⃗ ∂ t 2 \nabla^2 \vec{H} = \dfrac{\partial^2 \vec{H}}{\partial t^2} ∇ 2 H = ∂ t 2 ∂ 2 H ∇ 2 E ⃗ = ∂ 2 E ⃗ ∂ t 2 \nabla^2 \vec{E} = \dfrac{\partial^2 \vec{E}}{\partial t^2} ∇ 2 E = ∂ t 2 ∂ 2 E
Technique
Apply the vector identity ∇ × ( ∇ × A ⃗ ) = ∇ ( ∇ ⋅ A ⃗ ) − ∇ 2 A ⃗ \nabla\times(\nabla\times\vec A) = \nabla(\nabla\cdot\vec A) - \nabla^2\vec A ∇ × ( ∇ × A ) = ∇ ( ∇ ⋅ A ) − ∇ 2 A c = 1 c=1 c = 1
Solution
Use the standard identity, valid for any sufficiently smooth vector field A ⃗ \vec A A
∇ × ( ∇ × A ⃗ ) = ∇ ( ∇ ⋅ A ⃗ ) − ∇ 2 A ⃗ . ( ∗ ) \nabla\times(\nabla\times\vec A) = \nabla(\nabla\cdot\vec A) - \nabla^2\vec A. \tag{$\ast$} ∇ × ( ∇ × A ) = ∇ ( ∇ ⋅ A ) − ∇ 2 A . ( ∗ ) Wave equation for E ⃗ \vec E E
Take the curl of ∇ × E ⃗ = − ∂ H ⃗ ∂ t \nabla\times\vec E = -\dfrac{\partial\vec H}{\partial t} ∇ × E = − ∂ t ∂ H
∇ × ( ∇ × E ⃗ ) = − ∇ × ∂ H ⃗ ∂ t = − ∂ ∂ t ( ∇ × H ⃗ ) . \nabla\times(\nabla\times\vec E) = -\,\nabla\times\frac{\partial\vec H}{\partial t} = -\frac{\partial}{\partial t}\bigl(\nabla\times\vec H\bigr). ∇ × ( ∇ × E ) = − ∇ × ∂ t ∂ H = − ∂ t ∂ ( ∇ × H ) . (Space and time derivatives commute.) Substitute ∇ × H ⃗ = ∂ E ⃗ ∂ t \nabla\times\vec H = \dfrac{\partial\vec E}{\partial t} ∇ × H = ∂ t ∂ E
∇ × ( ∇ × E ⃗ ) = − ∂ ∂ t ( ∂ E ⃗ ∂ t ) = − ∂ 2 E ⃗ ∂ t 2 . \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}. ∇ × ( ∇ × E ) = − ∂ t ∂ ( ∂ t ∂ E ) = − ∂ t 2 ∂ 2 E . Now apply ( ∗ ) (\ast) ( ∗ ) ∇ ⋅ E ⃗ = 0 \nabla\cdot\vec E = 0 ∇ ⋅ E = 0
∇ ( ∇ ⋅ E ⃗ ⏟ = 0 ) − ∇ 2 E ⃗ = − ∂ 2 E ⃗ ∂ t 2 ⟹ − ∇ 2 E ⃗ = − ∂ 2 E ⃗ ∂ t 2 . \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}. ∇ ( = 0 ∇ ⋅ E ) − ∇ 2 E = − ∂ t 2 ∂ 2 E ⟹ − ∇ 2 E = − ∂ t 2 ∂ 2 E . Hence
∇ 2 E ⃗ = ∂ 2 E ⃗ ∂ t 2 . \nabla^2\vec E = \frac{\partial^2\vec E}{\partial t^2}. ∇ 2 E = ∂ t 2 ∂ 2 E . Wave equation for H ⃗ \vec H H
Take the curl of ∇ × H ⃗ = ∂ E ⃗ ∂ t \nabla\times\vec H = \dfrac{\partial\vec E}{\partial t} ∇ × H = ∂ t ∂ E
∇ × ( ∇ × H ⃗ ) = ∂ ∂ t ( ∇ × E ⃗ ) = ∂ ∂ t ( − ∂ H ⃗ ∂ t ) = − ∂ 2 H ⃗ ∂ t 2 , \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}, ∇ × ( ∇ × H ) = ∂ t ∂ ( ∇ × E ) = ∂ t ∂ ( − ∂ t ∂ H ) = − ∂ t 2 ∂ 2 H , using ∇ × E ⃗ = − ∂ H ⃗ ∂ t \nabla\times\vec E = -\dfrac{\partial\vec H}{\partial t} ∇ × E = − ∂ t ∂ H ( ∗ ) (\ast) ( ∗ ) ∇ ⋅ H ⃗ = 0 \nabla\cdot\vec H = 0 ∇ ⋅ H = 0
∇ ( ∇ ⋅ H ⃗ ⏟ = 0 ) − ∇ 2 H ⃗ = − ∂ 2 H ⃗ ∂ t 2 ⟹ ∇ 2 H ⃗ = ∂ 2 H ⃗ ∂ t 2 . \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}. ∇ ( = 0 ∇ ⋅ H ) − ∇ 2 H = − ∂ t 2 ∂ 2 H ⟹ ∇ 2 H = ∂ t 2 ∂ 2 H . Answer
∇ 2 E ⃗ = ∂ 2 E ⃗ ∂ t 2 , ∇ 2 H ⃗ = ∂ 2 H ⃗ ∂ t 2 . \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}.\;} ∇ 2 E = ∂ t 2 ∂ 2 E , ∇ 2 H = ∂ t 2 ∂ 2 H .