and multiplying (2·1) by i we get

iex/∂x₁ + ∂iey/∂x₂ + ∂iez/∂x₃ = iρ = ρ₄ ... ... (2·2)

Now substitute

mx = f₂₃ = -f₃₂ and iex = f₄₁ = -f₁₄

my = f₃₁ = -f₁₃ iey = f₄₂ = -f₂₄

mz = f₁₂ = -f₂₁ iez = f₄₃ = -f₃₄

and we get finally:—

f₁₂/∂x₂ + ∂f₁₃/∂x₃ + ∂f₁₄/∂x₄ = ρ₁ }

f₂₁/∂x₁ + ∂f₂₃/∂x₃ + ∂f₂₄/∂x₄ = ρ₂ } ... (3)

f₃₁/∂x₁ + ∂f₃₂/∂x₂ + ∂f₃₄/∂x₄ = ρ₃ }