Discussion
Under the Lorentz gauge condition, the electric potential V, is related to magnetic vector potential \(\vec{A}\) as :
Answer: B
The Lorentz gauge condition is \(\vec{\nabla}\cdot\vec{A}\) \(+ \mu_{0}\varepsilon_{0}\,\partial V/\partial t = 0\).
The Lorentz (Lorenz) gauge chooses the potentials so that the divergence of \(\vec{A}\) cancels the time derivative of V divided by \(c^{2}\). With \(c^{2} = 1/\mu_{0}\varepsilon_{0}\):
\[ \begin{aligned} &\vec{\nabla}\cdot\vec{A} \\ &+ \frac{1}{c^{2}}\frac{\partial V}{\partial t} = 0 \end{aligned} \]
In this gauge both V and \(\vec{A}\) satisfy inhomogeneous wave equations, and the condition is Lorentz invariant. Writing \(1/\mu_{0}\varepsilon_{0}\) instead multiplies by \(c^{2}\) where it should divide, the minus sign gives the wrong relative sign, and the second time derivative belongs to the wave equation, not the gauge condition.
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