I'm having a bit of trouble understanding how the gauge condition is found.
Consider the potentials V and →A and V′ and →A′ so that
→E=−→∇V−∂→A∂t=−→∇V′−∂→A′∂t →B=→∇×→A=→∇×→A′
The 2 sets of potentials are then related like this:
→A′=→A+→∇θ V′=V−∂θ∂t For a function θ.
The equations for the potentials are the following:
→∇2→A−εμ∂2→A∂t2=−μ→J+→∇(→∇.→A+εμ∂V∂t)
→∇2V−εμ∂2V∂t2=−ρε−∂∂t(→∇.→A+εμ∂V∂t)
They can be decoupled by choosing the right potentials so that (→∇.→A+εμ∂V∂t)=0
The above part is the theory in general.
Now to find the correct potentials where this is the case, let's say that this last part is true for potentials →A′ and V′, i.e.
→∇.→A′+εμ∂V′∂t=0
Filling in the relations between the 2 sets of potentials from earlier gives:
→∇.→A+∇2θ+εμ∂V∂t−εμ∂2θ∂t2=0
Yet my book tells me that this means that
∇2θ−εμ∂2θ∂t2=0
I don't see why this is the case.
Answer
You've just run into the fact that the Lorenz gauge is only a partial gauge choice; it does not uniquely specify the vector and scalar potentials. The condition given on θ ensures that, if your original A,V obeyed the Lorenz condition, then the new potentials A′,V′ will also obey the Lorenz condition.
Unique solutions for A,V arise only with additional conditions imposed (for instance, vanishing sufficiently fast at infinity).
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