# How do I derive the Lorenz gauge from the continuity equation?

I was reading my old electromagnetics book (Elements of Electromagnetics, by Sadiku, 3rd edition) and after the author explained what the Lorenz gauge is mathematically and why it is useful in decoupling the scalar electrical potential from the vector magnetic potential to produce the wave equations, the author made a side comment below eq. (9.52) that the Lorenz gauge $$\nabla \cdot \vec A = -\mu\varepsilon \dfrac{\partial V}{\partial t}$$ can be derived from the continuity equation.

I don't have a background in relativity. I am an electrical engineering student. So given these limitations, how would I go about deriving the Lorenz gauge from the continuity equation?

Note: I am sticking to the conventions of my book in writing equations.

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The author probably means that it makes some equations derived from the continuity equation look simpler (some terms drop out). By construction, you cannot 'derive' a gauge, since it is a matter of choice. –  Vibert Jun 9 '13 at 22:08
The exact words were "It can be shown that the Lorentz condition can be obtained from the continuity equation. Therefore our choice of [Lorentz gauge condition] is not arbitrary." I think the author's point was that unlike the Coulomb gauge condition, div(A)=0, as used in the case of magnetostatics, which is arbitrarily chosen to make the math simpler, the Lorenz gauge has some other condition that makes the Lorenz gauge condition preferable to others that may be imagined. –  Sophisticated Idiot Jun 9 '13 at 22:45
By cointinuity equation, you mean $\nabla \cdot j + \partial_t \rho = 0$? –  Muphrid Jun 9 '13 at 22:47
If you write Maxwells equations in terms of the scalar and vector potentials using the Lorenz condition, you get $\Box \phi = -\rho$; $\Box {\bf{A}} = -{\bf{J}}$ (ignoring $c$, $\epsilon$, $\mu$). Taking the time derivative of the first:$\Box \frac{d\phi}{dt} = -\frac{d\rho}{dt}$and the div of the second $\Box \nabla.{\bf{A}} = -\nabla.{\bf{J}}$ Adding these equations, we get the continuity equation on the RHS, and $\Box$(Lorenz condition) on the LHS. –  twistor59 Jun 10 '13 at 18:23
So if the Lorenz condition holds, the continuity equation must hold. Not sure how to go the other way though, which is what you wanted I think.... –  twistor59 Jun 10 '13 at 18:24

We are for simplicity going to use the special relativistic formulation of E&M. Hopefully OP can still understand enough to extract an answer to his question.

Assume (-,+,+,+) signature and assume for simplicity units where $c=\varepsilon_0=\mu_0=1$. The continuity equation

$$\tag{1} \partial_{\mu} j^{\mu}~=~0$$

is a consequence of Maxwell's equations

$$\tag{2} \partial_{\mu} F^{\mu\nu}~=~j^{\nu}$$

with sources $j^{\mu}=(\rho,\vec{J})$. The existence of a $4$-gauge potential $A^{\mu}=(V,\vec{A})$ renders the source-free Maxwell equations trivial.

$$\tag{3} \partial_{\mu} A^{\mu}~=~0$$

(or any other gauge-fixing condition: Coulomb, axial, temporal, etc, for that matter) can not be derived from Maxwell's equations or its consequences. The whole point of a gauge-fixing condition is to (at least partially) fix the gauge ambiguity

$$\tag{4} A_{\mu}~\longrightarrow ~ A_{\mu}+\partial_{\mu}\Lambda.$$

Ref. 1 claims below eq. (9.52) that the Lorenz gauge-fixing condition (9.50) follows from the continuity eq. (1):

[...] It can be shown that the Lorentz$^1$ condition can be obtained from the continuity equation; therefore, our choice of eq. (9.50) is not arbitrary. [...]

Ref. 1 uses the retarded $4$-potential

$$\tag{5} A^{\mu}(\vec{x},t)~=~\iiint_V \frac{d^3y}{4\pi} \frac{j^{\mu}(\vec{y},t-R)}{R}, \qquad R~:=~|\vec{x}-\vec{y}|,$$

for the gauge potential $A^{\mu}$, see Ref. 1 eqs. (9.53-55). Indeed, the Lorenz gauge-fixing condition (3) follows from the retarded formula (5) and the continuity eq. (1) via integration by part. In fact, Maxwell equations [in Lorenz gauge and with sources, aka. wave equations with sources, see Ref. 1 eqs. (9.51-52)]

$$\tag{6} \Box A^{\mu}~=~j^{\mu} , \qquad \Box ~:=~\partial_{\mu}\partial^{\mu}~=~ \partial^2_t - \Delta, \qquad \Delta ~:=~\vec{\nabla} \cdot\vec{\nabla},$$

follows from the retarded formula (5).

What Ref. 1 fails to state is, that the retarded formula (5) is itself a gauge-fixing choice, although admittedly a very natural choice.

References:

1. M.N.O. Sadiku, Elements of Electromagnetics, 2009.

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$^1$ Ref. 1 wrongly spells Lorenz with a "t", cf. the Wikipedia page.

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A potential point of confusion: I read your text just below your eqn (4) as stating that the ref claims the continuity equation can be obtained from the Lorenz condition, but you then quote the ref as claiming the converse (as in the question). I presume the two claims should agree (or am I missing something). –  Art Brown Jun 11 '13 at 5:34
@Art Brown: Thanks for noticing the typo. I updated the answer. –  Qmechanic Jun 11 '13 at 8:02