# What is the physical consequence of the Lorenz Gauge Term not equaling zero?

What happens to the physics of the electromagnetic field if the Lorenz gauge term does not equal to zero?

\begin{align} \partial_{\mu}A^{\mu} \neq 0 \end{align}

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The demand that the Lorenz term $\partial_\mu A^{\mu}=0$ is a gauge fixing procedure applied to classical Maxwell field theory, due to invariance under a transformation,

$$A_{\mu} \to A'_\mu = A_\mu + \partial_\mu \alpha(x)$$

where $\alpha(x)$ is any scalar function of spacetime. If the term were set to not necessarily be zero for all field configurations, then the physics remain unchanged. All you would have done is not gauge fix the theory. The purpose of the term was to remove artificial degrees of freedom. The reason we often use Lorenz gauge (specifically when quantizing Maxwell's theory) is because it has the added benefit of preserving Lorentz covariance, unlike - for example - Coulomb gauge.

The reason we can always pick a representative field $A_{\mu}$ for every physically inequivalent configuration such that $\partial_\mu A^{\mu}=0$ is because we need only make a gauge transformation with $\alpha(x)$ satisfying,

$$\partial_\mu \partial^\mu\alpha(x)=-f(x)$$

where $f(x)=\partial_\mu(A')^\mu$. The differential equation always has a solution, and hence we can always find an $A_\mu$ satisfying the Lorenz gauge.

Remark: Minor typo edited thanks to Qmechanic.

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Physically, nothing happens. A choice of gauge does not have a physical interpretation, since it just removes unphysical modes from the theory. One might as well use another choice of gauge in which the quantity you mention is not equal to zero. For a discussion of possible gauge choices, see my answer to this question.

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