Relationship between the continuity equation and the wave equation What exactly is the relationship between the continuity equation and the wave equation?
Suppose $J^\mu$ is a contravariant vector that satisfies the continuity equation $\partial_\mu J^\mu=0$. Let $J^\mu$ be defined by $J^\mu=\partial^\mu\varphi$, where $\varphi$ is some Lorentz scalar. Performing a simple substitution yields $\partial_\mu\partial^\mu\varphi=0$, or $\square\varphi=0$, where $\square$ is the D'Alembert operator. This seems to be a manifestation of the wave equation. Is this a correct derivation? If so, what is the physical interpretation behind it?
 A: The wave equation  can be written as $\nabla_i\nabla^i\varphi = 0$ where $\nabla$ is the Levi-Civita connection on Minkowski space, $\varphi$ does not need to be Lorentz invariant. $\partial_i\partial^i\varphi = 0$ is the Laplace equation on both Minkowski and Euclidean space since ordinary partial derivatives do not respect the metric.
In (pseudo)Riemannian geometry the covariant derivative $\nabla_i$ replaces partials $\partial_i$.
The Laplace-Beltrami operator 
$$\Delta \triangleq \nabla^i\nabla_i$$
is a common generalization of both the D'Alembertian and the Laplacian,
$$\Delta \varphi = 0$$
is the Laplace-Beltrami equation.
Your observation then generalizes to say that, for scalar field, a vanishing Laplace-Beltrami operator is the same as a divergence free contravariant gradient. This is universally valid since
$$\Delta \varphi = \nabla^i\nabla_i\varphi = g^{ij}\nabla_i\nabla_j\varphi= \nabla_i\nabla^i\varphi$$
It does not depend on the metric or coordinate system.
However, the precise geometric meaning of this depends on the metric.
In Euclidean spaces it means a function is harmonic iff it has a divergence free gradient since the Laplace-Beltrami operator is just the Laplacian.
In Minkowski space the Laplace-Beltrami operator is the D'Alembertian,
and the Laplace-Beltrami equation becomes the wave equation. In Minkowski space covariant derivatives are just ordinary partial derivatives, as in Euclidean space, since there is no curvature, making Christoffel symbols vanish.
Furthermore for scalars
$$\nabla_i \varphi = (d\varphi)_i = \partial_i\varphi$$
is true in all metrics.
However the contravariant derivative $\nabla^i\varphi$ is not the usual gradient vector since index raising depends on the the metric.
The gradient is therefore more naturally thought of as a covector or a 1-form.
Geometrically the Minkowski gradient vectors are the time reflections of what the Euclidean gradient vectors would be.
I do not have a good intuitive picture of contravariant derivatives and their divergences in general non Euclidean spaces though.
Note that for 
$$\varphi \triangleq t^2 + x^2$$
in Euclidean space
$$\nabla_i\nabla^i\varphi = 4$$
while in Minkowski space
$$\nabla_i\nabla^i\varphi = 0$$
making it a solution to the equation even though it is not Lorentz invariant.

A general vector field $J^i$ may satisfy $\nabla_i J^i = 0$, but still have non zero curl or in more generally $\nabla_{[i} J_{j]} \neq 0$. Such a field is not a gradient of a scalar field. So at least $\nabla_{[i} J_{j]} = 0$ is required. The Poincare lemma says this is sufficient for fields on contractible subsets of Euclidean space.
