Why does the irrotational term in the vector potential equation suggest it may cancel a corresponding piece of the current density? In Jackson's book about vector potential of Maxwell's equation:
$$\mathbf{\nabla^2A}- \frac{1}{c^2}\frac{\partial^2 \mathbf{A}}{\partial^2 t}=-\mu_0 \mathbf{J} + \frac{1}{c^2}\nabla{\frac{\partial \Phi}{\partial t}}(6.24)$$

Since it (i.e. The last term that involves the scalar potential) involves the
  gradient operator, it is a term that is irrotational, that is, has
  vanishing curl. This suggests that it may cancel a corresponding piece
  of the current density.

Although it indeed cancel the irrotational part of $\mathbf{J}$, but that is a result of the mathematical deduction that follows on. Why the book says "vanishing curl suggests that it may cancel a corresponding piece" in that place of the book?
 A: Jackson's intention is this: he is assuming that the reader knows about Helmholtz decomposition theorem: every "nice-enough" vector field can be written as sum of two unique decaying-at-infinity component fields: irrotational field (zero curl) and solenoidal field (zero divergence). This applies to electric current density $\mathbf J$ too, so for given $\mathbf J$, there are two fields such that $\mathbf J = \mathbf J_{irr} + \mathbf J_{sol}$ and $\nabla \times \mathbf J_{irr} = \mathbf 0$ and $\nabla\cdot \mathbf J_{sol} = 0$.
Then, he is making the reader notice that in the expression
$$
-\mu_0 \mathbf{J} + \frac{1}{c^2}\nabla{\frac{\partial \Phi}{\partial t}}
$$
the potential term has zero curl, so it is irrotational and maybe this term can totally or partially cancel the other irrotational term $-\mu_0\mathbf J_{irr}$. Of course, this cancellation may not in general happen: the terms may be of the same sign, and even if they are of opposite signs, they may not cancel each other completely, because $\mathbf J$ can have non-zero curl.
However, it turns out that if $\Phi$ is given by the Coulomb formula, then the cancellation is perfect and the right-hand side can be written simply as
$$
-\mu_0\mathbf J_{sol}
$$
thus making the wave equation source term dependent only on current density: electric potential and charge density are removed from the equation. The current density $\mathbf J_{sol}$ is an artificial construct (for example it is non-zero in space points where no actual current is flowing) but it does depend only on current density field $\mathbf J$.
A: OK I found it. The equation 6,24 applies only to the Coulomb gauge Jackson explains exactly what he means   in eqs 6.27 and 6.28. Did you turn over the page to read this?
