I am working through Griffith, and was confused by some of the restrictions he imposed on the potentials in introducing the Coulomb and Lorenz gauge.
It is easy to see that adding some arbitrary gradient field $\nabla\lambda$ to the vector potential and subtracting the derivative $\frac{\partial\lambda}{\partial t}$ from the scalar potential will not change the electric and magnetic fields. But following this introduction, he fixes the value of the divergence of A, not A itself, assuming at least without proof that there always exists a gradient field such that $\nabla\cdot\left(A+\nabla\lambda\right) = 0$. Is this always true? And furthermore, he never adds that $\nabla\lambda$ to any of the other occurrences of the vector potential in the equation, nor does he subtract the time derivative of the field from the scalar potential. If we're using the equation
$$\left(\nabla^2A -\mu_0\epsilon_0\frac{\partial^2 A}{\partial t^2}\right)-\nabla\left(\nabla\cdot A + \mu_0\epsilon_0\frac{\partial V}{\partial t}\right)=-\mu_0 J$$
To calculate the value of the vector potential, doesn't changing the value of A and the time dependence of V matter? Why are we not forced to solve for lambda?