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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?

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    $\begingroup$ Possible duplicate of Are gauge choices in electrodynamics really always possible? $\endgroup$ – ACuriousMind Jul 7 '16 at 14:40
  • $\begingroup$ Thanks. This answers my main question. However, I am still curious why we don't have to explicitly add $\nabla\lambda$ or subtract the derivative from A and V respectively. In other words, why am I not forced to solve for $\lambda$? $\endgroup$ – JAustin Jul 7 '16 at 15:25
  • $\begingroup$ What are going to do with an explicit solution $\lambda$? $\endgroup$ – Vladimir Kalitvianski Jul 7 '16 at 16:23
  • $\begingroup$ I was under the impression that the degree of freedom allows us to adjust the vector potential only by adding some factor $\alpha$ with curl zero and simultaneously subtracting the time derivative of lambda from the vector potential so E and B remain the same. Thinking about it more though, the actual numerical value of A and V don't matter, so I suppose the values of A and V after the gauge transformation will differ from the pre-transformation value by that factor, without affecting the E and B fields. I think I understand. $\endgroup$ – JAustin Jul 7 '16 at 16:42