Purcell in his book was deriving the vector potential $\bf A$ using $\text{curl}\;(\text{curl}\; \mathbf A)= \mu_0 \mathbf J\; .$
After some algebra, he came to this: $$-\frac{\partial^2 A_x}{\partial x^2}-\frac{\partial^2 A_x}{\partial y^2}- \frac{\partial^2 A_x}{\partial z^2} +\frac{\partial}{\partial x}\left(\frac{\partial A_x}{\partial x}+ \frac{\partial A_y}{\partial y}+\frac{\partial A_z}{\partial z}\right)= \mu_0 J_x.$$
Then he wrote:
[…] The quantity in parenthesis is the divergence of $\bf A\;.$ Now we have a certain latitude in the construction of $\bf A\; .$ All we care about is its curl; its divergence can be anything we like. Let us require that $$\text{div}\;\mathbf{A}= 0$$ ….
He gave the reason:
To see why we are free to do this, suppose we had an $\bf A$ such that $\text{curl}\;\mathbf A= \mathbf B,$ but $\text{div}\; \mathbf A= f(x,y,z)\ne 0.$ Treating $f$ like the charge density $\rho$ in electrostatic field , we obviously find a field $\bf F,$ the analogue of $\bf E,$ such that $\text{div}\; \mathbf F= f.$ But we know the curl of such a field is zero. Hence we could add $-\bf F$ to $A,$ making a new field with the correct curl and zero divergence.
I am having problem in understanding his reason.
First, shouldn't $\bf F$ be equal to $\bf A$ itself as the divergence of both is $f\;?$ Secondly, why should I add $-\bf F$ to $\bf A$ – wouldn't it nullify the function as both $\bf F$ & $\bf A$ are equal, isn't it?
Can anyone please help me understand his reasoning why we are free to take anything for $\text{div}\; \mathbf A\;?$
Best Answer
First, the physical thing we care about is $\vec B$. So we can do anything to $\vec A$ we like as long as we get the same $\vec B$. That is, we can do anything that doesn't change the curl of $\vec A$.
Now, suppose that $\vec \nabla\cdot\vec A = f$. Here's where Purcell neglects to stress what he means by "analogue of $\vec E$ in electrostatics" - the curl of an electrostatic field is zero! So $\vec F$ is a field with $\vec\nabla\cdot\vec F = f$, but $\vec \nabla\times\vec F = 0$. So $\vec A - \vec F$ now has zero divergence, but still the same curl.