Why Div of Magnetic vector potential=0, i.e. $\nabla \cdot A = 0$?

Question

I know it's somehow a silly question but I'm stuck!
In Griffith, after introducing vector potential $A$ for a magnetic field $B$, he says we are free to assume that $$\nabla \cdot A = 0.$$
I understand this is a way to simplify our life so that:
$$\nabla \cdot ( A=A0+\nabla \lambda) =0$$ $$\nabla^2 \lambda = -\nabla \cdot A0$$
And now we can treat it as poisson's equation.
BUT why we are free to do so?

Answer 1

It's Coulomb gauge condition.

When you introduce the vector potential from the Gauss law for the magnetic field

$$\nabla \cdot \mathbf{b} = 0 \qquad \rightarrow \qquad \mathbf{b} =: \nabla \times \mathbf{a} \ ,$$

you can easily prove that this is a definition up to an additive gradient of a scalar field, i.e $\mathbf{a}$ and

$$\mathbf{a} + \nabla \lambda \ ,$$

produce the same magnetic field.

Now, if you write the equations using potentials as the primary unknowns, you get a non-determined problem with an infinite number of solutions (even if they produce the same magnetic field). In order to get a determined system, you need to add the condition about the divergence.

Answer 2

This is based on the principle of electromagnetic gauge invariance. According to this principle only $\bf E$ and ${\bf B}$ are physical. This is inspired on the Lorentz force, $${\bf f} = q{\bf E} + q{\bf v}\times {\bf B} \,.$$ Since only the rotation of ${\bf B} = \nabla \times {\bf A}$ is relevant, you are then free to choose $\nabla \cdot {\bf A} = 0$. This called a gauge choice and this particular choice is the radiation gauge. Other choices are the Lorenz (no typo) gauge and the Coulomb gauge.

Answer 3

Beware that it is $\boldsymbol{B(r)},\boldsymbol{E(r)}$ that really play the most important part in our real life, so we should first pay attention to the practical value of $\boldsymbol{B} \;and\; \boldsymbol{E}$ and others come after.

Suppose we've already known $\boldsymbol{B(r)},\boldsymbol{E(r)}$ in the space and want to know $\boldsymbol{A(r)}$ and $\varphi(\boldsymbol{r})$ too. From the definition formula we know $\boldsymbol{B}=\nabla \times \boldsymbol{A}$ and $\boldsymbol{E}=-\nabla \varphi-\displaystyle{\frac{\partial\boldsymbol{A}}{\partial t}}$. Obviously, we can't uniquely determine $\boldsymbol{A} \;and \;\varphi$ by these two equations. Hence, in order to determine the exact value of $\boldsymbol{A}\; and \;\varphi$, we'll have to artificially "create" another equation, i.e. gauge, which helps us simplify calculation under specific circumstances and discuss further about their properties.

In electrostatics, defining the created equation as $\nabla \cdot\boldsymbol{A}=0$ would play its simplification role a lot. Because, from the Maxwell's equations we know: $$\nabla\times \boldsymbol{B}=\nabla\times(\nabla\times\boldsymbol{A})=\nabla(\nabla\cdot\boldsymbol{A})-\nabla^{2}\boldsymbol{A}=\mu \boldsymbol{J}$$ Plug in the created equation, then we have: $$\nabla^{2}\boldsymbol{A}=-\mu\boldsymbol{J}$$ We can easily notice that this equation take the similar form of $\nabla^{2}\varphi=-\displaystyle{\frac{\rho}{\varepsilon}}$, so we can calculate $\boldsymbol{A}$ in the same way we calculate $\varphi$. And this created equation is called Coulomb gauge.

However, if you don't like Coulomb gauge, you can also create another equation that will help uniquely determine the value of $\boldsymbol{A}$. As long as your equation doesn't affect the practical value of $\boldsymbol{B(r)}\;and\;\boldsymbol{E(r)}$, it will be a usable gauge (unmindful of whether it will even lead to harder calculations).

Answer 4

You can set the divergence of $\vec{A}$ to anything you like, by choosing an appropriate form of the gauge field $\lambda$.

In the Coulomb gauge you have shown that the choice is that $\nabla^2\lambda$ is equal to the negative divergence of $\vec{A}_0$.

You are free to make that choice since it has no effect on the measured electric or magnetic fields since, in addition to adding $\nabla \lambda$ to $\vec{A}_0$ you must subtract $\partial \lambda/\partial t$ from the scalar potential $\phi_0$. $$\nabla \times \vec{A} = \nabla \times (\vec{A}_0 + \nabla \lambda) = \nabla \times \vec{A}_0$$ $$-\partial_t \vec{A} -\nabla\phi = -\partial_t(\vec{A}_0 + \nabla \lambda) - \nabla(\phi_0 - \partial_t \lambda) = -\partial_t \vec{A}_0 -\nabla\phi_0\ .$$