I am having trouble reconciling the retarded potentials, with a possibility for a background homogenous solution to the EM field to exist.
In the Lorenz gauge $$\nabla \cdot \vec{A} = – \mu_0 \epsilon_0 \frac{\partial \phi}{\partial t},$$
Maxwell eqs. become
$$\nabla^2 \phi – \mu_0 \epsilon_0 \frac{\partial^2 \phi}{\partial t^2}= -\frac{\rho}{\epsilon_0}$$
$$\nabla^2 \vec{A} – \mu_0 \epsilon_0 \frac{\partial^2 \vec{A}}{\partial t^2}= -\mu_0 \vec{J}.$$
A set of potentials that satisfy these conditions are the retarded potentials.
$A(r,t)= \frac{\mu_0}{4\pi}\iiint \frac{\vec{J}(r',t_{r})}{|\vec{r}- \vec{r}'|} d^3 r'$
$\phi(r,t)= \frac{1}{4\pi\epsilon_{0}}\iiint \frac{\rho(r',t_{r})}{|\vec{r}- \vec{r}'|} d^3 r'$
Now as far as I'm aware, when $r = \infty$ $(\phi,\vec{A} = 0)$
Now, because
$\vec{E} = -\nabla \phi – \frac{\partial \vec{A}}{\partial t}$
$\vec{B} = \nabla × \vec{A}$
Now correct me if I'm wrong :
When $r =\infty$,
I would expect $\vec{E} = 0$, $\vec{B} = 0$
(This part is probably where I'm getting confused, as there probably is a non zero gradient and derivative,) ***
However I am aware that the electric and magnetic fields satisfy the inhomogenous wave equation, and thus any homogenous solution can be added. Meaning if one were to exist, E and B would not be zero near infinity?
But the retarded potentials seem to imply otherwise ( provided my conclusions are correct)
I remember asking a similar question before, and was told that jefimenkos equations are a PARTICULAR solution to maxwells equations, which is why there is no homogenous solution.
However I cannot find any particular solution to this problem in this 'derivation', as although YES, the retarded potentials are particular solutions to the inhomogenous wave equation, it shouldn't matter as any potential will leave the field invariant.
And if my *** Statement is correct, where do I input my homogenous solution? Or set the derivatives when solving for the electric and magnetic fields?
Edit:
For clarification, I am fully aware that I can add the homogenous solution to the retarded potential, and will still satisfy the inhomogenous wave equation.
My assertion isn't that we can't. I am stating that, adding the homogenous solution to the potentials doesn't change the fields. Remembering that when we add the homogenous solution onto $\vec{A}$, we also have to add a homogenous solution onto $\phi$, in order to satisfy the lorenz gauge condition. Maintaining the lorenz gauge condition, even adding the homogenous solution onto the potentials, does NOT give the electric and magnetic fields a homogenous wave equation solution. As the homogenous solution given to $\phi$ in order to maintain the lorenz gauge, cancels out with the homogenous solution added to $\vec{A}$, and I can prove it with a counter example. All potentials in the lorenz gauge when $\rho = 0$, give 0 fields, which YES is technically a solution to the homogenous wave equation, but it is not the full Solution.
The lorenz gauge In freespace,
$\nabla^2 \vec{A} = \mu_0 \epsilon_0 \frac{\partial^2 \vec{A}}{\partial t^2}$
Now when $\vec{A}$ is chosen to be 0, this represents a 0 homogenous solution added to the retarded potentials, for when we are not in freespace.
Let's add a homogenous solution onto the retarded potentials, aka, by choosing A to be non zero whilst also satisfying the homogenous wave equation.
A potential that satisfies the homogenous wave equation is –
$\vec{A} = cos(kx-\omega t)\hat i$
$\nabla \cdot \vec{A} = -k sin(kx-\omega t)$
In order to maintain the lorenz gauge condition.
$-k sin(kx-\omega t) = -\mu_0 \epsilon_{0} \frac{\partial \phi}{\partial t}$
$\phi = \frac{k}{\omega \mu_0 \epsilon_{0}}cos(kx-\omega t)$
$-\nabla \phi = \frac{k^2}{\omega \mu_0 \epsilon_0}sin(kx-\omega t) \hat i$
$-\frac{\partial \vec{A}}{\partial t} = -\omega sin(kx-\omega t)\hat i$
So let's find the field!
The coefficient in question is $\frac{k^2}{\omega \mu_0 \epsilon_0} – \omega$
$$\frac{k^2}{\omega \mu_0 \epsilon_0} – \omega = 0 $$
When the potential is chosen to be non zero in free space, there is still no EM wave, the lorenz gauge inheretly sets the homogenous E field to 0.
Meaning EVEN AFTER adding the homogenous solution to the retarded potentials, we STILL get DO NOT get a homogenous E field added to the fields produced by charges.
Best Answer
Let's set $c=1$. There really isn't anything special about retarded potentials - they are particular solutions of Maxwell's equations for the potential in the Lorenz gauge $$ \partial_\mu\partial^\mu A^\nu = J^\nu\tag{1}$$ (where $A^0 = \phi$ and $J^0=\rho$ - since the Lorenz gauge conditions is Lorentz-invariant, we can use covariant language here). Since eq. (1) is the inhomogeneous version of $$\partial_\mu\partial^\mu A^\nu = 0,\tag{2}$$ the general theory of inhomogeneous equations applies - if $A_\text{ret}^\mu$ is a particular of eq. (1) (for instance the usual retarded potentials) and $A_\text{hom}$ is any solution of eq. (2), then $A_\text{ret}^\mu + A_\text{hom}^\mu$ is a solution of eq. (1), too. The Lorenz gauge condition is linear as well: If $\partial_\mu A_\text{ret}^\mu = 0$ and $\partial_\mu A_\text{hom}^\mu = 0$, then $\partial_\mu(A_\text{ret}^\mu + A_\text{hom}^\mu) = 0$, too. In particular you can choose $A_\text{hom}^\mu$ to be the usual plane waves for EM waves in all of space that don't vanish at infinity.
This is exactly analogous to us solving the equations in terms of $\vec E$ and $\vec B$ and being able to add homogeneous solutions to one particular inhomogeneous solution - it's the same mathematical principle, and it's completely irrelevant that the equations are in terms of potentials and not fields now.