Relation between EM scalar potential and EM vector potential I am doing caculation using $\phi$ and $\mathbf{A}$, however I can't find the relation between $\phi$ and $\mathbf{A}$. For E and B we have
\begin{align}
\frac{{{d^2}E\left( x \right)}}{{d{x^2}}} + \frac{{{\omega ^2}}}{{{c^2}}}E\left( x \right) &= 0\\
\frac{{{d^2}B\left( x \right)}}{{d{x^2}}} + \frac{{{\omega ^2}}}{{{c^2}}}B\left( x \right) &= 0
\end{align}
in vacuum for 1-D case. Given $E(0) = E_0$ and $E'(0) = ik E_0$, the solution for E is $E(x) = E_0 e^{ikx}$. And we can find solution for B is $B(x) = \frac{E_0}{c} e^{ikx} $ for there is $|E|=c|B|$ in SI unit.
But I can't find the realtion between $A$ and $\phi$ similar to E and B for 
\begin{align}
\frac{1}{c^2}\frac{\partial^2\varphi}{\partial t^2} - \nabla^2{\varphi}  &= 0\\
\frac{1}{c^2}\frac{\partial^2\mathbf A}{\partial t^2} - \nabla^2{\mathbf A} &= 0
\end{align}
using Lorenz gauge.
What is the relation between $A$ and $\phi$ besides gauge condition?
 A: 
What is the relation between A and ϕ besides gauge condition?

Maxwell's equations in vacuum are, in terms of $\phi$ and $\mathbf{A}$:
$$\nabla^2\phi + \frac{\partial}{\partial t}\left(\nabla \cdot \mathbf{A}\right) = 0$$
$$\left(\nabla^2 \mathbf{A} -  \frac{1}{c^2}\frac{\partial^2\mathbf{A}}{\partial t^2}\right) - \nabla\left(\nabla \cdot \mathbf{A} +  \frac{1}{c^2}\frac{\partial\phi}{\partial t}\right) = 0 $$
Since $\frac{\partial\phi}{\partial t}$ and $\nabla \cdot \mathbf{A}$ are gauge degrees of freedom, it looks like the answer to your question is none.
A: If ${\bf A}$ satisfies a homogeneous wave equation of the form suggested, then
$${\bf A} = {\bf A_0} f({\bf k}\cdot {\bf r} - \omega t)$$
is a solution, with $\omega/k=c$.
In that case, I think that you can write
$$ \phi = \frac{c^2}{\omega} {\bf k}\cdot {\bf A} = \frac{c^2}{\omega} {\bf k}\cdot {\bf A_0}f,$$
which is also a solution to the homogeneous wave equation for $\phi$ and satisfies the Lorenz gauge. i.e.
$$\nabla \cdot {\bf A} = {\bf k}\cdot {\bf A_0} f'$$
and
$$-\frac{1}{c^2}\frac{\partial \phi}{\partial t} = -\frac{1}{c^2} \frac{c^2}{\omega}{\bf k}\cdot {\bf A_0}(-\omega f') = {\bf k}\cdot {\bf A_0}f'   $$
So your answer appears to be
$$ \phi = \frac{c^2}{\omega} {\bf k}\cdot {\bf A}$$
