Verma S.P. said:
The
Coulomb gauge (also known as the
transverse gauge) is much used in
physics and is defined by the gauge condition (more precisely, gauge fixing condition)
It is particularly useful for "semi-classical" calculations in quantum mechanics, in which the vector potential is
quantized but the Coulomb interaction is not.
The Coulomb gauge has a number of properties:
- The potentials can be expressed in terms of instantaneous values of the fields and densities (in SI units)
where ρ(r, t) is the electric charge density,...
- so i will request you to visit wiki and see how your length gauge is related to coulomb vector potential/or see griffiths book on electrodynamics-
First of all, let's get this straight, because it's obviously not correct. The vector potential is not simply instantaneous (this cannot be, because relativity forbids the observable quantities, i.e., the electromagnetic field to relate in an action-at-a-distance way with the sources, i.e., the current and charge densities).
So let's start with the Maxwell equations, using
\begin{equation}
\label{1}
F_{\mu \nu}=\partial_{\mu} A_{\nu} - \partial_{\nu} A_{\mu},
\end{equation}
which uses the homogeneous Maxwell equations, which are constraint equations on the electromagnetic field. So we can forget about them completely with this ansatz with the four-vector potential. The disadvantage is that the four-potential is not observable, because it is not completely determined by the dynamical equations of motion, which come from the inhomogeneous Maxwell equations, which read
\begin{equation}
\label{2}
\partial_{\mu} F^{\mu \nu} = \frac{1}{c} j^{\nu}.
\end{equation}
The reason is "gauge invariance". Indeed the field-strength tensor does not change, when we add an arbitrary four-gradient field to the four-potential, i.e.,
\begin{equation}
\label{3}
A_{\mu}'=A_{\mu}+\partial_{\mu} \chi.
\end{equation}
Thus, we have to constrain the four-potential by one additional condition, which at least partially "fixes the gauge", so that the resulting dynamical equations have unique solutions.
Common gauges are the Lorenz gauge, which has the advantage of being manifestly covariant or the Coulomb gauge, which has advanatages in some calculations and in quantizing the electromagnetic field, but the disadvantage of spoiling manifest Lorentz covariance. A naive Lorentz transformation usually leads to four-potentials in another gauge, if the gauge condition is not a four-tensor equation. Of course, the general Lorentz transformation of the four-potential is a naive Lorentz transformation of the four-potential as a vector field followed by an arbitrary gauge transformation. Expressed in the new coordinates, you can restore the chosen gauge also to be valid in the new frame, but that's not relevant here. We just work in a fixed inertial reference frame.
The Coulomb-gauge condition reads
\begin{equation}
\label{4}
\vec{\nabla} \cdot \vec{A}=0.
\end{equation}
Because this is not Lorentz covariant, it is simpler to use the 1+3-dimensional formalism. The fields read
\begin{equation}
\label{5}
\vec{E}=-\frac{1}{c} \partial_t \vec{A} -\vec{\nabla} \Phi, \quad \vec{B}=\vec{\nabla} \times \vec{A}.
\end{equation}
I've written ##A^0=\Phi##. The gauge transformation (\ref{3}) translates into
\begin{equation}
\label{6}
\Phi'=\Phi+\frac{1}{c} \partial_t \chi, \quad \vec{A}'=\vec{A}-\vec{\nabla} \chi.
\end{equation}
The inhomogeneous Maxwell equations read
\begin{equation}
\label{7}
\vec{\nabla} \times \vec{B}-\frac{1}{c} \partial_t \vec{E}=\frac{1}{c} \vec{j}, \quad \vec{\nabla} \cdot \vec{E}=\rho.
\end{equation}
Plugging in (\ref{5}) we get
$$\vec{\nabla} \times (\vec{\nabla} \times \vec{A}) + \frac{1}{c} \partial_t \left (\frac{1}{c} \partial_t \vec{A}+\vec{\nabla} \Phi \right )=\frac{1}{c} \vec{j}$$
and with the Coulomb gauge condition (\ref{4})
\begin{equation}
\label{8}
\Box \vec{A}=\frac{1}{c} (\vec{j}-\vec{\nabla} \partial_t \Phi)=\frac{1}{c} \vec{j}_{\perp}.
\end{equation}
Further we have
\begin{equation}
\label{9}
\vec{\nabla} \cdot \vec{E}=-\vec{\nabla} \left (\frac{1}{c} \vec{A}+\vec{\nabla} \Phi \right)=-\Delta \Phi=\rho.
\end{equation}
Thus we have to use the retarded propagator to solve (\ref{8}),
\begin{equation}
\label{10}
\vec{A}(t,\vec{x})=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x}' \frac{\vec{j}_{\perp}(t-|\vec{x}-\vec{x}'|/c,\vec{x}')}{4 \pi c |\vec{x}-\vec{x}'|}.
\end{equation}
For the scalar potential we get from (\ref{9})
\begin{equation}
\label{11}
\Phi(t,\vec{x})=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x}' \frac{\rho(t,\vec{x})}{4 \pi |\vec{x}-\vec{x}'|},
\end{equation}
which appears not to be causal. One can, however prove, that the fields ##(\vec{E},\vec{B})## from the found solution for the four-potential is a retraded solution with the physical sources ##\rho## and ##\vec{j}## (Jefimenko equations).
Of course, we must have charge conservation, which is a integrability constraint, resulting from gauge invariance,
\begin{equation}
\label{12}
\partial_t \rho + \vec{\nabla} \cdot \vec{j}=0,
\end{equation}
and we thus have
\begin{equation}
\label{13}
\vec{\nabla} \cdot \vec{j}_{\perp}=\vec{\nabla} \cdot \vec{j}-\partial_t \Delta \Phi=0,
\end{equation}
because of (\ref{9}).
Finally to the original question. I don't understand it at all, because, of course, just using
##\chi=\vec{x} \cdot \vec{A}##
in the gauge transformation (\ref{6}) yields for the magnetic field
$$\vec{A}'=\vec{A}-\vec{\nabla} (\vec{x} \cdot \vec{A}).$$
Now writing this in the Ricci calculus (in 3D Cartesian coordinates, where I write all vector indices as lower indices) we get
$$A_j'=A_j-\partial_j (x_k A_k)=-x_k \partial_j A_k \neq 0.$$