- #1

George444fg

- 26

- 4

\begin{equation}

\vec{A} = \begin{pmatrix}

g(t,x)\\

0\\

0\\

g(t,x)\\

\end{pmatrix}

\end{equation}

And I would like to know how it will transform under the Lorenz Gauge transformation. I know that the Lorenz Gauge satisfy

\begin{equation}

\nabla \cdot A +\frac{1}{c^2}\frac{\partial\phi}{\partial t}=0

\end{equation}

So by applying a gauge transformation to my original expression I obtain that:

\begin{equation}

\tilde{\vec{A}} = \begin{pmatrix}

g(t,x)+\frac{\partial f}{\partial t}(t,x,y,z)\\

\frac{\partial f}{\partial x}(t,x,y,z)\\

\frac{\partial f}{\partial y}(t,x,y,z)\\

g(t,x)+\frac{\partial f}{\partial z}(t,x,y,z)\\

\end{pmatrix}

\end{equation}

That implies that:

\begin{equation}

\frac{1}{c^2}\frac{\partial g}{\partial t}+\frac{1}{c^2}\frac{\partial^2 f}{\partial^2 t}+\frac{\partial^2 f}{\partial^2 x}+\frac{\partial^2 f}{\partial^2 y}+\frac{\partial^2 f}{\partial^2 z} =0

\end{equation}

This expression doesn't help me a lot fixing my gauge. Except in the case that I take the f(t) but then $\partial_t{f} = g(t,x)+const$. But then $\tilde{A}$ gives back a 0 magnetic and electric field which is impossible. Probably I do somewhere a mistake, could you please help me out find out how to solve it?