# 4-Current vector potential transformation under Gauge fixing

• I
• George444fg

#### George444fg

I am given an initial vector potential let's say:

\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?

• Kulkarni Sourabh

Here's the posting with LaTeX rendering:
I am given an initial vector potential let's say:

$$\begin{equation} \vec{A} = \begin{pmatrix} g(t,x)\\ 0\\ 0\\ g(t,x)\\ \end{pmatrix} \end{equation}$$

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?

Here's the posting with LaTeX rendering:
Its my bad, I meant that my original Gauge was in the form:

\begin{equation}
\vec{A} = \begin{pmatrix}
g(t+x)\\
0\\
0\\
g(t+x)\\
\end{pmatrix} = (\Phi, A)
\end{equation}

such that
\begin{equation}
\nabla \cdot A = g'(t+x) \Longrightarrow \Box \chi = g'(t+x)
\end{equation}

Now I apply the Lorenz Gauge to get a solution for A, in the Lorenz gauge.

Last edited:
I don't understand your notation. Please give your notation for the spacetime four-vector (standard is ##(x^{\mu})=(t,x,y,z)## (with ##c=1##) and also ##(A^{\mu})=(\Phi,\vec{A})##.

So to write down in the conventional form we have:

\begin{equation}
A^{\mu} = (f(t+y), f(t+y), 0, 0) = (\Phi, A)
\end{equation}

So you have
$$\partial_{\mu} A^{\mu}=\dot{f}(t+y)$$
If you now want a new four-potential such that the Lorenz gauge condition ##\partial_{\mu} A^{\prime '}=0## is fulfilled you make
$$A_{\mu}'=A_{\mu} + \partial_{\mu} \chi$$
and
$$\partial^{\mu} A_{\mu}' = \dot{f}(t+y)+\Box \chi=0 \; \Rightarrow \; \Box \chi=-\dot{f}(t+y),$$
which you can solve with, e.g., the retarded Green's function of the d'Alembert operator.