I 4-Current vector potential transformation under Gauge fixing

AI Thread Summary
The discussion centers on transforming a given vector potential under the Lorenz Gauge condition. The initial vector potential is expressed as a function of time and space, leading to a transformation that introduces a new function f. The resulting equations indicate that fixing the gauge is challenging, particularly when aiming to maintain non-zero electric and magnetic fields. The conversation highlights the necessity of clarifying notation and utilizing the d'Alembert operator to solve for the gauge function. Ultimately, the transformation process and gauge fixing require careful consideration of the underlying mathematical relationships.
George444fg
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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?
 
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Here's the posting with LaTeX rendering:
George444fg said:
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

$$\begin{equation}
\nabla \cdot A +\frac{1}{c^2}\frac{\partial\phi}{\partial t}=0
\end{equation}$$
George444fg said:
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?
 
vanhees71 said:
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.
 
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