4-Current vector potential transformation under Gauge fixing

In summary, the conversation discusses an initial vector potential in the form of a four-vector. The Lorenz Gauge transformation is applied to this vector potential, resulting in a new four-vector with a gauge condition that can be solved using the retarded Green's function of the d'Alembert operator.
  • #1
George444fg
26
4
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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  • #2
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?
 
  • #3
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:
  • #4
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})##.
 
  • #5
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}
 
  • #6
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.
 

Related to 4-Current vector potential transformation under Gauge fixing

What is the 4-current vector potential transformation under gauge fixing?

The 4-current vector potential transformation under gauge fixing refers to the procedure in electromagnetism where the vector potential \(A_\mu\) is modified by a gauge transformation to simplify the equations of motion or to satisfy certain conditions (gauge conditions). This is often done to make calculations more manageable or to impose certain physical constraints.

Why is gauge fixing necessary in electromagnetism?

Gauge fixing is necessary in electromagnetism to remove the redundancy in the description of the electromagnetic field. The potentials \(A_\mu\) are not uniquely determined by the physical fields \(E\) and \(B\); they can be transformed without changing the physical fields. Gauge fixing imposes additional conditions to uniquely determine the potentials, simplifying the mathematical treatment and ensuring a unique solution to the equations of motion.

What are common gauge choices used in gauge fixing?

Common gauge choices include the Lorenz gauge, where \(\partial^\mu A_\mu = 0\), and the Coulomb gauge, where \(\nabla \cdot \mathbf{A} = 0\). Each gauge has its advantages depending on the problem being solved. The Lorenz gauge is often used in relativistic formulations, while the Coulomb gauge is frequently used in non-relativistic quantum mechanics and in problems involving static fields.

How does gauge fixing affect the physical observables of a system?

Gauge fixing does not affect the physical observables of a system, such as the electric and magnetic fields. These fields are gauge-invariant, meaning they do not change under a gauge transformation. Gauge fixing only affects the potentials \(A_\mu\), which are used as intermediate steps in calculations but do not directly correspond to physical observables.

Can gauge fixing be applied in quantum field theory?

Yes, gauge fixing is crucial in quantum field theory (QFT). In QFT, gauge fixing is used to handle the redundant degrees of freedom associated with gauge symmetries, allowing for a well-defined quantization procedure. The Faddeev-Popov procedure is a common method used to introduce gauge fixing in the path integral formulation of QFT, ensuring that the theory remains consistent and free of divergences due to gauge redundancy.

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