Electromagnetic Potential Gauge Transformations

[Petar Mali]

\Delta\vec{A}-\frac{1}{c^2}\frac{\partial^2 \vec{A}}{\partial t^2}=-\mu_0\vec{j}

\Delta\varphi-\frac{1}{c^2}\frac{\partial^2 \varphi}{\partial t^2}=-\frac{1}{\epsilon_0}\rho

c=\frac{1}{\sqrt{\epsilon_0\mu_{0}}}

\epsilon_0=8,85\cdot 10^{-12}\frac{F}{m}

\mu_0=4\pi 10^{-7}T

div\vec{A}+\frac{1}{c^2}\frac{\partial \varphi}{\partial t}=0

\frac{\partial A_x}{\partial x}+\frac{\partial A_y}{\partial _y}+\frac{\partial A_z}{\partial z}+\frac{\partial (\frac{1}{c}\varphi)}{\partial (ct)}=0

A^{\mu}=(\vec{A},\frac{1}{c}\varphi)

A_{\mu}=g_{\mu\nu}A^{\nu}

A_{\mu}=(-\vec{A},\frac{1}{c}\varphi)

divA^{\mu}=0

Can I say from

A^{\mu}=(\vec{A},\frac{1}{c}\varphi)

and

A_{\mu}=(-\vec{A},\frac{1}{c}\varphi)

something more about gauge transformations of electromagnetic potentials

\varphi_0=\varphi-\frac{\partial f}{\partial t}

\vec{A}_0=\vec{A}+gradf

I think about - sign in first term -\frac{\partial f}{\partial t} and + sign in second term +gradf

Or

\varphi_0=\varphi+\frac{\partial f}{\partial t}

\vec{A}_0=\vec{A}-gradf

\Delta A^{\mu}-\frac{1}{c^2}\frac{\partial^2 A^{\mu}}{\partial t^2}=-\mu_0j^{\mu}

j^{\mu}=(j_x,j_y,j_z,c\rho)

 

[arkajad]

The gauge transformation of A_\mu is

A_\mu\mapsto A_\mu +\frac{\partial f}{\partial x^\mu}

or, alternatively, which is the same replacing f by -f:

A_\mu\mapsto A_\mu -\frac{\partial f}{\partial x^\mu}

All the rest you derive from this basic formula.

 

[Petar Mali]

@ arkajad

Thanks!

 

[Dale]

Can I say from

A^{\mu}=(\vec{A},\frac{1}{c}\varphi)

and

A_{\mu}=(-\vec{A},\frac{1}{c}\varphi)

something more about gauge transformations of electromagnetic potentials

I thought that only the Lorentz gauge condition was Lorentz covariant. So does the four-potential in a different gauge even transform like a tensor? I would think not.