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.