How do I determine the minus sign in gauge transformations?

[Petar Mali]

\vec{B}=rot\vec{A}

\vec{E}=-\frac{\partial\vec{A}}{\partial t}-grad\varphi


If I define

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

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

where

f=f(x,y,z,t)

I will get

\vec{B}=rot\vec{A}=rot\vec{\widetilde{\vec{A}}}

\vec{E}=-\frac{\partial\vec{A}}{\partial t}-grad\varphi=-\frac{\partial\widetilde{\vec{A}}}{\partial t}-grad\widetilde{\varphi}

But if I say

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

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

I wouldn't get that result. How I know how to take minus sign in this relations!

 

[CompuChip]

The transformation for A is clear, because you want B to stay the same and you simply exploit the fact that grad(rot F) = 0 for any vector field F.
So then you can simply define \tilde\phi = \phi + \delta, write out
-\frac{\partial \tilde{\vec A}}{\partial t} - \nabla \tilde\phi
and see what \delta has to be to cancel the extra term from the \tilde A-derivative so you get
-\frac{\partial \vec A}{\partial t} - \nabla \phi
back.

 

[Petr Mugver]

How I know how to take minus sign in this relations!

Another way to see it is with four vectors:

Define the (contravariant) four potential putting together the scalar and vector potential:

A^\mu=(\phi,\mathbf{A})

and define the (covariant) four derivative putting together the time derivative and the gradient:

\partial_{\mu}=(\partial_t,\nabla)

Then the gauge transformation is simply

A^\mu\longmapsto A^\mu+\partial^\mu f

and the fact that you have to raise the index of \partial_\mu yields the minus sign.