# Scalar and vector potential

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

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

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 wouldnt get that result. How I know how to take minus sign in this relations!

CompuChip
Homework Helper
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.

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.