# Gauge invariance for the electric field

We know that $$\boldsymbol{E}=-\nabla V-\frac{\partial\boldsymbol{A}}{\partial t}$$

$$\boldsymbol{B}=\nabla\times\boldsymbol{A}$$ But I see that the following changes do not change these fields: $$\boldsymbol{A}\to\boldsymbol{A}+\nabla f$$ $$V\to V-\frac{\partial f}{\partial t}$$

My question is about $$V$$, why is the partial derivative with respect to time does not change the electric field? why would $$\nabla\left(\frac{\partial f}{\partial t}\right)=0$$ I don't see that explained anywhere, and I fail to understand how that is so trivial for any function $$f$$ that depends on position and time.

• Have you tried to just plug the gauge-transformed potentials into the expression of the electric field and see what happens? Commented Apr 18, 2021 at 13:33
• I feel stupid for not thinking about the cancelling terms, thought there is a deeper reason for it. Well, thank you. Commented Apr 18, 2021 at 14:00

The zero is due to canceling the part from vector potential change and the scalar potential change in the electric field definition.

E = −∇V − ∂A/∂t

B = ∇ × A

A → A + ∇f

V → V − ∂f/∂t

When you transform E, you have to transform both the V and the A which occur in E (and note that mixed partial derivatives commute):

E → −∇(V − ∂f/∂t) − (∂/∂t)(A + ∇f) = −∇V + ∇(∂f/∂t) - ∂A/∂t - ∇(∂f/∂t) = E