# How to arrive at the identity between electric field $\vec{E}$ and magnetic vector potential $\vec{A}$?

We know that since $$\vec{\nabla}\cdot\vec{B}=\vec{0}$$, so $$\vec{B}$$ can be expresses as the curl of some special kind of vector function, i.e., $$\vec{B}=\vec{\nabla}\times\vec{A}$$.
Also, since always $$\vec{\nabla}\times\vec{E}=0,$$ thus $$\vec{E}$$ can be expressed as the gradient of some special scaler function. i.e, $$\vec{E}=-\vec{\nabla}V.$$

Now we have Faraday's equation, given as: $$\vec{\nabla}\times\vec{E}=-\frac{\partial\vec{B}}{\partial t},$$ which is applicable only when there is a changing magnetic field. So, $$\vec{\nabla}\times\vec{E}=0$$ applies when all fields are static, and the later applies in the non-static field scenario.

My question is, how can we arrive at the following equation: $$$$\vec{E}=-\vec{\nabla}V(\vec{r},t)-\frac{\partial}{\partial t}\vec{A}(\vec{r},t)$$$$

I can reason that this equation is true, since applying curl on both sides makes 1st term on the right zero, and second term gives partial derivative of magnetic field vector. So the result is nothing but Faraday's equation.

But I feel thats not the whole story.
Further, how to arrive at the said equation at first place? What concepts am I missing?

Contrary to your claim, $$\vec{\nabla} \times \vec{E}=0$$ does not hold "always", but only in the special case where $$\partial \vec{B}/ \partial t=0$$ can be assumed. Instead, being one of Maxwell's equations, $$\vec{\nabla} \times \vec{E}=-\partial \vec{B}/ \partial t$$ (in units where $$c=1$$) does indeed hold always.
Starting from the Maxwell equation $$\vec{\nabla} \cdot \vec{B} =0$$, one infers the existence of a vector field $$\vec{A}(t, \vec{x})$$ such that the magnetic field can be written as $$\vec{B} = \vec{\nabla} \times \vec{A}$$. This is a mathematical theorem holding under certain topological conditions being fulfilled here. Inserting this relation into the Maxwell equation $$\vec{\nabla} \times \vec{E}=-\partial \vec{B} /\partial t$$, one arrives at $$\vec{\nabla} \times (\vec{E} + \partial \vec{A} / \partial t)=0$$, which implies the existence of a scalar function $$V(t, \vec{x})$$ with $$\vec{E}+ \partial \vec{A} / \partial t= - \vec{\nabla} V$$ by a similar mathematical theorem (the minus sign is a convention).
We have thus found that the electric field $$\vec{E}$$ and the magnetic field $$\vec{B}$$ can be expressed in terms of the scalar potential $$V$$ and the vector potential $$\vec{A}$$ as $$\vec{E}= -\vec{\nabla} V - \frac{\partial \vec{A}}{\partial t}, \qquad \vec{B}= \vec{\nabla} \times \vec{A}.$$ Note that $$V$$ and $$\vec{A}$$ are not uniquely determined by $$\vec{E}$$ and $$\vec{B}$$ but only up to a possible "gauge transformation" $$V^\prime = V + \frac{\partial\Lambda}{\partial t}, \quad \vec{A}^{\, \prime} = \vec{A}- \vec{\nabla} \Lambda,$$ where $$\Lambda(t, \vec{x})$$ is an arbitrary scalar field.
• Is this $V$ and the "electric scaler potential" the same thing? Jan 16 at 10:25
• $V(t, \vec{x})$ is the scalar potential of the electromagnetic field (in general). In electrostatics / magnetostatics (where $\partial \vec{ B}/ \partial t=0$) it reduces to that what you might call "electric potential" with $\vec{E} = -\vec{\nabla} V(\vec{x})$. In this special case, you can always achieve a time-independent $V$ and a time-independent vector potential $\vec{A}$ in a suitable "gauge". Jan 16 at 10:38