General Definition of Potential in Circuits with Time-dependent Fields? Electric potential is a very common observable to measure (especially in electro-engineering fields). It is measured using a multimeter, and thought of as the energy per charge that one charged particle gains when it is moved to a certain point of reference . However, since in a setup with time-dependent magnetic fields the Electric field is not conservative anymore, this concept of voltage becomes ill-defined for this cases. 
Wikipedia proposes a solution to this Problem, which is to split the Electric field into 2 components, one being rotation-free, and then defining the potential to be just the Energy per charge that one gets uppon integrating this rotation-free component. 
The Formula Wikipedia gives for this is:
\begin{align}
V(\vec{x}) = \int_{\vec{x}_0}^{\vec{x}} d\vec{x}\cdot\left(\vec{E} + \frac{\partial \vec{A}}{\partial t}\right)
\end{align}
With $\vec{A}$ being the Vector-Potential that the magnetic Field stems from. Of course, calculated this way, the potential Wikipedia gives this way is nothing but the Skalar Potential $\phi$ of the electromagnetic field. 
My problem now is that this skalar potential is not unique, but can be altered via gauge transformations. This wouldn't be a problem if it was just about adding constants, but since gauge transformations give more freedom to choosing $\phi$, this does as well mean that the potential difference between two points is no longer defined. 
So my question is: What gauge do we choose to define the potential used in circuit models in electro-engineering? Is that even the most general definition of "potential"? Is there something as consensus over the definition of "potential" that is used in different fields and applications?
 A: With variable currents, there are very classic experiments that show that two voltmeters connected to identical terminals give different indications if the magnetic flux through the circuit of the voltmeter is not negligible.
https://www.youtube.com/watch?v=FUUMCT7FjaI  (Walter Lewin. min 39...)
This means that in general we must renounce the definition of tension as the variation of a quantity between two points. And so give up the potential difference.
We can recover the usual law in the case of lumped circuits elements. The generalization of potential difference is the circulation of the electric field on a path to be defined, without using the field potentials $V$ and $\overrightarrow{A}$. 
This is pretty well detailed in the book "Classical Electromagnetism in a Nutshell", Garg, chap 17.
He defines a generic lumped circuit element as a circuit limited by a box such that "the line integral of $\overrightarrow{E}$ along any curve C lying outside the box is assumed to be independent of the path" (p 418). It is an approximate concept. He then defines the voltage drop as the circulation of the electric field $V(t)=\int\limits_{C}{\overrightarrow{E}\overrightarrow{dl}}$ . With this definition, With these definitions, he finds the usual impedances of capacitor, inductance and resistance.
Sorry for my poor english.
A: 
What gauge do we choose to define the potential used in circuit models in electro-engineering? Is that even the most general definition of "potential"?

In electrical engineering, one rarely hears world "potential", but one always hears "voltage", even in situations where the system is pervaded by solenoidal electric field. The reason is simple: voltage is a very useful concept not only in electrostatics but also in AC circuits; it always means "difference of the Coulomb potential", for the simple reason the resulting voltage (difference of potential) is connected in the most simple way with actual distribution of electric charges in the system, and can be always measured with cleverly positioned probe wires that cancel effect of induced emf.
The solenoidal part of electric field, if present, is handled by a different concept - induced emf. So, both concepts - voltage (difference of Coulomb potential) and emf (integral of induced electric field) - coexist and are used in general situations.
A: I will try to clarify what seems tricky to me with the potential in variable regime. I hope I can be clear and without mistake ! Sorry for my poor english.
An ohmic portion of conductor at rest with length l and section s is considered. It obeys Ohm's law $\overrightarrow{j}=\gamma \overrightarrow{E}=\gamma \left( -\overrightarrow{\nabla }V-\frac{\partial \overrightarrow{A}}{\partial t} \right)$, with $\overrightarrow{j}=\frac{I}{s}\overrightarrow{t}$introducing the unitary vector tangent to the conductor.
If we integrate along the conductor $\underbrace{\left( \int\limits_{C}^{D}{\frac{1}{\gamma }\frac{dl}{s}} \right)}_{{{R}_{CD}}}I=\int\limits_{C}^{D}{\left( -\overrightarrow{\nabla }V-\frac{\partial \overrightarrow{A}}{\partial t} \right)\centerdot \overrightarrow{dl}}={{V}_{(C)}}-{{V}_{(D)}}+\underbrace{\int\limits_{C}^{D}{\left( -\frac{\partial \overrightarrow{A}}{\partial t} \right)\centerdot \overrightarrow{dl}}}_{{{e}_{CD}}}$
At first sight, it is a very attractive formula that generalizes the law of Ohm : ${{V}_{(C)}}-{{V}_{(D)}}={{R}_{CD}}I-{{e}_{CD}}$.
But there are several problems :
*) Both terms ${{V}_{(C)}}-{{V}_{(D)}}$ and ${{e}_{CD}}=\int\limits_{C}^{D}{\left( -\frac{\partial \overrightarrow{A}}{\partial t} \right)\centerdot \overrightarrow{dl}}$ depend on the choice of the gauge. We can modify them. Only the sum has a meaning independent of the gauge.
*) Even assuming that the static gauge can be extended, ${{V}_{(M)}}=\int{\frac{dq(P)}{4\pi {{\varepsilon }_{0}}PM}}$, a voltmeter does not generally measure the potential difference. In total, it is expected that the voltmeter gives us the indication ${{R}_{CD}}I$. And the distribution between ${{V}_{(C)}}-{{V}_{(D)}}$and ${{e}_{CD}}$ is not easy to unravel.
An example : a square circuit of side $2a$, of total resistance $R$ which surrounds an infinite solenoid which generates a variable magnetic field. The external magnetic field is zero and one does not have to worry about the position of the voltmeter wires until it does not surround the solenoid.

We define ${{e}_{ind}}=-S\frac{dB}{dt}$  . This is the total emf induced along the conductor. The intensity is simply obtained by writing ${{e}_{ind}}=-S\frac{dB}{dt}=RI$
I think we will agree on the following results:
If we connect a voltmeter on one side, it will indicate ${{U}_{1}}={{e}_{ind}}/4$ (case 1)
If you connect a voltmeter on a half-side, it will indicate ${{U}_{2}}={{U}_{3}}={{e}_{ind}}/8$ whatever the position of the voltmeter. (case 2 and case 3)
Do you think these voltages are simply the variation of the Coulomb potential?
The induced electric field generated by the solenoid is easy to calculate using the Maxwell Faraday equation and the symmetries $\oint{\overrightarrow{E}\cdot \overrightarrow{dl}}={{e}_{ind}}=-S\frac{dB}{dt}$ or  ${{\overrightarrow{E}}_{\text{ind}}}=\frac{{{e}_{\text{ind}}}}{2\pi r}\overrightarrow{{{e}_{\theta }}}$
The electric field within the metallic conductor is also obtained using the Maxwell Faraday equation. But this field is directed in the direction of the conducting wire. For example, on the vertical right side ${{\overrightarrow{E}}_{\text{conductor}}}=\frac{{{e}_{\text{ind}}}}{8a}\overrightarrow{{{e}_{z}}}$.
The difference comes from the electric field generated by the charges on the conductor's surface. They force the field in the right direction. This field is an electrostatic type field and it derives from a Coulomb potential.
We find the following component $Oz$ of this field by making the difference: ${{E}_{\text{zCoulomb}}}=\frac{{{e}_{\text{ind}}}}{8a}-\frac{{{e}_{\text{ind}}}}{2\pi r}\cos (\theta )=\frac{{{e}_{\text{ind}}}}{8a}\left( 1-\frac{8}{2\pi }\frac{{{a}^{2}}}{{{z}^{2}}+{{a}^{2}}} \right)$ This Coulomb field varies along the vertical conductor. 
The variation of the associated Coulomb potential is easy to calculate $V({{z}_{2}})-V({{z}_{1}})=-\int\limits_{{{z}_{1}}}^{{{z}_{2}}}{{{E}_{\text{zCoulomb}}}dz=}-\frac{{{e}_{\text{ind}}}}{8}{{\left[ \left( \frac{z}{a}-\frac{4}{\pi }\arctan \left( \frac{z}{a} \right) \right) \right]}_{{{z}_{1}}}}^{{{z}_{2}}}$
You can check that:
*) $V(a)-V(-a)=0$ so in the first case, the voltmeter does not measure the Coulomb potential difference at all.
*) $V(a/2)-V(-a/2)\ne 0$ (case 3) while $V(0)-V(-a)=0$ (case 2)
So, in these two last cases, the voltmeter indicates the same result ${{U}_{2}}={{U}_{3}}={{e}_{ind}}/8$ but the two contributions of the Coulomb potential are very different.
Rather long text ! Hope there is no mistake and it can help !
