# 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?

• Note that it is $\partial_t \mathbf A$ which is divergenceless. The integrand contains also total electric field, which in general, has non-zero divergence. – Ján Lalinský Feb 8 at 21:56
• I corrected myself. I meant to write "rotation-free", because the component of the field that stems from the vector potential is effectively subtracted from the general electric Field. – Quantumwhisp Feb 8 at 22:00

## 3 Answers

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.

• The question is "what does voltmeter mesure ?" it's the name of an old article of the AJP I read nearly 40 years ago. The answer was : the circulation of the electric field along the circuit of the voltmeter. If the measurement of the "difference of potential" between two points A and B does not depend only on these two points, we can not express it as a difference $f (A) -f (B)$ ? – Vincent Fraticelli Feb 8 at 19:38
• Prof. Lewin is going against standard engineering practice/terminology. It is true that the voltmeters will show different results, but this will always be the case if the probe leads are allowed to capture net induced emf from solenoidal electric fields (which always exist somewhere). If the voltmeter leads are positioned to negate the effect of the electromotive force, voltage can be measured (difference of Coulomb potential). – Ján Lalinský Feb 8 at 19:38
• What real voltmeter in real circuit measures depends on positioning of its probe wires and the type of emf present. Sometimes the circulation is zero (such as when the voltage source of the circuit is Volta cell or a battery of them) so voltmeter can measure difference of Coulomb potential (provided electrochemical potentials at the contacts are taken into consideration). Other times, when current is due to induced emf, it measures that induced emf, either by measuring tiny current flowing throught the voltmeter or by creating an actual voltage across it leads and measuring field effect. – Ján Lalinský Feb 8 at 19:51
• Another book where the issue of defining voltage for lumped elements is treated in detail is R. M. Fano, L. J. Chu, R. B. Adler, Electromagnetic Fields, Energy, and Forces. – Massimo Ortolano Feb 8 at 19:56
• @Vincent Fraticelli: It seems to me that by this definition, the measured voltage is exactly the difference in the coulomb-pottential (which is the potential generated only by the charges, and which is the same as the skalar-potential in the coulomb-gauge)- would you agree on that? – Quantumwhisp Feb 9 at 12:06

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 !

• The voltmeter won't show some fixed part of net emf for the main circuit such as $e_{ind}/4$ for case 1, which you suggest. What the voltmeter will show will depend on how the wires of the voltmeter probes are positioned. By aligning them with the square side, the voltmeter reading can be made 0 which is correct value of voltage (difference of potential) if the source of B does not have any concentrated charges in it. By increasing the area of the voltmeter circuit, the voltmeter ceases to measure voltage and shows rather emf in the voltmeter circuit, which can be made arbitrary large. – Ján Lalinský Feb 9 at 18:41
• I do not think that the position of the voltmeter has an influence on this measurement since the external magnetic field is null : no change of flux if you move the probes. Simply, the probes do not have to completely surround the solenoid, which is supposed. The situation is not cylindrically symmetrical for the circuit. This is the reason why charges appear on the conductor. – Vincent Fraticelli Feb 9 at 19:01
• Oh, I see that you have conveniently chosen an infinite solenoid so there is no magnetic flux through the voltmeter circuit. In that case, the voltmeter will always show 0 no matter the position of probe wires, because the voltmeter circuit experiences zero emf. In this special case the reading is also valid as voltage, because there is no source of electric potential variation, all points on the conducting square have the same potential. – Ján Lalinský Feb 9 at 19:22
• I disagree with you on this point. But we probably are too far away to make the experiment. I choose a geometry rather schematic to make the computation easy, but would you say that if I put four equal resistors around the magnetic core of a transformer, and a voltmeter connected to one resistor, the voltage indicated by the voltmeter would be zero ? – Vincent Fraticelli Feb 9 at 21:21
• It would be interesting to do this experiment with a solenoid, I'll try to do that later next week with my friend who has an oscilloscope. With magnetic core of a transformer, the geometry is more complicated, but if the wires are placed so net induced emf in the voltmeter circuit is zero, then the voltmeter should indicate potential difference across the resistor. For symmetrical placement of the resistor with respect to the core, that should be zero, even if there is current running through the resistor. – Ján Lalinský Feb 9 at 21:31

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.

• Could the downvoter explain his problems with this answer? I find it rather useful. – Quantumwhisp Feb 8 at 20:13
• I second @Quantumwhisp 's request. I'd like to learn something new here if possible. – Ján Lalinský Feb 8 at 21:48
• Could you ellaborate on how exactly induced emf and voltage (talking about concepts) coexist? Are there situations in which they interfere, or are simultaneously used? – Quantumwhisp Mar 13 at 16:47
• @Quantumwhisp emf is a more general concept, that has several different instances ( chemical emf, motional emf, induced emf, thermoelectric emf...) Voltage, in my mind, is specific to difference of electric potential. Elaborating on this could be several pages, so maybe you can ask a new question on this. – Ján Lalinský Mar 13 at 18:55