# Electric potential vs electromagnetic potential questions

We were all taught about the electric potential $$V$$, which is defined up to a constant, and can be measured with a voltmeter or an oscilloscope. On the other hand, in electromagnetism are defined the scalar potential $$\varphi$$ and the vector potential $$\mathbf A$$, and there holds everywhere inside and outside the wires$$\nabla \varphi = - {\mathbf E} - {\partial {\mathbf A}\over \partial t}.$$

I always believed that the electric potential $$V$$ coincides with the electromagnetic scalar potential $$\varphi$$ inside the electric wires, and that's what is measured by a voltmeter or an oscilloscope. But thinking about, I get in trouble with that.

Question 1: The electromagnetic potential $$\varphi$$ is defined only up to a gauge. So, if the electric potential $$V$$ between two points $$A$$ and $$B$$ coincides with $$\varphi(B)-\varphi(A)$$, what is it? Let assume it's $$V = \int_{A}^B {\mathbf E}\cdot d{\ell}$$ (curvilinear integral along the wire between $$A$$ and $$B$$). Then we have $$V = \int_A^B (\nabla \varphi - {\partial {\mathbf A}\over \partial t})\cdot d \ell = \varphi(B) - \varphi(A) - \int_A^B {\partial {\mathbf A}\over \partial t}\cdot d \ell.$$ This works for steady currents, where there exists a gauge for which $$A$$ does not depend on $$t$$ (hence the integral vanishes). But for quickly varying currents like what can be measured with an oscilloscope (say 500 MHz), this fails: there is no electromagnetic potential that fulfills $$\varphi(B)-\varphi(A) = \int_A^B {\mathbf E}\cdot d\ell$$. So, what is the electric potential V?

question 2: I get even more in troubles when induction is involved: Consider a single loop solenoid, of surface $$S$$, with no resistance, subject to a uniform varying magnetic field of amplitude $$B(t)$$ normal to the loop. The two terminals of the solenoid are not connected, or more precisely are connected to an oscilloscope whose input impedance is very high (10 Mega ohm say). We are more or less taught that the voltage measured by the oscilloscope is $$V(t) = S{\partial B\over \partial t}$$. But it suffices to have a look at the demonstration of Faraday's law to see that all what is proved is that $$\oint {\mathbf E}\cdot d\ell = - {\partial B\over \partial t},$$ where the integral is the curvilinear integral over a spatial closed loop. By what magics do we deduce that the potential between the two terminals of the electrical wire loop is of the aforementioned form? (I expect a relatively mathematical justification).

question 3: To bring things to an apocalyptic state, I ask about the following problem, taken from the Wikipedia article "electromotive force". As in the previous question, assume we have an electric wire loop subject to a normal uniform varying magnetic field $$B(t)$$. But now, we assume that one half of the loop has a resistance $$R_1=100\ \Omega$$, and the other half has a resistance $$R_2 = 200\ \Omega$$. I believe we can suppose the induced current $$I(t)$$ is the same all along the loop, so, at the two terminals of the resistors , we have on one hand $$V = 100\ I$$, and on the other hand $$V = -200\ I$$. The electric potential is even not well defined in this case. I'm puzzled.

Q1: You got confused by using total electric field, whose integral can't uniquely define electric potential, because it depends on the path.

Electric potential (also in AC circuits with very high frequency, and even radiating circuits) is defined as integral of the conservative part of electric field. In other words, it is any function $$\varphi$$ of position, for which

$$-\nabla \varphi = \mathbf E_C.$$

The conservative part of electric field can be defined via the Helmholtz decomposition theorem.

https://en.wikipedia.org/wiki/Helmholtz_decomposition

This definition of potential is still ambiguous, because if $$\varphi$$ is electric potential, then $$\varphi + C$$ where $$C$$ is constant independent of position, is also electric potential.

This can be fixed and value of potential can be made unique. Let $$G$$ be the point where the electric potential is put by definition to be zero (often called "ground"). Then the unique value of electric potential at any point $$a$$ can be expressed as $$\varphi(a) = \int_a^G\mathbf E_C \cdot d\mathbf s.$$

Why using $$\mathbf E_C$$ instead of $$\mathbf E$$? Electric potential can't be defined as integral of total electric field, because in general this integral depends on the path.

We can introduce drop of potential when going from point $$a$$ to point $$b$$, and we can express it using $$\mathbf E_C$$ this way:

$$\varphi_a - \varphi_b = \int_a^b \mathbf E_C\cdot d\mathbf s.$$

Letter $$V$$ is better used for this drop of potential on two terminal devices ($$V$$ for voltage). In AC circuit theory, $$V = RI$$ for resistor, $$V = LdI/dt$$ for inductor, $$V = Q/C$$ for capacitor, and $$V = -\mathscr{E}$$ for voltage source of electromotive force $$\mathscr{E}$$.

This is why the KVL is valid in AC circuits; it is just rephrasing of the conservativeness of $$\mathbf E_C$$: $$\oint \mathbf E_C \cdot d\mathbf s = 0.$$

Q2: Potential drop on an inductor is $$LdI/dt$$ provided the only EMF active in the inductor is the self-induction EMF, i.e. there must not be any external source of EMF (no moving magnets, no other changing currents nearby the inductor). Then it can be derived from the Faraday law, and the zero ohmic resistance of the perfect inductor, that potential drop on the inductor equals minus induced EMF.

How? In perfect inductor body, there must not be net electric field (because of zero resistance). Hence integral of induced electric field must be the same but opposite sign to the integral of the conservative electric field. The first is the induced EMF $$-LdI/dt$$, so the second is $$LdI/dt$$.

Q3: In this case the wire experiences external EMF, and the standard formulae for potential drops like $$V = RI$$ are not applicable. Potential can be still uniquely defined (see above), and thus potential drop when going from $$a$$ to $$b$$ is defined, but this potential drop is not related to $$RI$$, because the electric current is not purely due to conservative field, but also due to induced non-conservative field, which potential drop does not capture.

Electric current in this scenario can be determined from the original Kirchhoff's second circuital law: sum of emfs acting on a closed conductive path equals sum of terms $$R_k I_k$$ over all parts of the conductive path. So we have

$$\text{total emf} = -\frac{d\Phi_B}{dt} = R_1I + R_2I$$ because currents in both parts of the loop are equal.

• thx for the effort. But how do you define the "conservative part" of the electric field? I think it is important this point be included inside you answer, otherwise the answer would be superficial. Commented May 26, 2022 at 13:26
• I've added reference to Helmholtz's theorem which in general can be used to define the conservative part of the field. Commented May 26, 2022 at 13:31
• thx. OK for q1. There remains two points a bit unclear: how to show that an induced E-field is solenoidal (that's what you say when you remove the induced field from the Ec field). And second, how to show that the potential drop of an inductor is $L dI/dt$ if the potential cannot be defined here (since the inductor acts on itself by induction, which prevents using potentials anyway). The usual proof only shows that it's the integral of the total field along the solenoid. (I think I will mark this question as the accepted answer, so, you may want to clarify these points inside your answer). Commented May 26, 2022 at 13:59
• Induced field is solenoidal field by definition $\mathbf E_i = \mathbf E - \mathbf E_C$ since this field has zero divergence everywhere. And potential can and is defined everywhere in inductor, since it is based on $\mathbf E_C$. I address this in my answer: potential drop has to be $LdI/dt$ on terminals of a perfect inductor in order for the conservative field integral to cancel out the induced electric field integral for path following the coils (emf) $-LdI/dt$. Commented May 26, 2022 at 23:48

Building on the answer of Ján Lalinský and on further research, I add here my own answer to the question.

question 1: It turns out that the electric potential $$V$$ measured at some point of the wires identifies, up to a constant, to the electromagnetic potential $$\varphi$$ under the Coulomb gauge, which vanishes at infinity (this last condition specifies $$\varphi$$ uniquely in the whole space). It can also be defined, up to a constant, as the integral of the conservative part of the electric field; this is the irrotational part of the E-field in the Helmholtz decomposition.

To see why these two definitions are equivalent, let $$\varphi$$ be the potential under the Coulomb gauge $$\nabla\cdot {\mathbf A}=0$$. This last condition implies that $$\mathbf A$$ is a solenoidal field: $${\mathbf A} = \nabla \times \mathbf F$$. We have Maxwell equation $${\mathbf E} = -\nabla \varphi - {\mathbf \partial_t A} = -\nabla \varphi - \partial_t (\nabla \times {\mathbf F}) = -\nabla \varphi - \nabla \times (\partial_t \mathbf F),$$ hence $$\nabla \varphi$$ is the conservative part of $$\mathbf E$$.

Unfortunately, there is some cheating here, because the "conservative part of the E-field" is not uniquely defined by Helmholtz decomposition which needs not be unique, even if it is specified as some "ground" point. Indeed, two possible functions $$V_1$$ and $$V_2$$ which agree at some point are possible: it suffices to choose an harmonic function $$\Lambda$$ equal to $$0$$ at the specified point, and to set $$V_2 = V_1+\Lambda$$. Then it is easily seen that if $$\mathbf E = \nabla V_1 + \nabla\times \mathbf F_1$$, then $$\mathbf E = \nabla V_2 + \nabla \times (\mathbf F_1- \mathbf F_2)$$, where $$\nabla \times \mathbf F_2 = \nabla \Lambda$$ (which is possible since $$\nabla^2\Lambda = 0$$ by definition).

So, it is probably better to define the electric potential $$V$$ as above, as the electromagnetic Coulomb potential $$\varphi$$ which vanishes at infinity, up to a constant.

question 2 and 3: Assume we have a filiform two terminal device $$D$$ of some shape. The device may be moving or deforming, so, we shall denote by $$\mathbf V(z)$$ the speed of the device at (curvilinear) coordinate $$z$$. Let $$\rho(z, t)$$ be the moving charge linear density at some point $$z$$ of $$D$$, and $$\mathbf v$$ its curvilinear speed. So, the total speed of the moving charges inside the conductor is $$\mathbf v + \mathbf V$$. We define the quantity $$V_T$$ by $$V_T = \int_D (\mathbf E + \mathbf V\times \mathbf B) \cdot d\vec\ell.$$ Pay attention that $$V_T$$ should not be seen as a potential between two points, but as an electromotive propensity along a (filiform) device.

Now, let us compute the electrical power transmitted by the electromagnetic field to $$D$$. Notice that only the component of the electromagnetic force parallel to the curvilinear speed $$\mathbf v$$ of the charges in the device provides electrical work to the device, hence $$P = \int_D d\mathbf F\cdot \mathbf v = \int_D (\rho\, d\ell )\,[{\mathbf E + (\mathbf v+\mathbf V)\times \mathbf B}]\cdot \mathbf v.$$ Since $$\mathbf v$$ is orthogonal to $$\mathbf v\times \mathbf B$$, this expression simplifies to $$P = \int_D \rho\, (\mathbf E + \mathbf V\times \mathbf B)\cdot \mathbf v \,d\ell = \int_D(\mathbf E + \mathbf V \times \mathbf B)\cdot \mathbf j \, d\ell,$$ where we have denoted by $$\mathbf j$$ the current density with respect to the conductor (the curvilinear current density component of the total current density). If we assume that the current is uniform inside $$D$$, which is a valid assumption as far as the wave length of the electric wave is large with respect to the dimension of the device, then $$||\mathbf j|| = I$$, $$\mathbf j = I{d\vec \ell \over d\ell}$$ and $$P = I\int_D (\mathbf E + \mathbf V \times \mathbf B)\cdot d\vec\ell = I V_T.$$ This is valid for every filiform two terminal devices.

On the other hand, one of the most basic law of electrical engineering is that the power transmitted to a two terminal device is $$P = VI$$, where $$V$$ is the electric potential between the two terminals. We see that this is valid only whenever $$V = V_T$$, that is, with Maxwell equation $$E = -\nabla \varphi - \partial_t \mathbf A$$, whenever $$\partial_t\mathbf A = 0$$ (under the Coulomb gauge). Since $$\mathbf B = \nabla \times \mathbf A$$, this implies $$\partial_t \mathbf B = 0.$$ The meaning of this equation is :

Electric laws of the form $$\Delta V = f(I)$$, related to two terminal devices, are valid only whenever there is no varying magnetic field along the device.

Conversely, if this condition holds, then $$\nabla \times \partial_t \mathbf A = 0$$, hence $$\partial_t A$$ is irrotational: $$\partial_t\mathbf A = \nabla \mathbf F$$. Coulomb gauge $$\nabla\cdot \mathbf A = 0$$ implies $$\nabla^2 \mathbf F = 0.$$
Observe that $$\mathbf E$$ vanishes at infinity (since the electric wires are contained in a bounded domain). From the definition of $$\varphi$$ and Maxwell equation $$\mathbf E = -\nabla \varphi - \partial_t \mathbf A,$$ it follows that $$\partial_t \mathbf A = \nabla \mathbf F$$ vanishes at infinity. Hence $$\mathbf F$$ is constant (uniqueness of the solution of Laplace equations with Neuman boundary conditions). Hence $$\partial_t \mathbf A = 0$$. We see that the above condition is also sufficient. In general, laws of the form $$V = f(I)$$ for two terminal devices have to be replaced by $$V_T = f(I),$$ and one has to be careful that $$V_T$$ is not a potential between two points, that would be independant of the path joining those points, but an "electromotrive force" along a given path, which depends also on the shape of the path. If the path is oriented, then $$V_T$$ is defined as above if the orientation is opposite to the current flow, and is defined as its opposite otherwise. With these conventions, everything holds true by replacing everywhere $$V$$ by $$V_T$$, except for the path dependence. Nevertheless, at those zones of the circuit where there is no magnetic induction, the electromotive force $$V_T$$ along the path joining two points $$A$$ and $$B$$ is equal to $$V(B) - V(A)$$, as explained above. So everything works, if one is careful to isolate those zones where there is some magnetic induction and to give them a special treatment.

This works in synergy with induction loops (inductor etc.), where all what is specified is the electromotive force as a function of the flux $$\Phi$$: $$V_T = \partial_t \Phi.$$

So, the apparent paradox in question 3 can be solved as follows: $$V_T = \partial_t \Phi = V_T({\rm path_1}, R_1) + V_T({\rm path_2}, R_2) = R_1 I + R_2 I = (R_1 + R_2)I.$$ Hence $$I = V_T/(R_1+R_2)$$.

At the terminals of the loop, and for any inductor in general, the magnetic induction is weak and we have $$V_T(\mathrm{along\ any\ path\ joining\ the\ terminals}) = \Delta V$$ (assuming the other paths joining the terminals lie in induction free zones).

• $V_{emf}$ is a confusing notation; it is better, to prevent maintaining the widespread confusion on these issues, to use symbol $\mathscr{E}$ for EMF, $\mathscr{E}_i$ for induced EMF, and reserve the symbol $V$ to drops of potential. Commented May 30, 2022 at 12:42
• @Ján Lalinský. Good Idea. Commented May 30, 2022 at 12:44
• However, induced EMF $\mathscr{E}_i$ is not integral of total electric field, but integral only of the induced field $\mathbf E_i$, i.e. solenoidal part of electric field. Commented May 30, 2022 at 12:44
• Your $V_{emf}$ is total impressed electric electromotive force (not a standard term), including the effects of induced EMF and potential drop. In ideal conductor, including ideal inductor, this is zero. But in real conductors it is not. Commented May 30, 2022 at 12:46
• So, what do you suggest for the integral of the total electric field? Commented May 30, 2022 at 12:46