# Why do we have the negative on the RHS of $\Delta \mathscr{l} E_{t1} - \Delta \mathscr{l} E_{t2} = -\Delta \mathscr{l} M_s$?

I am currently studying the textbook Microwave Engineering, fourth edition, by David Pozar. Section Fields at a General Material Interface of chapter 1.3 FIELDS IN MEDIA AND BOUNDARY CONDITIONS says the following:

For the tangential components of the electric field we use the phasor form of (1.6), $$\oint_C \bar{E} \cdot d\bar{l} = -j\omega \int_S \bar{B} \cdot d\bar{s} - \int_S \bar{M} \cdot d \bar{s}, \tag{1.33}$$ in connection with the closed contour $$C$$ shown in Figure 1.7. In the limit as $$h \to 0$$, the surface integral of $$\bar{B}$$ vanishes (because $$S = h \Delta \mathscr{l}$$ vanishes). The contribution from the surface integral of $$\bar{M}$$, however, may be nonzero if a magnetic surface current density $$\bar{M}_s$$ exists on the surface. The Dirac delta function can then be used to write $$\bar{M} = \bar{M}_s \delta(h), \tag{1.34}$$ where $$h$$ is a coordinate measured normal from the surface. Equation (1.33) then gives $$\Delta \mathscr{l} E_{t1} - \Delta \mathscr{l} E_{t2} = -\Delta \mathscr{l} M_s,$$ or $$E_{t1} - E_{t2} = -M_s, \tag{1.35}$$ which can be generalised in vector form as $$(\bar{E}_2 - \bar{E}_1) \times \hat{n} = \bar{M}_s. \tag{1.36}$$ (1.6) is given as follows:

Applying Stokes' theorem (B.16) to (1.1a) gives $$\oint_C \bar{\mathcal{E}} \cdot d\bar{l} = - \dfrac{\partial}{\partial{t}} \int_S \bar{\mathcal{B}} \cdot d \bar{s} - \int_S \bar{\mathcal{M}} \cdot d \bar{s}, \tag{1.6}$$ which, without the $$\bar{\mathcal{M}}$$ term, is the usual form of Faraday's law and forms the basis for Kirchhoff's voltage law.

Chapter 1.2 Maxwell's Equations introduces Maxwell's equations as follows:

The general form of time-varying Maxwell equations, then, can be written in "point," or differential, form as

$$\nabla \times \overline{\mathcal{E}} = \dfrac{-\partial{\overline{\mathcal{B}}}}{\partial{t}} - \overline{\mathcal{M}}, \tag{1.1a}$$ $$\nabla \times \overline{\mathcal{H}} = \dfrac{\partial{\overline{\mathcal{D}}}}{\partial{t}} + \overline{\mathcal{J}}, \tag{1.1b}$$ $$\nabla \cdot \overline{\mathcal{D}} = \rho, \tag{1.1c}$$ $$\nabla \cdot \overline{\mathcal{B}} = 0 \tag{1.1d}$$ The MKS system of units is used throughout this book. The script quantities represent time-varying vector fields and are real functions of spatial coordinates $$x$$, $$y$$, $$z$$, and the time variable $$t$$. These quantities are defined as follows:

$$\overline{\mathcal{E}}$$ is the electric field, in volts per meter $$(\text{V}/\text{m})$$.
$$\overline{\mathcal{H}}$$ is the magnetic field, in empires per meter $$(\text{A}/\text{m})$$.
$$\overline{\mathcal{D}}$$ is the electric flux density, in coulombs per meter squared ($$\text{Coul}/\text{m}^2$$).
$$\overline{\mathcal{B}}$$ is the magnetic flux density, in webers per meter squared ($$\text{Wb}/\text{m}^2$$).
$$\overline{\mathcal{M}}$$ is the (fictitious) magnetic current density, in volts per meter $$(\text{V}/\text{m}^2)$$.
$$\overline{\mathcal{J}}$$ is the electric current density, in amperes per meter squared ($$\text{A}/\text{m}^2$$).
$$\rho$$ is the electric charge density, in coulombs per meter cubed ($$\text{Coul}/\text{m}^3$$).

Equation (1.33) then gives $$\Delta \mathscr{l} E_{t1} - \Delta \mathscr{l} E_{t2} = -\Delta \mathscr{l} M_s,$$ or $$E_{t1} - E_{t2} = -M_s, \tag{1.35}$$ which can be generalised in vector form as $$(\bar{E}_2 - \bar{E}_1) \times \hat{n} = \bar{M}_s. \tag{1.36}$$

Why do we have the negative on the RHS of $$\Delta \mathscr{l} E_{t1} - \Delta \mathscr{l} E_{t2} = -\Delta \mathscr{l} M_s$$? It seems to me that it has something to do with having $$E_{t1} - E_{t2}$$, and that $$E_{t2} - E_{t1}$$ instead would have been positive ($$\Delta \mathscr{l} E_{t2} - \Delta \mathscr{l} E_{t1} = \Delta \mathscr{l} M_s$$). But I don't understand why the order of subtraction for the electric fields matters here. What does it matter if we subtract the electric field of medium 1 from the electric field of medium 2 or vice-versa? What leads to the sign change on the RHS of the equation?

• Just some thoughts, but could the problem be within choice/convention? I think the negative initially arises from Maxwell's equations, and the author chooses to subtract 2 from 1 to preserve the convention of negative M, but then generalizes the result in vector form in eqn 1.36. So from 1.36, what happens if we apply the right hand rule? M points out of page. But what if we then swap the order of n and E? M points into the page. In the book, eqn 1.40c shows this generalized result. Mar 28, 2022 at 0:11
• @bleuofblue Yes, I think that could be it. When you say "preserve the convention of negative $M$," are you referring to (1.1a)? Mar 29, 2022 at 19:00
• Yes I'm referring to 1.1a, the general form of Maxwell-Faraday eqn including the magnetic current density term. Mar 29, 2022 at 23:04
• @bleuofblue Hmm, ok. If you'd like, feel free to post an answer elaborating on your comment. It seems that the bounty will lapse without an answer. Mar 29, 2022 at 23:13

The result of the fictitious magnetic current density $$\bar{\mathcal{M}}$$ being negative stems from how they arrive at equation 1.35. Ultimately Stokes' theorem is applied to Faraday's law, which can be written with the magnetic current density to bring symmetry to Maxwell's equations. This is due to the fact that magnetic monopoles are nonexistent in reality, otherwise the equations would already exhibit symmetry. We can see that the sign of this fictitious quantity always extends $$B$$ and comes directly from Faraday's law, in that: $$\nabla \times E = -\frac{\partial B}{\partial t}-M \space\space\rightarrow\space\space -\nabla \times E = \frac{\partial B}{\partial t}+M$$
Equation 1.36 then is given in vector form, where the direction (sign) of $$\bar{\mathcal{M}}$$ is determined based on the cross product of the net $$E$$ field and vector $$\vec{n}$$. We can see that the sign of $$\bar{\mathcal{M}}$$ will flip depending on if we cross $$\vec{n}$$ with $$E$$ or vise versa.
• Sorry, I was trying to add depth to the comment, but I guess I made it more muddled. Essentially I'm saying that the eqn 1.1a can be written with a negative curl, and positive RHS, or positive curl with negative RHS - both are valid. The book chose the form with the positive curl, and from this choice the $M$ is negative, and remains negative until eq1.36 where in vector form, the cross product of $\vec{n}$ and $E$ will determine the sign of $M$. Mar 30, 2022 at 23:23