# Imhomegenous Wave Equation: Possible Born and Wolf Error

My concern involves the following lines from Principle of Optics (Born and Wolf 7th ed, 60 year anniverary, Section 1.2 pg 11). I fail to derive these equations 5,6 exactly.

\begin{align} \nabla^2E -\frac{\varepsilon\mu}{c^2}\partial^2_t E+\nabla(\ln\mu)\times(\nabla \times E)+\nabla(E\cdot\nabla(\ln\varepsilon))&=0 \tag{5}\\ \nabla^2H -\frac{\varepsilon\mu}{c^2}\partial^2_t H+\nabla(\ln\varepsilon)\times(\nabla \times H)+\nabla(H\cdot\nabla(\ln\mu))&=0 \tag{6} \end{align}

No matter how many times I do the derivation I keep getting the same result (which differ from the text).

The discrepancy starts when you go from \begin{align} \frac{\varepsilon}{c^2}\partial^2_t E+\nabla\times\left(\frac{1}{\mu}\nabla \times E\right)&=0 \tag{2}\\ \end{align} which arises from taking the curl of Maxwell's equations. Expansion via $$\nabla\times(u \mathbf v)=u\nabla\times \mathbf v+\nabla u \times \mathbf v\\ \nabla \times \nabla \times \mathbf v = \nabla (\nabla \cdot \mathbf v)-\nabla^2 \mathbf v$$

and the other Maxwell's equations

$$\nabla \cdot (\varepsilon E)=0 \\ \nabla \cdot (\mu H) = 0 \\$$

\begin{align} \nabla^2E -\frac{\varepsilon\mu}{c^2}\partial^2_t E-\nabla(\ln\mu)\times(\nabla \times E)+\nabla(E\cdot\nabla(\ln\varepsilon))&=0 \tag{mine}\\ \nabla^2H -\frac{\varepsilon\mu}{c^2}\partial^2_t H-\nabla(\ln\varepsilon)\times(\nabla \times H)+\nabla(H\cdot\nabla(\ln\mu))&=0 \tag{mine} \end{align}

where the 3rd term signs are flipped relative to the equations in the text. I've checked older.

I've checked another text , Mathematical Theory of Optics (Luneberg) which seems to have an identical result but it reference Born and Wolf so for all I know it just copies it.

To me the $$\nabla^2 E$$ term simply has to have the opposite sign of the $$\nabla(\ln\mu)\times(\nabla \times E)$$ term.

I suspect I made an error somewhere in my derivation but I just can't see where it could be. I've spent few hours and can't seem to find where I could have made the sign error.

I also tried searching the book website for an errata

Any help is appreciated.

• just checked it, B&W is right: it is $+$. you may have forgotten the $-$ in Faraday's law. Commented Mar 10 at 23:01
• Did you also get equation (2) above? I started with (2) in B+W. Did you derive from scratch? Commented Mar 11 at 0:06
• took the curl of Faraday's eq, and expanded it $\nabla \times (\nabla \times E)=-\nabla \times \mu \dot H$, took the time diff of Maxwell-Ampere $\nabla \times \dot H = -\mu \ddot E$ and $\nabla \cdot \epsilon E=0$, that is all. Commented Mar 11 at 0:13
• Thanks for the help, I resolved my sign error. It was related to how $\nabla \ln \mu= -\mu \nabla(\frac{1}{\mu})$ I missed the negative sign there. A silly mistake on my part. Commented Mar 11 at 0:51

$$\nabla \ln \mu= -\mu \nabla(\frac{1}{\mu})$$