- #1

PhDeezNutz

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- Homework Statement
- This is more so a conceptual question rather than a specific homework problem. I always thought that the polarization vector (i.e. direction of the ##\vec{E}## - field denoted as ##\vec{\varepsilon}##) in an EM wave was a Euclidean three vector but apparently I am wrong because in one equation in my book they mention the complex conjugate ##\vec{\varepsilon}##. I'm using Zangwill's "Modern Electrodynamics" btw.

The specific part I'm confused about is

Call this equation 1

##\frac{d \sigma_{scatt}}{d \Omega} = \frac{\langle \frac{dP}{d \Omega} \rangle}{\frac{1}{2} \varepsilon_0 c E_0^2} = r^2 \frac{\vec{E}_{rad}^2}{\vec{E}_{0}^2} = \left| \vec{f} \left( \vec{k} \right) \right|^2##

"If our interest is a scattered electric field with a particular polarization ##\vec{\varepsilon}##, the differential scattering cross section (above) generalizes to"

Call this equation 2

##\left. \frac{d \sigma_{scatt}}{d \Omega} \right|_{\vec{\varepsilon}} = r^2 \frac{\left|\vec{\varepsilon}^* \cdot \vec{E}_{rad} \right|^2}{\left| \vec{E}_0\right|^2} = \left| \vec{\varepsilon}^* \cdot \vec{f}\left( \vec{k} \right) \right|^2##

Again I always thought ##\vec{\varepsilon}## was a 3D euclidean vector and I'm trying to reconcile what it would mean for it to be complex.

- Relevant Equations
- The differential scattering cross section is defined as

##\frac{d \sigma_{scatt}}{d \Omega} = \frac{\langle \vec{S}_{rad} \rangle \cdot r^2 \hat{r}}{\left| \langle \vec{S}_{inc} \rangle \right|}##

In the far field

##\vec{E} = \vec{E}_{inc} + \vec{E}_{rad} = E_0 \left[ \hat{e}_0 e^{i \vec{k}_0 \cdot \vec{r}} + \frac{e^{ikr}}{r} \vec{f} \left( \vec{k} \right)\right]e^{-i \omega t}##

I guess I will show my work for substantiating equation 1 and hopefully by doing so someone will be able to point out where I could generalize.

##\langle \vec{S}_{rad} \rangle = \frac{1}{2 \mu} \mathfrak{R} \left( \vec{E}_{rad} \times \vec{B}^*_{rad}\right) = \frac{1}{2 \mu} \mathfrak{R} \left( \vec{E}_{rad} \times \left( \hat{r} \times \vec{E}_{rad}^*\right) \sqrt{\mu \varepsilon} \right) = \sqrt{\frac{\varepsilon}{\mu}} \mathfrak{R} \left(\vec{E}_{rad} \times \left( \hat{r} \times \vec{E}_{rad}^* \right) \right) = \sqrt{\frac{\varepsilon}{mu}} \mathfrak{R} \left( \hat{r} \left(\vec{E}_{rad} \cdot \vec{E}_{rad}^* \right) - \vec{E}_{rad}^* \left( \hat{r} \cdot \vec{E}_{rad} \right) \right) = \sqrt{\frac{\varepsilon}{\mu}} \frac{E_0^2 \left|\vec{f} \left(\vec{k} \right) \right|^2}{r^2} \hat{r}##

##\langle \vec{S}_{inc} \rangle = \frac{1}{2 \mu} \mathfrak{R} \left( \vec{E}_{inc} \times \vec{B}_{inc}^*\right) = \sqrt{\frac{\varepsilon}{\mu}} \mathfrak{R} \left( \hat{r} \left( \vec{E}_{inc} \cdot \vec{E}_{inc}^*\right) \right) = \sqrt{\frac{\varepsilon}{\mu}} E_0^2##

Therefore

##\frac{d \sigma_{scatt}}{d \Omega} = \left| \vec{f} \left( \vec{k} \right) \right|^2 ##

##\langle \vec{S}_{rad} \rangle = \frac{1}{2 \mu} \mathfrak{R} \left( \vec{E}_{rad} \times \vec{B}^*_{rad}\right) = \frac{1}{2 \mu} \mathfrak{R} \left( \vec{E}_{rad} \times \left( \hat{r} \times \vec{E}_{rad}^*\right) \sqrt{\mu \varepsilon} \right) = \sqrt{\frac{\varepsilon}{\mu}} \mathfrak{R} \left(\vec{E}_{rad} \times \left( \hat{r} \times \vec{E}_{rad}^* \right) \right) = \sqrt{\frac{\varepsilon}{mu}} \mathfrak{R} \left( \hat{r} \left(\vec{E}_{rad} \cdot \vec{E}_{rad}^* \right) - \vec{E}_{rad}^* \left( \hat{r} \cdot \vec{E}_{rad} \right) \right) = \sqrt{\frac{\varepsilon}{\mu}} \frac{E_0^2 \left|\vec{f} \left(\vec{k} \right) \right|^2}{r^2} \hat{r}##

##\langle \vec{S}_{inc} \rangle = \frac{1}{2 \mu} \mathfrak{R} \left( \vec{E}_{inc} \times \vec{B}_{inc}^*\right) = \sqrt{\frac{\varepsilon}{\mu}} \mathfrak{R} \left( \hat{r} \left( \vec{E}_{inc} \cdot \vec{E}_{inc}^*\right) \right) = \sqrt{\frac{\varepsilon}{\mu}} E_0^2##

Therefore

##\frac{d \sigma_{scatt}}{d \Omega} = \left| \vec{f} \left( \vec{k} \right) \right|^2 ##