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Understanding the dipole model for Rayleigh scattering

  1. May 24, 2013 #1
    Hello.

    I am currently studying scattering theory in detail for my BSc thesis, and I'm starting with Rayleigh scattering. I'm following Scattering of electromagnetic waves: theories and applications by L Tsang, J A Kong and K-H Ding, which is pretty much what I like in a science textbook: descriptive but concise. I'm stuck deriving equation 1.2.1, however, and I've looked at other books and websites for hours with no result. The following are the notes I've gathered (mainly from that book, with a few additions from others and my own calculations); some of the calculations may be a little long, but I'm including every step so you may see which things I understand and which ones I don't (as well as exactly where everything comes from):

    Consider a plane wave with wavelength [itex]\lambda[/itex] incident on a particle of size (maximum distance between points within the particle) [itex]D\ll\lambda[/itex]. Say the electric permittivity of the (homogeneous) medium surrounding the particle is [itex]\epsilon_m[/itex]. Omitting the time dependence [itex]e^{-i\,\omega\,t}[/itex], we may write the incident electric field as
    [tex]\vec{E}_i=\vec{E}_o\,e^{i\,\vec{k}\cdot\vec{r}}.[/tex]
    Because the particle is much smaller than the wavelength, we may consider the wave's electric field to be constant in the region of space occupied by the particle. The wave, then, induces an oscillatory electric dipole in the particle with dipole moment [itex]\vec{p}=q\,\vec{d}[/itex], where [itex]\vec{d}[/itex] is the vector from the dipole's negative charge [itex]-q[/itex] to its positive charge [itex]q[/itex].

    A dipole's electric potential at [itex]\vec{r}[/itex] is given by
    [tex]\phi(\vec{r})=\frac{q}{4\,\pi\,\epsilon_m}\left[\frac{1}{|\vec{r}-\vec{r}_+|}-\frac{1}{|\vec{r}-\vec{r}_-|}\right],[/tex]where [itex]\vec{r}_\pm=\pm\frac{1}{2}\vec{d}[/itex] is the position of charge [itex]\pm q[/itex] and the dipole is centred at the origin.

    Say the two charges lie on the [itex]z[/itex] axis, so that the angle between [itex]\vec{r}[/itex] and [itex]\vec{d}=\vec{r}_+-\vec{r}_-[/itex] is [itex]\theta[/itex]. We write
    [tex]|\vec{r}-\vec{r}_+|=\left|\vec{r}-\frac{1}{2}\vec{d}\right|=\left[\left(\vec{r}-\frac{1}{2}\vec{d}\right)\cdot\left(\vec{r}-\frac{1}{2}\vec{d}\right)\right]^{1/2}=\left[r^2-\vec{r}\cdot\vec{d}+\frac{d^2}{4}\right]^{1/2}=r\left[1-\frac{d}{r}\cos(\theta)+\frac{d^2}{4\,r^2}\right]^{1/2},[/tex][tex]|\vec{r}-\vec{r}_-|=r\left[1+\frac{d}{r}\cos(\theta)+\frac{d^2}{4\,r^2}\right]^{1/2},[/tex]whereby
    [tex]\phi(\vec{r})=\frac{q}{4\,\pi\,\epsilon_m\,r}\left[\left(1-\frac{d}{r}\cos(\theta)+\frac{d^2}{4\,r^2}\right)^{-1/2}-\left(1+\frac{d}{r}\cos(\theta)+\frac{d^2}{4\,r^2}\right)^{-1/2}\right].[/tex]
    Now, for small [itex]x[/itex] we have
    [tex](1+x)^n=1+n\,x+\frac{n(n-1)}{2}x^2+\mathcal{O}(x^3).[/tex]By setting [itex]x:=\mp\frac{d}{r}\cos(\theta)+\frac{d^2}{4\,r^2}[/itex] and [itex]y:=\frac{d}{r}[/itex], in the far-field limit ([itex]r\gg d[/itex]) we have
    [tex]\phi(\vec{r})=\frac{1}{4\,\pi\,\epsilon_m\,r}\left[\left(1+\frac{d}{2\,r}\cos(\theta)-\frac{d^2}{8\,r^2}+\frac{3\,d^2}{8\,r^2}\cos^2(\theta)+\mathcal{O}(y^3)\right)-\left(1-\frac{d}{2\,r}\cos(\theta)-\frac{d^2}{8\,r^2}+\frac{3\,d^2}{8\,r^2}\cos^2(\theta)+\mathcal{O}(y^3)\right)\right].[/tex]Eliminating everything of order [itex]y^2[/itex] and greater, we finally obtain
    [tex]\phi(\vec{r})\approx\frac{q\,d\,\cos(\theta)}{4\,\pi\,\epsilon_m\,r^2}=\frac{q\,\vec{d}\cdot\hat{r}}{4\,\pi\,\epsilon_m\,r^2}=\frac{\vec{p} \cdot\hat{r}}{4\,\pi\,\epsilon_m\,r^2},[/tex]where [itex]\hat{r}=\frac{1}{r}\vec{r}[/itex].

    The electric field radiated by the dipole is
    [tex]\vec{E}_s=-\nabla\phi=-\frac{\partial\phi}{\partial r}\hat{r}-\frac{1}{r}\frac{\partial\phi}{\partial\theta}\hat{\theta}-\frac{1}{r\,\sin(\theta)}\frac{\partial\phi}{\partial\varphi} \hat{\varphi}.[/tex]
    Here I run into the first of three problems: is the "real" physical object produced by the dipole the potential or the field? In other words, at which step do I introduce the oscillatory nature ([itex]e^{i\,k\,r}[/itex], where the scalar product [itex]\vec{k}\cdot\vec{r}[/itex] is simply [itex]k\,r[/itex] because the dipole's far-field electric potential and electric field are approximately those of a spherical wave)? If I introduce it from the beginning, in the potential, then
    [tex]\phi=\frac{p\,\cos(\theta)}{4\,\pi\,\epsilon_m\,r^2}e^{i\,k\,r},[/tex][tex]-\frac{\partial\phi}{\partial r}=\frac{(2-i\,k\,r)\,p\,\cos(\theta)}{4\,\pi\,\epsilon_m\,r^3}\,e^{i\,k\,r};[/tex]if I introduct it into the electric field only, then
    [tex]\phi=\frac{p\,\cos(\theta)}{4\,\pi\,\epsilon_m\,r^2},[/tex][tex]-\frac{\partial\phi}{\partial r}=\frac{2\,p\,\cos(\theta)}{4\,\pi\,\epsilon_m\,r^3}\,e^{i\,k\,r}.[/tex]
    Regardless of which expression for the radial component of [itex]\vec{E}_s[/itex] is correct, the other two components (omitting the exponential) are given by
    [tex]-\frac{1}{r}\frac{\partial\phi}{\partial\theta}=\frac{p\,\sin(\theta)}{4\,\pi\,\epsilon_m\,r^3},[/tex][tex]-\frac{1}{r\,\sin(\theta)}\frac{\partial\phi}{\partial\varphi}=0,[/tex]so
    [tex]\vec{E}_s=\frac{p}{4\,\pi\,\epsilon_m\,r^3}e^{i\,k\,r}\left[A\,\cos(\theta)\,\hat{r}+\sin(\theta)\,\hat{\theta}\right],[/tex]where [itex]A[/itex] equals either [itex]2-i\,k\,r[/itex] or [itex]2[/itex] (whichever is correct).

    Now, [itex]\vec{p}=(\vec{p}\cdot\hat{r})\,\hat{r}+(\vec{p}\cdot\hat{\theta})\,\hat{\theta}=p\,\cos(\theta)\,\hat{r}-p\,\sin(\theta)\,\hat{\theta}[/itex] ([itex]\vec{p}[/itex] has no [itex]\varphi[/itex] component), so
    [tex]\vec{E}_s=\frac{p}{4\,\pi\,\epsilon_m\,r^3}e^{i\,k\,r}\left[(A+1)\,\cos(\theta)\,\hat{r}-\cos(\theta)\,\hat{r}+\sin(\theta)\,\hat{\theta}\right]=\frac{1}{4\,\pi\,\epsilon_m\,r^3}e^{i\,k\,r}\left[(A+1)\,(\vec{p}\cdot\hat{r})\,\hat{r}-\vec{p}\right].[/tex]
    Finally, for any three vectors [itex]\vec{a},\vec{b},\vec{c}[/itex] the following identity is true: [itex](\vec{a}\cdot\vec{c})\,\vec{b}-(\vec{a}\cdot\vec{b})\,\vec{c}=\vec{a}\times(\vec{b}\times\vec{c})[/itex]. By writing [itex]\vec{p}=(\hat{r}\cdot\hat{r})\,\vec{p}[/itex] and using this identity, we obtain
    [tex]\vec{E}_s=\frac{1}{4\,\pi\,\epsilon_m\,r^3}e^{i\,k\,r}\left[A\,(\vec{p}\cdot\hat{r})\,\hat{r}+\hat{r}\times(\hat{r}\times\vec{p}) \right].[/tex]
    Herein lie my second and third problems. The second problem is that the electric field radiated by a dipole never has a radial component, regardless of whether the dipole oscillates or not, whereas the last equation above implies that there is a radial component with amplitude [itex]\frac{A}{4\,\pi\,\epsilon_m\,r^3}[/itex]. The third problem is that this expression for [itex]\vec{E}_s[/itex] is not equal to that given by equation 1.2.1 of the book I am using, which is
    [tex]\vec{E}_s=-\frac{k^2}{4\,\pi\,\epsilon_m\,r}e^{i\,k\,r}\,\hat{r}\times(\hat{r} \times\vec{p}).[/tex]
    Any help would be greatly appreciated. :)
     
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  3. May 24, 2013 #2

    Jano L.

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    Gold Member

    The last formula refers to "wave" field of the oscillating dipole, which is almost the total field at great distances. If you are interesting in this field, you cannot base the calculation on the Coulomb potential alone. That will give you only electrostatic field - hence the ##1/r^3## dependence.

    You can solve Maxwell's equations for one point-like particle instead, and then the dipole field is just superposition of two such solutions, one from the positive, one from the negative particle. Try to get books by Jackson and Landau&Lifgarbagez, they show how this can be done.
     
  4. May 25, 2013 #3
    Thank you, Jano. I'll take a look at Landau and Lifgarbagez's Electrodynamics of continuous media, since I cannot digest Jackson's book.
     
  5. May 25, 2013 #4

    Jano L.

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    Try also their other book "The Classical Theory of Fields", it is probably better for what you are trying to learn.
     
  6. May 25, 2013 #5
    All, right, I shall, thank you. =)
     
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