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Solution to EM wave

  1. Jan 15, 2009 #1

    KFC

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    I know that in free space, the general solution of the wave equation about electric field is of sine and cosine form. One can also write it in complex form as

    [tex]E = E_0 \exp(i\vec{k}\cdot\vec{r} - i\omega t)[/tex]

    I have two queations about this solution

    1) If consider the polarization, how should I rewrite the solution?

    2) In the presence of polarization, will the angular frequency [tex]\omega[/tex] depend on polarization?
     
  2. jcsd
  3. Jan 15, 2009 #2

    KFC

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    I just find the solution involves polarization in another textbook, it reads

    [tex]\vec{E} = \tilde{E} \exp(i\vec{k}\cdot\vec{r} - i\omega t)[/tex]

    where [tex]\tilde{E}[/tex] is complex amplitude represent the amplitude and the polarization. The book read: for finding the general multi-mode solution, we have to add up all possible solutions of above form

    [tex]\sum_{k}\sum_p\tilde{E}_{k,p} \exp(i\vec{k}\cdot\vec{r} - i\omega_{k, p} t)[/tex]

    Now the complex amplitude depends on wavenumber k and p (polarization I guess). Well, I don't understand why a 'p' there? If we need to sum over p, so what values of p is allowed? In the book, it said p can be 1 or 2, but why is that? The last question is about the frequency, why it also depends on p?
     
  4. Jan 16, 2009 #3
    That is because the EM wave is a transverse wave. The direction of the wave propagation is "k". So the E vector is in the plane perpendicular to the "k" vector. So E can be given in a basis of two vectors in the plane, representing polarization.
    Explicitly:
    Let [tex]\textbf{n}[/tex] be a unit vector parallel to the direction of wave propagation, that is parallel to [tex]\textbf{k}[/tex]. This means that we can write [tex]\textbf{k}=k\textbf{n}[/tex] where k is the absolute value of the k vector.

    Now lets take a unit vector in the plane perpendicular to the k vector, that is let:

    [tex]\textbf{k}\cdot \textbf{e}^1(\textbf{n})=0[/tex]

    We want an orthonormal basis in this plane. But we know that we can generate a unit vector orthogonal to e^1 via the cross product, so let:

    [tex] \textbf{e}^2(\textbf{n})=\textbf{n}\times\textbf{e}^1(\textbf{n})[/tex]

    Now since the E vector is in this plane, it can be given as a linear combination of the above two unit vectors. So we have for E:

    [tex]\textbf{E}(\textbf{r},t)=A_1(\textbf{k})\textbf{e}^1(\textbf{n})e^{i(\textbf{k}\textbf{r}-\omega(\textbf{k})t)}+A_2(\textbf{k})\textbf{e}^2(\textbf{n})e^{i(\textbf{k}\textbf{r}-\omega(\textbf{k})t)}[/tex]

    Using summation notation:

    [tex]\textbf{E}(\textbf{r},t)=\sum_{p=1}^2A_p(\textbf{k})\textbf{e}^p(\textbf{n})e^{i(\textbf{k}\textbf{r}-\omega(\textbf{k})t)}[/tex]

    So the polarization state can be represented by the vector (A_1,A_2). This is called the Jones vector. hence the p index.
     
  5. Jan 16, 2009 #4

    KFC

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    Thanks a lot. Very clear :)

    But in your explanation, it seems that the angular frequency only depends on k but not polarization (so no p index in [tex]\omega[/tex])?

    By the way, if it is not plane wave, do we still have something like polarization?
     
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