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I Do non-monochromatic "waves" exist in dispersive media?

  1. Apr 29, 2016 #1
    Hi.

    Is the superposition of two different monochromatic waves in a dispersive medium still a wave (i.e. a solution of a wave equation) if the phase velocity is not the same? Since the wave equation contains the phase velocity, the two individual waves are solutions of different wave equations. Is there a wave equation that is solved by the superposition?
     
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  3. Apr 29, 2016 #2

    blue_leaf77

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    When more than one waves are present, it's more intuitive to write the wave equation in frequency domain
    $$
    \nabla^2 E(\mathbf{r},\omega) + \frac{\omega^2}{c^2}\epsilon(\omega)E(\mathbf{r},\omega) = 0
    $$
    where ##\epsilon(\omega)## is the medium's permittivity.
     
  4. Apr 29, 2016 #3
    But isn't that the same problem? In the time domain the phase velocity depends on the frequency, now the permittivity does. Still, waves of different frequency are solutions to different wave equations.
     
  5. Apr 29, 2016 #4

    vanhees71

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    It's still a linear PDE (the Helmholtz equation), and thus you can apply Fourier decomposition, i.e., you write
    $$E(\vec{r},\omega)=\int_{\mathbb{R}^3} \frac{\mathrm{d}^3 \vec{k}}{(2 \pi)^3} \tilde{E}(\vec{k},\omega) \exp(\mathrm{i} \vec{k} \cdot \vec{x}).$$
    Plugging this into the equation leads to the dispersion relation
    $$k=|\vec{k}|=\frac{\omega}{c} \sqrt{\epsilon(\omega)}.$$
    This leads to
    $$E(\vec{r},\omega)=\int_{\Omega} \mathrm{d}^2 \Omega A(\omega,\vartheta,\varphi) \exp(\mathrm{i} k \vec{n} \cdot \vec{x})|_{k=\omega \sqrt{\epsilon(\omega)}/c},$$
    where ##\vec{n}=(\cos \varphi \sin \vartheta,\sin \varphi \sin \vartheta,\cos \vartheta)## and ##\mathrm{d}^2 \Omega=\mathrm{d} \vartheta \mathrm{d}\varphi \sin \vartheta##, ##\vartheta \in \{0,\pi \}##, ##\varphi \in \{0,2 \pi \}##, and ##A## is an arbitrary function. It can be evaluated from some initial condition (in the time domain).
     
  6. May 3, 2016 #5
    So does that mean that ##c## in the wave equation (in the time domain) is not to be regarded as a constant but as a function of the frequency, ##c(\omega)##?
     
  7. May 3, 2016 #6

    blue_leaf77

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    Yes, for a non-magnetic medium, the wave equation reads
    $$
    \nabla^2 E(\mathbf{r},t) - \frac{\epsilon(\omega)}{c^2} \frac{\partial^2}{\partial t^2}E(\mathbf{r},t) = 0
    $$
     
  8. May 4, 2016 #7

    vanhees71

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    Here you mix the frequency and time domains, and the above equation is not correct for a general wave packet but only for a monochromatic mode. The correct equation is the Helmholtz equation for ##\tilde{E}(\omega,\vec{r})## in #2. In the time domain you usually have a retarded integral between ##D## and ##E##
    $$D(t,\vec{x})=\int_{\mathbb{R}} G_{\text{ret}}(t,t') E(t',\vec{x}),$$
    and this is only the special case, where you neglect spatial dispersion. For details, see vol. VIII of Landau&Lifshitz (Macroscopic E&M).
     
  9. May 4, 2016 #8

    blue_leaf77

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    My bad, you are right. The equation in time domain should have been
    $$
    \nabla^2 E(\mathbf{r},t) - \frac{1}{c^2} \frac{\partial^2}{\partial t^2}\int R(t-t')E(\mathbf{r},t') \, dt' = 0
    $$
    where ##R(t) = FT[\epsilon(\omega)]## is the so-called medium's response function, typically required to satisfy the condition ##R(t<0)=0##. This requirement has a physical meaning that the contribution to the present's field due to the field in the future is zero - causality principle. I think this function is what you meant by ##G(t)##. Basically, the convolution ##\int R(t-t')E(\mathbf{r},t') \, dt' ## is just the FT of ##\epsilon(\omega)E(\mathbf{r},\omega)##.
    The wave equation in #6 is only approximately true when the permittivity varies slowly within the frequency range of the spectrum of the field, or equivalently when the field's spectrum is sufficiently narrow.
     
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