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Different expressions for the polarization

  1. Oct 3, 2011 #1
    Hi

    People usually write the (total) polarization like this

    [tex]P(t) = \epsilon_0(\chi^{(1)}E(t)+chi^{(2)}E(t)^2+\ldots)[/tex]

    where χ is the susceptibility. But in my book I see they write it like this

    [tex]
    P(t) = \varepsilon _0 \frac{1}{{2\pi }}\int\limits_{ - \infty }^\infty {\chi ^{(1)} (t)E(t' - t)dt} + \varepsilon _0 \frac{1}{{4\pi \pi }}\int\limits_{ - \infty }^\infty {\chi ^{(2)} (t_1 ,t_2 )E(t' - t_1 )E(t' - t_2 )dt_1 dt_2 + }
    \ldots[/tex]

    I'm not quite sure what the difference is between these two expressions. Do they apply to different situations?

    Thanks in advance.

    Niles.
     
  2. jcsd
  3. Oct 3, 2011 #2

    vanhees71

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    Let's stick to linear-response theory first and consider normal media (no plasma), where temporal dispersion is sufficient to describe the polarization. Also let's stick to homogeneous media (say in thermal equilibrium). Then you impose an electric field on the medium and disturb this state of the charges inside the medium. Then, if this field is small compared to the inner bindings of the electrons to the ions, the response of the medium is linear, and the polarization can be written with help of the retarded Green's function as

    [tex]\vec{P}(t,\vec{x})=\int_{\mathbb{R}} \mathrm{d} t' \chi(t-t') \vec{E}(t',\vec{x}).[/tex]

    Here, [itex]\chi(t-t') \propto \Theta(t-t')[/itex] is the retarded Green's function.

    Now you can express the Green's function and the electric field with help of its Fourier transform via

    [tex]\chi(t-t')=\int_{\mathbb{R}} \frac{\mathrm{d} \omega}{(2 \pi)} \tilde{\chi}(\omega) \exp[-\mathrm{i} \omega(t-t')][/tex]

    and

    [tex]\vec{E}(t,\vec{x})=\int_{\mathbb{R}} \frac{\mathrm{d} \omega}{(2 \pi)} \tilde{\vec{E}}(\omega,\vec{x}) \exp(-\mathrm{i} \omega t).[/tex]

    The only restriction to [itex]\tilde{\chi}[/itex] in order to be retarded is that it is an entire function in the upper complex [itex]\omega[/itex] plane. Then the convolution integral wrt. time becomes a simple multiplication in freqency space:

    [tex]\tilde{\vec{P}}(\omega,\vec{x})=\tilde{\chi}( \omega ) \tilde{\vec{E}}(\omega,\vec{x}).[/tex]
     
  4. Oct 3, 2011 #3
    Thanks for taking the time to reply.

    Aren't you missing an ε0/2∏?




    Two things:

    1) Why does χ have to be entire in order to be retarded? I can't quite see that point.
    2) Do you have a reference to this derivation?

    Again, thanks.

    Best,
    Niles.
     
    Last edited: Oct 3, 2011
  5. Oct 3, 2011 #4
    Can I please ask a moderator to perhaps move this thread to the homework section?
     
  6. Oct 4, 2011 #5

    vanhees71

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    All constants are included in the symbol [itex]\chi[/itex]. It depends on the system of units you use. In SI units the dielectric constant of the vacuum is usually taken out of [itex]\chi[/itex], and then you have to write [itex]\epsilon_0 \chi[/itex]. In physics the SI is not a very intuitive system of units, and I prefer Heaviside-Lorentz units (which are rationalized Gauss units), where [itex]\epsilon_0=\mu_0=1[/itex].

    To answer your second question, just look at

    [itex]\chi(t,t')=\int_{\mathbb{R}} \frac{\mathrm{d} \omega}{2 \pi} \tilde{\chi}(\omega) \exp[-\mathrm{i} \omega (t-t')]. [/itex]

    You can evaluate this integral by using Cauchy's integral theorem (or the theorem of residues) by closing the integration path with a large semicircle in the upper or lower [itex]\omega[/itex]-half plane. In order to avoid divergences from the exponential function, you have to close the path in the upper (lower) plane, for [itex]t-t'<0[/itex] ([itex]t-t'>0[/itex]). In order to have [itex]\chi(t,t') \propto \Theta(t-t')[/itex], thus there must not be singularities of [itex]\tilde{\chi}(\omega)[/itex] in the upper [itex]\omega[/itex]-half plane, i.e., it should be an entire function there.

    This you find in nearly any textbook on theoretical electromagnetism or optics. One very good source are Sommerfeld's lectures on theoretical physics (vol. III and particularly volume IV).
     
    Last edited: Oct 4, 2011
  7. Oct 4, 2011 #6
    Thanks, I'll have to sit down with my complex analysis book and go through your arguments. It's been a while since I've done these things. I'll post a reply, when I understand it (or if I mess it up).

    Thanks for now.
     
  8. Oct 10, 2011 #7
    I understand the argument now, thanks for simplifying it. If we go past linear response theory, and look at the n'th order susceptibility, then am I correct to say that the same property has to apply to [itex]\chi^{(n)}\propto \Theta(t-t_1) \Theta(t-t_2)\ldots \Theta(t-t_n)[/itex], i.e. that [itex]\chi^{(n)}[/itex] has to be entire in the upper complex (multidimensional?!) plane?

    Best,
    Niles.
     
    Last edited: Oct 10, 2011
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