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Plane wave in Minkowski space-time

  1. Mar 21, 2012 #1

    Wox

    User Avatar

    The classical expression of a plane electromagnetic wave (electric part)

    [tex]
    \bar{E}(t,\bar{x})=\bar{E}_{0}e^{i(\bar{k}\cdot \bar{x}-\omega t)}
    [/tex]

    looks a lot like the basis function of the Fourier decomposition in Minkowski space-time

    [tex]
    \bar{E}(\bar{w})=\int_{-\infty}^{\infty}\hat{\bar{E}}(\bar{\nu})e^{2\pi i\ \eta(\bar{\nu},\bar{w})}d\bar{\nu}
    [/tex]

    where [itex]\bar{E}\colon\mathbb{R}^{4}\to\mathbb{R}^{4}[/itex] and [itex]\bar{w}=(ct,\bar{x})[/itex]. If I write [itex]\bar{\nu}=\frac{\nu}{c}(1,\bar{n})[/itex] with [itex]\left\|\bar{n}\right\|=1[/itex] then we get

    [itex]\eta(\bar{\nu},\bar{w})=\frac{\nu}{c}\bar{n}\cdot \bar{x}-\nu t=\frac{1}{2\pi}(\bar{k}\cdot \bar{x}-\omega t)[/itex] and

    [tex]
    \bar{E}(\bar{w})=\int_{-\infty}^{\infty}\hat{\bar{E}}(\bar{\nu})e^{ i\ (\bar{k}\cdot \bar{x}-\omega t)}d\bar{\nu}
    [/tex]

    and for a monochromatic wave

    [tex]
    \bar{E}(\bar{w})=\bar{E}(ct,\bar{x})=\hat{\bar{E}}(\bar{\nu})e^{ i\ (\bar{k}\cdot \bar{x}-\omega t)}
    [/tex]

    which is close to the classical expression, but not exactly. So the point is, I feel Fourier decomposition in Minkowski space and the classical plane wave are related, but I'm not sure how. Can someone clarify?
     
  2. jcsd
  3. Mar 21, 2012 #2
    Hi Vox,

    Fourier decomposition in both cases is implemented by taking the inner product of each possible frequency component with each frequency component of the actual wave. That works because all frequency components are mutually orthogonal (when evaluated over a suitably long time interval). But the definition of an inner product is very different in Minkowski space compared with a classical setting (Euclidean geometry).

    The definition of the inner product for Minkowski space is described here:

    http://en.wikipedia.org/wiki/Minkowski_space#The_Minkowski_inner_product
     
    Last edited: Mar 21, 2012
  4. Mar 21, 2012 #3

    Wox

    User Avatar

    I used the proper Minkowskian inner product in [itex]\eta(\bar{\nu},\bar{w})=...[/itex] assuming signature (-+++), while [itex]\bar{k}\cdot \bar{x}[/itex] is the Euclidean inner product.

    I don't know what you mean by "in both cases". Consider the Fourier decomposition of any map [itex]\bar{E}\colon\mathbb{R}^{4}\to\mathbb{R}^{4}[/itex] then the basis functions of the Fourier decomposition (frequency components) are [itex]\hat{\bar{E}}(\bar{\nu})e^{2\pi i\ \eta(\bar{\nu},\bar{w})}[/itex]. In the context of electrodynamics, [itex]\bar{E}[/itex] is an electric field and the basis functions look very much like plane waves. My question is: are these plane waves and if yes, how can you show this? (my attempt in the original post didn't work).

    When I say "looks like", I mean that [itex]\bar{k}\cdot \bar{x}-\omega t[/itex] is the Minkowskian inner product of [itex](t,\bar{x})[/itex] and [itex](\omega,\bar{k})[/itex], although I'm not sure what the last four-vectors represents.
     
    Last edited: Mar 21, 2012
  5. Mar 21, 2012 #4
    Sorry. I'm not sure how you would even define a plane in Minkowski space. There seems to be some ideas expressed in mathematical literature but they look a bit esoteric to me.
     
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