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How does the formula C-6 come about in the following image?

  1. Feb 23, 2016 #1
    tann.JPG
     
  2. jcsd
  3. Feb 23, 2016 #2

    Simon Bridge

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    Via the principle of superposition ... where did it lose you?
    C-6 is just summing up all the independent solutions already found...
     
  4. Feb 23, 2016 #3

    dextercioby

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    Do you know anything about Fourier transformation and how you can use it for PDEs?
     
  5. Feb 23, 2016 #4
    I meant why does the superposition look like this? i.e. a Fourier transform?
     
  6. Feb 23, 2016 #5
    I do know about Fourier transforms but am not familiar with their relation to PDEs
     
  7. Feb 24, 2016 #6

    jtbell

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    A superposition of two plane waves looks like this: $$\psi(\vec r, t) = g_1 e^{i(\vec k_1 \cdot \vec r - \omega_1 t)} + g_2 e^{i(\vec k_2 \cdot \vec r - \omega_2 t)}$$
    A superposition of infinitely many (but still "countable" 1,2,3...) plane waves looks like this: $$\psi(\vec r, t) = \sum_{j=0}^\infty g_j e^{i(\vec k_j \cdot \vec r - \omega_j t)}$$
    Suppose the ##\omega_j## and the ##\vec k_j## can be related by ("fitted to") a continuous function ##\omega(\vec k)##. Likewise for the ##g_j## and the ##\vec k_j##, with another function ##g(\vec k)##. Then we can rewrite this slightly as $$\psi(\vec r, t) = \sum_{j=0}^\infty g(\vec k_j) e^{i(\vec k_j \cdot \vec r - \omega(\vec k_j) t)}$$
    Finally, suppose we have a superposition of an "uncountably" infinite number of waves with a continuous (not discrete) distribution of values of ##\vec k##. In this case the sum becomes an integral: $$\psi(\vec r, t) = \int g(\vec k) e^{i(\vec k \cdot \vec r - \omega(\vec k) t)} d^3 k$$
    This is a sort of "volume integral" in 3-dimensional ##\vec k##-space. The extra constant in front of your equation C-6 is basically a sort of normalization factor.
     
    Last edited: Feb 24, 2016
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