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General solution to the wave equation of electromagnetic field

  1. Nov 10, 2014 #1
    Suppose that we have the four-vector potential of the electromagnetic field, [texA^i[/tex]

    The wave equation is given by $$(\frac {1}{c^2} \frac {\partial^2}{\partial t^2}-\nabla^2) A^i=0$$

    Now the solution, for a purely spatial potential vector, is given by

    $$\mathbf{A}(t, \mathbf{x})=\mathbf{a_k} \exp{i(\pm \omega_{\mathbf{k}}t-\mathbf{k}.\mathbf{x}}); \mathbf{k}.\mathbf{a}=0$$

    To find the general solution we write the superposition as

    $$\mathbf{A}(t, \mathbf{x})=\int (\mathbf{f(k)}\exp{i( \omega_{\mathbf{k}}t-\mathbf{k}.\mathbf{x}})+\mathbf{g(k)}\exp{-i( \omega_{\mathbf{k}}t+\mathbf{k}.\mathbf{x}})) \frac {d^3 \mathbf{k}}{(2 \pi)^3}$$

    My question is that where this $$\frac {d^3 \mathbf{k}}{(2 \pi)^3}$$

    comes from? Shouldn't it be $$d^3\mathbf{x}$$
     
  2. jcsd
  3. Nov 10, 2014 #2

    DrClaude

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    Staff: Mentor

    It comes from the density of states in ##\mathbf{k}##-space. If you were to integrate over ##d^3\mathbf{x}##, you wouldn't get the vector potential as a function of ##\mathbf{x}##, would you?
     
  4. Nov 10, 2014 #3
    Why we do not say that $$\mathbf{A}(t, \mathbf{x})=\sum_{\mathbf{f(k)}\mathbf{g(k)}} (\mathbf{f(k)}\exp{i( \omega_{\mathbf{k}}t-\mathbf{k}.\mathbf{x}})+\mathbf{g(k)}\exp{-i( \omega_{\mathbf{k}}t+\mathbf{k}.\mathbf{x}})) $$

    Isn't superposition simply given by the above sum? Why and how we transform to the integral?
     
  5. Nov 10, 2014 #4

    dextercioby

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    Science Advisor
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    Isnt' k a continuous variable? How could you sum after its continuous values?
     
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