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I've come across a problem with finding the energy stored in time/frequency electric field. In space/time we have (taking $\epsilon = 1$)

$$ Energy = \frac{1}{2} \int_V |\mathbf{E}(\mathbf{x},t)|^2 \;d^3x $$

But, I presume that the formula is different for a frequency-dependent electric field. I've searched Griffiths and Jackson but can't quite find what I'm looking for...

I've also tried to Fourier transform my expression for the electric field back to space/time, but my electric field is a fairly gruesome expression - I couldn't FT it easily. I was hoping to compute the energy from $E(\mathbf{x},\omega)$ via computational integral, once I find an expression for the energy.

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Something is odd about this question. What you can calculate is energy at any given time. So the LHS, ie the energy must be function of time. I think it is meaningless to calculate energy as a function of Omega. What is a very sensible question is to calculate energy as a function of time in terms of Fourier transform of position variables. I am doubtful, so I will wait for someone to look into it. – Prathyush Oct 28 '12 at 0:28

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I do not think it depends on frequency. Why should it? E.g., if the $\vec E$ is sinusoidal $$ \vec E(\vec x, t)=\vec E_0\sin(\vec k\cdot \vec x-\omega t) $$ you just have $$ \mbox{Power}= \int d^3x E_0^2 \sin(\vec k\cdot \vec x-\omega t)^2 $$

As Crazy Buddy pointed out, if you want the Energy you have to integrate over time as well $$ \mbox{Energy}= \int_0^T dt \int d^3x E_0^2 \sin(\vec k\cdot \vec x-\omega t)^2 $$

However, i point out that that formula is true only if the volume is infinite (i.e., you integrate with all coordinates going from $-\infty$ to $\infty$). Otherwise, you should also add the contribution given by integrating the Poynting vector on the boundary of the volume.

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It doesn't make sense to say energy contained 'per frequency'. If you're integrating over time and space, you will get the total energy in the field. $$E = \int_{0}^{T}dt\int_{-\infty}^{\infty}d^{3}x |E(x)|^{2} $$

But if you take a fourier transform, you'll have to take one on both sides. So the quantity on the left will be the fourier transform of the energy. I don't really know what that means, but that's what you'll get. You cannot take a fourier transform of one side of an equation alone.

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Why can't you take a Fourier transform of one side alone? FT is just a change of basis - it completely preserves your function. In quantum terms: $$ \tilde{f}(p) = \langle p | f \rangle = \int_{-\infty}^{\infty} \langle p | x \rangle \langle x | f \rangle \; dx = \int_{-\infty}^{\infty} \frac{e^{-ipx}}{(2\pi\hbar)^{1/2}} f(x) \;dx $$ All I did was inserted the identity and got $\tilde{f}(p)$ - the Fourier transform of $f$. – alexvas Oct 30 '12 at 6:39
But if you notice, when you changed basis, you got the fourier transform of $f(x)$. Meaning, when you went from x to p basis, your function became the fourier transform of itself. Which is the same in this case with energy. When you go from time to frequency, your energy will become the fourier transform of itself. The physical content of the function is untouched, you're right. But the form of the function will change. – Kitchi Oct 30 '12 at 7:21

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