# Transforming electric field in frequency domain to intensity in frequency domain

I'm currently struggling to convert the electric field to the intensity in the frequency domain.

In principle it seems like I need to do the following: $$I(\omega)=\mathcal{F}[I(t)]=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}dt|\mathcal{R}(E(t))|^2e^{-i\omega t}$$ where $$E(t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}d\omega E(\omega)e^{i\omega t}$$ with the given electric field $$E(\omega)$$ in the frequency domain.

This is in the first place not really doable analytically and also not very easy numerically. That's why I wanted to ask if there is a better way to do this transformation.

It's actually remarkably easy. No need to go into the time domain. Use Ohm's law and the impedance of free space, $$Z_0$$ (about 377Ω).

$$I(\omega)= E(\omega)^2 /Z_0$$

• Okay, but what is then my electric field spectrum consists of deltas? Delta squared is always a bit weird. Commented Sep 9, 2022 at 11:43

Taking square of EM signal in time domain is a pointwise product leading to a convolution in the Fourier domain: F(x×x)=F(x)∗F(x)

As EM signals are mostly sinosuidal, trigonometric identities defines the change in frequency e.g:

$$cos^2x=1/2(1+cos2x)$$ $$sin^2x=1/2(1−cos2x)$$ or

From the above formulae, one can see that:

Trignometric conversions explains the peaks shifts in freq domain. Scalar components yield frequency peaks at reference point