Mathematical Methods of Physics/Fourier transforms

Fourier transforms are used in practically every field of physics. They are used to transform a variable into the space of another which is either easier or more informative (for instance, one may be able to express an electromagnetic wave in terms of the electric field strength - but for real-life measurements we want to express the wave as a sum of several frequencies (called spectral-decomposition))

This is similar in usage to the Laplace transform - but is much better suited to functions with some oscillating behavior.

$\tilde{f}(s)=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}f(t)e^{-2\pi ist}dt$ and $f(t)=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\tilde{f}(s)e^{2\pi ist}ds$

(Note that some textbooks will have different factors in front of these integrals)

== Examples

1. Consider an electromagnetic wave $E(t)=E_{0}sin({\omega }_{0}t)$. Determine the intensity of the wave as a function of the frequency $\omega$. (It should be obvious to any physics major that this wave has only one frequency - so we can use that fact to check that our final result matches that expectation)

Applying the Fourier Transform integral : $\tilde{E}(\omega )=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{E}_{0}sin({\omega }_{0}t)e^{-2\pi i\omega t}dt$

To do this we will need to remember a few identities,

• $sin(x)=\frac{e^x-e^{-x}}{2i}$
• $e^a{e}^b=e^{a+b}$
• $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{i(x-x^{\prime })t}dt=\delta (x-x^{\prime })$

Using the first and second identities, $\tilde{E}(\omega )=\frac{E_0}{2i}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}[e^{2\pi i({\omega }_0+\omega )t}+e^{2\pi i({\omega }_0-\omega )t}]dt$

From here we can see that each integral is simply a delta function and we reduce it to $\tilde{E}(\omega )=\frac{E_0}{2i}[\delta (2\pi ({\omega }_0+\omega ))+\delta (2\pi ({\omega }_0-\omega ))]$

This tells us that the only frequencies with a non-zero amplitude are ${\omega }_0$ and $-{\omega }_0$. But as physicists we know that frequencies are not negative, so this matches the result expected (i.e. there is only one frequency present)