# Fourier transform for the wave vector dervation problem

• fisher garry
In summary, the conversation discusses using Fourier transforms to solve the Maxwell equations and the challenges of obtaining a transform for the wave vector and position. The steps for obtaining the Fourier series are provided and the conversation concludes with an integration of the series, leading to the use of a limit as L goes to infinity and the introduction of a new variable, g(k), resulting in the final equation of f(x) = 1/2pi * integral from -infinity to infinity of g(k)e^(ikx).

#### fisher garry

Below is my walkthrough of a Fourier transform. My problem is that I want to do all the similar steps for a Fourier transform between position x and the wave vector k. That is working on a solution of the maxwell equations. The maxwell equations has many possible solutions for example:

$$cos(kx-\omega t)+isin(kx - \omega t)$$

But I don't know how to start up with it in a sum of cos and sin and end up with a transform. One of the problems that I see is that I don't have any denumerator like L that can go to infinity. I have added all the steps below because I need a directly point to where I must change something in order to get the Fourier transform for the wave vector and position. Look for the bold part below for my main problem to get to the problem with the L, Just scroll down there to get directly to the question. The rest of the text is just to avoid overlooking any linear additions and arrangments that might be the problem.

$$\int_{-L}^{L}\sin\frac{2m\pi}{L}t\, dt=0 \tag A$$

$$\int_{-L}^{L}\cos\frac{2m\pi}{L}t\, dt=0$$

$$\int_{-L}^{L}\frac{1}{2}\left[\cos\frac{2n \pi}{L}t\sin\frac{2m\pi}{L}t\right] \,dt$$

Using standard trigonometric relations:

\begin{align}
\frac{1}{2}\int_{-L}^{L}(\frac{2m
\pi}{L}-\frac{2n
\pi}{L})t + sin(\frac{2m
\pi}{L}+\frac{2n
\pi}{L})t dt\\
\end{align}

\begin{align}
\frac{1}{2}\int_{-L}^{L}sin(\frac{2
\pi}{L}(m-n))t + sin(\frac{2
\pi}{L}(m+n))t dt\\
\end{align}

For m=n

\begin{align}
\frac{1}{2}\int_{-L}^{L} 0 + sin(\frac{2
\pi}{L}(2m))t dt=0\\
\end{align}

$$m\neq n$$ We have the same integral as in (A) which is zero for integers m,n

\begin{align}
\int_{-L}^{L}\frac{1}{2}[sin\frac{2n
\pi}{L}tsin\frac{2m\pi}{L}t] dt\\
\end{align}

Using standard trigonometric relations:

\begin{align}
\frac{1}{2}\int_{-L}^{L}cos(\frac{2m
\pi}{L}-\frac{2n
\pi}{L})t - (\frac{2m
\pi}{L}+\frac{2n
\pi}{L})t dt \ \ \ (B) \\
\end{align}

\begin{align}
\frac{1}{2}\int_{-L}^{L}cos(\frac{2
\pi}{L}(m-n))t - (\frac{2m
\pi}{L}(m+n))t dt\\
\end{align}

for $$m \neq n$$ we get sine when it is inegrated and it is 0 for all + and - of integers m,n

when m=n m+n becomes 0 as well for the same reason as when $m \neq n$

but for m-n

\begin{align}
\frac{1}{2}\int_{-L}^{L}cos(\frac{2
\pi}{L})t dt=\frac{1}{2}\int_{-L}^{L}1 dt=L\\
\end{align}

since cos0=1

\begin{align}
\int_{-L}^{L}\frac{1}{2}[cos\frac{2n
\pi}{L}tcos\frac{2m\pi}{L}t] dt\\
\end{align}

Using standard trigonometric relations:

\begin{align}
\frac{1}{2}\int_{-L}^{L}cos(\frac{2m
\pi}{L}+\frac{2n
\pi}{L})t + cos(\frac{2m
\pi}{L}-\frac{2n
\pi}{L})t dt\\
\end{align}

for similar reasoning as in (B) $$m \neq n$$ everything is 0

for m=n

\begin{align}
\frac{1}{2}\int_{-L}^{L}cos(\frac{2m
\pi}{L}-\frac{2n
\pi}{L})t dt = \frac{1}{2}\int_{-L}^{L} 1 dt\\
\end{align}

The Fourier series are given by:

$$f(x)=\sum_{=0}^{\infty}a_n'cos(nx)$$ + $$\sum_{n=0}^{\infty}b_n'sin(nx)$$

$$f(x)= \frac{1}{2}a_0 +$$ $$\sum_{=1}^{\infty}a_ncos(nx)$$ + $$\sum_{n=1}^{\infty}b_nsin(nx) \ \ \ (C)$$

Multipying with $$cos\frac{k\pi x}{L}$$

$$f(x)cos\frac{k\pi x}{L}=cos\frac{k\pi x}{L} \frac{1}{2}a_0 +$$ $$\sum_{=1}^{\infty}cos\frac{k\pi x}{L}a_ncos\frac{n\pi x}{L}$$ + $$\sum_{n=1}^{\infty}cos\frac{k\pi x}{L}b_nsin\frac{n\pi x}{L} \ \ \ (C)$$

Putting integration signs on both sides and using trigonometric relations above we obtain

$$\int_{-L}^{L}f(x)cos\frac{k\pi x}{L} dx=\int_{-L}^{L}cos\frac{k\pi x}{L} \frac{1}{2}a_0 dx +$$ $$\int_{-L}^{L}\sum_{=1}^{\infty}cos\frac{k\pi x}{L}a_ncos\frac{n\pi x}{L} dx$$ + $$\int_{-L}^{L}\sum_{n=1}^{\infty}cos\frac{k\pi x}{L}b_nsin\frac{n\pi x}{L} dx\ \ \$$

$$\int_{-L}^{L}cos\frac{k\pi x}{L} \frac{1}{2}a_0 dx =0$$

$$\int_{-L}^{L}\sum_{n=1}^{\infty}cos\frac{k\pi x}{L}b_nsin\frac{n\pi x}{L} dx=0$$

$$a_k=\frac{1}{L}\int_{-L}^{L} f(x)cos\frac{k\pi x}{L} dx (D)$$

if we instad multiply with $$sin\frac{k\pi x}{L}$$ and doing the same I obtain

$$b_k=\frac{1}{L}\int_{-L}^{L} f(x)sin\frac{k\pi x}{L} dx (E)$$

Normal complex number relation:

$$e^{in\frac{\pi x}{L}}=cos\frac{n\pi x}{L} + isin\frac{n\pi x}{L}$$

from symmetry:

$$e^{-in\frac{\pi x}{L}}=cos\frac{n\pi x}{L} - isin\frac{n\pi x}{L}$$

addition and subtraction from theese two equations gives:

$$e^{in\frac{\pi x}{L}}+e^{-in\frac{\pi x}{L}}=2cos x$$

$$e^{in\frac{\pi x}{L}}-e^{-in\frac{\pi x}{L}}=2isin x$$

$$a_ncos\frac{n\pi x}{L} + b_nsin\frac{n\pi x}{L}=\frac{a_n}{2}(e^{in\frac{\pi x}{L}}+e^{-in\frac{\pi x}{L}})+\frac{b_n}{i2}(e^{in\frac{\pi x}{L}}-e^{-in\frac{\pi x}{L}})$$

using $$\frac{1}{i}=-i$$

$$a_ncos\frac{n\pi x}{L} + b_nsin\frac{n\pi x}{L}=(\frac{a_n}{2}-\frac{b_n}{i2})e^{in\frac{\pi x}{L}}+(\frac{a_n}{2}+\frac{b_n}{i2})e^{-in\frac{\pi x}{L}})$$

$$c_n=\frac{a_n}{2}-\frac{b_ni}{2} \\\ c_n*=\frac{a_n}{2}+\frac{b_ni}{2}$$

We can now rewrite the Fourier series:

$$f(x)= \frac{1}{2}a_0 +$$ $$\sum_{=1}^{\infty}a_ncos\frac{n\pi x}{L}$$ + $$\sum_{n=1}^{\infty}b_nsin\frac{n\pi x}{L} \ \ \ =\frac{1}{2}a_0 +$$ $$\sum_{=1}^{\infty}c_ne^{in\frac{\pi x}{L}}+c_n*e^{-in\frac{\pi x}{L}})$$

if we let the last part go from $$-\infty$$ to -1 we can redifine the name of the constants keeping in mind that $$e^0=1$$ for $$c_0=\frac{1}{2}a_0$$ and get:

$$f(x)= \sum_{-\infty}^{\infty}c_ne^{in\frac{\pi x}{L}} \ \ \ (F)$$

using (D):

$$c_0=\frac{a_k}{2}=\frac{1}{2L}\int_{-L}^{L} f(x) dx (G)$$

$$c_n=\frac{1}{2}(a_n-ib_n)=\frac{1}{2L}\int_{-L}^{L} f(x)cos\frac{k\pi x}{L} - if(x)sin\frac{k\pi x}{L} dx$$

$$c_n=\frac{1}{2}(a_n-ib_n)=\frac{1}{2L}\int_{-L}^{L} f(x)e^{-\frac{k\pi x}{L}} dx \ \ \ (H)$$

$$c_n*=\frac{1}{2}(a_n+ib_n)=\frac{1}{2L}\int_{-L}^{L}f(x) e^{\frac{k\pi x}{L}} dx$$

by letting n=-1,-2,-3..

$$c_n*=\frac{1}{2}(a_n+ib_n)=\frac{1}{2L}\int_{-L}^{L}f(x) e^{-\frac{k\pi x}{L}} dx \ \ \ (I)$$

similar as for the discrete case we get by adding (G), (H) and (I):

$$c_n=\frac{1}{2L}\int_{-L}^{L}f(x) e^{-\frac{k\pi x}{L}} dx \ \ \ for -\infty<n<\infty$$

using (F) we get

$$f(x)= \sum_{-\infty}^{\infty}\frac{1}{2L}\int_{-L}^{L} e^{-\frac{k\pi x}{L}}e^{in\frac{\pi x}{L}} \ \ \$$

Here is where L goes to infinity don't know what to do for the wave vector

Letting $$L\rightarrow \infty$$

$$f(x)= \lim_{L\to \infty}\sum_{n=-\infty}^{\infty}\frac{1}{2L}\int_{-L}^{L}f(x') e^{-\frac{k\pi x}{L}}e^{in\frac{\pi x}{L}} \ \ \$$

$$\frac{\delta k}{2\pi}=\frac{1}{L}$$

$$f(x)= \lim_{\delta k\to 0}\sum_{n=-\infty}^{\infty}\frac{\delta k}{2 \pi}\int_{-\frac{\pi}{\delta k}}^{\frac{\pi}{\delta k}}f(x') e^{-in \delta k x}e^{in\delta k x} \ \ \$$

$$g(k)=f(x') e^{-ikx'} dx'\ \ \$$

$$k=n\delta k$$

And then add the sigma sum like an integral

$$f(x)= \frac{1}{2 \pi}\int_{-\infty}^{\infty}g(k)e^{ik x} \ \ \$$

Last edited:
Z
fisher garry said:
LLsin2mπLtdt=0(A) \int_{-L}^{L}\sin\frac{2m\pi}{L}t\, dt=0

It seems as if you are "going twice around". In my experience, the correct integrals would be: $\int_{-L}^{L}sin(\frac{m\pi}{L}t )dt = 0$ etc.

Yes I do get that it would be one round to much. Thank you for sorting that out. I have found a Fourier transform online that seems to be related to the wavelength of an EM wave bt I have ran into a problem there as well:

How can they let h go to 0 when $h=\frac{2 \pi}{\lambda}$
when they do Fourier transforms for Heisenberg for example where they look at EM waves that has very small $\lambda$

This is the site I am referring to:
http://www.jpoffline.com/physics_docs/y2s4/cvit_ft_derivation.pdf