# I Fourier analysis and the sinusoidal plane wave

1. Jun 12, 2017

### BacalhauGT

hey

So Fourrier transform is

$f(t) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} F(\omega) e^{i \omega t} d\omega$

with

$F(\omega) = \int_{-\infty}^{\infty} f(t) e^{-i \omega t} dt$

Question 1 - The fourier mode for the continuous case is $\frac{1}{2 \pi} F(\omega) e^{i \omega t}$, is it right?

So a function can be written as a superposition of Fourier modes.

Question 2 - Why does the sinusoidal plane wave is always considered in the form

$\phi(x,t) = A \exp [i (k x - \omega t)]$ ?

I mean, Should it be $\phi(x,t) = A \exp [i (kx + \omega t)]$ ? Because this is the form of two dimensional fourier mode for x and t.

So using this, it makes sense for me analysing the linear equations using a Fourier mode, because since it is linear, any function can be given by a sum of Fourier modes, so we can see what happens to one frequency and k.

Thank you

2. Jun 13, 2017

### vanhees71

All the signs and factors in front of the Fourier transformation are arbitrary, but it's good to stick to conventions of your field. You have the sign for the time-frequency FT opposite to what physicists are used to. I've seen this convention in some engineering textbook once. That said concerning question 1, it's correct.

Question 2 adresses the Fourier transform in time and space for the wave equation (setting the phase velocity of the waves to 1 for convenience)
$$(\partial_t^2-\partial_x^2) \phi(x,t)=0.$$
The plane wave you've given first is the one describing a wave moving in positive $x$-direction, i.e., setting the phase constant, $k x-\omega t=\text{const}$ implies that
$$x=\frac{\omega}{k} t + \text{const}.$$
For the same reason your 2nd mode describes a wave moving in the negative $x$-direction. A general wave is a superposition of both (and a superposition over $\omega$).

Here, btw. you use the physicists' sign convention (i.e., a $-$ in front of time in your Fourier mode). Here you can write the solutions of the wave equation as
$$\phi(x,t)=\frac{1}{2 \pi} \int_{\mathbb{R}} \mathrm{d} \omega \left \{A(\omega) \exp[-\mathrm{i} \omega (t-x)] + B(\omega) \exp[-\mathrm{i} \omega (t+x)] \right \},$$
where I used the wave equation to get $k=\pm \omega$.

Since the wave equation is 2nd order in time you need to give two initial conditions to get a unique solution, i.e., you need to give as initial conditions
$$\phi(0,x)=\phi_0(x), \quad \dot{\phi}(0,x)=\psi_0(x).$$
You can think about, how to determine $A(\omega)$ and $B(\omega)$ given these initial conditions!

3. Jun 13, 2017

### BacalhauGT

Thank you. But please let me clarify one thing. I still dont get why the signs of the Fourier transformation are arbitrary. Can you explain me?

So Fourrier transform is

$f(t) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} F(\omega) e^{i \omega t} d\omega$ with $F(\omega) = \int_{-\infty}^{\infty} f(t) e^{-i \omega t} dt$

can i write:

$f(t) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} F(\omega) e^{-i \omega t} d\omega$ with $F(\omega) = \int_{-\infty}^{\infty} f(t) e^{+i \omega t} dt$

Please check if thats the reason: This is because since the domain is $-\infty$ to $\infty$, so integrating in $\omega$ covers the positive and negative $\omega$. So making $\omega \rightarrow -\omega'$:

$f(t) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} F(\omega) e^{i \omega t} d\omega = \frac{1}{2 \pi} \int_{\infty}^{-\infty} F(-\omega') e^{i -\omega' t} - d\omega' =\frac{1}{2 \pi} \int_{-\infty}^{\infty} F(-\omega') e^{i -\omega' t} d\omega' = \frac{1}{2 \pi} \int_{-\infty}^{\infty} F'(\omega') e^{i -\omega' t} d\omega'$

Is this correct?

So:

$f(t) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} F(\omega) e^{ \pm i \omega t} d\omega$ with $F(\omega) = \int_{-\infty}^{\infty} f(t) e^{\mp i \omega t} dt$

4. Jun 13, 2017

### I like Serena

The signs are indeed arbitrary and so is the place where we put the 2pi normalization constant. That's indeed because we can just as easily use $\omega'=-\omega$, and we can put the 2pi also in the other equation, or divide a $\sqrt{2\pi}$ over the 2 of them.

However, the first equation that you have called the fourier transform is usually called the inverse fourier transform. And the 2nd equation is usually called the fourier transform. Together they form a fourier transform pair.

Last edited: Jun 13, 2017
5. Jun 13, 2017

### BacalhauGT

thank you. It was a distraction

6. Jun 14, 2017

### BacalhauGT

btw, can anyone recommend me a book for fourier analysis with applications in physics? thank you

7. Jun 14, 2017

### Dr Transport

8. Jun 14, 2017