Fourier analysis and the sinusoidal plane wave

[BacalhauGT]

hey

So Fourier 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

 

[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!

 

[BacalhauGT]

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!

Thank you. But please let me clarify one thing. I still don't get why the signs of the Fourier transformation are arbitrary. Can you explain me?

So Fourier 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 that's 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 $F(\omega) = \int_{-\infty}^{\infty} f(t) e^{\mp i \omega t} dt$

 

[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.

 

[BacalhauGT]

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.

thank you. It was a distraction

 

[BacalhauGT]

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

 

[Dr Transport]

https://www.amazon.com/dp/0821868896/?tag=pfamazon01-20

I used the first edition and consistently go back to it as a reference...

 

[BacalhauGT]

https://www.amazon.com/dp/0821868896/?tag=pfamazon01-20

I used the first edition and consistently go back to it as a reference...

Thank you!