# Convolution and fourier transform puzzle

1. Oct 29, 2006

### quasar987

My notes (from a physics course) justifies the following equality by invoking the convolution thm:

$$\int_{-\infty}^{+\infty} \chi(\omega)\vec{E}_0(\omega)e^{-i\omega t}d\omega=\int_{-\infty}^{+\infty} \chi(\tau)\vec{E}_0(\tau-t)d\tau$$

From a mathematical standpoint (i.e. without reference to what the functions chi and E are), is the above some consequence of the convolution thm? The convolution thm says that $\mathcal{F}(f*g)=\mathcal{F}(f)\cdot\mathcal{F}(g)$ I dont see how this applies here. For one, there are two integrals mutilplied by one another on the RHS of the convolution thm. And for another, the RHS of my equation is simply the convolution of chi and E; not its fourier transform or anything. Anyone sees?

And what if I were to add that

$$\vec{E}_0(\omega)=\int_{-\infty}^{+\infty} \vec{E}(T)e^{i\omega T}dT$$ ?

Notice the 0 subscript on the LHS but not in the integral; the difference is meant to represent two dinscting functions [of course one is just the "fourier coefficients" of the other]

Final hypothesis: perhaps my professor had another "convolution thm" in mind when he wrote that?

P.S. For clarity, I have not kept track of all the 2pi's popping up now and then in the above equations.

Last edited: Oct 29, 2006
2. Oct 29, 2006

### quasar987

I added a 0 suscript to the RHS of the first equatiom too. Sorry about that.

3. Oct 29, 2006

### StatusX

Is that also supposed to be $\chi_0(\omega)$? If so, then this is a direct consequence of the fourier transform identity you mentioned: just take the inverse fourier transform of both sides. If not, it's obviously false: just take $\chi=1$. Also, the usual notation for the fourier transform of f is F or $\tilde f$. The f0 notation doesn't make sense to me.

Last edited: Oct 29, 2006
4. Oct 29, 2006

### quasar987

mmmh, I'm confused nowon who has a 0 and who doesnt. In my notes there are no zeroes anywhere.

The zero subscipt notation is from the fact that these E in there are electromagnetic waves. Typically, a single electromagnetic wave is noted

$$\vec{E}=\vec{E}_0e^{-kz-\omega t)$$

so the natural extension to a wave packet is to say that we are going to superpose functions of the type

$$\vec{E}=\vec{E}_0(\omega)e^{-kz-\omega t)$$

leading to the notation E_0 for the fourier coefficients. (I thought you were a physics student StatusX, am I mistaken?)

Last edited: Oct 29, 2006
5. Oct 29, 2006

### quasar987

on which chi would you put a 0 subscript?

6. Oct 29, 2006

### StatusX

I am a physics student, and I have seen that notation for a plane wave. It makes sense there because then $\vec E_0$ is a constant vector. But I've never seen that subscript carried over to a fourier trasnform. It's always the tilde notation (eg, griffiths).

The equation only makes sense if both functions on the LHS are fourier transforms of the corresponding functions on the RHS.

7. Oct 29, 2006

### quasar987

There's still a mistake in the sign of the exponent as far as I can tell.

Let $\chi$ and $\vec{E}$ be functions of $\tau$. Then

$$\mathcal{F}[\chi*\vec{E}(t)]=\sqrt{2\pi}\mathcal{F}[\chi]\mathcal{F}[\vec{E}]=\sqrt{2\pi}\chi(\omega)\vec{E}(\omega)$$

$$\Rightarrow \frac{1}{\sqrt{2\pi}}\chi*\vec{E}(t)=\mathcal{F}^{-1}[\chi(\omega)\vec{E}(\omega)]$$

where is it implicit that when the argument is omega, we're talking about the fourier transform of the function. Writing explicitely the last equation

$$\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}\chi(\tau)\vec{E}(t-\tau)d\tau = \frac{1}{2\pi}\int_{-\infty}^{+\infty}\chi(\omega)\vec{E}(\omega)e^{i\omega t}d\omega$$

(the 2pi's don't match those in my notes either, in my notes, I have

$$\frac{1}{2\pi}\int_{-\infty}^{+\infty} \chi(\tau)\vec{E}(\tau-t)d\tau=\int_{-\infty}^{+\infty} \chi(\omega)\vec{E}(\omega)e^{-i\omega t}d\omega$$)

Last edited: Oct 29, 2006
8. Oct 29, 2006

### StatusX

The inverse transform is usually the one with the minus i, but this is just convention. Also, please try to use different symbols for a function and its fourier transform. tau and omega are just dummy variables, so, eg:

$$\int_{-\infty}^\infty \chi(\tau) d\tau = \int_{-\infty}^\infty \chi(\omega) d\omega$$

9. Oct 29, 2006

### quasar987

duh :P

I want to thank you StatusX for your assistance.

Are you a grad student?

10. Oct 29, 2006

### StatusX

No, I'm a senior. I actually just made a thread about applying to grad schools.

11. Oct 29, 2006

### quasar987

You might not have an answer to that but...

Why are you so freakishly good at math?

12. Oct 29, 2006

### StatusX

Thanks, I appreciate being called a freak. Maybe it's because I was born down river from a nuclear power plant and down wind from a chemical weapons testing ground.

13. Oct 29, 2006

### quasar987

I have a natural talent for diplomacy. :uhh:

14. Oct 31, 2006

### quasar987

If I take it from the top and try to recreate the result of my prof by doing every step in between...

We know that given a monochromatic plane wave of frequency $\omega_0$ propagating in a dielectric material in the z direction, the resulting polarization density function of the material is

$$\vec{P}(z,t)=\epsilon_0\chi_e(\omega_0)\vec{E}_0e^{i(k_R(\omega_0)z-\omega_0 t)}$$

Now, if we shoot a wave packet

$$\vec{E}(z,t)=\int_{-\infty}^{+\infty}\tilde{\vec{E}}_0(\omega)e^{i(k_R(\omega)z-\omega t)}d\omega$$

(happy? I put a tilde ) into the material, the resulting polarization should be

$$\vec{P}(z,t)=\int_{-\infty}^{+\infty}\epsilon_0 \chi_e(\omega)\tilde{\vec{E}}_0(\omega)e^{ik_R(\omega)z}e^{-i\omega t}d\omega$$

Define

$$\tilde{\vec{E}}_0(z,\omega)\equiv \tilde{\vec{E}}_0(\omega)e^{ik_R(\omega)z}$$

Then by the convolution thm,

$$\vec{P}(z,t)=\int_{-\infty}^{+\infty}\epsilon_0 \tilde{\chi_e}(\tau)\vec{E}(z,t-\tau)d\tau$$

where

$$\tilde{\chi_e}(\tau)=\mathcal{F}^{-1}[\chi_e(\omega)]=\int_{-\infty}^{+\infty}\chi_e(\omega) e^{-i\omega \tau}d\omega$$

and

$$\vec{E}(z,\tau)=\mathcal{F}^{-1}[\vec{E}_0(z,\omega)]=\int_{-\infty}^{+\infty}\tilde{\vec{E}}_0(\omega)e^{i(k_R(\omega)z-\omega \tau)}d\omega$$

is just the wave packet I sent in.

Does that look good or is the presence of the z not legal and I have to set it to 0 everywhere?

Last edited: Oct 31, 2006
15. Oct 31, 2006

### StatusX

Yea, that looks about right, although I think there are some factors of 2pi missing. z is fine - you can just think of it as fixed throughout the calculation, then when you're done the result is true for all z.

16. Oct 31, 2006

Awesome.

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