# Fourier Transform of Polarization

1. Feb 12, 2017

### David92

1. The problem statement, all variables and given/known data
The problem is from an optics text, however I believe the problem to be a mathematical one.

I'm trying to take the Fourier transform of
P(t) = ε0∫ X(t-τ)E(τ) dτ which should equal
P(ω) = ε0X(ω)E(ω)
where ε0 is a constant
X is the susceptibility
E is the electric field

2. Relevant equations
P(t) = ε0∫ X(t-τ)E(τ) dτ <-starting point

Fourier transforms:
P(ω) = ∫ P(t)eiωt dt
P(t) = 1/2π ∫ F(ω)e-iωt

P(ω) = ε0X(ω)E(ω) <- end point (final solution)

3. The attempt at a solution
First off not sure if convolution is part of the answer where
P(t) = ε0∫ X(t-τ)E(τ) dτ also equals
P(t) = ε0∫ X(τ)E(t-τ) dτ
but I will start my attempt with the first equation

If I plug the P(t) time transforms into the left hand side I get

1/2π ∫ F(ω)e-iωt dω = ε0∫ X(t-τ)E(τ) dτ

now multiply by eiω'tdt and integrate over all time

1/2π ∫ F(ω)e-iωt dω ∫ eiω'tdt= ε0∫ X(t-τ)E(τ) dτ ∫ eiω'tdt

where the left hand side can be arranged to give

1/2π ∫ F(ω) dω ∫ e-i(ω-ω')tdt= ε0∫ X(t-τ)E(τ) dτ ∫ eiω'tdt

using ∫ e-i(ω-ω')tdt = 2πσ(ω-ω') <- Dirac Delta function

∫ F(ω)σ(ω-ω') dω = ε0∫ X(t-τ)E(τ) dτ ∫ eiω'tdt <-Dirac Delta function 'picks out' ω'

F(ω') = ε0∫ X(t-τ)E(τ) dτ ∫ eiω'tdt

and now I'm stuck, unless my approach is wrong to begin with.
I also know the electric field is the real part of eiωt
so the last equation could be rewritten

F(ω') = ε0∫ X(t-τ)eiωτ dτ ∫ eiω'tdt

Though I am not sure if this helps.

2. Feb 12, 2017

### blue_leaf77

It can be rearranged to give
$$F(\omega') = \epsilon_0 \int E(\tau)\ d\tau \left( \int X(t-\tau)e^{i\omega' t} dt \right)$$
The integral in the parentheses is the Fourier transform of $X(t-\tau)$. Using the shifting property of Fourier transform you should be able to find its relation with the transform of $X(t)$. If you can get the idea of this part, the rest should be pretty straight forward.

3. Feb 13, 2017

### vela

Staff Emeritus

Just wanted to mention you could have gotten to this point without all that work. Just plug in
$$P(t) = \varepsilon_0 \int X(\tau-t)E(t)\,d\tau$$ into
$$F(\omega) = \int P(t)e^{i\omega t}\,dt.$$

4. Feb 13, 2017

### David92

Thank you blue leaf for pointing me towards the shifting property of Fourier transforms, and vela you are certainly right I just didn't notice. This does save some time!

As for the shifting property I still have a few questions about it,
is the below true?

∫ X(t-τ) eiω't dt = ∫ X(u) eiω'(u+τ) du
where u = t - τ and this then equals

eiω'τ ∫ X(u) eiω'u = eiω'τ X(ω)

This turns F(ω'), from my previous post, into

F(ω') = ε0 X(ω)∫ E(τ) eiω'τ

Does this mean

∫ E(τ) eiω'τ dτ = E(ω) ?