Fourier Transform of Polarization

[David92]

Homework Statement

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

Homework Equations

P(t) = ε₀∫ X(t-τ)E(τ) dτ <-starting point

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

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

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 e^iω'tdt and integrate over all time

1/2π ∫ F(ω)e^-iωt dω ∫ e^iω'tdt= ε₀∫ X(t-τ)E(τ) dτ ∫ e^iω'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 ∫ e^iω'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 ∫ e^iω'tdt <-Dirac Delta function 'picks out' ω'

F(ω') = ε₀∫ X(t-τ)E(τ) dτ ∫ e^iω'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 e^iωt
so the last equation could be rewritten

F(ω') = ε₀∫ X(t-τ)e^iωτ dτ ∫ e^iω'tdt

Though I am not sure if this helps.

 

[blue_leaf77]

F(ω') = ε0∫ X(t-τ)E(τ) dτ ∫ eiω'tdt

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.

 

[vela]

F(ω') = ε₀∫ X(t-τ)E(τ) dτ ∫ e^iω'tdt

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
$$F(\omega) = \int P(t)e^{i\omega t}\,dt.$$

 

[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) e^iω't dt = ∫ X(u) e^iω'(u+τ) du
where u = t - τ and this then equals

e^iω'τ ∫ X(u) e^iω'u = e^iω'τ X(ω)

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

F(ω') = ε₀ X(ω)∫ E(τ) e^iω'τ dτ

Does this mean

∫ E(τ) e^iω'τ dτ = E(ω) ?