Subsections

• Definitions
• Correlation Function
• Relaxation in Fourier Space
• Example: Velocity Autocorrelation Function
• Example: Diffusion

Spectral Representation

Definitions

We applied Fourier transform for space-dependent problems--and we were successful. Let us do the same for time-dependent problems:  

$$x_\omega = \int_{-\infty}^{\infty} x(t)e^{i\omega t}\,dx$$ (1)

Inverse transform:

$$x(t) = \int_{-\infty}^{\infty} x_\omega e^{-i\omega t}\,\frac{d\omega}{2\pi}$$

Correlation Function

The product of x's is

$$\left\langle x(t)x(t')\right\rangle=\left\langle \iint_{-\in... ...\omega't')}\, \frac{d\omega\,d\omega'}{(2\pi)^2}\right\rangle$$

Or, since only x's are random,

$$\left\langle x(t)x(t')\right\rangle=\iint_{-\infty}^{\infty}... ...{-i(\omega t +\omega't')}\, \frac{d\omega\,d\omega'}{(2\pi)^2}$$

Symmetry with respect to time: $\left\langle x(t)x(t')\right\rangle$ depends only on t-t'. This means that in the integral $\omega t +\omega't'\to\omega(t-t')$. This means that $\omega'=-\omega$.

Conclusion:

$\left\langle x_{\omega}x_{\omega'}\right\rangle$ should be proportional to $\delta$-function $\delta(\omega+\omega')$

Definition:

Let  

$$\left\langle x_{\omega}x_{\omega'}\right\rangle=2\pi\varphi_\omega\delta(\omega+\omega')$$ (2)

Then

$$\varphi(t)=\int_{-\infty}^{\infty} \varphi_\omega e^{-i\omega t}\,\frac{d\omega}{2\pi}$$

We see that $\varphi_\omega$ is indeed the Fourier transform of $\varphi(t)$. This is exactly the function measured in inelastic scattering, relaxation experiments (NMR, ESR, dielectric spectroscopy, shear rheometry,...). Inverse:

$$\varphi_\omega = \int_{-\infty}^{\infty} \varphi(t)e^{i\omega t}\,dx$$

If t=0, we obtain

$$\varphi(0)=\int_{-\infty}^{\infty} \varphi_\omega \,\frac{d\omega}{2\pi}$$

But $\varphi(0) = \left\langle x^2(0)\right\rangle=kT/\gamma$. We obtained:

$$\int_{-\infty}^{\infty}\varphi_\omega \,\frac{d\omega}{2\pi} = \frac{kT}{\gamma}$$

Relaxation in Fourier Space

We obtained for correlation function

$$\varphi(t)=\frac{kT}{\gamma}e^{-\lambda \left\lvert t\right\rvert}$$

Take Fourier transform:  

$$\varphi_{\omega} = \int_{\infty}^{\infty}\frac{kT}{\gamma}e... ...ght\rvert}\,dt = \frac{2\lambda kT}{\gamma(\lambda^2+\omega^2)}$$ (3)

This is a Lorentzian!

Relaxation with one characteristic time is Lorentzian.

Many characteristic times--more complex formulae.

Example: Velocity Autocorrelation Function

Consider a large particle surrounded by small particles (a colloidal particle among molecules). Equation for average velocity:

$$\frac{d\left\langle \mathbf{v}\right\rangle}{dt}=-\lambda\left\langle \mathbf{v}\right\rangle$$

with friction coefficient $\lambda$. For a spherical particle with radius r and mass m in liquid with viscosity $\eta$ 

$$\lambda=6\pi\eta r/m$$ (4)

The average square

$$\left\langle \mathbf{v}^2(0)\right\rangle = \frac{3kT}{m}$$

We immediately obtain autocorrelation function:

• In the real time:

  $$\left\langle \mathbf{v}(0)\mathbf{v}(t)\right\rangle=\frac{3kT}{m}e^{-\left\lvert \lambda\right\rvert t}$$

• After Fourier transform:  

  $$\left\langle \mathbf{v}(0)\mathbf{v}(t)\right\rangle_{\omega} = \frac{3\lambda kT}{m(\lambda^2+\omega^2)}$$ (5)

Equations (3) and (5) are special cases of fluctuation-dissipation theorem.

In particular, if $\omega=0$

$$\begin{gathered} \varphi_{\omega=0} = \frac{2 kT}{\gamma\la... ...\frac{3kT}{m\lambda} = \frac{kT}{2\pi\eta mr} \end{gathered}$$

Example: Diffusion

Consider particles with concentration c. Diffusion equation with diffusion coefficient D:

$$\frac{\partial c(\mathbf{r},t)}{\partial t} = D\nabla^2 c(\mathbf{r},t)$$

Fourier transform in space:

$$c(\mathbf{r},t) = \sum_{\mathbf{k}} c_{\mathbf{k}}e^{i\mathbf{k}\mathbf{r}}$$

We obtain:

$$\frac{\partial c_{\mathbf{k}}(t)}{\partial t} = -D\mathbf{k}^2 c_{\mathbf{k}}(t)$$

Let us now consider equilibrium. In equilibrium the fluid is homogeneous. For $\mathbf{k}\ne0$:

$$\left\langle c_{\mathbf{k}}\right\rangle=0,\quad \left\lang... ...ne\mathbf{k}'\\ kT/\gamma,&\mathbf{k}=\mathbf{k}' \end{cases}$$

with $\gamma=\partial^2A/\partial c^2$.

We immediately obtain:

$$\left\langle c_{\mathbf{k}}(0)c_{\mathbf{k}}(t)\right\rangle=\frac{kT}{\gamma}e^{-D\mathbf{k}^2t}$$

Or, in Fourier space

$$S(\mathbf{k},w) = \frac{2D\mathbf{k}^2 kT}{\gamma(D^2\mathbf{k}^4+\omega^2)}$$

This is dynamic structure factor.