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KuboDistributions

< Kubo relaxation functions | Index | A simple prototype model of dynamical relaxation? >

Assume a Gaussian distribution of Kubo Toyabe second moments {$\Delta$}

{$ P(\Delta)=\frac 1 {\sigma\sqrt{2\pi}} e^{- \frac 1 2 \left(\frac{\Delta-\mu}{\sigma}\right)^2}$}

Then the average relaxation function {$G(\sigma,\mu;t)$} may be obtained directly from the KT relaxation functions {$G(\Delta;t)$}

{$G(\Delta;t)= \frac 1 3 \left( 1+2(1-\Delta^2t^2)e^{- \frac 1 2 \left(\frac{\Delta}{\sigma}\right)^2}\right) $}

as

{$G(\sigma,\mu;t)= \int_{-\infty}^{\,\,\,\,\,\,\,\,\,\,\,\,\infty} G(\Delta;t)P(\Delta)d\Delta $}

The integral splits into three terms, the first being

{$\frac 1 3 \int_{-\infty}^{\,\,\,\,\,\,\,\,\,\,\,\,\infty} P(\Delta)d\Delta = \frac 1 3$}

The second term is

{$\frac 2 3 \frac 1 {\sigma\sqrt{2\pi}}\int_{-\infty}^{\,\,\,\,\,\,\,\,\,\,\,\,\infty} e^{- \frac 1 2 \left(\frac{\Delta}{\sigma}\right)^2} e^{- \frac 1 2 \left(\frac{\Delta-\mu}{\sigma}\right)^2}d\Delta $}

The exponent in the integral is {$-\frac 1 2 \left[\left(\frac{\Delta}{\sigma}\right)^2 +\left(\frac{\Delta-\mu}{\sigma}\right)^2 \right]=-a\Delta^2+b\Delta-c$} with

{$\begin{eqnarray*} a&=&\frac 1 {2\sigma^2} (1+\sigma^2t^2) \\ b &=& \frac \mu {\sigma^2} \\ c & =& \frac {\mu^2}{2\sigma^2} \end{eqnarray*}$}

According to Wolfram

{$ \int_{-\infty}^{\,\,\,\,\,\,\,\,\,\,\,\,\infty} e^{-ax^2+bx} dx=\sqrt{\frac \pi a} e^{\frac {b^2} {4a}} $}

Therefore the second term is

{$\frac 2 3 \frac 1 {\sigma\sqrt{2\pi}}e^{-\frac {\mu^2}{2\sigma^2}}\sqrt{\frac \pi a} e^{\frac {b^2} {4a}} = \frac 2 3 e^{-\frac {\mu^2}{2\sigma^2}}\,e^{\frac {\mu^2} {2(1+\sigma^2t^2)}}\,\sqrt{\frac 1 {1+\sigma^2t^2}} $}

The third term is

{$-\frac {2t^2} 3 \frac 1 {\sigma\sqrt{2\pi}}\int_{-\infty}^{\,\,\,\,\,\,\,\,\,\,\,\,\infty} \Delta^2 e^{- \frac 1 2 \left(\frac{\Delta}{\sigma}\right)^2} e^{- \frac 1 2 \left(\frac{\Delta-\mu}{\sigma}\right)^2}d\Delta $}

According to Wolfram

{$ \int_{-\infty}^{\,\,\,\,\,\,\,\,\,\,\,\,\infty} x^2 e^{-ax^2+bx} dx=\sqrt{\frac \pi a} e^{\frac {b^2} {4a}} \left[ \frac {b^2} {4a^2} + \frac 1 {2a}\right]$}

Therefore the third term is {$\begin{eqnarray*}-\frac {2t^2} 3 \frac 1 {\sigma\sqrt{2\pi}}e^{-\frac {\mu^2}{2\sigma^2}}\sqrt{\frac \pi a} e^{\frac {b^2} {4a}} \left[ \frac {b^2} {4a^2} + \frac 1 {2a}\right] &=& \\ -\frac {2} 3 e^{-\frac {\mu^2}{2\sigma^2}}\,e^{\frac {\mu^2} {2(1+\sigma^2t^2)}}\,\sqrt{\frac 1 {1+\sigma^2t^2}} \left[ \frac {\mu^2t^2 + \sigma^2t^2 +\sigma^4t^4} {(1+\sigma^2t^2)^2} \right]\end{eqnarray*}$} The sum of the three terms is {$G(\sigma,\mu;t)= \frac 1 3 \left[ 1 + 2 e^{-\frac {\mu^2}{2\sigma^2}}\,e^{\frac {\mu^2} {2(1+\sigma^2t^2)}}\,\sqrt{\frac 1 {1+\sigma^2t^2}} \frac {1 + \sigma^2t^2 - \mu^2t^2 } {(1+\sigma^2t^2)^2} \right] $}

Resulting average function with width of the distribution σ ranging from 0.3 to 3 μs-1

< Kubo relaxation functions | Index | A simple prototype model of dynamical relaxation? >