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If the period is much smaller than the half-life, ie. the degradation rate is much smaller than the frequency : $latex \Omega \gg \gamma $ , we can make the following simplifications.
$latex \displaystyle{\Large \arctan\left(\frac{\Omega}{\gamma}\right) \sim \frac{\pi}{2} }$ $latex \displaystyle{\Large \frac{1}{2}\arctan\left(\frac{\omega}{\gamma}\right) \sim \frac{\pi}{4} }$ (angle to avoid self-coupling, $latex \displaystyle{\Large \theta_k }$ ) $latex \displaystyle{ \Large R_A(\Omega_j, x-x_j) \sim \frac{\alpha}{2 \sqrt{D\Omega}} \exp^{- \sqrt{\frac{\Omega}{2D}}|x-x_j|}}$ $latex \displaystyle{ \Large \theta_A(\Omega_j, x-x_j) \sim -(\frac{\pi}{4}+\sqrt{\frac{\Omega}{2D}}|x-x_j|) } $ Let us define two more useful parameters : $latex \displaystyle{\Large \rho= \sqrt{\frac{2D}{\Omega}} }$ (this describes the distance travelled by one particule during a period ) $latex \displaystyle{\Large M= \frac{\alpha}{\sqrt{D\Omega}} }$ (this is a coupling strength associated to the shortest time scale, $latex \displaystyle{\Large \Omega }$ ) Reporting this in the self-consistent equation for the frequency, one finds :
$latex \displaystyle{\Large \Omega=\omega + \frac{M}{2}Im \left( \sum_{j=-\frac{N}{2}}^{\frac{N}{2}}e^{-\frac{|j|}{\rho}}e^{i \left( -(\frac{\pi}{4}+\frac{|j|}{\rho})+\frac{\pi}{4}\right ) } e^{i\beta j} \right) = \omega+\frac{M}{2} Im \left( \sum_{j=-\frac{N}{2}}^{\frac{N}{2}}e^{-\frac{|j|}{\rho}(1+i)} e^{i\beta j} \right) = \omega+\frac{M}{2} Im \left(\sum_{j=0}^{\frac{N}{2}}e^{-j\left(\frac{(1+i)}{\rho}-i \beta \right)} + \sum_{j=1}^{\frac{N}{2}}e^{-j\left(\frac{(1+i)}{\rho}+i \beta \right)} \right)} $
Using again the formula for the convergent geometric sum, one finds :
$latex \displaystyle{\Large \Omega=\omega + \frac{M}{2}Im \left( \frac{1}{1-e^{-\left(\frac{1+i}{\rho}-i \beta \right)}}+\frac{1}{1-e^{-\left(\frac{1+i}{\rho}+i \beta \right)}}-1\right) } $
To find the allowed frequencies, we then have to find the zeroes of the function :
$latex \displaystyle{\Large F(\Omega)=\Omega -\omega - \frac{\alpha}{2\sqrt{D\Omega}}Im \left( \frac{1}{1-e^{-\left((1+i)\sqrt{\frac{\Omega}{2D}}-i \beta \right)}}+\frac{1}{1-e^{-\left((1+i)\sqrt{\frac{\Omega}{2D}}+i \beta \right)}}\right) } $
Limit for $latex \displaystyle{\Large \Omega\to\infty }$
$latex \displaystyle{\Large \frac{1}{1-e^{-\sqrt{\frac{\Omega}{2D}}}e^{-i\sqrt{\frac{\Omega}{2D}}+i\beta}}+\frac{1}{1-e^{-\sqrt{\frac{\Omega}{2D}}}e^{-i\sqrt{\frac{\Omega}{2D}}-i\beta}} \sim_{\Omega\to 0} 2(1+e^{-\sqrt{\frac{\Omega}{2D}}}\cos(\beta)e^{-i\sqrt{\frac{\Omega}{2D}}}) } $ and the imaginary part vanishes
$latex \displaystyle{\Large F(\Omega) \sim_{\Omega\to\infty} \Omega } $
Limit for $latex \displaystyle{\Large \Omega\to 0 }$
If $latex \displaystyle{\Large \beta \ne 0 }$ $latex \displaystyle{\Large \frac{1}{1-e^{-(1+i)\sqrt{\frac{\Omega}{2D}}+i\beta}} \sim \frac{1}{1-e^{i(\beta-\sqrt{\frac{\Omega}{2D}})}(1-\sqrt{\frac{\Omega}{2D}})} \sim \frac{1-\sqrt{\frac{\Omega}{2D}}e^{i(\beta-\sqrt{\frac{\Omega}{2D}})}}{1-e^{i(\beta-\sqrt{\frac{\Omega}{2D}})}} }$
Then $latex\displaystyle{\Large F(\Omega) \sim_{\Omega\to 0} \omega+ \frac{\alpha}{2\sqrt{D\Omega}} Im \left( \frac{1}{1-e^{i(\beta-\sqrt{\frac{\Omega}{2D}})}}+ \frac{1}{1-e^{i(\beta+\sqrt{\frac{\Omega}{2D}})}} -\sqrt{\frac{\Omega}{2D}}e^{-i\sqrt{\frac{\Omega}{2D}}}2\cos(\beta)\right)}$
In the case where $latex \displaystyle{\Large \beta }$ is pretty big, when $latex \displaystyle{\Large \Omega\to 0 }$ the imaginary part of all the terms goes to 0 like a sine function, so the whole function behaves like $latex \displaystyle{\Large F(\Omega) \sim \Omega-cst } $
In the case where $latex \displaystyle{\Large \beta }$ is small and positive, there should be a singularity at $latex \displaystyle{\Large \beta = \sqrt{\frac{\Omega}{2D}}}$
In this case, if we write $latex \displaystyle{\Large \beta }$ as $latex \displaystyle{\Large \sqrt{\frac{\Omega}{2D}}+\epsilon }$ then one of the terms behaves like : $latex \displaystyle{\Large \frac{1}{1-e^{i \epsilon}} \sim \frac{i}{\epsilon} }$
The summation of the geometric series is not allowed for $latex \displaystyle{\Large \epsilon = 0 }$, so this function does not describe well this region, but there is a sudden increase that leads to a new solution ? (more details needed)
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