Timeline for Question about graph Ginzburg-Landau equation
Current License: CC BY-SA 3.0
11 events
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| Nov 4, 2017 at 23:21 | comment | added | sleeve chen | Oh, I see. The last question is about 3., how to obtain the picture "mexican hat" you mentioned? (By Taylor expansion which gives the linear equation for perturbation and then plot it?) | |
| Nov 4, 2017 at 21:55 | comment | added | Carlo Beenakker | you have to look at the left-hand-side of the equation, it is the term $\dot{\theta}_i$ that acquires an extra term. | |
| Nov 4, 2017 at 18:43 | comment | added | sleeve chen | However, applying your argument to phase equation, $\sin(\theta_i-\theta_k)$ remains unchanged also: $\sin(\theta_k+\delta\theta-\theta_i-\delta\theta)=\sin(\theta_i-\theta_k)$. Could you please provide a complete comparison on these? | |
| Nov 4, 2017 at 12:07 | comment | added | Carlo Beenakker | of course not: $\theta_k\mapsto\theta_k+\delta\theta$, $\theta_i\mapsto\theta_i+\delta\theta$, so the difference $\theta_k-\theta_i$ remains unchanged. | |
| Nov 4, 2017 at 10:09 | comment | added | sleeve chen | But based on your argument, in the amplitude equation the value of the term $\cos(\theta_k-\theta_i-\delta \theta)$ will also be changed due to $\delta \theta$. | |
| Nov 4, 2017 at 10:02 | comment | added | Carlo Beenakker | it has nothing to do with the order $\theta_k-\theta_i$ or $\theta_i-\theta_k$; it just that $\delta\theta=\omega t+\gamma$ depends on time, so $(d/dt)(\theta_i+\delta\theta)=\dot{\theta}_i+\omega$; so the phase equation gets an offset $\omega$ which the amplitude equation does not. | |
| Nov 4, 2017 at 2:07 | comment | added | sleeve chen | To 1., I am still confused that for amplitude eq it is $\theta_k-\theta_i$ and for the phase eq it is $\theta_i-\theta_k$. Why one is unchanged and the other one is changed by $\omega$? To 3., how to obtain the picture you mentioned? (By Taylor expansion which gives the linear equation for perturbation and then plot it?) Sorry for many questions. Thanks! | |
| Nov 3, 2017 at 22:07 | comment | added | Carlo Beenakker | I clarified 1; concerning 3, the typical picture is the mexican hat --- amplitude perturbations are radial and damped, while phase fluctuations are azimuthal and undamped. | |
| Nov 3, 2017 at 22:03 | history | edited | Carlo Beenakker | CC BY-SA 3.0 |
added 109 characters in body
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| Nov 3, 2017 at 21:10 | comment | added | sleeve chen | To 1., can I use the same argument to say that the phase equation is unchanged and so phase dynamic is invariant? I am confused about 3., how to view it? Thanks! | |
| Nov 3, 2017 at 10:50 | history | answered | Carlo Beenakker | CC BY-SA 3.0 |
