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Appendix A: derivation WKB equations in presence of a condensate

Let us split $ \phi$ into its real and imaginary parts $ a = \Re \phi$ and $ b = \Im \phi $. Then the equation ([*]) splits into two coupled equations

$\displaystyle \partial_{t}a$ $\displaystyle + \triangle b + 2\vec{v}\cdot\nabla a
 +\frac{\nabla\varrho}{\varrho}\cdot\nabla b 
 +\rho(2 a b + b(a^2+b^2))
 = 0,$ (37)
$\displaystyle \partial_{t}b$ $\displaystyle - \triangle a +2a\varrho+2\vec{v}\cdot\nabla b
 -\frac{\nabla\varrho}{\varrho}\cdot\nabla a 
 + {\varrho}(3 a^2+b^2 +a(a^2+b^2))
 = 0,$ (38)

where we have used the fact that $ \frac{\nabla\psi_{0}}{\psi_{0}} =
\frac{\nabla\varrho}{2\varrho}+\frac{i}{2} \vec{v} $, which follows from ([*]).

Gabor transforming our two coupled equations ([*]) and ([*]) and using Taylor series to represent large-scale quantities,

$\displaystyle \varrho(\vec{x}_{0}) = \varrho(\vec{x}) +
 (\vec{x}_{0}-\vec{x})\cdot\nabla\varrho(\vec{x}) +
 O(\varepsilon^{2}),$    

we find

$\displaystyle \partial_{t}\hat{a} + \triangle \hat{b} 
 + \frac{\nabla \varrho}{\varrho}\cdot \nabla \hat{b}
 + \vec{v}\cdot\nabla\hat{a}$ $\displaystyle +{\cal G} \left[\rho(2 a b + b(a^2+b^2))\right]
 = 0,$ (39)
$\displaystyle \partial_{t}\hat{b} - \triangle \hat{a} 
 - {\bf }\frac{\nabla \varrho}{\varrho}\cdot \nabla \hat{a}
 + 2{\bf }\vec{v}\cdot\nabla\hat{b}$ $\displaystyle + 2\varrho\hat{a} 
 + 2i{\bf }\nabla\varrho\cdot\partial_{k}\hat{a}$    
  $\displaystyle + {\cal G}\left[( \varrho(3 a^2+b^2 +a(a^2+b^2)))\right] 
 = 0.$ (40)

Where $ {\cal G}[f(x)]$ is the Gabor transform of $ f(x)$. We have kept only $ O(\varepsilon)$ terms and neglected the $ O(\varepsilon^{2})$ and higher order terms. For generality, we have kept the nonlinear term.



Subsections
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Next: The order - Up: text Previous: Summary
Dr Yuri V Lvov 2007-01-23