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Calculations of $\Sigma ({\bf k},\omega )$

In the one-loop approximation expression for $\Sigma ({\bf k},\omega )$ has the form
\begin{displaymath} \Sigma({\bf k},\omega)= \Sigma_{a1}({\bf k},\omega) + \Sigma_{a2}({\bf k},\omega) + \Sigma_{a3}({\bf k},\omega) \,, \end{displaymath} (79)

where $\Sigma_j({\bf k},\omega)$ is given by (A1-A3). Our goal here is to analyze these expressions in one-pole approximation, by substituting in it ``one-pole'' $n({\bf k},\omega)$ and $G({\bf k},\omega)$ from (4.11) and (4.16). In the resulting expression one can perform the integration over $\omega$ analytically. The result is
    $\displaystyle \Sigma({\bf k},\omega)=\int \frac{d^3 k_1 d^3 k_2}{(2\pi)^3}$ (80)
  $\textstyle \times$ $\displaystyle \Bigg( \frac{\vert V({\bf k}_2,{\bf k},{\bf k}_1)\vert^2 \delta({... ..._2)\big]} {\omega+\omega({\bf k}_1)-\omega({\bf k}_2) +i ( \gamma_1+ \gamma_2)}$  
  $\textstyle +$ $\displaystyle \frac{\vert V({\bf k}_0,{\bf k}_1,{\bf k}_2)\vert^2 \delta({\bf k... ... {\omega-\omega({\bf k}_1)-\omega({\bf k}_2)+i ( \gamma_1+ \gamma_2)}\Bigg) \ .$  

Next we introduce $\Sigma({\bf k})=\Sigma({\bf k},\omega_*)$ , with $\omega_*$ given by (4.14) and consider (4.18) in the limit of small $\gamma$, which allows us to perform analytically integrations over perpendicular components of wavevectors. The result for the damping frequency $\gamma(k)$ may be represented in the following form (for details see Appendix B):
$\displaystyle \gamma(k)=\frac{A^2 k^2}{4 \pi c }\int_{1/L}^\infty {n(q) q^2 d q} \simeq \frac{A^2 k^2 }{4\pi c } N(\Omega) \ .$     (81)

We introduced here cut-off for small $k$ at $1/L$, where $L$ is the size of the box. We also introduced ``the density of the number of particles'' $N(\Omega)$ in the solid angle according to
\begin{displaymath} N(\Omega)=\int k^2 n({\bf k}) d k \ , \end{displaymath} (82)

such that the total number of particles
\begin{displaymath} N=\int N(\Omega)d\Omega\ . \end{displaymath} (83)

After substituting $A$ from (B11), one has the following estimate for $ \gamma({\bf k})$:
\begin{displaymath} \gamma({\bf k})\simeq k^2 N(\Omega)/\rho_0\, , \end{displaymath} (84)

Consider now $\Sigma'({\bf k})\equiv {\rm Re \Sigma({\bf k})}$. It follows from (B12) that

$\displaystyle \Sigma({\bf k})$ $\textstyle =$ $\displaystyle \frac{A^2}{4 \pi ^2 c } \int d q \int _0 d y q^2 n(q) \frac{y}{y^2+\Gamma_{k12}^2}$ (85)
    $\displaystyle \times \left[ (k^2 + 2 k q + q^2) - (k^2 - 2 k q + q^2 ) \right]$  
  $\textstyle \simeq$ $\displaystyle \frac{A^2 k } { \pi^2 c^2}\int d q \int_0^{y_{\rm max}} \frac{y d y }{y^2 + \Gamma_{k12}^2} \left[ c q^3 n(q) \right] \,,$  

where $\Gamma_{k12}=\gamma (k)+\gamma (k_1)+\gamma (k_2)$ is the ``triad interaction'' frequency. One may evaluate the integral with respect to $y$ as
\begin{displaymath} \L (q)=\ln{\frac{y_{\rm max}}{\Gamma_{kkq}}} \simeq \ln{\frac{c k^2}{q \gamma({\bf k})}}. \end{displaymath} (86)

After substituting $ \gamma({\bf k})$ from (4.22), one has
\begin{displaymath} \L (q)\propto \ln{\rho_0\c/q N(\Omega)} \end{displaymath} (87)

The main contribution to the integral (4.23) over $q$ comes from the infrared region $q\simeq 1/L$. It gives the estimate,
\begin{displaymath} \Sigma'({\bf k}) = \frac{A^2 k}{\pi^2 c^2}\L E(\Omega) \end{displaymath} (88)

where we have defined the density of the wave energy in solid angle as
\begin{displaymath} E(\Omega)= \int \omega_0(k)n(k) k^2 d k\ . \end{displaymath} (89)

This value relates to $N(\Omega)$ as follows:
\begin{displaymath} E(\Omega)\simeq \frac{ c }{L}N(\Omega) \ . \end{displaymath} (90)

Equation (4.26) together with the expression (B11) for $A$ may be written as
\begin{displaymath} \Sigma'({\bf k})\simeq c k \epsilon \ln{1/\epsilon}\ , \end{displaymath} (91)

where
\begin{displaymath} \epsilon\simeq E(\Omega)/\rho_0 c^2 \ , \end{displaymath} (92)

is the dimensionless parameter of nonlinearity, the ratio of energy of acoustic turbulence and the density of thermal energy of media $\rho_0 c^2 \simeq \tilde n T$, where $\tilde n$ is the concentration of atoms.

Equation (4.22) for $ \gamma({\bf k})$ may be written in a similar form

\begin{displaymath} \gamma({\bf k})\simeq c k (k L) \epsilon\ . \end{displaymath} (93)

One can see that
\begin{displaymath} \frac{ \gamma({\bf k})}{\Sigma'({\bf k})}\propto \frac{k L} {\ln{1/\epsilon}}\ . \end{displaymath} (94)

It means that for large enough inertial interval
\begin{displaymath} \gamma({\bf k}) \gg\Sigma'({\bf k}) \end{displaymath} (95)

and one may neglect the nonlinear corrections $\Sigma'({\bf k})$ to the frequency with respect to the damping of the waves $ \gamma({\bf k})$. That shows that our above calculations of $\Sigma({\bf k})$ is self-consistent. Later we also will take into account only damping $ \gamma({\bf k})$ in the expressions for the Green's functions taking $\omega(k)=\omega_0(k)=c k$.
next up previous
Next: Calculations of . Up: One-loop approximation Previous: One-loop approximation
Dr Yuri V Lvov 2007-01-17