DispersiveWiki:Sandbox

Welcome to the sandbox! Please feel free to edit this page as you please by clicking on the "edit" tab at the top of this page. Terry 14:58, 30 July 2006 (EDT)

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this is the sandbox.

Gradient Expansion

Given a nonlinear equation as

$\Box \phi +\lambda \phi ^{3}=0$

one can always build a gradient expansion by assuming $\lambda \rightarrow \infty$ as

Duality in perturbation theory

In this section we will show how a duality principle holds in perturbation theory showing how to derive a strong coupling expansion with the leading order ruled by an adiabatic dynamics in order to study the evolution of a physical system. We consider the following perturbation problem

$\partial _{t}u=L(u)+\lambda V(u)$

being $\lambda$ an arbitrary ordering parameter: As is well known an expansion parameter is obtained by the computation of the series itself. The standard approach assume the limit $\lambda\rightarrow 0$ and putting

$u=u_{0}+\lambda u_{1}+\ldots$

one gets the equations for the series

Failed to parse (syntax error): {\displaystyle \partial_t u_0 = L(u_0) \\ }

Failed to parse (syntax error): {\displaystyle \partial_t u_1 = L'(u_0)u_1 + V(u_0) \\ }

$\vdots$

where a derivative with respect to the ordering parameter is indicated by a prime. We recognize here a conventional small perturbation theory as it should be. But the ordering parameter is just a conventional matter and so one may ask what does it mean to consider $L(u)$ as a perturbation instead with respect to the same parameter. Indeed one formally could write the set of equations

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \partial_t v_0 = V(v_0) \\ \partial_t v_1 = V'(v_0)v_1 + L(v_0) \\ \vdots }

where we have interchanged $L(u)$ and $V(u)$ and renamed the solution as $v$. The question to be answered is what is the expansion parameter now and what derivative the prime means. To answer this question we rescale the time variable as $\tau = \lambda t$ into eq.(\ref{eq:eq1}) obtaining the equation

$\lambda \partial _{\tau }u=L(u)+\lambda V(u)$

and let us introduce the small parameter $\epsilon=\frac{1}{\lambda}$. It easy to see that applying again the small perturbation theory to the parameter $\epsilon\rightarrow 0$ we get the set of equations (\ref{eq:set}) but now the time is scaled as $t/\epsilon$, that is, at the leading order the development parameter of the series will enter into the scale of the time evolution producing a proper slowing down ruled by the equation

$\epsilon \partial _{t}v_{0}=V(v_{0})$

that we can recognize as an equation for adiabatic evolution that in the proper limit $\epsilon\rightarrow 0$ will give the static solution $V(u_0)=0$. We never assume this latter solution but rather we will study the evolution of eq.(\ref{eq:lead}). Finally, the proof is complete as we have obtained a dual series

$u=v_{0}+{\frac {1}{\lambda }}v_{1}+\ldots$

by simply interchanging the terms for doing perturbation theory. This is a strong coupling expansion holding in the limit $\lambda\rightarrow\infty$ dual to the small perturbation theory $\lambda\rightarrow 0$ we started with and having an adiabatic equation at the leading order.

It is interesting to note that, for a partial differential equation, we can be forced into a homogeneous equation because, generally, if we require also a scaling into space variables we gain no knowledge at all on the evolution of a physical system. On the other side, requiring a scaling on the space variables and not on the time variable will wash away any evolution of the system. So, on most physical systems a strong perturbation means also a homogeneous solution but this is not a general rule. As an example one should consider fluid dynamics where two regimes dual each other can be found depending if it is the Eulerian or the Navier-Stokes term to prevail. In general relativity things stay in a way to get a homogeneous equation at the leading order. The reason for this is that products of derivatives or second order derivatives in space coordinates are the only elements forming the Einstein tensor beside time dependence.