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Why is the solution of the phi^6 potential not a soliton?

  1. May 8, 2015 #1
    1. The problem statement, all variables and given/known data

    Consider a theory with a [itex]\phi^6[/itex]-scalar potential:

    [itex]\mathcal{L} = \frac{1}{2}(\partial_\mu\phi)^2-\phi^2(\phi^2-1)^2.[/itex]

    Why is the solution to the equation of motion not a soliton?

    2. Relevant equations

    [itex]\phi''=\frac{\partial V}{\partial\phi}[/itex]

    3. The attempt at a solution

    [itex]
    \phi'\phi''=\phi'\frac{dV}{d\phi}\\
    \frac{d}{dx}\left(\frac{\phi'^2}{2}\right)=\frac{dV}{dx}\\
    \phi'=\pm \sqrt{2V}\\
    \phi'=\phi-\phi^3\\
    \Rightarrow\phi(x)= \frac{e^x}{\sqrt{e^{2x}-C_1}}-\frac{e^{-x}}{\sqrt{e^{-2x}-C_2}}[/itex]

    Yet the solution on the last line is not a soliton. Why is that so?
     
  2. jcsd
  3. May 13, 2015 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
  4. May 13, 2015 #3
    More generally, one has

    [itex]\mathcal{L} = \frac{1}{2}(\partial_\mu\phi)^2-(\phi^2-\epsilon)(\phi^2-1)^2[/itex]

    in which case there is no soliton, since the energy, defined as [itex]E=\int_{-\infty}^{+\infty}dx [\frac{\phi'^2}{2}+V(\phi)][/itex] is such that

    [itex]\lim_{t\to +\infty}E(t) = -\infty[/itex]

    for [itex]\epsilon\neq 0[/itex] and hence is topologically unstable.
     
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