The equation of motion of a real scalar field $\phi(x,t)$ in 1+1 dimension in an arbitrary potential $V(\phi)$ is given by $$\frac{\partial^2\phi}{\partial t^2}-\frac{\partial^2\phi}{\partial x^2}+\frac{dV}{d\phi}=0.\tag{1}$$

To obtain the soliton solution, one first assumes $\phi$ is time-independent so that it satisfies $$\frac{\partial^2\phi}{\partial x^2}=\frac{dV}{d\phi}\tag{2}$$ and the energy is finite i.e., $E<\infty$. For a potential of the form $V(\phi)=\frac{\mu^2}{2}\phi^2+\frac{\lambda}{4}\phi^4$, Eq. (2) has the following time-independent, finite energy solution $$\phi(x)=\frac{\mu}{\sqrt \lambda}\tanh\left(\frac{\mu x}{\sqrt{2}}\right).\tag{3}$$ The existence of such time-independent solutions means that it is possible to find a frame in which $\phi$ is time-independent. Because the distribution $\phi(x)$ in (3) is time-independent, the coordinate $x$, too, is time-independent i.e., $\dot{x}=0$ implying that the distribution isn't moving along $x-$axis. This is why the frame in which $\phi$ is time-independent is called the soliton's rest frame. To obtain moving solution one makes a Lorentz boost to obtain $$\phi(x,t)=\frac{\mu}{\sqrt \lambda}\tanh\left(\frac{\mu\gamma(x-\beta ct)}{\sqrt{2}}\right)\tag{4}$$ where $\beta=\frac{v}{c},$ and $\gamma=(1-\beta^2)^{-1/2}$. So it's like a pulse moving along $x$-axis without changing its shape.

Let O be a centre of the solition (4) (where $\phi=0$). At time $t=0$, the location of the centre O be $x=0$, and at time $t$ (when the pulse moved forward) the centre occupies a location $x=\beta ct$. Does it enable us to define the soliton velocity? If yes, how? Is it $dx/dt=c\beta$?

  • $\begingroup$ Much better, although there is still some notational confusion: It appears you use $x$ both for the spatial variable and for something like the location of the soliton, since $\dot{x}$ doesn't make sense otherwise. Your actual question is therefore unclear: For a general solution, what do you actually mean by "x is time-independent"? $\endgroup$ – ACuriousMind Oct 13 '17 at 10:37
  • $\begingroup$ If a wave pulse $f(x-vt)$ moves, its phase velocity is given by the velocity with which any constant phase point on the waveform moves, and is given by $\frac{dx}{dt}=v$. In exactly the same manner, can we not define $dx/dt$ as the phase velocity of the soliton where $x$ is any point on the soliton pulse? @ACuriousMind $\endgroup$ – SRS Oct 13 '17 at 11:03
  • $\begingroup$ 1. You're still overloading and misusing notation: The "$x$" in your $\mathrm{d}x/\mathrm{d}t$ is not the same as your $x$ in $f(x - vt)$ (one is a free variable, the other the location of a specific point), please try not to do that as it is confusing and provide an actual definition of what you call $\dot{x}$ ("phase velocity" only makes sense for waves). Furthermore, the l.h.s. of your eq. (4) should be $\phi(x,t)$, not $\phi(x)$. 2. Why do you feel the need to point out that $\dot{\phi}\neq 0$ may happen for a general solution? It's also non-zero for a solution like eq. (3). $\endgroup$ – ACuriousMind Oct 13 '17 at 12:29
  • $\begingroup$ 1. Let P be a point on a wave $f(x-vt)$. If at time $t$, the location of the point P be $x$, and if at time $t+dt$ (when the pulse moved forward) the same point P occupies a location $x+dx$. Then equating the phases at these two times, we have $dx/dt=v$ in the limit $dt\to 0$, which is the phase velocity. Similarly, at $t=0$, the centre O of the soliton $\phi(x,t)$ was at $x=0$ and after time $t$, the centre O shifts to $x=\beta t$. Am I still thinking wrong? 2. Why do you say $\dot{\phi}=0$ for (3) when it is a time-independent solution? @ACuriousMind $\endgroup$ – SRS Oct 13 '17 at 13:41
  • $\begingroup$ I have changed the question because I thought I think I need to understand the idea of soliton velocity first. @ACuriousMind $\endgroup$ – SRS Oct 13 '17 at 13:52

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