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In 3+1 dimensions with signature +1 -1 -1 -1, $$ \mathcal{L}= \frac{1}{2}\partial^\mu\phi\partial_\mu\phi -\phi^2/2 -\phi^4/4$$ field equation: $$\square\phi+\phi+\phi^3=0$$ (check this) $$\square=\partial^2_t-\nabla^2$$ Note that it is not a Mexican hat. I guess this haven't been solved exactly before but, somebody have shown or discarded that there could be soliton solutions or at least solitary waves? |
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The solution for the soliton in a $\phi^4$ model is given by making a field $\phi$ which depends only on x and t, and is independent of any other spatial dimensions. This is a classical one dimensional problem. When the mass-squared parameter is negative, then the soliton appears. It is the solution to the equation $$ \partial_x^2 \phi + \phi - \phi^3 = 0 $$ Where x is rescaled to absorb $\mu^2$, and $\phi$ is rescaled to absorb $\lambda$. The solution is gotten by using a version of conservation of energy, which works here because the above is a second order differential equation, which looks just like the motion of a particle in a potential $$ V(\phi) = {1\over 2} \phi^2 - {1\over 4} \phi^4$$ Note that this is the inverted field potential appearing in the Lagrangian. The solution for $\phi$ has a x-conservation of x-energy, because if you call x "time", then the second order equation turns into Newton's laws for a one-dimensional motion. The conserved quantity is $$ {1\over 2} (\partial_x \phi)^2 + V(\phi) = E$$ For the soliton solution, $\phi$ should go to the vacuum solution at $x=\pm\infty$. The two vacua are the two minima of the original potential, the places where $$ \phi - \phi^3 = 0$$ or $$\phi = \pm 1 $$ The potential at these field values gives the energy, because the field gradient has to go to zero at infinty. This makes the x-energy 1/4 at infinity. The conservation of x-energy then tells you the field gradient $$ (\partial\phi(x))^2 + \phi^2 - {1\over 2} \phi^4 = 1/2 $$ or that $$ {1\over (\phi^2-1) } (\partial_x \phi) = \pm t+C $$ or $$ \tanh^{-1} \phi = t+C $$ Which gives the standard $\phi^4$ domain wall soliton $$ \phi(x) = \tanh(t+C)$$ This solution is a particle in 1d (1+1), a line in 2d (2+1), a domain wall in 3d (3+1), and in general, a d-1 dimensional object in d dimensions. |
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I would like to give you another solution to your equation that was recently published (see here). The equation $$\partial_x^2\phi+\mu_0^2\phi+\lambda\phi^3=0$$ admits the exact solution $$\phi(x)=\pm\sqrt{\frac{2\mu^4}{\mu_0^2 + \sqrt{\mu_0^4 + 2\lambda\mu^4}}}{\rm sn}\left(p\cdot x+\theta,\sqrt{\frac{-\mu_0^2 + \sqrt{\mu_0^4 + 2\lambda\mu^4}}{-\mu_0^2 - \sqrt{\mu_0^4 + 2\lambda\mu^4}}}\right) $$ being sn a Jacobi elliptic function, $\theta$ and $\mu$ two integration constants and provided that $$ p^2=\mu_0^2+\frac{\lambda\mu^4}{\mu_0^2+\sqrt{\mu_0^4+2\lambda\mu^4}}. $$ This has the appearance of a dispersion relation and $p$ is a quasi-momentum as happens in solid state with quasi-particles in a strong interacting environment. As pointed out before, this is not a soliton solution. |
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