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When solitons collide

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Adapted from notes I copied during a course I attended

Consider a 1+1 dimensional scalar field \phi(x,t) where x corresponds to the space dimension, and t corresponds to the time dimension. The Lagrangian and the Hamiltonian energy density for the field are given by,

\mathcal{L}=\frac{1}{2} \partial_{\mu}\phi \partial^{\mu}\phi - U(\phi)

\mathcal{H}=\frac{1}{2}(\partial_0 \phi)^2 + \sum_i\frac{1}{2}(\partial_i \phi)^2 + U(\phi)

Here, the Einstein summation convention is used, where repeated indices are summed over. \partial_{\mu} is used to denote the four-vector (\frac{1}{c}\frac{\partial}{\partial t}, \nabla)

The evolution of this scalar field is governed by the Euler-Lagrange equation,

\frac{\partial \mathcal{L}}{\partial \phi} = \partial_{\mu}\left(\frac{\partial \mathcal{L}}{\partial \left( \partial_{\mu}\phi \right)}\right)

Plugging in Lagrangian for the scalar field I’ve written above, the E-L equation reduces to,

\Box \phi = -\frac{\partial U(\phi)}{\partial \phi}

where, \Box is the D’Alembertian. ie, \Box \phi = \phi_{tt}  - \phi_{xx}

A specific example, where the solutions of the E-L equation have interesting properties, is given by the potential,

U(\phi)=\frac{\lambda}{2} \left(\phi^2-a^2\right)^2

The lowest energy solutions which satisfy the E-L equations are known as the classical vaccum. In this specific example, they are given by

\phi(x)=a
\phi(x)=-a

You can see that these solutions correspond to the minimum of the potential. Additionally, the total energy is also minimized by these solutions, because the contribution of the space and time derivatives to the Kinetic Energy density term is zero.

Apart from these vacuum solutions, there exist finite energy solutions of the E-L equations for this potential. From the criteria that the energy must be finite when the Hamiltonian energy density is integrated over x, such a solution must tend to the classical vacuua solutions at the extremities.

Such a solution is known as the kink solition, and is given by,

\phi(x)=a \tanh{\mu x}

where, \mu^2=\lambda a^2

You can check that this solution satisfies the Euler-Lagrange equation by plugging it in. Note that the solution is time-independent. Furthermore, when x \rightarrow \pm \infty, \phi \rightarrow \pm a. Therefore, the solution interpolates between the classical vacuua solutions.

To understand why it’s called a kink soliton, let’s plot its functional form, and the corresponding Energy density. For the plots below, I’ve set a=1, \lambda=1

Plot of the kink soliton solution

Plot of the kink soliton solution

Hamiltonian density for the kink-soliton solution

Energy density for the kink-soliton solution

As you can see, the solution tends to the vaccum solution (here, a=1) at either end and the energy density appears as a stationary wave packet which is like a kink in a rope.

Now, there are certain global symmetries that Lagrangian has, which are absent in the kink soliton solution. For instance, if we set x \rightarrow -x, the Lagrangian in Eq. (1), and consequently the corresponding action, is invariant. However, the solution \phi(-x) \rightarrow -\phi(x), is a new solution and interpolates the other way (ie, \phi \rightarrow \mp a as x \rightarrow \pm \infty. This is called the anti-kink solition, and has the same energy as the kink soliton.

There are other symmetries where the Lagrangian is invariant which give rise to new solutions. For example, a spatial translation will give a solution \phi_{new}=\phi(x-x_0), which shifts the location of the center of the kink. A lorentz transformation x \rightarrow \gamma(x-vt) will give a new time dependent solution, \phi_{new}=\phi\left(\gamma(x-vt)\right), where \gamma=1/\sqrt{1-v^2/c^2}. This corresponds to a kink which starts out at the origin and moves to the right with a constant velocity.

Therefore, the most general form of the kink soliton can be written as,

\phi_K(x,t) = a \tanh{\mu(x-x_0-vt)}

This corresponds to a soliton which is initially centered at x_0 and moves with a velocity v to the right. The anti-kink soliton corresponding to this is simply -\phi_K(x,t)

Now, consider a box which has both a kink and an anti-kink solition placed away from each other and with velocities in opposite directions. What happens to them as they approach each other? They can’t pass each other, otherwise they’d violate the boundary condition (remember that the kink and anti-kink solitons tend to different vacuum solutions at either end). So you’d expect them to bounce off each other.

In fact, we can even numerically test this, because we know that they are governed by the Euler-Lagrange equation. If \phi(x,t) represents both the kink and anti-kink soliton, then initially, one has

\phi(x,t) = \left\{\begin{array}{c} a\tanh{\mu(x+x_0-vt)} \quad x \leq 0 \\ -a\tanh{\mu(x-x_0+vt)} \quad x \geq 0 \end{array}\right\}

then, the E-L equation can be rewritten as two first order differential equations

\left(\begin{array}{c} \partial_t \phi(x,t) \\ \partial_t \pi(x,t) \end{array}\right) = \left(\begin{array}{c} \pi(x,t) \\ \partial_x^2\phi(x,t) - \frac{\partial U(\phi)}{\partial \phi}\end{array}\right)

where, \pi(x,t) = \partial_t \phi(x,t) is the momentum density

To solve the system of equations above, I discretized the spatial derivative using the central difference method, and used a fourth-order Runge Kutta solver with corresponding intial and boundary condition discussed above. You can see how the kink and anti-kink soliton bounce off each other in the video below.

Kinky, ain’t it?

Wikipedia has a gallery of images on solitons for the Sine-Gordon model, where a Sine potential is used instead of the quartic potential. There are also a lot of nice websites on the fantastic properties that such non-linear PDE’s exhibit.

See Also
Sine-Gordon Solitons and Soliton Collisions:
Pendulum Model

Soliton-Lab Art Gallery

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Written by parseval

November 30, 2008 at 6:20 am

Posted in physics, science, videos

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