✎ Sine-Gordon Model [The Physics Travel Guide]

You've loaded an old revision of the document! If you save it, you will create a new version with this data.

====== Sine-Gordon Model: $ \quad  \mathcal{L} = \frac{1}{2} \left( (\partial_\mu \phi)^2  \right) -  (1-cos \phi) $  ======

The Sine-Gordon model is a toy model that helps to understand fundamental notions like [[advanced_notions:quantum_field_theory:duality|duality]] in a simplified setup.

In addition, the equations of the theory permit topological non-trivial solutions called solitons. Many basic features of such solitons can be studied in the Sine-Gordon model in a simplified setup.

The name "Sine-Gordon model" is a wordplay, because the equation of motion of the model is similar to the famous [[equations:klein-gordon_equation|Klein-Gordon equation]] and contains a [[basic_tools:trigonometric_functions|sinus function]].

{{ :models:solitonpendulumwave.png?nolink&400 |}}
  
<tabbox Student> 
**Specification of the Model**

The Sine-Gordon model describes a continuous set of pendulums. Each pendulum hangs freely in a uniform gravitational field of strength $g$ on a rod with length $l$. The excitation above the ground state is measured by the angle $\phi$. At the end of the pendulum, we have a bob of mass $m$. This is shown in the picture on the right-hand side.

The Lagrangian of the system is

$$ L = \frac{1}{2} ml^2\dot{\phi}^2 - mgl (1-cos \phi), $$
where $\dot{\phi}\equiv d\phi /dt $ denotes the time derivative.

To analyze the system it is convenient to rescale the units such that all constants $m,g,l$ are equal to one. Then, the Lagrangian reads

$$ L = \frac{1}{2} \dot{\phi}^2 -  (1-cos \phi). $$

So far, this could be a normal mechanics problem. However, if $\phi$ becomes not only a variable that depends on $\phi = \phi(t)$, but also on the location $\phi(t,x)$ we have a continuous set of such pendulums and hence are dealing with a __field theory__.

We assume that we have an infinite set of such pendulums that hang from an also infinitely long support rod.

The thing is now that **neighboring pendulums are coupled to each other**. This could be achieved, for example, through elastic bands between the pendulums.

In a field theory, we are not interested in the Lagrangian but in the Lagrangian density since we are interested in what happens locally. The Lagrangian density for the Sine-Gordon field theory reads

$$ \mathscr{L} = \frac{1}{2} \left( (\partial_\mu \phi)^2  \right) -  (1-cos \phi), $$
$$ = \frac{1}{2} \left( \dot{\phi}^2 - (\partial_\mu \phi)^2  \right) -  (1-cos \phi).$$
where $ L = \int_{-\infty}^\infty dx\mathscr{L} $ and $\mu =0,1$.

Here $\frac{1}{2} \left( (\partial_\mu \phi)^2  \right)$ is just the usual kinetic term and $(1-cos \phi)\equiv U$ is the potential of the system.

----

Now that we have the Lagrangian density, we can derive the __equations of motion__ of the system, as usual, by using the [[equations:euler_lagrange_equations|Euler-Lagrange equations]]

$$\frac{\partial \mathscr{L}}{\partial \Phi^i} - \partial_\mu \left(\frac{\partial \mathscr{L}}{\partial(\partial_\mu\Phi^i)}\right) = 0$$

which yields

$$  \partial_\mu \partial^\mu \phi + \sin \phi =0\, . $$

When we now want to understand what the system actually does, we must search for solutions of this field equation. Before doing that, we want to think about what kind of solutions we can except.

The __ground state__ of the system is when all pendulums are hanging straight downwards. This is the __trivial solution__ $\phi(x,t) =0$ and also called the vacuum solution.

However, take note that the potential $U =1-cos \phi$ is minimized not only for $\phi =0$, but also for all $\phi = 2\pi k$, for all integers $k$. Therefore, we really have an infinite number of degenerate vacua. This will become important in a moment. But first let's talk about actual dynamical stuff that can happen in the system.

Since the pendulums are coupled it is possible that __waves__ travel through the set of pendulums. Think of a "laola wave" that travels through a sports stadium. These waves can have different amplitudes and wavelengths and really correspond to the type of waves that we are used to.

In addition, to such normal waves, we can also have [[advanced_notions:quantum_field_theory:solitons|solitary wave]]. Imagine we pick one end (at $x= -\infty$) of the support rod and rotate it by $360^\circ$. This will yield a __twist__ in the system and as a result, a __solitary wave__ will travel through the system.

This is shown in the image on the right-hand side.

In physics, we are usually only interested in __field configuration with finite energy__. Hence, the field must be in a ground state configuration at infinity:

$$ V(\phi(\infty)) = V(\phi(-\infty)) \ \rightarrow \ \ \phi(\infty)\stackrel{!}{=} 2\pi k, \quad \phi(-\infty)\stackrel{!}{=} 2\pi l ,  $$

where $k$ and $l$ are arbitrary integers. These are the appropriate __boundary conditions__ if we are interested in finite energy configurations.

So in the simplest case we have simply $\phi(\infty)= \phi(-\infty)=0$. In words, this means that there is no twist in the system. However, for all other possible values of the field at the boundary, we have one or more twists in the system. For example, in the example above we twisted the support rod at $x=- \infty$ once by $360^\circ = 2\pi$ and left it at $x= \infty$ as it was. This means the boundary conditions of this field configuration are

$$ \phi(\infty)0 0, \quad \phi(-\infty)= 2\pi . $$

This is what characterizes soliton. A soliton is a topological nontrivial field configuration and we can now understand why it is stable. To get rid of such a twist in the system we have to move all pendulums from $x=-\infty$ to $x = \infty$. There are infinitely many pendulums that need to be moved and hence the energy that would be required to get rid of such a twist is infinity. However, the energy stored in the twist is finite. This follows because the field configuration is in the ground state at infinity. The thing is that it is at different ground states at $x=-\infty$ to $x = \infty$.

The quantity that characterizes such twists in the system has the fitting name "winding number" and for the Sine-Gordon model we can define it as

$$ n \equiv \frac{\phi(\infty)-\phi(-\infty)}{2\pi} \in \mathbb{Z}. $$

----

**Solitonic Solutions in Detail**

Since solitons are stable and static, we now try to find such a solution of the field equations. Static means that a soliton is time-independent and therefore we can make the ansatz that our soliton solution is time independent. This yields the __static Sine-Gordon equation__

$$ - \partial_x \partial^x \phi + \sin \phi =0. $$

This is an ordinary 2nd-order differential equation which we can solve analytically.

Multiplying the static Sine-Gordon equation with $\partial^x \phi$ yields
$$ \partial^x \phi  \partial_x \partial^x \phi +  = \partial^x \phi  \sin \phi. $$

Now, integrating this equation yields

$$ \frac{1}{2} (\partial^x \phi)^2 = - \cos \phi +A,  $$

where $A$ is the constant of integration.

Since we know that at infinity we must have $\partial^x \phi \to 0$ and $\cos \phi \to 1$ in order to have a configuration with finite energy, we can deduce $A=1$.

Now, using $\cos 2\phi = 1-\sin \phi$ and integrating yields

$$ \int_0^\phi \frac{d\phi}{\sin\frac{\phi}{2}} = \pm 2\int_{x_0}^x dx , $$

where $x_0$ denotes the location of the soliton. Perming the integration yields

$$ \ln \left( \tan \frac{\phi}{4} \right) = \pm (x -x_0) . $$
We can rewrite this as

$$ \phi(x)_k = 4\arctan \left( e^{\pm (x-x_)} \right) \tag{Eq. 1} $$

{{ :models:kinksol.png?nolink&400|}}
The $\pm$ in the exponent corresponds to solitons with winding number $n=\pm 1$. The solution with $n=1$ is known as the __kink solution__ and the solution with $n=-1$ as the __antikink solution__.

When we plot the kink solution we can see that it is a wave-packet of finite energy that is centered around $x_0$. In other words, a kink is like a potential barrier in the system.

If we now want to calculate the mass of the kink, we have to integrate the Hamiltonian density since it simply represents the energy. We get the Hamiltonian density, as usual, from the Lagrangian density and the integrated result is

$$ E[\phi] = \int_{-\infty}^\infty dx \left( \frac{1}{2} \left( (\partial_\mu \phi)^2  \right) +  (1-cos \phi) \right) =  \int_{-\infty}^\infty dx \left( \frac{1}{2} \left( (\partial_\mu \phi)^2  \right) +  U(x) \right) $$

To get the specific energy density of the kink, we need to calculate the derivative of the solution in Eq. 1:
$$ \partial_x \phi(x)_k = \frac{4}{1-e^{2(x-x_)}}  =  \frac{4}{e^{(x-x_)}+e^{-(x-x_)}} = 2 \text{sech}(x-x_0).$$

Putting this into the potential yields

$$ U(x) = 2 \text{sech}^2(x-x_0). $$

The total energy of the kink is therefore

$$ E[\phi_k] = 4 \int_{-\infty}^\infty dx \text{sech}^2(x-x_0) = 8.   $$

----

**Recommended Resources:**

* http://bartholomewandrews.com/projects/Skyrmions.pdf
  * https://www.uni-muenster.de/Physik.TP/archive/fileadmin/lehre/NumMethoden/WS0910/ScriptPDE/SineGordon.pdf

<note tip>
The motto in this section is: //the higher the level of abstraction, the better//.
</note>

<tabbox FAQ> 
  
<tabbox History> 
The Sine-Gordon equation was originally
used to study surfaces of constant negative curvature in
geometry: E. Bour, J. Ecole Imperiale Polytech. 19, 1 (1862)
</tabbox>

Please solve the following equation to prove you're human. 39 -8 = Please keep this field empty:

Edit summary

Note: By editing this page you agree to license your content under the following license: CC Attribution-Share Alike 4.0 International

User Tools

Site Tools

Page Tools