How Does the WZW Model Simplify the Action Integral?

In summary: A_\mu \right) = -\frac{1}{2}\int_0^1 \gamma_{\mu\tau}\frac{dx^\mu}{d\tau}\frac{dx^\tau}{d\tau}d\tau$. This is exactly the first expression in the WZW model, which can be written as$I = \frac{1}{2}\int_0^1 \gamma_{\alpha\beta}\frac{dx^\alpha}{d\tau}\frac{dx^\beta}{d\tau}d\tau$. Thus, we can see that the two expressions are equivalent.
  • #1
Jim Kata
197
6
Hi I'm trying to understand very basic aspects of the WZW model, and would appreciate any help. The main thing i don't get is how you replace

[tex]
I = \frac{1}
{2}\int\limits_0^1 {\gamma _{\alpha \beta } } \frac{{dx^\alpha }}
{{d\tau }}\frac{{dx^\beta }}
{{d\tau }}d\tau
[/tex]

with
[tex]
I = \frac{{ - k}}
{4}\int\limits_0^1 {tr} (g^{ - 1} \frac{d}
{{d\tau }}g)^2 d\tau
[/tex]


for some group g

My thoughts are I don't know how the metric enters and the connection doesn't. For example consider

[tex]
{\mathbf{g}} = P\exp \int\limits_0^1 {\frac{{dx^\mu }}
{{d\tau }}} {\mathbf{A}}_\mu d\tau
[/tex]

then

[tex]
{\mathbf{g}}^{ - {\mathbf{1}}} \frac{d}
{{d\tau }}{\mathbf{g}} = - \frac{{dx^\mu }}
{{d\tau }}{\mathbf{A}}_\mu
[/tex]

so the connection enters but not sure how the metric does cos I'm not familiar with any identity that says

[tex]
tr\left( {{\mathbf{A}}_\mu {\mathbf{A}}_\tau } \right) = \gamma _{\mu \tau }
[/tex]
 
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  • #2
Thanks in advance.The two expressions are equivalent because they both describe the same action. The first expression is written in terms of a metric tensor, $\gamma_{\alpha\beta}$, and the components of the velocity vector, $\frac{dx^{\alpha}}{d\tau}$. The second expression is written in terms of a group element, $g$, and its inverse, $g^{-1}$. The connection between these two expressions comes from the fact that the group element can be parameterized in terms of the connection. In other words, we can write the group element as$g = P\exp \int_0^1 \frac{dx^\mu}{d\tau}A_\mu d\tau$. Here, $P$ is the path-ordering operator, which ensures that the exponential is properly ordered according to the path parameter, $\tau$. Taking the derivative of this expression with respect to $\tau$, we find that$g^{-1}\frac{dg}{d\tau} = -\frac{dx^\mu}{d\tau}A_\mu$. This is the connection between the two expressions.To see how the metric enters into the equation, let us take the trace of the above expression. We have$\mathrm{tr}\left( g^{-1}\frac{dg}{d\tau} \right) = -\mathrm{tr}\left( \frac{dx^\mu}{d\tau}A_\mu \right)$. By definition, the trace of a matrix is invariant under change of basis. In other words, we can choose any basis we like, and the trace will remain the same. In particular, we can choose a basis in which the components of the connection are expressed in terms of the metric tensor. In this basis, we have$\mathrm{tr}\left( \frac{dx^\mu}{d\tau}A_\mu \right) = \gamma_{\mu\tau}\frac{dx^\mu}{d\tau}\frac{dx^\tau}{d\tau}$. Therefore, we have$-\mathrm{tr}\left( \frac{dx^\mu
 
  • #3



Hi there,

The WZW model is a conformal field theory that describes the dynamics of a certain type of field theory called a sigma model. In this model, the fields are maps from a two-dimensional space (usually called the worldsheet) to a target space, which can be a group manifold. The action of the WZW model is given by the integral of a certain quantity called the WZW term, which is what you have written above.

To understand the replacement you mentioned, it is helpful to first understand the WZW term in a bit more detail. The WZW term is a measure of the curvature of the target space, and it is given by the trace of the product of two matrices: the inverse of the metric on the target space (which is usually denoted by g) and the derivative of the group element g with respect to the worldsheet coordinates. This is what you have written in your first equation.

Now, the replacement you mentioned is a simplification that is often used in calculations involving the WZW model. It is based on a mathematical identity known as the Maurer-Cartan equation, which relates the derivative of a group element to the group element itself and the group's connection (the A's in your second equation). Using this identity, you can show that the WZW term can be written in terms of the connection and its derivative, without explicitly involving the metric. This is what is done in your second equation.

In summary, the replacement you mentioned is a simplification that is often used in calculations involving the WZW model, and it is based on a mathematical identity known as the Maurer-Cartan equation. I hope this helps clarify things a bit!
 

Related to How Does the WZW Model Simplify the Action Integral?

1. What is the WZW model?

The WZW model is a theoretical framework in physics that describes the interactions between particles known as gauge bosons. It is based on a mathematical structure called a Lie algebra and is used to study the behavior of elementary particles and their symmetries.

2. How does the WZW model relate to other theories in physics?

The WZW model is closely related to other theories in physics, such as quantum field theory and string theory. It has been used to understand the dynamics of particles in high energy collisions, and it has also been used to study the dynamics of strings in string theory.

3. What are the basic principles of the WZW model?

The WZW model is based on three main principles: gauge symmetry, conformal symmetry, and chiral symmetry. Gauge symmetry refers to the invariance of physical laws under certain transformations, conformal symmetry refers to the invariance of physical laws under scale transformations, and chiral symmetry refers to the invariance of physical laws under certain transformations that flip the direction of time.

4. What are the applications of the WZW model?

The WZW model has many applications in theoretical physics, such as describing the behavior of particles in high energy collisions, understanding the properties of quantum field theories, and studying the dynamics of strings in string theory. It has also been used in condensed matter physics to study the behavior of strongly correlated systems.

5. How can I learn more about the WZW model?

There are many resources available for learning more about the WZW model, such as textbooks, research papers, and online courses. It is recommended to have a strong background in mathematics and physics before delving into the complexities of the WZW model. Additionally, consulting with experts in the field and participating in discussions and conferences can also help deepen understanding of the model.

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