# Is My Localized Yang-Mills Theory Solution Correct?

• A
• Antarres
In summary, the conversation discusses the localization of internal symmetries and the introduction of a covariant derivative to maintain invariance of the Lagrangian. The transformation properties of the quantity K^mu are determined to be -K^mu*theta, with this solution being deemed correct but the reasoning behind it remaining uncertain.
Antarres
While revising Yang-Mills theory, I have stumbled upon a certain problem, which I solved in a somewhat trivial way so I would like to check whether my reasoning is correct.

Let's say we have a multicomponent matter field ##\{\phi^m(x)\}## which transforms according to some Lie group ##G## of internal symmetry:
$$\phi'(x) = \phi(x) + \delta_0\phi(x)$$
$$\delta_0\phi(x) = \theta^aT_a\phi(x) \equiv \theta\phi(x)$$
where ##\theta^a## are parameters, ##T_a## are generators, and ##a## is multiplet index.
We localize this internal symmetries by ##\theta^a \rightarrow \theta^a(x)##. In order to maintain invariance of the Lagrangian, we introduce the covariant derivative:
$$\nabla_\mu \phi(x) = (\partial_\mu + A_\mu)\phi(x) \qquad A_\mu \equiv A^a_\mu T_a$$
where ##A_\mu## is the introduced gauge field.
Now when we localize the symmetry and introduce the covariant derivative in such a way that keeps Lagrangian invariant, we have that the covariant derivative of the field transforms according to the rule(by definition):
$$\delta_0\nabla_\mu\phi(x) = \theta\nabla_\mu\phi(x)$$
And we have that equations of motion can be written in covariant form:
$$\frac{\partial \mathcal{L}}{\partial \phi} - \nabla_\mu\frac{\partial\mathcal{L}}{\partial\nabla_\mu\phi} =0$$

We define the following quantity:
$$K^\mu = \frac{\partial\mathcal{L}}{\partial\nabla_\mu\phi}$$
and we're looking for the transformation properties of this quantity. I have solved it by saying that the quantity ##K^\mu\nabla_\mu\phi## must be gauge invariant, that is:
$$\delta_0(K^\mu\nabla_\mu\phi) = 0$$
from which it is easy to find that ##\delta_0K^\mu = -K^\mu\theta##. This is the correct solution, however I'm not sure whether my argument is correct. I have a gut feeling that it must be correct, but in that case I don't see why it would rigorously be true, although it seems trivially true for quadratic Lagrangian.

So it would be good if someone would look at this and point it out if I did this correctly. Thanks.

Antarres

vanhees71
It looks correct to me.

Antarres

## 1. What is Yang-Mills theory?

Yang-Mills theory is a mathematical framework used to describe the interactions between elementary particles. It is a quantum field theory that combines the principles of quantum mechanics and special relativity.

## 2. What is the significance of Yang-Mills theory in physics?

Yang-Mills theory is a fundamental theory in particle physics that helps explain the strong nuclear force, one of the four fundamental forces in nature. It is also an important component of the Standard Model of particle physics.

## 3. How does Yang-Mills theory differ from other quantum field theories?

Yang-Mills theory is unique in that it describes the interactions between particles through the exchange of other particles, known as gauge bosons. This is in contrast to other quantum field theories that use scalar fields or spinor fields to describe interactions.

## 4. What is the role of symmetry in Yang-Mills theory?

Symmetry plays a crucial role in Yang-Mills theory, as it is based on the principle of gauge symmetry. This means that the mathematical equations of the theory remain unchanged when certain transformations are applied, allowing for a more elegant and consistent description of particle interactions.

## 5. How has Yang-Mills theory been tested and validated?

Experimental evidence for Yang-Mills theory has been found through high-energy particle collider experiments, such as the Large Hadron Collider, which have confirmed the existence of the predicted gauge bosons. The theory has also been successfully used to make predictions about particle interactions that have been confirmed through experiments.

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