Do non-Abelian gauge fields takes the same value in the Lie subalgebra

In summary, the conversation discusses the Lie algebra and the field strength tensor for the $\mathrm{SU(2)}$ gauge fields. It is clarified that the field strength tensor lies in the subalgebra spanned by the unbroken generators and the gauge fields in the first two terms on the R.H.S. also lie in the same subalgebra. The confusion arises regarding the gauge fields in the third term, but it is concluded that they must also lie in the subalgebra. In the case of only one unbroken generator, the third term vanishes and the remaining terms reduce to the electromagnetic field strength tensor.
  • #1
AlbertEi
27
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Do the $\mathrm{SU(2)}$ gauge fields takes the same value in the Lie subalgebra spanned by the $\mathrm{SU(2)}$ field strength tensor?

I will try to clarify my questions. Define the Lie algebra as:

\begin{equation}
[t^a,t^b] = \varepsilon_{abc}t^c
\end{equation}

where $\varepsilon$ is the usual Levi-Civita symbol. The field strength tensor:

\begin{equation}
F_{\mu \nu} = \partial_\mu A_\nu - \partial_\nu A_\mu + [A_\mu , A_\nu]
\end{equation

In components:

\begin{equation}
F_{\mu \nu}^a t^a = \partial_\mu A_\nu^a t^a - \partial_\nu A_\mu^a t^a + A_\mu^b A_\nu^c \varepsilon_{abc}t^a
\end{equation}

Now I will skip a lot of details, but basically in the Higgs vacuum $F_{\mu \nu}$ lies in the subalgebra spanned by the unbroken generators (in my example there is only one unbroken generator). I will denote the unbroken generators as $t^A$ (i.e. the group index of the unbroken generator is denoted by a capital letter), therefore:

\begin{equation}
\Rightarrow F_{\mu \nu}^A t^A = \partial_\mu A_\nu^A t^A - \partial_\nu A_\mu^A t^A + A_\mu^b A_\nu^c \varepsilon_{Abc} t^A
\end{equation}

Clearly, the gauge fields in the first two terms on the R.H.S. lie in the subalgebra. But I'm confused about the last term. To me it seems that those gauge field do not necessary live in the same subalgebra. However, the sources that I read seem to suggest that I'm wrong. If anybody can help me, then that would be greatly appreciated.

P.s. I've recently learned Latex, so I hope that I have used it correctly.

Edit: It seems that something is going wrong with my formulas, but I'm not sure how to fix it. I hope people can still understand what I've written.

Thanks!
 
Last edited:
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  • #2
Ok, I think I know where I went wrong in my thought process, the first two terms on the R.H.S. must be in the same subalgebra as the field strength tensor, and the gauge fields in the third term are the same gauge fields. Therefore, they must also lie in the subalgebra. However, if we only have one unbroken generator, then that mean that the last term vansishes and so it reduces to the electromagnetic field strength tensor. Is that correct?
 

FAQ: Do non-Abelian gauge fields takes the same value in the Lie subalgebra

1. What are non-Abelian gauge fields?

Non-Abelian gauge fields are mathematical constructs used in theoretical physics to describe the interactions between particles. They are a type of gauge field that do not obey the commutative property, meaning that their values at different points cannot be swapped without changing the overall result.

2. What is a Lie subalgebra?

A Lie subalgebra is a subset of a larger Lie algebra, which is a mathematical structure that describes the algebraic properties of a non-commutative group. In the context of non-Abelian gauge fields, the Lie subalgebra represents a subgroup of the larger group of gauge transformations that preserves the value of the gauge field.

3. How do non-Abelian gauge fields take values in a Lie subalgebra?

Non-Abelian gauge fields take values in a Lie subalgebra through a process known as gauge fixing. This involves choosing a specific set of parameters that define the gauge transformation, which then determines the values of the gauge field in the Lie subalgebra.

4. Are the values of non-Abelian gauge fields the same in all Lie subalgebras?

No, the values of non-Abelian gauge fields can vary in different Lie subalgebras. This is because the choice of gauge fixing parameters can affect the specific values of the gauge field within the Lie subalgebra.

5. What is the significance of the values of non-Abelian gauge fields in the Lie subalgebra?

The values of non-Abelian gauge fields in the Lie subalgebra can have physical implications, as they determine the strength and nature of the interactions between particles. In addition, the mathematical structure of the Lie subalgebra can provide insights into the underlying symmetries and properties of the system being studied.

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