# Showing the equivalence of lagrangians

• dingo_d
In summary, the conversation discusses a Lagrangian and two different ways to express it. The first is in terms of a field and its covariant derivative, while the second expands the covariant derivative and includes weak isospin and hypercharge operators. The speaker is seeking help in showing that the kinetic part is equivalent in both expressions.
dingo_d

## Homework Statement

I have a lagrangian written as:

$\mathcal{L}_H = \text{Tr}\left[\,(D_\mu \Phi)^\dagger D^\mu \Phi\right] - \mu^2 \text{Tr}\left[\,\Phi^\dagger \Phi\right] - \lambda (\text{Tr}\left[\,\Phi^\dagger \Phi\right])^2$

Where the field is:

$\Phi \equiv \frac{1}{\sqrt{2}}(i \sigma_{2} \phi^*, \phi) = \frac{1}{\sqrt{2}} \begin{pmatrix}\phi^{0*} & \phi^+ \\-\phi^- & \phi^0\end{pmatrix} \;,$

and

$D_\mu \Phi \equiv \partial_\mu \Phi + i g \frac{\sigma^{i}}{2} W^{i}_\mu \Phi - i \frac{g'}{2} B_\mu \Phi \sigma_3 \;.$

And I need to show that that is the same as:

$\mathcal{L}_H = \left[(\partial_\mu+ig{\bf{T}\cdot W_\mu}+ig'\frac{Y}{2}B_\mu)\phi\right]^\dagger\left[(\partial_\mu+ig{\bf{T}\cdot W_\mu}+ig'\frac{Y}{2}B_\mu)\phi\right]-\mu^2(\phi^\dagger\phi)-\lambda(\phi^\dagger\phi)^2$.

Now, I have easily shown that the potential part really is the same, by explicitly doing matrix multiplication, and showing that the expression with the trace is the same as the one without, but how to show the kinetic part? What to do here?

Any help is welcome :)

You haven't specified the weak isospin $\mathbf{T}$ and hypercharge operator $Y$, but I think you will find that they are precisely defined so that this equivalence follows.

## 1. What is a Lagrangian?

A Lagrangian is a mathematical function that describes a physical system in terms of its position and velocity. It is used in the field of classical mechanics to determine the equations of motion for a system.

## 2. How do you show the equivalence of Lagrangians?

To show the equivalence of Lagrangians, one must demonstrate that the two Lagrangians result in the same equations of motion for a given system. This can be done through mathematical manipulation and substitution.

## 3. Why is it important to show the equivalence of Lagrangians?

Showing the equivalence of Lagrangians is important because it allows us to use different Lagrangians to describe the same physical system. This can provide a deeper understanding of the system and can sometimes simplify calculations.

## 4. Are there any limitations to showing the equivalence of Lagrangians?

Yes, there are limitations to showing the equivalence of Lagrangians. In some cases, it may be impossible to show the equivalence due to the complexity of the system or the limitations of our mathematical understanding.

## 5. How does the equivalence of Lagrangians relate to the principle of least action?

The principle of least action states that a physical system will follow the path of least action, which is determined by the Lagrangian. Showing the equivalence of Lagrangians ensures that the same path of least action is determined, regardless of which Lagrangian is used.

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