Partial and Covariant derivatives in invarint actions

In summary, the conversation discusses the invariance of the integral I under transformations and the question of whether the same is true for a covariant derivative. It is mentioned that in nonabelian field theory, the g[A_{\mu},\varphi] term may or may not vanish within the integral. It is also mentioned that in cases of supersymmetry, the answer to the question often includes a covariant derivative. However, without enough information about the local properties, it is not possible to determine if the action is invariant.
  • #1
AlphaNumeric
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It's physics based but actually a maths question so I'm asking it here rather than the physics forums.

[tex]I = \int \mathcal{L}\; d^{4}x[/tex]

I is invariant under some transformation [tex]\delta_{\epsilon}[/tex] if [tex]\delta_{\epsilon}\mathcal{L} = \partial_{\mu}X^{\mu}[/tex] for some function/tensor/field thingy [tex]X^{\mu}[/tex]. This I've no problem with.

However, is the same true for a covariant derivative? If [tex]\delta_{\epsilon}\mathcal{L} = D_{\mu}X^{\mu}[/tex] where [tex]D_{\mu}\varphi = \partial_{\mu}\varphi + g[A_{\mu},\varphi][/tex], as you get in nonabelian field theory. Is the action still invariant? Obviously the [tex]\partial_{\mu}[/tex] part of [tex]D_{\mu}[/tex] represents no problem but I don't know if the [tex]g[A_{\mu},\varphi][/tex] term vanishes or not within the integral.

I've been doing some supersymmetry and a number of times I've got the answer the question has asked to find plus a covariant derivative of something. :cry: If I just got a mess of terms I'd know I'm way off, but the fact everything collects nicely into a covariant derivative makes me feel I'm at least on the right track.

Thanks for any help :)
 
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  • #2
If ##[A_\mu,X^\mu]=0## we are done, otherwise we just do not have enough information to draw a conclusion. The reason is, that ##\partial_\mu## is a local property, and ##D_\mu## connects this local property with another one ##[A_\mu,X^\mu]## at another location. So ##\mathcal{L}## could still be invariant or equally not.
 

Related to Partial and Covariant derivatives in invarint actions

1. What is the difference between partial and covariant derivatives?

Partial derivatives are taken with respect to one variable while holding all other variables constant. In contrast, covariant derivatives take into account the variation of all variables and their relationships to each other. They are used in the study of manifolds and coordinate systems.

2. How are partial and covariant derivatives used in invariant actions?

Partial and covariant derivatives are used in invariant actions to express the invariance of physical laws under coordinate transformations. By taking derivatives with respect to different variables, we can determine how a physical quantity changes under a change of coordinates.

3. Can partial and covariant derivatives be used interchangeably?

No, partial and covariant derivatives are not interchangeable. They have different mathematical properties and are used for different purposes. Invariant actions require the use of both types of derivatives to fully describe the invariance of a physical system.

4. How do partial and covariant derivatives relate to the principle of least action?

The principle of least action states that the path taken by a system between two points is the one that minimizes the action (or energy) of the system. Partial and covariant derivatives are used to determine the equations of motion for a system, which can then be used to find the path that minimizes the action.

5. Are there any practical applications of partial and covariant derivatives in physics?

Yes, partial and covariant derivatives have many practical applications in physics. They are used in the study of relativity, quantum mechanics, electromagnetism, and many other fields. They are essential for understanding the behavior of physical systems and form the mathematical basis for many physical laws and theories.

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