Commutator between covariant derivative, field strength

In summary, the conversation is about proving the equation ∂μFμ\nu + ig[Aμ, Fμ\nu] = [Dμ,Fμ\nu] using the definition Dμ = ∂μ + igAμ. The individual is having trouble with the term Fμ\nu∂μ and is trying to show that it is nil. They mention using a smooth function, which is nil at infinity, to demonstrate this. However, they ask for clarification on whether their approach is correct and if not, for the correct answer and explanation. The response states that their approach is correct and should be sufficient for a physics question.
  • #1
oliveriandrea
9
0
Hello,
i try to prove that
∂μFμ[itex]\nu[/itex] + ig[Aμ, Fμ[itex]\nu[/itex]] = [Dμ,Fμ[itex]\nu[/itex]]
with the Dμ = ∂μ + igAμ

but i have a problem with the term Fμ[itex]\nu[/itex]∂μ ...
i try to demonstrate that is nil, but i don't know if it's right...

Fμ[itex]\nu[/itex]∂μ [itex]\Psi[/itex] = [itex]\int[/itex] (∂[itex]\nu[/itex]Fμ[itex]\nu[/itex]) (∂μ[itex]\Psi[/itex]) + [itex]\int[/itex] Fμ[itex]\nu[/itex]∂μ∂[itex]\nu[/itex] [itex]\Psi[/itex] = (∂[itex]\nu[/itex]Fμ[itex]\nu[/itex]) [[itex]\Psi[/itex] ]∞∞ - [itex]\int[/itex][itex]\Psi[/itex]∂μ∂[itex]\nu[/itex]Fμ[itex]\nu[/itex] = 0

with [itex]\Psi[/itex] a smooth function, nil at infinity

if it's wrong please do you post the right answers? and why it is wrong...
thank you
 
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  • #2
This is right. Of course, if you need it mathematically accurate, you have to think about, how fast you test function has to go to 0 at infinity, but if it's a physics question, what you did should be sufficient.
 
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Likes 1 person
  • #3
Thank you (yes it's physics question)
 

FAQ: Commutator between covariant derivative, field strength

1. What is a commutator between covariant derivative and field strength?

The commutator between covariant derivative and field strength is a mathematical expression that represents the failure of two operations, the covariant derivative and the field strength, to commute or produce the same result regardless of the order in which they are performed. It is an important quantity in the study of gauge theories and is used to calculate the curvature of a gauge field.

2. How is the commutator between covariant derivative and field strength calculated?

The commutator between covariant derivative and field strength is calculated using the Lie bracket, which is a mathematical operation that measures the failure of two vector fields to commute. In the context of gauge theories, this bracket is used to calculate the curvature of the gauge field, which is related to the commutator by a proportionality constant.

3. What is the physical significance of the commutator between covariant derivative and field strength?

The commutator between covariant derivative and field strength is significant in the study of gauge theories because it represents the non-commutativity of the gauge fields. This non-commutativity is a fundamental aspect of gauge theories and allows for the existence of gauge bosons, which are the particles responsible for mediating the strong, weak, and electromagnetic forces.

4. How does the commutator between covariant derivative and field strength relate to gauge transformations?

The commutator between covariant derivative and field strength is closely related to gauge transformations. In fact, the commutator is a gauge-invariant quantity, meaning that it remains unchanged under gauge transformations. This property is important in the construction of gauge theories and plays a crucial role in the understanding of their physical implications.

5. Can the commutator between covariant derivative and field strength be extended to higher dimensions?

Yes, the commutator between covariant derivative and field strength can be extended to higher dimensions. In fact, it is a fundamental concept in differential geometry and is used to study the curvature of higher-dimensional spaces. In the context of gauge theories, this extension is crucial for understanding the behavior of gauge fields in higher-dimensional spacetimes.

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