Constant of motion, Maxwell's equations

In summary, we are given the Faraday Tensor F_{ab} and a killing vector field \xi ^{a}. If the lie derivative \mathcal{L} _{\xi }F_{ab} = 0, then we can show that F_{ab}\xi ^{b} = \triangledown _{a}\varphi for some smooth scalar field \varphi. We can also show that I = mu^{a}\xi _{a} + q\varphi is a constant of motion for a charged particle with mass m, charge q, and 4-velocity u, subject to the Lorentz force. This is done by showing that \triangledown _{u}I =
  • #1
WannabeNewton
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Homework Statement


Let [itex]F_{ab}[/itex] be the Faraday Tensor and [itex]\xi ^{a}[/itex] a killing vector field. Suppose that the lie derivative [itex]\mathcal{L} _{\xi }F_{ab} = \xi ^{c}\triangledown _{c}F_{ab} + F_{cb}\triangledown _{a}\xi ^{c} + F_{ac}\triangledown _{b}\xi ^{c} = 0[/itex]. Show that [itex]F_{ab}\xi ^{b} = \triangledown _{a}\varphi [/itex] for some smooth scalar field [itex]\varphi [/itex] and that [itex]I = mu^{a}\xi _{a} + q\varphi [/itex] is a constant of motion for a charged particle of mass m, charge q, and 4 - velocity u.

Homework Equations


Maxwell's equations: [itex]\triangledown ^{a}F_{ab} = 0[/itex], [itex]\triangledown _{[a}F_{bc] } = 0[/itex]. Equations of motion for charged particle / Lorentz force: [itex]mu^{a}\triangledown _{a}u^{b} = qF^{b}_{c}u^{c}[/itex].

The Attempt at a Solution


In order to show that [itex]F_{ab}\xi ^{b} = \triangledown _{a}\varphi [/itex] for some smooth scalar field [itex]\varphi [/itex], we want to show that the one - form field [itex]F_{ab}\xi ^{b}[/itex] is closed i.e. [itex]\triangledown _{a}(F_{bc}\xi ^{c}) - \triangledown _{b}(F_{ac}\xi ^{c}) = 0[/itex]. We find that [itex]\triangledown _{a}(F_{bc}\xi ^{c}) - \triangledown _{b}(F_{ac}\xi ^{c}) = (\xi ^{c}\triangledown _{a}F_{bc} - \xi ^{c}\triangledown _{b}F_{ac} - F_{cb}\triangledown _{a}\xi ^{c} - F_{ac}\triangledown _{b}\xi ^{c})[/itex]. Using the second of Maxwell's equations, [itex]\triangledown _{[a}F_{bc] } = 0[/itex], we find that [itex]\triangledown _{a}(F_{bc}\xi ^{c}) - \triangledown _{b}(F_{ac}\xi ^{c}) = (-\xi ^{c}\triangledown _{c}F_{ab} - F_{cb}\triangledown _{a}\xi ^{c} - F_{ac}\triangledown _{b}\xi ^{c}) = 0[/itex] because this is the expression of the lie derivative which we are told vanishes. Thus we can conclude that [itex]F_{ab}\xi ^{b} = \triangledown _{a}\varphi [/itex] for some smooth scalar field [itex]\varphi [/itex]. In order to show that [itex]I = mu^{a}\xi _{a} + q\varphi [/itex] is a constant of motion along the worldline of this charged particle subject to the Lorentz force, we must show [itex]\triangledown _{u}I = u^{b}\triangledown _{b}I = 0[/itex]. We find that [itex]u^{b}\triangledown _{b}I = u^{b}\triangledown _{b}(mu^{a}\xi _{a} + q\varphi ) = mu^{b}u^{a}\triangledown _{b}\xi _{a} + mu^{b}\xi _{a}\triangledown _{b}u^{a} + qu^{b}\triangledown _{b}\varphi [/itex]. Since [itex]u^{b}u^{a}[/itex] is symmetric in the indices and [itex]\triangledown _{b}\xi _{a}[/itex] is anti - symmetric in the indices, on account of [itex]\xi [/itex] being a killing field, their contraction must vanish so we get [itex]u^{b}\triangledown _{b}I = mu^{b}\xi _{a}\triangledown _{b}u^{a} + qu^{b}\triangledown _{b}\varphi = mu^{b}\xi _{a}\triangledown _{b}u^{a} + qu^{b}F_{bc}\xi ^{c} = mu^{b}\xi _{c}\triangledown _{b}u^{c} - qu^{b}F_{cb}\xi ^{c} = \xi _{c}(mu^{b}\triangledown _{b}u^{c} - qu^{b}F_{b}^{c}) = 0[/itex]. Can anyone check if this is the right way to do it because I'm not too sure; it seems fine however I never had to use anywhere the first of Maxwell's equations [itex]\triangledown ^{a}F_{ab} = 0[/itex] even though we are given it. Maybe it is superfluous information for the purposes of this problem? Thanks again!
 
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  • #2
It looks ok! Maybe the unused equation was given for sake of completness, letting you decide if it will be used or not.
 
  • #3
Ok thank you mate!
 

1. What is the constant of motion in physics?

The constant of motion in physics refers to a quantity that remains unchanged during the course of a physical process or event. It is a fundamental concept in classical mechanics and is essentially a conserved quantity, meaning it is the same before and after a physical process, such as a collision or motion of a system.

2. What are Maxwell's equations and why are they important?

Maxwell's equations are a set of four equations that describe the fundamental laws of electromagnetism. They were formulated by James Clerk Maxwell in the 19th century and have since been widely accepted and used in the study of electromagnetism. These equations are important because they provide a mathematical framework for understanding the behavior of electric and magnetic fields, and have led to many technological advancements such as radio, television, and wireless communications.

3. How do Maxwell's equations relate to the constant of motion?

Maxwell's equations are derived from the principle of conservation of energy, which is a type of constant of motion. This principle states that energy can neither be created nor destroyed, but can only be transformed from one form to another. The equations describe the relationship between electric and magnetic fields, which are forms of energy, and their conservation throughout a physical process.

4. Can the constant of motion be violated?

No, the constant of motion cannot be violated. It is a fundamental principle in physics and has been rigorously tested and confirmed through numerous experiments. Any apparent violation of a constant of motion can be explained by factors such as external forces or errors in measurement.

5. How are Maxwell's equations used in practical applications?

Maxwell's equations are used in a wide range of practical applications, including the design and operation of electronic devices, communication systems, and medical equipment. They are also essential in the development of new technologies such as solar panels, MRI machines, and particle accelerators. Understanding and applying Maxwell's equations has greatly advanced our understanding of electromagnetism and revolutionized many industries.

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