Help with Reducing an Equation into Jacobi Identity Form

Click For Summary

Homework Help Overview

The discussion revolves around reducing the equation \(\partial_\mu {*} F^{\mu \nu} = 0\) into the Jacobi identity form \(\partial_\lambda F_{\mu \nu} + \partial_\mu F_{\lambda \nu} + \partial_\nu F_{\lambda \mu} = 0\). The subject area pertains to tensor calculus and differential forms in the context of physics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the meaning of the '*' operator and its implications for the equation. There is an exploration of the relationship between the dual operator and the form of \(F_{\mu \nu}\). Some participants question the interpretation of indices and how they relate to the Jacobi identity.

Discussion Status

The discussion is ongoing, with participants sharing insights and clarifications about the notation and mathematical concepts involved. Some guidance has been offered regarding the interpretation of the dual operator and its application to the equation.

Contextual Notes

There is mention of previous problems and forms of \(F_{\mu \nu}\) that may influence the current understanding. Participants express uncertainty about specific notations and their meanings, indicating a need for further clarification.

kreil
Science Advisor
Insights Author
Messages
665
Reaction score
68

Homework Statement


Reduce the equation [tex]\partial_\mu {*} F^{\mu \nu} = 0[/tex] into the following form of the Jacobi Identity:

[tex]\partial_\lambda F_{\mu \nu} + \partial_\mu F_{\lambda \nu} + \partial_\nu F_{\lambda \mu} = 0[/tex]

The Attempt at a Solution



I can't figure out what the '*' is supposed to be. My first thought was that it was a typo and is meant to signify a dot product, but the partial derivative is not a vector, so I don't see how this could be the case.

At any rate, this problem seems straightforward but I could use some help getting started.

Thanks for your thoughts.
 
Physics news on Phys.org
hi kreil! :smile:

(have a curly d: ∂ and a mu and an nu: µν :wink:)

the clue is in the alteration of the position of the indices …

this is ∂µ(*Fµν) :wink:

(see http://en.wikipedia.org/wiki/Hodge_dual" )
 
Last edited by a moderator:
Thanks Tim, I figured it was an operator of some sort but my notes/book were not immediately helpful in figuring out what it was.

In a previous problem we used the following form for F,

[tex]F_{\mu \nu} = \partial_\mu A_\nu - \partial_\nu A_\mu[/tex]

I'm having trouble seeing how the dual operator works on this form of F. Assuming a more general form, is the following true?

*[tex]F^{\mu \nu} = \left ( F^\mu e_\mu + F^\nu e_\nu + F^\lambda e_\lambda \right ) e_\mu e_\nu e_\lambda = F_{\nu \lambda} + F_{\mu \lambda} + F_{\mu \nu}[/tex]
 
Last edited:
hi kreil! :smile:

i've never really grasped the index way of doing it :redface:

i always think of it this way …

i see Fµν as a wedge product Exx∧t + Eyy∧t + Ezz∧t + Bxy∧z + Byz∧x + Bzx∧y

then for "curl" i wedge-multiply by ∂∧ = ∂xx∧ + ∂yy∧ + ∂zz∧ + ∂tt∧,

and use the equivalence x∧y∧z → *t, y∧z∧t → *x, z∧x∧t → *y, x∧y∧t → *z

(the jacobi identity is written in triple-wedges x∧y∧z etc)

i seem to remember there's a fairly good description of this in (surprisingly) Misner Thorne and Wheeler "Gravitation"
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
5K
  • · Replies 1 ·
Replies
1
Views
5K
  • · Replies 30 ·
2
Replies
30
Views
8K
Replies
10
Views
2K
  • · Replies 2 ·
Replies
2
Views
5K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K