Dual of Maxwell tensor as gauge curvature

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please see http://camoo.freeshell.org/dual_gauge.pdf" [Broken]
Thanks
Laura
 
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jtbell
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Hi, did you know that we have inline LaTeX capabilities here? Enter your code between [ tex ] and [ \tex ] tags for standalone equations, or [ itex ] and [ /itex ] for inline equations. In either case, remove the extra spaces between the brackets. I had to add them here so the tags would show up as plain text instead of being treated as actual tags.

You can also try clicking on the sigma icon in the message-composition window, although I suspect you'll find it easier to write the code directly, because you apparently did it for that PDF file.

This will make it more convenient for readers than downloading a PDF file. Also, they can quote your equations if necessary which makes it easier to reply.
 
Like so:

The following comment in a book was puzzling to me : ”Recall the Hodge dual of the Maxwell
tensor F. We could imagine a ’dual’ U(1) gauge connection that has F as its bundle curvature
rather than F.”

But [itex]F_ab[/itex] satisfies d[itex]F_ab[/itex] = 0, which means that [itex]F_ab[/itex] can be expressed as a derivative of a 1-form:
[tex]F_ab = \frac{\partial{_A}}{\partial{_b}}-\frac{\partial{A_a}}{\partial{x^b}}.[/tex]
and the vector potential [itex]A_a[/itex] gives you a gauge connection

[tex]\nabla_a= \frac{\partial}{\partial{x^a}} - ieA_a[/tex]

And so on.

So [itex]F_ab[/itex] is the
curvature of the connection,

[tex]\nabla_a \nabla{_b}-\nabla_a \nabla_b -ie =-ie \left(\frac{\partial{A_b}}{\partial{x^a}}-\frac{\partial A_b}{\partial x^b}\right)[/tex]
And so on.
 
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Hi, did you know that we have inline LaTeX capabilities here?
I know, but I use the built-in Latex for simple things, not for long complicated posts.
Laura
 
Part 2

But

[tex]d^*F_{ab}=-\frac{*j}{\epsilon 0}[/tex]

But how can *F be a gauge curvature when d*F[itex]\neq[/itex] 0? If it were in empty
space you could come up with a 1-form [itex]Z_a[/itex] so that *[itex]F_{ab}=\frac{\partial {Z_b}}{\partial{x^a}}-\frac {\partial{Z_a}}{\partial{x^b}}[/itex] , that would give
you an alternate bundle connection. Who knows what it would be a connection for!
Is the empty-space version of the Maxwell equations what you’d be using at the quantum
level? Maybe all the interactions are being described explicitly so you wouldn’t be using a
charge-current vector. What do you think?
Well there's the whole question anyway, parts 1 & 2. I don't know how hard pdf formatting of latex is?
 
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So the book's statement is a statement about electromagnetic duality. That the source free or empty space maxwell's equations remain unchanged when you exchange [tex]E[/tex] and [tex]B[/tex], or equivalently, [tex]F[/tex] and [tex]*F[/tex]. Note that [tex]*F[/tex] is only a 2-form in 4 dimensions, so the statement is only true in 4. In this case, [tex]*F[/tex] is the curvature of a connection corresponding to the electromagnetic dual (electric monopoles become magnetic monopoles, etc.)

There's some interest in this for Yang-Mills theories and in other contexts. Anyway I think the point is that it has to be source free. Or, approximately source free (say, in some non-simply connected region). See for example the Dirac monopole.
 
jtbell
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I use the built-in Latex for simple things, not for long complicated posts.
Can't you simply paste the LaTeX code that you have to prepare anyway, in between the tags? Or does that not work for some reason?
 
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There's some interest in this for Yang-Mills theories and in other contexts. Anyway I think the point is that it has to be source free. Or, approximately source free (say, in some non-simply connected region). See for example the Dirac monopole.
Yes but the book didn't say anything about the field being source-free. Is my guess right, that in the application where the dual connection would be used, I guess on wavefunctions, that you wouldn't be including the local sources in the Maxwell equations, because you are looking at things at too small a scale to consider charge density, current etc.? He might be leaving this unsaid - but it seems like an obligatory aspect of considering the dual of the Maxwell tensor as a curvature. I spent a lot of time thinking about whether there'd be some way to get around that restriction, some other way of defining the covariant derivative that would give the dual as the curvature. But nothing really came up.
Laura
 

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