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electromagnetic fields

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Thread replies: 6
Thread images: 2

What is it exactly?
>>
Fucking magic 'n shiet
>>
>>9172561
They're a property of the universe, we just so happen to model and call them electromagnetic fields.
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>>9172589

Are magnetic fields made of photons?
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>>9172561
They're [math]U(1)[/math] gauge fields on [math]\mathbb{R}^3[/math]. Let [math]\mathcal{G}
= \operatorname{Map}(X,G)[/math] be the space of smooth maps from a compact Lie group [math]G[/math] to a manifold [math]X[/math], then the set of gauge fields is given by the pullback of the representation [math]\rho : G \rightarrow V[/math] of [math]\mathcal{G}[/math] onto the vector bundle [math]V \rightarrow X[/math] over [math]X[/math]. This induces a representation of the connection [math]A \in \rho^*\Omega^1(X) \otimes \mathfrak{g}[/math] on the Lie algebra [math]\mathfrak{g}[/math] of [math]G[/math] from connections on the principle [math]G[/math]-bundle [math]G\rightarrow P \rightarrow X[/math], and these connections are the gauge fields, while the representations [math]\rho^*\mathcal{G}[/math] are the gauge transformations.
In the case of [math]G=U(1), X =\mathbb{R}^3[/math] and the unitary representation [math]U(1) \rightarrow S^1[/math], the set of gauge fields is given by [math]A_\mu dx^\mu, \mu = 0.\dots,3[/math] where [math]A_\mu:\mathbb{R}^3\rightarrow \mathbb{R}[/math] is a smooth function. The gauge transformations are [math]e^{ig(x)}[/math] where [math]g:\mathbb{R}^3\rightarrow \mathbb{R}[/math] is also smooth. The curvature (or the representation thereof) of the principle [math]G[/math]-bundle is then given by [math]F = dA[/math], which is the electromagnetic tensor [math]F_{\mu\nu} =
\partial_\mu A_\nu - \partial_\nu A_\mu =
\frac{1}{2}(\delta_{0\mu} - \delta_{\mu 0}){E_\mu} - \epsilon^{\mu\nu\lambda\tau}B_{\lambda\tau}[/math] where [math]\epsilon^{\mu\nu\lambda\tau}[/math] is the totally antisymmetric tensor of rank 4.
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>>9172720
There are typos but I'm too lazy to fix them.
Thread posts: 6
Thread images: 2


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