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 Exercise [19.10] 
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Joined: 03 Jun 2010, 15:18
Posts: 136
Post Exercise [19.10]
Attached my solution.


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07 Sep 2010, 13:55

Joined: 12 Jul 2010, 07:44
Posts: 154
Post Re: Exercise [19.10]
I'd just like to add something to this, because it confused me for a little while:

\xi \in \mathrm{U}(1) \subset \mathbb{C}

i.e. \xi = e^{i\theta(\boldsymbol x)}, \theta\in\mathbb{R}

Since \xi is restricted to U(1), this means that derivatives of \xi have to have a particular form. For example, along a path parameterised by x, we have:

\frac{\mathrm{d}\xi}{\mathrm{d}x}<br />\ =\ \frac{\mathrm{d}e^{i\theta(x)}}{\mathrm{d}x}<br />\ =\ i\frac{\mathrm{d}e^{i\theta}}{\mathrm{d}(i\theta)}.\frac{\mathrm{d}\theta}{\mathrm{d}x}<br />\ =\ i\xi.\frac{\mathrm{d}\theta}{\mathrm{d}x}

In general, then, derivatives of a U(1) field \xi must have the form:

i\xi\ \times \ (real\mathrm{-}valued\ quantity)


The bundle (guage) connection applied to \xi:

\nabla_a\xi\ \equiv\ \frac{\partial\xi}{\partial x^a}\ -\ ieA_a\xi

clearly fits this pattern, when e \in \mathbb{R} and \boldsymbol A is a real-valued one-form, so it does indeed define a valid derivative on U(1).


19 Mar 2012, 12:23

Joined: 12 Jul 2010, 07:44
Posts: 154
Post Re: Exercise [19.10]
Note that although the spacetime fibre bundle \mathbb{M} \times U(1) is unique, as is the electromagnetic potential A_a (or alternatively the Maxwell tensor F_{ab}), the bundle connection \nabla_a is different for each particle, as it depends on the particle's charge, e.

(Alternatively we could instead regard each particle or particle-type as defining a completely different set of U(1) fibres over the same base Minkowski space \mathbb{M}, with a specific bundle connection then being unique to each such fibre set)...

This is in contrast to the metric (Riemannian) connection of ยง14.7, which can be regarded as an intrinsic property of spacetime, and doesn't vary depending on what you are applying it to!


19 Mar 2012, 12:45
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