Aethelwulf Posted February 1, 2015 Report Share Posted February 1, 2015 Knowing [math]\Delta = \gamma_{\mu} \partial^{\mu}[/math] One can introduce this next equation as the square root of the spacetime interval [math]i \gamma^{\mu} \Delta \gamma_{\mu} R = 1[/math] Repeating indices cancel out [math]i \gamma^{\mu} \gamma_{\mu} \partial^{\mu} \gamma_{\mu} R = 1[/math] [3] so what we really have is [math]i \gamma^{\mu}\partial^{\mu} R = 1[/math] Which is the square root of the space time interval. In an antisymmetric product, the spin matrices is given as [math]2 \sigma_{\mu \nu} = \gamma_{\mu} \gamma_{\nu} - \gamma_{\nu} \gamma_{\mu}[/math] This is redundant view in a geometric algebraic-sense, and... Hestenes of course mentions this in his own work. The unit two form is [math]dx \wedge dy[/math] This term is contracted under the sigma, so it already implies the term [math]\sigma_{\mu \nu}[/math]. The reason lies within making sure for instance, [math]\sigma_{\mu} \sigma_{\mu}[/math] is [math]0[/math] and not [math]1[/math]. This means it has a property which allows it to square to a negative [math]-1[/math], indeed, applying this rule we find a rotation in the imaginary axis [math]i^2 \gamma^{\mu} \partial^{\mu} R = -1[/math] It's almost like a special case of a wick rotation. Quote Link to comment Share on other sites More sharing options...
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