elfmotat Posted February 4, 2013 Share Posted February 4, 2013 This came up in another forum, and as of yet we weren't able to come up with an answer. Someone posted the following pdf about KK theory: http://www.weylmann.com/kaluza.pdf . In it, the Lorentz force law is derived from the 5D geodesic equation. What was puzzling us was the appearance of an additional term: [math]\frac{d^2x^\lambda }{ds^2}+\Gamma^\lambda_{\mu \nu }\frac{dx^\mu}{ds} \frac{dx^\nu}{ds}=-kF^\lambda_{~\mu}\frac{dx^\mu}{ds} \frac{dx^4}{ds}-\frac{1}{2}kA_\nu F^\lambda_{~\mu} \frac{dx^\mu}{ds} \frac{dx^\nu}{ds}[/math] This is troubling for two reasons: (1) there's no experimental evidence to suggest deviations from the traditional Lorentz force law, and (2) it is directly dependent on the four-potential which suggests it is not gauge-invariant and that the potential is directly observable. Any insight would be appreciated. Link to comment Share on other sites More sharing options...
elfmotat Posted February 5, 2013 Author Share Posted February 5, 2013 (edited) I was able to resolve this with an another (anonymous) individual. (1) The factor of 1/2 in front of the second term is incorrect. (2) it turns out that the second term on the RHS of the equation isn't gauge-invariant because the first term isn't either! A gauge transformation is equivalent to a change in the 5th coordinate, so the first term on the RHS becomes: [math]k F^{\lambda}_{~\mu} \frac{dx^{\mu}}{ds} \frac{dx^4}{ds} \rightarrow kF^\lambda_{~\mu} \frac{dx^\mu}{ds} \left(\frac{dx^4}{ds}+\frac{d \zeta}{ds}\right)[/math] The second term becomes: [math]k A_{\nu}F^{\lambda}_{~\mu}\frac{dx^{\mu}}{ds} \frac{dx^{\nu}}{ds} \rightarrow k (A_{\nu}-\partial_{\nu} \zeta)F^{\lambda}_{~\mu} \frac{dx^{\mu}}{ds} \frac{dx^{\nu}}{ds} = kA_{\nu}F^{\lambda}_{~\mu} \frac{dx^{\mu}}{ds} \frac{dx^{\nu}}{ds}-kF^{\lambda}_{~\mu} \frac{dx^{\mu}}{ds} \frac{d \zeta}{ds}[/math] So the entire RHS is actually gauge-invariant. (3) The Noether conserved momentum about the "curled up" 5th dimension turns out to be: [math]k\left (\frac{dx^4}{ds}+A_\mu \frac{dx^\mu}{ds} \right )[/math] So if we identify this conserved quantity with electric charge, we obtain the standard Lorentz force law with no additional term. Edited February 5, 2013 by elfmotat Link to comment Share on other sites More sharing options...
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