im wondering if there is a general geodesic equation that describes the forces and how they act. For example I started off with the original nieve derivation of the geodesic equation: $$\frac{d}{d\tau}e_\tau=\frac{d}{d\tau}\left(\frac{dx^\mu}{d\tau}e_\mu\right)=0$$ Using the product rule $$\frac{d}{d\tau}\left(\frac{dx^\mu}{d\tau}e_\mu\right)=\frac{d^2x^\mu}{d\tau^2}e_\mu+\frac{dx^\mu}{d\tau}\frac{de_\mu}{d\tau}$$ Im not too sure on this next step, however if we imagine an internal space controlled by a paramater $\theta$ can we do the chain rule on this to give $$\frac{d^2x^\mu}{d\tau^2}e_\mu+\frac{dx^\mu}{d\tau}\frac{de_\mu}{d\tau}=\frac{d^2x^\mu}{d\tau^2}e_\mu+\frac{dx^\mu}{d\tau}\frac{d\theta}{d\tau}\frac{de_\mu}{d\theta}$$ Rewriting gives the following $$\frac{d^2x^\mu}{d\tau^2}e_\mu+\frac{dx^\mu}{d\tau}\frac{d\theta}{d\tau}\frac{de_\mu}{d\theta}=\frac{d^2x^\mu}{d\tau^2}e_\mu+\frac{d\theta}{d\tau}\frac{de_\mu}{d\theta}\frac{dx^\mu}{d\tau}$$ Where the $\frac{d e_\mu} {d\theta}$ is how the basis vectors change when moving along the internal space. So this can be interpreted as a connection. $$\frac{d e_\mu} {d\theta}=F_\mu^\alpha e_\alpha$$ Plugging this back into the equation gives $$\frac{d^2x^\mu}{d\tau^2}e_\mu+\frac{d\theta}{d\tau}F_\mu^\alpha \frac{dx^\mu}{d\tau}e_\alpha=0$$ Rewriting the summed indicies $$\left(\frac{d^2x^\mu}{d\tau^2}+\frac{d\theta}{d\tau}F_\alpha^\mu\frac{dx^\alpha}{d\tau}\right)e_\mu=0$$ $$\frac{d^2x^\mu}{d\tau^2}+\frac{d\theta}{d\tau}F_\alpha^\mu\frac{dx^\alpha}{d\tau}=0$$ If $\frac{d\theta}{d\tau}= -\frac{q}{m}$ Then we get the following. $$\frac{d^2x^\mu}{d\tau^2}-\frac{q}{m}F_\alpha^\mu\frac{dx^\alpha}{d\tau}=0$$ Which is the form of an charged particle moving in an electromagnetic field Is this derivation valid and am I allowed to do that suspect step?

  • $\begingroup$ $\tau$ is proper time ? What is $e_\tau$ and $e_\mu$ ? Not sure if this helps. $\endgroup$
    – Kurt G.
    Mar 31 at 18:42
  • $\begingroup$ Sorry I should of clarified $e_\tau$ is the basis vector moving along the proper time and $e_\mu$ is the positional basis vector $\endgroup$ Mar 31 at 18:46


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