- 59

- 0

Assume that in the inertial (Minkowski) lab frame the observer is seen to be orbiting with constant angular velocity [itex]\omega[/itex] at constant radius [itex]r[/itex]. Then the worldline of the observer is given by [itex]\mathcal{P}_0 = \left(\gamma\tau, r\cos(\gamma\omega\tau), r\sin(\gamma\omega\tau), 0)[/itex] where [itex]\tau[/itex] is the proper time of the observer and [itex]\gamma\equiv(1-r^2\omega^2)^{-1/2}[/itex]. I'm working in [itex]c=1[/itex] units with metric signature (+---). This lets us find the 4-vectors

[tex]u^\mu = \partial_\tau\mathcal{P}_0 = \gamma\left(1, -\omega r\sin(\gamma\omega\tau), \omega r\cos(\gamma\omega\tau), 0\right)[/tex]

[tex]a^\mu = \partial_\tau u^\mu = -\gamma^2\omega^2r\left(0, \cos(\gamma\omega\tau), \sin(\gamma\omega\tau), 0\right)[/tex]

Now, we should have that [itex]u^\mu[/itex] satisfies the FWT DE such that [itex]\partial_\tau u^\mu = \left(u^\mu a^\nu - u^\nu a^\mu\right)u_\nu[/itex]. For simplicity, consider 1-component. Then looking at the two sides separately gives us

[tex]\partial_\tau u^1 = -\gamma^2\omega^2 r\cos(\gamma\omega\tau)[/tex]

and

[tex]\left(u^1 a^\nu - u^\nu a^1\right)u_\nu = \left(a^0u^1-a^1u^0\right)u^0 - \left(a^1u^1-a^1u^1\right)u^1 - \left(a^2u^1-a^1u^2\right)u^2 - \left(a^3u^1-a^1u^3\right)u^3[/tex]

[tex] = \gamma^4\omega^2r\cos(\gamma\omega\tau) - \gamma^4\omega^4r^3\cos(\gamma\omega\tau) = \gamma^2\omega^2r\cos(\gamma\omega\tau) = -\partial_\tau u^1.[/tex]

It's that final minus sign that shouldn't be there.... I've checked all 16 components of the tetrad and they are all yield the LHS equal to the negative of the RHS of the DE (on occaision because both sides are zero). Anybody have any ideas?

Thanks,

Mike