\(\theta\) is \(60^o\). After the rotation, the \(t_{c}\) component of \(t_{now}\), \(t^{'}_c\), is orthogonal to \(t_c\) and parallel to \(t_g\). \(t^{'}_c\) has no effect on \(t_c\) and so becomes invisible, optically and to radar. This suggest that our sight is a charge phenomenon. This rotation is equivalent to rotating \(t_c\) by \(90^o\) in the \(t_g\)-\(t_c\) plane, but we simply don't have a grip on this angle.
After this rotation, \(t_T\) is invariant, there is no temperature change, but \(t^{'}_c\) is now \(t_g\) and \(t^{'}_g\) is now \(-t_c\). Gravity and charge have swapped roles and the result is all sort of weird effects. Maybe it is better to rotate,
\(t_c\) \(90^o\) towards \(t_T\), with the \(t_g\) axis held invariant. The problem is \(t_{now}\) is always rotated with respect to \(OA\), the projection of \(t_{now}\) on the \(t_T\)-\(t_c\) plane. Maybe, \(t_g\) can be reduced by increasing gravity \(g\), \(t_{now}\) collapses onto \(OA\) and then be rotated by \(90^o\).
Gravity is then reduced to normal after the phase shift. \(t^{'}_c\) is now \(t_T\) and \(t^{'}_T\) is now \(-t_{c}\). Temperature and charge swapped roles. Since the electrical/electronic instruments are on the \(t_T\)-\(t_c\) plane in the first place, maybe that is acceptable. With gravity invariant, nobody gets struck in the walls.
Theoretically.
