Rotating Al Again

From the post "Thank You, Al",  apparently $t_c$ is not rotated by rotating $t_{now}$ directly.  The projection of $t_{now}$ on the $t_T$-$t_c$ plane is rotated through $\theta$ about the origin as shown in the diagram below. An electrical/electronic instrument produces heat and manipulates charges, so it is on the $t_T$-$t_c$ plane.  The phase shift rotated is as measured and experienced by the instrument.

$\theta$ 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.