Brownian Motion, Short Time Travel, O particles

Then we come to the mistake I have made,

What are these?

We experience $t_c$, $t_g$ and $t_T$ as a whole, just as in any direction space is just $x$; then $t_c$, $t_g$ and $t_T$ are undifferentiated and so are these particles.  However, if our sight is a charge phenomenon (along $t_c$ only) then $o_{t_T}$($x$,$t_c$,$t_g$) and $o_{t_g}$($x$,$t_c$,$t_T$) can impart energy along $t_c$ and maybe slow/speedup $p_{t_c}$ and induce invisibility.  The notation $o_{t_x}$ is used here.

These O particles can still affect the space dimensions via the space-time energy oscillations in photons.   And can couple energy directly onto the time dimension.  Indirectly, they require a photon or a basic particle to be detected in space.  In the case of the left most O particle in the diagram, $p_{t_T}$($x_1$,$x_2$,$t_g$), $p_{t_T}$($x_1$,$x_2$,$t_c$), $p_{t_c}$($x_1$,$x_2$,$t_g$) or $p_{t_g}$($x_1$,$x_2$,$t_c$); photons with energy oscillating in either $t_g$ or $t_c$ time dimension and a space dimension, or a stationary basic particle where the first entry inside the parenthesis is replaced with a time dimension and has energy oscillation between $t_c$ or $t_g$ and a space dimension.  That is, $q$($t_T$,$x$,$t_g$), $T$($t_c$,$x$,$t_g$), $T$($t_g$,$x$,$t_c$) and $g$($t_T$,$x$,$t_c$).

OK?!