\(\cfrac{\partial^2\psi}{\partial\,t^2}=c^2\cfrac{\partial^2\psi}{\partial\,x^2}\)
And to go where no man has gone before,
\(\cfrac{\partial^2\psi}{\partial\,t^2}=\cfrac{\partial\,x_1}{\partial\,t}\cfrac{\partial\,x_2}{\partial\,t}\cfrac{\partial^2\psi}{\partial\,x_1\partial\,x_2}\)
I have made a mistake in previous posts!
In the case where the wave is travelling in the time dimension, \(t_g\) at speed \(c\), in the wave \(\psi\) varies with space \(x\),
\(\cfrac{\partial}{\partial\,t_g}\left\{\cfrac{\partial\psi}{\partial\,x}\right\}\)
Obviously,
\(\cfrac{\partial\,t_g}{\partial\,t}.\cfrac{\partial}{\partial\,t_g}\left\{\cfrac{\partial\psi}{\partial\,x}\right\}=\cfrac{\partial}{\partial\,t}\left\{\cfrac{\partial\psi}{\partial\,x}\right\}\)
and
\(\cfrac{\partial\,x}{\partial\,t}.\cfrac{\partial\,t_g}{\partial\,t}.\cfrac{\partial}{\partial\,t_g}\left\{\cfrac{\partial\psi}{\partial\,x}\right\}=\cfrac{\partial}{\partial\,t}\left\{\cfrac{\partial\psi}{\partial\,t}\right\}=\cfrac{\partial^2\psi}{\partial\,t^2}\)
since,
\(\cfrac{\partial\,x}{\partial\,t}=c\)
and the wave is at light speed down the time dimension, \(t_g\),
\(\cfrac{\partial\,t_g}{\partial\,t}=c\)
as such
\(\cfrac{\partial^2\psi}{\partial\,t^2}=\cfrac{\partial\,x}{\partial\,t}.\cfrac{\partial\,t_g}{\partial\,t}.\cfrac{\partial}{\partial\,t_g}\left\{\cfrac{\partial\psi}{\partial\,x}\right\}=c.c\cfrac{\partial}{\partial\,t_g}\left\{\cfrac{\partial\psi}{\partial\,x}\right\}=c^2\cfrac{\partial^2\psi}{\partial\,t_g\partial\,x}\)
which is the wave equation for a wave travelling down the time dimension \(t_g\). And when we substitute for specific time \(t\), for example \(t_g\), we obtain,
\(\cfrac{\partial^2\psi}{\partial\,t^2_g}=ic\cfrac{\partial^2\psi}{\partial\,x\partial\,t_g}\) --- (*)
using
\(t_c=\cfrac{1}{\sqrt{2}}t.e^{-i\pi/4}\)
\(t_g=\cfrac{1}{\sqrt{2}}t.e^{+i\pi/4}\),
\(t_c=t_ge^{-i\pi/2}\)
\(t_g=it_c\)
\(\cfrac{\partial\,t_g}{\partial\,t_c}=i\)
and
\(\cfrac{\partial\,x}{\partial\,t_c}=\cfrac{\partial\,x}{\partial\,t_g}=ic\)
where \(i\) is used to express orthogonality explicitly. The details are in the post "Not A Wave, But Work Done!" dated 18 Nov 14.
The problem is at \(\cfrac{\partial\,x}{\partial\,t}=c\). The particle is also at light speed in space, as would be the case of electrons in orbit around a nucleus, but not a stationary charge.
If \(\cfrac{\partial\,x}{\partial\,t}=0\) there is no wave. So what about the electrostatic field around a stationary charge then? Luckily, we are always in motion; not relative motion between two interacting particles but absolute motion of the particles involved.
Notice that in expression (*), only one of the two dimensions between which energy is oscillating is explicitly in the expression. This is because the oscillation is fully defined by what happens on just one of these two dimensions. So, in the case where another space dimension \(x\) has been replaced by a time dimension, expression (*) still applies. As long as one space dimension \(x\) remains as part of the oscillation pair, expression (*) is valid.
So the new particles in the post "Wrong Wrong Wrong", has the same wave equation as the above. It is just (*). And all derivations based on (*) is still valid.
This lead us to the wave equation for photons. Although one of the dimensions with oscillatory energy is replaced with a time dimension, the remaining space dimension \(x\) of the pair together with the fact that this wave is travelling in space at light speed \(c\), means that the wave equation is just,
\(\cfrac{\partial^2\psi}{\partial\,t^2}=\cfrac{\partial\,x_1}{\partial\,t}\cfrac{\partial\,x_2}{\partial\,t}\cfrac{\partial^2\psi}{\partial\,x_1\partial\,x_2}=c^2\cfrac{\partial^2\psi}{\partial\,x^2}\)
