What about this?
\(\cfrac{\partial\,t_g}{\partial\,t}=c\)
\(\cfrac{\partial\,x}{\partial\,t}=c\)
So,
\(\cfrac{\partial\,x}{\partial\,t_g}=\cfrac{\partial\,t}{\partial\,t_g}.\cfrac{\partial\,x}{\partial\,t}=1\)
Not quite! We have to consider the fact that \(x\) is orthogonal to \(t_g\) and in general,
\(\cfrac{\partial\,t_g}{\partial\,t}\neq c\)
In fact, the expression,
\(\cfrac{\partial\,x}{\partial\,t_g}\)
means change in \(x\) for every change in time \(t_g\) (\(t_g\) being the independent variable), just as
\(\cfrac{\partial\,x}{\partial\,t}\)
means change in \(x\) for every change in time \(t\) except we have to consider the fact that,
\(i\Delta t_g= \Delta x\)
\(c\) per change in \(t_g\) along \(x\) must be multiplied by \(i\) to reflect the fact that \(t_g\) is orthogonal to \(x\).
\(ic.\Delta t_g= \Delta x\)
and so,
\(ic= \lim\limits_{\Delta t_g\to0}\cfrac{\Delta x}{\Delta t_g}=\cfrac{\partial\,x}{\partial\,t_g}\)
We are now at time speed \(c\) does not mean we cannot achieve light speed in space. If we are constrained by energy conservation, as the case of a body at constant velocity, free motion, enters a conservative gravitational field, speeds up in space slows down time.
\(v_t^2+v_s^2=c^2\) --- (*)
where \(c\) is a constant, \(v_t\), time speed and \(v_s\), speed in space.
And at light speed \(c\), in space, time speed is zero. When we are not constrained by energy conservation and energy is pumped into the body, space speed increases without regard to time speed! Einstein was working in a conservative gravitational field, ie accelerating under gravity without external energy input, as such total K.E is conserved including K.E along the time dimension (as in expression *).
