\(r.\Delta \theta = k.\Delta t_c\)
\( \omega _{ c }=\cfrac { \partial \, \theta }{ \partial \, t_{ c } } =\cfrac{k}{r} \)
\( \omega =\omega _{ c }\cfrac { \partial \, t_{ c } }{ \partial \, t } =\cfrac { \partial \, \theta }{ \partial \, t_{ c } } \cfrac { \partial \, t_{ c } }{ \partial \, t } =\cfrac{k}{r} \cfrac { \partial \, t_{ c } }{ \partial \, t } \)
\( \left( \omega -\cfrac { k }{ r } \right) \cfrac { \partial \, t_{ c } }{ \partial \, t } =0\)
As, \(r\rightarrow\infty\), \(k\ne0\)
\( \omega\cfrac { \partial \, t_{ c } }{ \partial \, t } =0\)
since,
\( \omega\ne0\)
\(\cfrac { \partial \, t_{ c } }{ \partial \, t } =0\)
This implies that as time stretches out, time speed is zero.
when \(r\ne\infty\),
\(2\pi r=kt_{\lambda}\)
\(2\pi rf=kt_{\lambda}f=k\cfrac { \partial \, t_{ c } }{ \partial \, t }\)
\(\cfrac { \partial \, t_{ c } }{ \partial \, t }=\cfrac { 2\pi rf }{ k }=\cfrac { r\omega }{ k }=\cfrac{c}{k}\)
What about,
\(t_{ c }=2\pi r\)
\(\cfrac{t_c}{t}= \cfrac { 2\pi r }{ t } =c\)
\( \cfrac { \partial \, t_{ c } }{ \partial \, t } =2\pi \cfrac { \partial \, r }{ \partial \, t }=c \)
This is wrong because \(r\) for a given time speed is fixed. Time \(t_c\), will always coil around \(x\) at a fixed \(r\), gvien \(\cfrac{\partial\,t_c}{\partial\,t}\),
\(\cfrac{\partial\,r}{\partial\,t}=0\)
\(r\) does not increase and the time coil does not expand as time progresses.
Time is in a helix with a phase associated with its position on its circular path along its axis of travel. If time returns to its original phase at a distance \(t_{\lambda}\) ahead, after time period \(T\),
\(2\pi r=kt_{\lambda}\)
\(\cfrac { 2\pi r }{c}=T=\cfrac{k}{c}t_{\lambda}\)
\(t_{\lambda}=\cfrac{c}{k}T\)
so, time linear speed \(v_t\) is,
\(v_t=\cfrac{c}{k}\)
Time speed \(v_t\), is not light speed, \(c\).
