The recurrence theorem of the Frenet formulas

Abel Cavaşi

Hi all!

Studying the Frenet formulas I have concluded that they are recursive. More specifically, using the trigonometric form of the Frenet formulas (form which you can find boring details on my blog), we proved the following

Theorem: If there is a right trihedron of the n order
[tex](\vec{T}_{n},\;\vec{N}_{n},\;\vec{B}_{n})[/tex]
that satisfies the Frenet formulas of the n order, written in the trigonometric form
[tex]\left\{\begin{array}{l}\dot{\vec{T}}_n=\omega_n \sin{\theta_n} \vec{N}_n\\\dot{\vec{N}}_n=\omega_n(-\sin\theta_n\vec{T}_n+\cos\theta_n\vec{B}_n)\\\dot{\vec{B}_n}=-\omega_n\cos\theta_n\vec{N}_n\end{array}\right.,[/tex]
then there is still a right trihedron of the n +1 order
[tex]\left\{\begin{array}{l}{\vec{T}}_{n+1}=\cos{\theta_n} \vec{T}_{n}+\sin \theta_{n} \vec{B}_{n}\\{\vec{N}}_{n+1}=-\sin \theta_{n} \vec{T}_{n}+\cos \theta_{n} \vec{B}_{n}\\{\vec{B}}_{n+1}=-\vec{N}_{n}\end{array}\right.[/tex]
that satisfying, in turn, the Frenet formulas of the n+1 order written also in the trigonometric form
[tex]\left\{\begin{array}{l}\dot{\vec{T}}_{n+1} = {\omega_{n+1}}\sin \theta_{n+1}{\vec{N}}_{n+1}\\{\dot{\vec{N}}}_{n+1}={\omega}_{n+1}(-\sin{\theta_{n+1}}{\vec{T}}_{n+1}+\cos{\theta_{n+1}}{\vec{B}}_{n+1})\\\ dot{\vec{B}}_{n+1}=-\omega_{n+1} \cos \theta_{n+1}{\vec{N}}_{n+1}\end{array}\right.,[/tex]
where
[tex]\theta_{n+1}=\arctan\frac{\dot{\theta}_{n}}{\omega_{n}}[/tex]
and
[tex]\omega_{n+1}=\sqrt{{\dot{\theta}_{n}}^{2}+\omega_{n}^{2}}[/tex] .

Demonstration: Through relations
[tex]\theta_{n+1}=\arctan\frac{\dot{\theta}_{n}}{\omega_{n}}[/tex]
and
[tex]\omega_{n+1}=\sqrt{{\dot{\theta}_{n}}^{2}+\omega_{n}^{2}}[/tex]
we have that
[tex]\sin\theta_{n+1}=\frac{\tan\theta_{n+1}}{\sqrt{1+ \tan^2 {\theta}_{n+1}}} = \frac {\frac {\dot{\theta}_n} {{\omega}_n}} {\sqrt{1+ \frac{{\dot{\theta}}^2} {\omega_{n}^{2}}}} = \frac{\dot{\theta}_{n}}{\sqrt{\omega_{n}^{2}+{\dot{\theta}_{n}}^{2}}}=\ frac{\dot\theta_n}{\omega_{n+1}}[/tex] ,
so
[tex]\dot{\theta}_{n}=\omega_{n+1}\sin\theta_{n+1}[/tex] .

We also have
[tex]\cos\theta_{n+1}=\sqrt{1-\sin ^{2}\theta_{n+1}}=\sqrt{1-\frac{{\dot{\theta}_{n}}^{2}}{\omega_{n}^{2}+{\dot{\theta}_{n}}^{2}}}=\ frac{\omega_{n}}{\omega_{n+1}}[/tex] ,
whence
[tex]\omega_{n}=\omega_{n+1}\cos\theta_{n+1}[/tex] .

Now, we derive the unit vectors of the trihedron of the n+1 order
[tex]\left\{\begin{array}{l}{\vec{T}}_{n+1}=\cos\theta_{n}\vec{T}_{n}+\sin\t heta_{n}\vec{B}_{n}\\{\vec{N}}_{n+1}=-\sin\theta_{n}\vec{T}_{n}+\cos\theta_{n}\vec{B}_{n}
\\{\vec{B}}_{n+1}=-\vec{N}_{n}\end{array}\right.[/tex]
and we obtain
[tex]\left\{\begin{array}{l}{\dot{\vec{T}}}_{n+1} = -\dot {\theta}_{n} \sin \theta_{n}\vec{T}_{n}+ \cos \theta \dot{\vec{T}}_{n}+ \dot{\theta}_{n} \cos \theta_{n} \vec{B}_{n}+ \sin \theta_{n} \dot{\vec{B}}_{n} \\{\dot{\vec{N}}}_{n+1} = -\dot{\theta}_{n}\cos\theta_{n}\vec{T}_{n}-\sin\theta_{n}\dot{\vec{T}}_{n}-\dot{\theta}_{n}\sin\theta_{n}\vec{B}_{n}+ \cos \theta_{n} \dot{\vec{B}}_{n}\\{\dot{\vec{B}}}_{n+1} = -\dot{\vec{N}}_{n}=-\omega_{n}(- \sin \theta_{n}\vec{T}_{n}+ \cos \theta_{n} \vec{B}_{n})\end{array}\right.[/tex] .

Replacing
[tex]\dot{\vec{T}}_{n}=\omega_{n}\sin\theta_{n}\vec{N}_{n}[/tex]
and
[tex]\dot{\vec{B}}_{n}=-\omega_{n}\cos\theta_{n}\vec{N}_{n}[/tex] ,
we obtain
[tex]\left\{\begin{array}{l}{\dot{\vec{T}}}_{n+1}=\dot{\theta}_{n}(-\sin\theta_{n}\vec{T}_{n}+\cos\theta_{n}\vec{B}_{n})\\{\dot{\vec{N}}}_{ n+1} = -\dot{\theta}_{n}(\cos \theta_{n} \vec{T}_{n}+ \sin \theta_{n} \vec{B}_{n})-\omega_{n} \vec{N}_{n}\\{\dot{\vec{B}}}_{n+1} = -\dot{\vec{N}}_{n} = -\omega_{n}(-\sin \theta_{n} \vec{T}_{n}+ \cos \theta_{n} \vec{B}_{n})\end{array}\right.[/tex] .

But, from the definition of the unit vectors of the high order, we know that
[tex]\left\{\begin{array}{l}{\vec{T}}_{n+1}=\cos\theta_{n}\vec{T}_{n}+\sin\t heta_{n}\vec{B}_{n}\\{\vec{N}}_{n+1}=-\sin\theta_{n}\vec{T}_{n}+\cos\theta_{n}\vec{B}_{n}\\{\vec{B}}_{n+1}=-\vec{N}_{n}\end{array}\right.[/tex] ,
so
[tex]\left\{\begin{array}{l}{\dot{\vec{T}}}_{n+1}=\dot{\theta}_{n}\vec{N}_{n +1}\\{\dot{\vec{N}}}_{n+1}=-\dot{\theta}_{n}\vec{T}_{n+1}+\omega_{n}\vec{B}_{n+1}\\{\dot{\vec{B}}}_ {n+1}=-\dot{\vec{N}}_{n}=-\omega_{n}\vec{N}_{n+1}\end{array}\right.[/tex] .

Because
[tex]\dot{\theta}_{n}=\omega_{n+1}\sin\theta_{n+1}[/tex]
and
[tex]\omega_{n}=\omega_{n+1}\cos\theta_{n+1}[/tex] ,
finally result that
[tex]\left\{\begin{array}{l}\\{\dot{\vec{T}}}_{n+1} = {\omega}_{n+1} \sin \theta_{n+1} {\vec{N}}_{n+1} \\{\dot{\vec{N}}}_{n+1}={\omega}_{n+1}(- \sin \theta_{n+1} {\vec{T}}_{n+1}+\cos \theta_{n+1}{\vec{B}}_{n+1}) \\{\dot{\vec{B}}}_{n+1}=-\omega_{n+1}\cos \theta_{n+1}{\vec{N}}_{n+1}\end{array}\right.[/tex] ,

qed.


I know that each of you are busy with lots of interesting problems, but I still like my answer, if you can, two important questions related to this theorem which I find rather unnoticed:

-1). Is it correct? It is so well made and well proven?
-2). Given that this theorem is a result about the important physical phenomenon called „mechanical movement”, what kind of impact do you think that have it in Physics? Consequently, it has any value, revolutionary?

Thank you very much for your attention!

p.s. Excuse my English, but I translated with Google translate.