# Frenet-Serret Formulas

• Apr 28th 2009, 02:29 AM
Kiwi_Dave
Frenet-Serret Formulas
Let $\displaystyle \kappa=\frac{\delta \overline T}{\delta s}\cdot \overline N$ and $\displaystyle \tau=\frac{\delta \overline N}{\delta s}\cdot \overline B$. Assume in turn that each of the intrinsic derivatives of $\displaystyle \overline T, \overline N, \overline B$ are some linear combination of the unit vectors $\displaystyle \overline T, \overline N, \overline B$ and hence derive the Frenet-Serret formulas of differential geometry.

I am sure thus must be easy but I cannot see it!

If $\displaystyle \frac{\delta \overline T}{\delta s}=a \overline T + b \overline N + c \overline B$ then taking the dot product of both sides with N and using the first equation given produces $\displaystyle \frac{\delta \overline T}{\delta s}=a \overline T + \kappa \overline N + c \overline B$ If I claim to know that N is orthogonal to T then I can get the first equation, that is $\displaystyle \frac{\delta \overline T}{\delta s}=\kappa \overline N$

But when I move onto the next equation $\displaystyle \frac{\delta \overline N}{\delta s}=a_2 \overline T + b_2 \overline N + c_2 \overline B$ then taking the dot product of both sides with B and using the second equation given produces $\displaystyle \frac{\delta \overline N}{\delta s}=a_2 \overline T + b_2 \overline N + \tau \overline B$.

I cannot see how to get from here to the required equation:
$\displaystyle \frac{\delta \overline N}{\delta s}=-\kappa \overline T + \tau \overline B$

Or the third equation:
$\displaystyle \frac{\delta \overline B}{\delta s}=-\tau \overline N$
• May 3rd 2009, 09:59 AM
Rebesques
Quote:

I am sure thus must be easy but I cannot see it!

What happens when you differentiate a unit vector?(Nerd)
• May 5th 2009, 01:21 AM
Kiwi_Dave
Differentiating a unit vector we get:

$\displaystyle \frac{\delta \overline T}{\delta s}\cdot \overline T=\frac{\delta \overline N}{\delta s}\cdot \overline N=\frac{\delta \overline B}{\delta s}\cdot \overline B=0$

So this helps me to reduce my three equations to:

$\displaystyle \frac{\delta \overline T}{\delta s}=\kappa \overline N + c_1 \overline B$

$\displaystyle \frac{\delta \overline N}{\delta s}=a_2 \overline T + \tau \overline B$.

$\displaystyle \frac{\delta \overline B}{\delta s}=a_3 \overline T + b_3 \overline N$

But how do I show that $\displaystyle c_1=a_3=0$ and $\displaystyle a_2=-\kappa, b_3=-\tau$?
• May 5th 2009, 07:00 AM
Jester
Quote:

Originally Posted by Kiwi_Dave
Let $\displaystyle \kappa=\frac{\delta \overline T}{\delta s}\cdot \overline N$ and $\displaystyle \tau=\frac{\delta \overline N}{\delta s}\cdot \overline B$. Assume in turn that each of the intrinsic derivatives of $\displaystyle \overline T, \overline N, \overline B$ are some linear combination of the unit vectors $\displaystyle \overline T, \overline N, \overline B$ and hence derive the Frenet-Serret formulas of differential geometry.

I am sure thus must be easy but I cannot see it!

If $\displaystyle \frac{\delta \overline T}{\delta s}=a \overline T + b \overline N + c \overline B$ then taking the dot product of both sides with N and using the first equation given produces $\displaystyle \frac{\delta \overline T}{\delta s}=a \overline T + \kappa \overline N + c \overline B$ If I claim to know that N is orthogonal to T then I can get the first equation, that is $\displaystyle \frac{\delta \overline T}{\delta s}=\kappa \overline N$

But when I move onto the next equation $\displaystyle \frac{\delta \overline N}{\delta s}=a_2 \overline T + b_2 \overline N + c_2 \overline B$ then taking the dot product of both sides with B and using the second equation given produces $\displaystyle \frac{\delta \overline N}{\delta s}=a_2 \overline T + b_2 \overline N + \tau \overline B$.

I cannot see how to get from here to the required equation:
$\displaystyle \frac{\delta \overline N}{\delta s}=-\kappa \overline T + \tau \overline B$

Or the third equation:
$\displaystyle \frac{\delta \overline B}{\delta s}=-\tau \overline N$

Given $\displaystyle \frac{d\vec{T}}{ds} = \kappa \vec{N}$ and $\displaystyle \frac{d \vec{B}}{ds} = - \tau \vec{N}$,

if we consider

$\displaystyle \vec{N} = \vec{B} \times \vec{T}$ then

$\displaystyle \frac{d \vec{N}}{ds} = \frac{d \vec{B}}{ds} \times \vec{T} + \vec{B} \times \frac{d \vec{T}}{ds} = - \tau \vec{N} \times \vec{T} + \vec{B} \times (\kappa \vec{N}) = \tau \vec{B} - \kappa \vec{T}$

as required.