1. ## Frenet-Serret Formulas

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

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

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

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

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

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

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

What happens when you differentiate a unit vector?

3. Differentiating a unit vector we get:

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

So this helps me to reduce my three equations to:

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

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

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

But how do I show that $c_1=a_3=0$ and $a_2=-\kappa, b_3=-\tau$?

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

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

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

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

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

Or the third equation:
$\frac{\delta \overline B}{\delta s}=-\tau \overline N$
Given $\frac{d\vec{T}}{ds} = \kappa \vec{N}$ and $\frac{d \vec{B}}{ds} = - \tau \vec{N}$,

if we consider

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

$\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.

,

,

,

,

,

,

,

,

,

,

,

,

,

,

# derive the frenet serret formula

Click on a term to search for related topics.