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Thread: Frenet-Serret Formulas

  1. #1
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    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:
    $\displaystyle \frac{\delta \overline B}{\delta s}=-\tau \overline N$
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  2. #2
    Super Member Rebesques's Avatar
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    I am sure thus must be easy but I cannot see it!


    What happens when you differentiate a unit vector?
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  3. #3
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    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 $\displaystyle a_2=-\kappa, b_3=-\tau$?
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  4. #4
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    Quote Originally Posted by Kiwi_Dave View Post
    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.
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