1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Differential geometry Frenet

  1. Jul 4, 2014 #1
    even γ: I-> R ^ 2 curve parameterized as to arc length (single speed) with curvature k (s)> 0 and torsion τ(s)> 0. I want to write the γ(s) as a combination of n(s), t(s), b(s). these are the types of Frenet.


    the only thing i know is that the types of Frenet are t(s)=γ'(s) , b(s)=t(s)Xn(s) and n(s)=t'(s)/k(s)
    but how i will write γ(s) with this types? is there any way or type to do this?
     
  2. jcsd
  3. Jul 5, 2014 #2
    Maybe it could help if you told us what you are trying to do. Are you trying to solve a specific problem? Also, did you intend to write ##\gamma : I \rightarrow \mathbb{R}^2##, as in a planar curve?
     
  4. Jul 5, 2014 #3
    my problem is:even γ: I-> R ^3 curve parameterized as to arc length (single speed) with curvature k (s)> 0 and torsion τ(s)>0. we assume that the γ is at the surface sphere with center the origin. Show that for any s we have:
    γ(s)=-(1/k(s))*n(s) + (k'(s)/(k^2(s)*τ(s)))*b(s)

    and he gives us a hint: write the γ(s) as a combination of n(s), t(s), b(s). these are the types of Frenet
    and γ:I->R^3
     
  5. Jul 5, 2014 #4
    i think that if i find a way to combine γ with the types of frenet and derivative them i will find the problem
     
  6. Jul 5, 2014 #5
    You are meant to write ##\gamma(s) = a_1(s)t(s) + a_2(s)n(s) + a_3(s)b(s)##, for unknown scalar functions ##a_i(s)##. You can immediately conclude that one of the ##a_i(s)## is zero identically. Which one and why? The other two can be determined by differentiating the remaining expression.
     
  7. Jul 5, 2014 #6
    a1(s) will be zero? and after that with differentiation i will conclude to the the expression i wrote?
     
  8. Jul 5, 2014 #7
    Don't ask. Try it! And if you run into more trouble, come back here, show us what you have done and we will give you more help.
     
  9. Jul 5, 2014 #8
    i tried to the derivative and i found this :
    γ'(s)=a2(s)n'(s)+a2'(s)n(s)+a3'(s)b(s)+a3(s)b'(s) , we know that γ'(s)=0 beacuse we are on the surface of sphere , after that i tried to replace n'(s)=-k(s)t(s)+τ(s)b(s) and the other types?
     
  10. Jul 5, 2014 #9

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    Why will ##a_1(s)## be zero?

    Why would ##\gamma^\prime(s) = 0##? I don't see what the surface of the sphere has to do with this.

    Do you know the Frenet formulas?
     
  11. Jul 5, 2014 #10
    because γ(s)*γ(s)=c where c is a constant and then 2γ(s)γ'(s)=0
     
  12. Jul 5, 2014 #11

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    OK, how does that answer my two questions?
     
  13. Jul 5, 2014 #12
    ##\gamma'(s) = 0## identically would mean that there is no motion and the curve consists of just one point. Clearly this is not true in our case, since we are allowed to move around as long as we stay on the surface of the sphere.
     
  14. Jul 5, 2014 #13
    i think that a1(s)=0 because i supposed that for γ'(s)=0 we know that t(s)=γ'(s) and then t(s)=0 and a1(s)=0
     
  15. Jul 5, 2014 #14
    but if we dont have γ'(s)=0 if we differentiate the γ(s)=a1(s)t(s)+a2(s)n(s)+a3(s)b(s) with what γ'(s) will be equal?
     
  16. Jul 5, 2014 #15
    That is not correct and it is not the reason ##a_1(s)## is zero. You wrote above that ##\gamma(s) \cdot \gamma'(s) = 0##, which is correct. Do you see how you could use this to conclude that ##a_1(s)## must be zero?
     
  17. Jul 5, 2014 #16
    no... can you tell me why? i focused on this that i said prior
     
  18. Jul 5, 2014 #17
    So ##\gamma(s) = a_1(s)t(s) + a_2(s)n(s) + a_3(s)b(s)## and ## \gamma'(s) \cdot \gamma(s) = 0 ##, where ##\gamma'(s) = t(s)## by definition.

    By the way, I hope you understand that ##\gamma(s)## and ##\gamma'(s)## are vector-valued: ##\gamma(s) = (\gamma_1(s), \gamma_2(s), \gamma_3(s)), \, \gamma'(s) = (\gamma'_1(s), \gamma'_2(s), \gamma'_3(s))## and that by ##\gamma(s) \cdot \gamma'(s)## I mean the scalar product of these vectors.
     
  19. Jul 6, 2014 #18
    if i differentiate i find this γ'(s)=a2(s)n'(s)+a2'(s)n(s)+a3'(s)b(s)+a3(s)b'(s) beacuse γ'(s)=t(s) i will replace this in this equation? and after that i am trying to replace the followings:
    b(s)=t(s)xn(s) , n(s)=t'(s)/k(s), t'(s)=k(s)n(s), n'(s)=-k(s)t(s)+τ(s)b(s), b'(s)=-τ(s)n(s) ?
     
  20. Jul 6, 2014 #19
    Use the last two of these and then scalar multiply the expression ##\gamma'(s) = ...## with some suitable vector.
     
  21. Jul 6, 2014 #20
    i replaced the n'(s) and b'(s) but why we must multiply with a suitable vector , for what reason? i am a little confused
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Differential geometry Frenet
  1. Differential Geometry (Replies: 0)

  2. Differential Geometry (Replies: 1)

  3. Differential geometry (Replies: 18)

  4. Differential geometry (Replies: 1)

Loading...