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Help with Frenet-Serret formula

  1. Sep 8, 2009 #1
    I'm learning vector analysis at the moment and having a bit of trouble grasping what's being put in front of me.

    My book (Theoretical Physics by Joos) explains it as such:

    I understand that dt/ds is perpendicular to t, but the problem I have is where Joos says matter-of-factly that |dt| = [tex]d\phi[/tex]. I simply cannot see how he is coming by that result, geometrically or algebraically. Can anyone point me in the right direction? What am I missing?
     
    Last edited: Sep 8, 2009
  2. jcsd
  3. Sep 10, 2009 #2
    i can see why this is confusing. It seems to be very old fashioned.

    I think he trying to say this. On any curve that is the trajectory of a point that is moving at constant speed one can approximate the motion in the limit as uniform circular motion. In uniform circular motion the only parameter is the radius of the circle. The reciprocal of the radius is the acceleration of the particle. Geometrically the acceleration is the curvature if the particle is moving at unit speed.

    In an arbitrary curve nearby tangents are close to being tangent to this instantaneous circle of uniform motion - this osculating circle - and so their perpendiculars intersect ,in the limit, in its center.

    So in the limit |ds/dt| since it is just the acceleration is the reciprocal of the radius of this osculating circle.

    I just also realized - you get the osculating circle by taking circles that pass by three nearby points and taking the limit as the points are moved towards each other (keeping the middle one fixed). Three points determine a circle uniquely - so in the limit you get some limiting circle - this is the osculating circle.
     
    Last edited: Sep 10, 2009
  4. Sep 10, 2009 #3

    Astronuc

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    Staff: Mentor

    Starting with [itex]s = \rho \phi[/itex]

    the [itex]ds = \rho d\phi[/itex].

    Then let ρ = 1, so [itex]ds = 1 d\phi = d\phi[/itex].


    dt is the infinitesimal or differential represented by ds = [tex]1\,d\phi\,\hat{t}\,=\,d\phi\,\hat{t}[/tex] where

    [tex]\hat{t} = \hat{\phi},\,or\,\bold{t}\,=\,\bold{\phi}, [/tex] the unit vector perpendicular to [tex]\hat{r}[/tex], the radial unit vector.
     
  5. Sep 10, 2009 #4
    Thanks a lot for the help! Now I understand it both intuitively and mathematically.
     
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