Help with Frenet-Serret formula

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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:

Since this vector (the unit tangent t) is always of unit length, its derivative must always be perpendicular to t, and so must be a vector in the normal plane to the curve. But this derivative, being the vector difference of two consecutive tangent vectors, must lie in the osculating plane formed by the latter, and so its direction is that of the principal normal, which direction we designate by the unit vector n. In order to calculate the magnitude of the vector dt/ds (s is the arc length), we note that the curve, in the neighbourhood of two consecutive tangents, corresponding to three neighbouring points, may be replaced by the circle of curvature, whose centre M is determined by the intersection of the perpendiculars to the two consectutive tangents. The angle [tex]d\phi[/tex] between these tangents is the same as that between the two radii of the circle of curvature. If [tex]\rho[/tex] is the radius of this circle then, in the limit,

[tex]ds = \rho d\phi[/tex].

On the other hand, |dt| = [tex]d\phi[/tex] whence

|dt/ds| = 1/[tex]\rho[/tex].

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?
 
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  • #2
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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?

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.
 
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  • #3
Astronuc
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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.
 
  • #4
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Thanks a lot for the help! Now I understand it both intuitively and mathematically.
 

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