Chapter 2: Space Curves
Section 2.7: Frenet-Serret Formalism
Prove Theorem 2.7.1.
Let curve C1 be defined by R1s, for which the TNB-frame consists of the vectors T1,N1,B1.
Let curve C2 be defined by R2s, for which the TNB-frame consists of the vectors T2,N2,B2.
By hypothesis, both curves have κs as curvature and τs as torsion.
By translation and rotation, align the curves so that the TNB-frames at s=0 coincide. Define
and by the calculations in Table 2.7.10(a), Q′s=0, so Q=c, constant. This constant can be determined by evaluating Q at s=0 where T1=T2=T0,N1=N2=N0,B1=B2=B0. At this point (since T0,N0,B0, are all unit vectors)
so c=3. It is now possible to conclude that T1=T2,N1=N2,B1=B2 for s≥0. The reasoning is as follows.
T1·T2=T1 T2 cosθ1 = 1⋅1⋅cosθ1
where θ1 is the angle between T1 and T2. Applying the same thinking to N1·N2 and B1·B2 leads to
The only way this can be true is if θ1=θ2=θ3=0 so that each cosine has the value 1.
The essential step has been to obtain T1s=T2s so that R1/s=R2/s and R1s=R2s+A. But at s=0, R10=R20, so A=0 and R1s=R2s.
=κ N1·T2+κ N2·T1
=−τ N1·B2+−τ N2·B1
=−κ T1+τ B1·N2+−κ T2+τ B2·N1
= −κ T1·N2+τ B1·N2−κ T2·N1+τ B2·N1
κ N1·T2+T1·N2−T1·N2−N1·T2+τ N2·B1+N1·B2−B2·N1−B1·N2=0
Table 2.7.10(a) Details showing that Q′=0
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