Frenet Serret Formulas

  • #1
Hi, I'm trying to derive the Frenet Serret Formulas, but I am having trouble to understand why, after some checking, that the derivative of binormal vector is:

[itex]\frac{d\hat{b}}{ds}=-\tau\hat{n}[/itex]

I understand that, [itex]\hat{t}\wedge\frac{d\hat{n}}{ds}\parallel\hat{n}[/itex] and [itex]|\frac{d\hat{b}}{ds}|=\tau[/itex], but why the negative sign? Isn't it equally possible that it has a positve sign?
 

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  • #3
Charles Link
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Since ## \tau ## is simply a constant of proportionality, I believe the choice is arbitrary, but whatever sign that is used for ## \tau ## needs to be consistent with the other Frenet equation involving ## \tau ##. The equations with their choice of the sign of ## \tau ## is apparently somewhat standard. It's a somewhat specialized topic, but I think you might find most of the textbooks use the same sign convention.
 
  • #4
lavinia
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Hi, I'm trying to derive the Frenet Serret Formulas, but I am having trouble to understand why, after some checking, that the derivative of binormal vector is:

##\frac{d\hat{b}}{ds}=-\tau\hat{n}##

I understand that, ##\hat{t}\wedge\frac{d\hat{n}}{ds}\parallel\hat{n}[/itex] and [itex]|\frac{d\hat{b}}{ds}|=\tau##, but why the negative sign? Isn't it equally possible that it has a positive sign?

##\frac{d\hat{b}}{ds}= \frac{d\hat{t×n}}{ds}= \frac{d\hat{t}}{ds}×\hat{n} + d\hat{t}×\frac{d\hat{n}}{ds} = 0 + \hat{t}×κ\hat{b} = κ\hat{t}×\hat{t}×\hat{n}##

Now use the cross product identity ##T×T×N = T(T⋅N) - N(T⋅T)##
 
Last edited:

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