If $B$ denotes the binormal vector, $S$ denotes the arc length, $\tau$ denotes torsion, and $N$ denotes principal normal vector, then,

$$\frac{dB}{dS} = -\Bigg|\frac{dB}{dS}\Bigg|N = -\tau N$$

So if $\tau$ is the magnitude of $\frac{dB}{dS}$, i.e. a modulus value, then how can it be negative? It is written here that "if torsion is positive, the curve "turns" to the side to which binormal vector points. If torsion is negative, the curve "turns" to the opposite side."


1 Answer 1


The definition of the torsion of the question, i.e. $$\frac{dB}{dS} = -\Bigg|\frac{dB}{dS}\Bigg|N = -\tau N$$ is not correct.

What is correct is that

$$\tau = -\frac{dB}{dS} \cdot N.$$ With this correct definition, the torsion can be positive as well as negative. See wikipedia for additional details.

  • $\begingroup$ But one of the Frenet–Serret formulas is $\frac{dB}{dS} = -\tau N$. And if you compare that with $\frac{dB}{dS} = -\Big|\frac{dB}{dS}\Big|N$, we get $\tau = \Big|\frac{dB}{dS}\Big|$ Why is that? $\endgroup$
    – Sasikuttan
    Feb 12, 2023 at 9:55
  • $\begingroup$ @Curiouserandcuriouser Indeed if $\frac{dB}{dS} = -\Bigg|\frac{dB}{dS}\Bigg|N$, then $\tau = \Big|\frac{dB}{dS}\Big|$. The problem is that the equation $\frac{dB}{dS} = -\Bigg|\frac{dB}{dS}\Bigg|N$ is not correct. Where did you got that? $\endgroup$ Feb 12, 2023 at 9:56
  • $\begingroup$ In my textbook (B.S. Grewal) it says "arc rate of turning of the binormal (i.e. the magnitude of $\frac{dB}{dS}$) is called torsion of the curve and is denoted by $\tau$". Also, shouldn't $\frac{dB}{dS}$ be parallel but opposite to $N$? $\endgroup$
    – Sasikuttan
    Feb 12, 2023 at 10:03
  • $\begingroup$ Then this equation was invented by you? $\endgroup$ Feb 12, 2023 at 10:39
  • $\begingroup$ I don't understand. If torsion is the magnitude of $\frac{dB}{dS}$, shouldn't be it equal to $\Big|\frac{dB}{dS}\Big|$? $\endgroup$
    – Sasikuttan
    Feb 12, 2023 at 10:43

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