# Torsion definition [closed]

Torsion is defined as $$\tau=-\frac{\mathrm{d}\vec{B}}{\mathrm{d}s}\cdot \vec{N}$$. Why is it not just $$\tau=\frac{\mathrm{d}\vec{B}}{\mathrm{d}s}\cdot \vec{N}$$? Could someone please show me a diagram showing $$\frac{\mathrm{d}\vec{B}}{\mathrm{d}s}$$, $$\vec{N}$$, $$\vec{T}$$, and $$\vec{B}$$ as a particle under torsion moves along a path?

• Some books (e.g., doCarmo) will actually have your formula without the negative. This is a matter of convention, although I believe 95% of us use the definition with the negative. If you want to see a picture, look on p. 12 of my differential geometry text, linked in my profile. Jan 23, 2020 at 23:27

## 1 Answer

I can't provide a diagram, but can present the relevant equations which show the origins of the equation

$$\tau = -\dfrac{d \vec B}{ds} \cdot \vec N = -\dot{\vec B} \cdot \vec N. \tag 1$$

I assume the reader is familiar with the first Frenet-Serret equation

$$\dot{\vec T} = \kappa \vec N, \tag 2$$

where

$$\Vert \vec N \Vert = 1 \tag 3$$

and

$$\kappa > 0; \tag 4$$

since

$$\vec T \cdot \vec T = 1, \tag 5$$

we of course have

$$\dot{\vec T} \cdot \vec T + \vec T \cdot \dot{\vec T} \Longrightarrow 2 \dot{\vec T} \cdot \vec T \Longrightarrow \dot{\vec T} \cdot \vec T = 0$$ $$\Longrightarrow \kappa \vec N \cdot \vec T = 0 \Longrightarrow \vec N \cdot \vec T = 0; \tag 5$$

$$\vec N \cdot \vec T = 0 \tag 6$$

yields, in light of (1),

$$\dot{\vec N} \cdot \vec T + \vec N \cdot \dot{\vec T} = 0 \Longrightarrow \dot{\vec N} \cdot \vec T + \vec N \cdot \dot{\vec T} = 0$$ $$\Longrightarrow \dot{\vec N} \cdot \vec T + \vec N \cdot \kappa \vec Nb\Longrightarrow \dot{\vec N} \cdot T = -\kappa, \tag 7$$

which shows that the component of $$\dot{\vec N}$$ along $$\vec T$$ is $$-\kappa$$. Now by (3),

$$\vec N \cdot \vec N = \Vert \vec N \Vert^2 = 1 \Longrightarrow \dot{\vec N} \cdot \vec N + \vec N \cdot \dot{\vec N} = 0$$ $$\Longrightarrow 2 \dot{\vec N} \cdot \vec N = 0 \Longrightarrow \dot{\vec N} \cdot \vec N = 0, \tag 8$$

and so $$\dot{\vec N}$$ has no component along $$\vec N$$ itself; since we are operating in $$\Bbb R^3$$, it is possible there is another component of $$\dot{\vec N}$$ normal to both $$\vec T$$ and $$\vec N$$; this motivates us to define

$$\vec B = \vec T \times \vec N; \tag 9$$

it follows from (3), (5) and (6) that

$$\Vert B \Vert = 1, \tag{10}$$

and from the definition of cross product that

$$\vec T \cdot \vec B = \vec N \cdot \vec B = 0; \tag{11}$$

thus $$\vec T$$, $$\vec N$$ and $$\vec B$$ form an orthonormal triad in $$\Bbb R^3$$ and hence we may complete the description of $$\dot{\vec N}$$ by specifying its component along $$\vec B$$; thus the torsion $$\tau$$ enters by means of the equation

$$\dot{\vec N} = -\kappa \vec T + \tau \vec B, \tag{12}$$

which further implies

$$\dot{\vec N} \cdot \vec B = \tau \vec B \cdot \vec B = \tau; \tag{13}$$

now (1) follows by differenting the second equation in (11), for

$$\vec N \cdot \vec B = 0 \Longrightarrow \dot{\vec N} \cdot \vec B + \vec N \cdot \dot{\vec B} = 0$$ $$\Longrightarrow \dot{\vec N} \cdot \vec B = -\vec N \cdot \dot{\vec B}; \tag{14}$$

combining this with (13) yields

$$\dot{\vec B} \cdot \vec N = -\tau, \tag{15}$$

which is essentially (1).

• \overrightarrow{N}. etc., may provide a better formatting: $\overrightarrow{N}$. Jan 23, 2020 at 23:24
• @Pythagoras: thanks I'll consider it. Cheers! Jan 23, 2020 at 23:25
• I actually wrote years ago a Mathematica script that will show the frame moving in real time along an arbitrary regular curve. Jan 23, 2020 at 23:37
• Robert, happy to send the notebook if you email me. Same to @DevrimA. Jan 23, 2020 at 23:40
• Yes, my profile is up-to-date, even if I'm out of date! Jan 24, 2020 at 0:59