0
$\begingroup$

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?

$\endgroup$
1
  • 1
    $\begingroup$ 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. $\endgroup$ Jan 23, 2020 at 23:27

1 Answer 1

0
$\begingroup$

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).

$\endgroup$
9
  • 1
    $\begingroup$ \overrightarrow{N}. etc., may provide a better formatting: $\overrightarrow{N}$. $\endgroup$
    – Pythagoras
    Jan 23, 2020 at 23:24
  • 1
    $\begingroup$ @Pythagoras: thanks I'll consider it. Cheers! $\endgroup$ Jan 23, 2020 at 23:25
  • 1
    $\begingroup$ I actually wrote years ago a Mathematica script that will show the frame moving in real time along an arbitrary regular curve. $\endgroup$ Jan 23, 2020 at 23:37
  • 1
    $\begingroup$ Robert, happy to send the notebook if you email me. Same to @DevrimA. $\endgroup$ Jan 23, 2020 at 23:40
  • 1
    $\begingroup$ Yes, my profile is up-to-date, even if I'm out of date! $\endgroup$ Jan 24, 2020 at 0:59

Not the answer you're looking for? Browse other questions tagged .