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I'm trying to show that (1) $T'\times T'' = k^2(kB +\tau T)$

$T' = \kappa N$, from Frenet Serret

$T'' = \kappa'N + N'\kappa$, but the algebra didn't follow when I tried to substitute this on the Left hand side, of (1) above

Buddy Holly
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    It seems you need to address the question of how $N'$ relates to $B$ and $T$. When I derive things I often find it helpful to write $V = c_1T+c_2N+c_3B$ and use dot-products to select the values of $c_1,c_2,c_3$. For example, $c_1 = V \cdot T$. We can use the orthonormality of the Frenet frame to make nice calculations. – James S. Cook Dec 08 '12 at 03:26

2 Answers2

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First using the frenet serret equation $$N' = -kT + \tau B,$$ substitute it into $T''$ to get $$T'' = k'N - k^2 T + k\tau B$$ so $$T' \times T'' = kN \times (-k'N - k^2T + kτB) = -kk'(N \times N) - k^3(N \times T) + k^2τ(N \times B)$$

If you unsure how I got to this point, here's a link to the properties of the cross product

Since $\{T,N,B\}$ is an orthonormal basis (meaning each vector is of unit length and each vector is orthogonal to one another), $$N \times N = 0,\quad T \times N = B$$ which implies $$N \times T = -B,\quad N \times B = T.$$

Going back to our equation we get

$$T' \times T'' = k^3B + k^2\tau T = k^2(kB + \tau T)$$

DreamAR
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Kevin
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  • Thank you so much for your explanation. Can you please explain the sign convention with why NxB = T, is there a systematic way of determining the sign? Thanks – Buddy Holly Dec 10 '12 at 05:13
  • The reason that N x B = T is because they form an orthonormal basis meaning that T, N, and B are orthogonal to each other. The easiest way to imagine this is picture this is to say that T lies on the x-axis, N the y-axis, and B the z-axis. You'll notice that all 3 vectors are orthogonal to one another. Now the cross product is a tool for computing a vector orthogonal to two vectors. So if you cross N and B you'll get a vector parallel to T and because all three vectors are of unit length and meet at the same point, N x B will in fact equal T. – Kevin Dec 10 '12 at 06:00
  • Since it's possible to get two orthogonal vectors to any two vectors, a convention was needed to determine what way to choose. If you go back to my previous example where T, N, and B lie on the x, y, and z axis respectively, you'll notice that both the z+ and z- are orthogonal to x and y. Now with our orthonormal basis {T,N,B} we set it up like this so that going left to right gives us a positive vector (T X N = B, N x B = T, and B X T = N) and going right to left gives us a negative vector (B X N = -T, N X T = - B, and T X B = -N). – Kevin Dec 10 '12 at 06:08
  • Now in general, a x b = c, the sign is detemined by angle between the two vectors. If you recall, a x b = |a||b|sinx where x is the angle between them measured counterclockwise. Assuming x can only make one full circular rotation, if x is less than 180 degrees or pi then a x b will be positive. If x is greater than 180 degrees (pi), it'll be negative. A simple trick to this is called the right hand rule and involves using your right hand to determine the direction of c. However I suggest going to youtube to learn how to use it. It's very simple just hard to explain over text. – Kevin Dec 10 '12 at 06:29
  • Thanks for the explanation, can you also pls help with the other differential geometry question on "christoffel symbols" I have asked? Somehow no response. Thx a lot – Buddy Holly Dec 10 '12 at 07:36
  • I don't know much about christoffel symbols sorry. I'm taking differential geometry too right now but my professor ran out of time in the semester and briefly talked about the gauss equations. Wish I could, they look interesting but I haven't studied them yet. – Kevin Dec 11 '12 at 03:19
  • no problem. thx again – Buddy Holly Dec 11 '12 at 07:32
1

Not really much to add after Kevin's rightfully accepted answer (+ 1), but the derivation is so cute I wanted to write it up for myself.

The first Frenet-Serret equation,

$\dot T = \kappa N, \tag 1$

may be differentiated with respect to the arc-length $s$ to give

$\ddot T = \dfrac{d}{ds}(\kappa N) = \dot \kappa N + \kappa \dot N \tag 2$

the second Frenet-Serret equation,

$\dot N = -\kappa T + \tau B, \tag 3$

may be substituted into (2):

$\ddot T = \dot \kappa N + \kappa (-\kappa T + \tau B) = \dot \kappa N -\kappa^2 T + \kappa \tau B; \tag 4$

we take the cross product of (1) and (4):

$\dot T \times \ddot T = \kappa N \times (\dot \kappa N -\kappa^2 T + \kappa \tau B) = \kappa \dot \kappa N \times N - \kappa^3 N \times T + \kappa^2 \tau N \times B$ $= - \kappa^3 N \times T + \kappa^2 \tau N \times B; \tag 5$

$B = T \times N \Longrightarrow N \times T = -B, \; N \times B = T; \tag 6$

thus (5) yields

$\dot T \times \ddot T = \kappa^3 B + \kappa^2 \tau T = \kappa^2 (\kappa B + \tau T), \tag 7$

as per request. QED.

Robert Lewis
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