MHB How Do Frenet Equations Relate Torsion to Time Derivatives of Position?

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The discussion focuses on deriving an expression for torsion using Frenet equations and time derivatives of the position vector. The key equations include the definitions of the Frenet frame and the relationship between torsion and the derivatives of the position vector. Participants explore how to manipulate the Frenet equations to express torsion in terms of time derivatives, ultimately leading to the formula involving the cross product of velocity and acceleration vectors. There is uncertainty about specific steps in the derivation, particularly in transitioning between different forms of the equations. The conversation highlights the complexity of relating geometric properties of curves to their dynamic behavior through calculus.
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Using Frenet equations, find an expression for the torsion in terms of time derivatives of the position vector.

The Frenet Equations are
\begin{align*}
\frac{d\hat{\mathbf{u}}}{ds} &= \frac{1}{\rho}\hat{\mathbf{n}}\\
\frac{d\hat{\mathbf{b}}}{ds} &= -\frac{1}{\tau}\hat{\mathbf{n}}\\
\frac{d\hat{\mathbf{n}}}{ds} &= \frac{1}{\tau}\hat{\mathbf{b}} - \frac{1}{\rho}\hat{\mathbf{u}}
\end{align*}

The torsion in terms of time derivatives is
\begin{align*}
\tau &= \frac{\dot{\mathbf{r}}\times \ddot{\mathbf{r}}\cdot\ddot{\mathbf{r}}}
{\lvert\dot{\mathbf{r}}\times \ddot{\mathbf{r}}\rvert^2}
\end{align*}

I am not sure how to go from the equations to answer though.
 
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Couple of comments:

1. It seems to me that some authors define
$$\frac{d \mathbf{b}}{ds}=- \tau \hat{\mathbf{n}}.$$

2. In either case, you can take this equation, and dot both sides into $\hat{\mathbf{n}}$ to solve for $\tau$. Then plug in from there.
 
Ackbach said:
Couple of comments:

1. It seems to me that some authors define
$$\frac{d \mathbf{b}}{ds}=- \tau \hat{\mathbf{n}}.$$

2. In either case, you can take this equation, and dot both sides into $\hat{\mathbf{n}}$ to solve for $\tau$. Then plug in from there.

So \(\hat{\mathbf{n}}\cdot\frac{d\hat{\mathbf{b}}}{ds} = \hat{\mathbf{n}}\cdot \left(\frac{d\hat{\mathbf{u}}}{ds} \times\hat{\mathbf{n}} + \hat{\mathbf{u}}\times \frac{d\hat{\mathbf{n}}}{ds}\right) = -\frac{1}{\tau}\), and since \(\frac{d\hat{\mathbf{u}}}{ds} = \frac{1}{\rho}\hat{\mathbf{n}}\), so the first term is zero due to \(\frac{1}{\rho} \hat{\mathbf{n}}\times \hat{\mathbf{n}} = 0\).

Then we are left with
\[
\hat{\mathbf{n}}\cdot \hat{\mathbf{u}}\times \frac{d\hat{\mathbf{n}}}{ds} = -\frac{1}{\tau}
\]

I am not sure how I am going to recover the time derivatives of \(r\) from the above expression.

I suppose I am making some progress now:

Torsion is defined as
\(-\frac{1}{\tau}\hat{\mathbf{n}} = \frac{d\hat{\mathbf{b}}}{ds}\).
Let's dot both side by \(\hat{\mathbf{n}}\).
\begin{align}
\frac{1}{\tau} &= -\hat{\mathbf{n}}\cdot \frac{d\hat{\mathbf{b}}}{ds}\\
&= -\rho\frac{d^2\mathbf{r}}{ds^2}\cdot \left(\hat{\mathbf{u}}\times
\rho\frac{d\hat{\mathbf{u}}} {ds}\right)
\end{align}
Now let's write \(\frac{d^2\mathbf{r}} {ds^2}\) as
\begin{align}
\frac{d^2\mathbf{r}}{ds^2} &= \frac{d^2\mathbf{r}} {dt^2}
\left(\frac{dt} {ds}\right)^2\\
&= \frac{1} {v^2}\ddot{\mathbf{r}}
\end{align}
and substitute back into equation above.
\begin{align}
&= -\frac{\rho^2} {v^2}\ddot{\mathbf{r}} \cdot \left(\frac{\dot{\mathbf{r}}}{v}\times
\frac{d\hat{\mathbf{u}}} {ds}\right)\\
&= ?\\
&= -\ddot{\mathbf{r}} \rho\cdot \left(\dot{\mathbf{r}}
\times\rho \ddot{\mathbf{r}}\right)_t\\
&= -\ddot{\mathbf{r}}\rho \cdot\left(\dot{\mathbf{r}}
\times\rho \dddot{\mathbf{r}}\right)\\
&= \frac{(\dot{\mathbf{r}} \times\ddot{\mathbf{r}}) \cdot
\dddot{\mathbf{r}}}
{\lvert \dot{\mathbf{r}} \times\ddot{\mathbf{r}} \rvert^2}
\end{align}

What is the ? step?
 
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