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[SOLVED] Frenet equation and torsion

dwsmith

Well-known member
Feb 1, 2012
1,673
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.
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,197
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.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
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.
[HR][/HR]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 in to 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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