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[SOLVED] Fernet-Serrat equations and vector calculus

dwsmith

Well-known member
Feb 1, 2012
1,673
I have shown the first two equality and I am working on the showing the 1st equals the 3rd.

\begin{alignat*}{4}
\frac{1}{\rho}\hat{\mathbf{{n}}} &= \frac{d\hat{\mathbf{{u}}}}{ds}
&{}= \frac{\dot{\hat{\mathbf{{u}}}}}{\dot{s}}
&{}= \left((\dot{\mathbf{r}} \cdot\dot{\mathbf{r}})\ddot{\mathbf{r}} -
(\dot{\mathbf{r}}\cdot\ddot{\mathbf{r}}) \dot{\mathbf{r}}\right)
\frac{1}{\lvert\dot{r}\rvert^4}
\end{alignat*}

[HR][/HR]

$$
\frac{1}{\rho}\hat{\mathbf{{n}}} = \frac{\dot{\hat{\mathbf{{u}}}}}{\dot{s}}
$$
We know that $\mathbf{v} = \frac{ds}{dt}\frac{dr}{ds}$ where $\dot{s} = v$ and $\hat{\mathbf{u}} = \frac{dr}{ds}$.

So $\mathbf{v} = v\hat{\mathbf{u}}\iff \dot{\hat{\mathbf{u}}} = \frac{1}{v}\frac{d\mathbf{v}}{dt}$.

Then $\frac{\dot{\hat{\mathbf{{u}}}}}{\dot{s}} = \frac{1}{v^2}\frac{d\mathbf{v}}{dt}$.

I know that
$$
\frac{d\mathbf{v}}{dt} = \frac{dv}{dt}\hat{\mathbf{u}} + \frac{v^2}{\rho}\hat{\mathbf{n}}.
$$
Then $\frac{\dot{\hat{\mathbf{{u}}}}}{\dot{s}} = \frac{1}{v^2}\frac{dv}{dt}\hat{\mathbf{u}} + \frac{1}{\rho}\hat{\mathbf{n}}$.
Therefore, $\frac{1}{v^2}\frac{dv}{dt}\hat{\mathbf{u}} = 0$ but how do I show that this is $0$?
 
Last edited:

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,774
So $\mathbf{v} = v\hat{\mathbf{u}}\iff \dot{\hat{\mathbf{u}}} = \frac{1}{v}\frac{d\mathbf{v}}{dt}$.
You need the product rule here.
That is:
$$\mathbf{v} = v\hat{\mathbf{u}} \Rightarrow \dot{\hat{\mathbf{u}}} = \frac{1}{v}\frac{d\mathbf{v}}{dt} - \frac{\dot v}{v^2}\mathbf v$$

The additional term cancels at the end.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
You need the product rule here.
That is:
$$\mathbf{v} = v\hat{\mathbf{u}} \Rightarrow \dot{\hat{\mathbf{u}}} = \frac{1}{v}\frac{d\mathbf{v}}{dt} - \frac{\dot v}{v^2}\mathbf v$$

The additional term cancels at the end.
How do I show the final equality? Convert to the vector triple product? Use Levi-Civita?

So I wrote the last term as $\dot{\mathbf{r}}\times\ddot{\mathbf{r}}\times\dot{\mathbf{r}}$ and used the fact that $\ddot{\mathbf{r}} = \dot{\mathbf{r}}\times\mathbf{c} \times\mathbf{r}$ where $\mathbf{c}$ is a constant vector.
\begin{align}
\dot{\mathbf{r}}\times\ddot{\mathbf{r}} \times\dot{\mathbf{r}} &=
\dot{\mathbf{r}}\times\dot{\mathbf{r}}\times \mathbf{c} \times\mathbf{r}\times\dot{\mathbf{r}}\\
&= \mathbf{c}\times\mathbf{r}\times\dot{\mathbf{r}}\\
&= \dot{\mathbf{r}}\times\mathbf{c}\times\mathbf{r}\\
&= \ddot{\mathbf{r}}
\end{align}
Then $\ddot{\mathbf{r}} = \left(\frac{ds}{dt}\right)^2\frac{d^2\mathbf{r}}{ds^2} = \frac{v^2}{\rho}\hat{\mathbf{n}}$.
Finally, $\frac{1}{\rho}\hat{\mathbf{n}} = \frac{v^2}{\lvert v\rvert^4\rho} \hat{\mathbf{n}} = \frac{1}{\rho} \hat{\mathbf{n}}$
 
Last edited: