MHB Fernet-Serrat equations and vector calculus

AI Thread Summary
The discussion focuses on proving the equality between the first and third expressions in the context of Fernet-Serrat equations and vector calculus. The key steps involve expressing velocity and its derivatives in terms of unit vectors and applying the product rule. It is established that the term involving the derivative of velocity cancels out, leading to a simplified form. The final equality is demonstrated by relating acceleration to curvature and confirming that both sides of the equation match. The conclusion reinforces the validity of the derived relationships in vector calculus.
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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*}



$$
\frac{1}{\rho}\hat{\mathbf{{n}}} = \frac{\dot{\hat{\mathbf{{u}}}}}{\dot{s}}
$$
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 $0$?
 
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dwsmith said:
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.
 
I like Serena said:
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, $\frac{1}{\rho}\hat{\mathbf{n}} = \frac{v^2}{\lvert v\rvert^4\rho} \hat{\mathbf{n}} = \frac{1}{\rho} \hat{\mathbf{n}}$
 
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