5

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}{\dot{r}^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 $0$?

Jyrki Lahtonen
  • 133,153
dustin
  • 8,241
  • 1
    er, if you've shown the first 2 equalities doesn't that immediately imply that the 1st equals the third? – Scaramouche Aug 30 '13 at 19:08
  • 1
    Are you sure that $\dot{\hat{\mathbf{u}}} = \frac{1}{v}\frac{d\mathbf{v}}{dt}$ ? : pay attention to the product rule ... – Tony Piccolo Aug 31 '13 at 14:50

1 Answers1

2

There is a mistake in the computation of $\dot {\hat {\mathbf u}}$.
Using the product rule, you have$$\dot {\hat {\mathbf u}}=-\frac {\dot v}{v^2}\mathbf v + \frac 1{v}\dot {\mathbf v}=-\frac {\dot v}{v}\hat{\mathbf u} + \frac 1{v}\dot {\mathbf v}$$and therefore $$\frac {\dot {\hat {\mathbf u}}}{v}=\frac 1{v^2}(-\dot v\, \hat{\mathbf u}+\dot {\mathbf v})=\frac 1{v^2}\left(-\dot v\, \hat{\mathbf u}+\dot v\, \hat{\mathbf u}+ \frac {v^2}{\rho}\hat {\mathbf n}\right)==\frac 1{\rho}\hat {\mathbf n}$$

Tony Piccolo
  • 4,426