# ϕ-component of equation of moition: proving a relationship [closed]

I have the following velocity vector (in spherical polars): $$\textbf{v} = u \hat{\textbf{r}} + v_{\phi}\hat{\boldsymbol\phi}$$ Where $u(r) = u$ and $v_{\phi} (r) = v_{\phi}$. $\theta =0$ in this system.

I am now required to prove the following: \begin{eqnarray} \rho(\textbf{v}\cdot\nabla)\textbf{v}=\rho \frac{u}{r} \frac{d}{dr}(r v_{\phi}). \end{eqnarray} The RHS is just part of the material derivative in the macroscopic equation of motion - and that is what I am ultimately calculating (the equation of motion). I am, though, only required to find the $\phi$-component of this (the RHS term) although I don't quite understand the mathematical implication of this.

Anyway, this is how I've been solving it (wrongly).

$\nabla$ in spherical coordinates is: $$\nabla = \frac{\partial}{\partial r} \hat{\boldsymbol r}+ \frac{1}{r} \frac{\partial}{\partial \theta}\hat{\boldsymbol\theta} +\frac{1}{r \sin \theta}\frac{\partial}{\partial \phi}\hat{\boldsymbol\phi}$$.

That is: \begin{eqnarray} (\textbf{v} \cdot \nabla) &=& (u \hat{\textbf{r}} + v_{\phi}\hat{\boldsymbol\phi})\cdot(\frac{\partial}{\partial r} \hat{\boldsymbol r}+ \frac{1}{r} \frac{\partial}{\partial \theta}\hat{\boldsymbol\theta} +\frac{1}{r \sin \theta}\frac{\partial}{\partial \phi}\hat{\boldsymbol\phi}) \\ &=& u \frac{\partial}{\partial r} + \underbrace{\frac{v_{\phi}}{r \sin \theta}\frac{\partial}{\partial \theta}}_{A}. \end{eqnarray} Note that term $A$ is equal to zero because there is no system dependence on $\phi$ and $\theta = 0$.

That is: $$(\textbf{v} \cdot \nabla) = u \frac{\partial}{\partial r}.$$ Therefore: \begin{eqnarray} \rho(\textbf{v} \cdot \nabla)\textbf{v} &=& \rho u \frac{\partial}{\partial r}(u \hat{\textbf{r}} + v_{\phi}\hat{\boldsymbol\phi}) \\ &=& \underbrace{\rho u \frac{\partial u}{\partial r}\hat{\textbf{r}}}_{B} + \rho u\frac{\partial v_{\phi}}{\partial r}\hat{\boldsymbol\phi} \end{eqnarray} I'm assuming now, because of the need for only the $\phi$-component, term $B$ is neglected, i.e.: $$\rho(\textbf{v}\cdot\nabla)\textbf{v} = \rho u\frac{\partial v_{\phi}}{\partial r}\hat{\boldsymbol\phi}$$ Which is not right.

This is relatively simple and it's irritated me hugely, and any help would be much appreciated.

Thanks.

## closed as off-topic by ACuriousMind♦, Kyle Kanos, John Rennie, JamalS, BernhardApr 5 '15 at 10:48

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## 1 Answer

I don't know the context, but I assume there is some sort of typo in the "correct" equation. I think it should read $$(\rho\nabla_\mathbf{v}\mathbf{v})_\phi=\rho\frac{u}{r}\frac{\mathrm{d}}{\mathrm{d}{r}}(rv_\phi)$$ To prove this, we use the formula found here for spherical coordinates. It gives $$(\nabla_\mathbf{v}\mathbf{v})_\phi=\frac{uv_\phi}{r}+u\frac{\mathrm{d}v_\phi}{\mathrm{d}r}=\frac{u}{r}\frac{\mathrm{d}}{\mathrm{d}{r}}(rv_\phi)$$ as was to be shown.