So this one might be a little simple for some of you but I was hoping I could get all the nuts and bolts needed to show this for myself.

I have the following relationship, which makes use of the the material derivative:

$$ (\vec{A}\cdot{\nabla})\vec{r}=\vec{A} $$

I am needing to show this result in spherical polar coordinates.

Now, I don't want to be vague in what I have so far, but I really have very little.

I've started with $\vec{r}$ in spherical polar coordinates being:

$$ \vec{r}=x\mathbf{\hat{e}}_x+y\mathbf{\hat{e}}_y+z\mathbf{\hat{e}}_z $$

Firstly, is this right? It doesn't feel right. Would the following be more appropriate:

$$ \vec{r}=\sin{\theta}\cos{\phi}\mathbf{\hat{e}}_x+\sin{\theta}\sin{\phi}\mathbf{\hat{e}}_y+\cos{\theta}\mathbf{\hat{e}}_z $$

Secondly, the $\nabla$ operator in spherical polar coordinates, I have given as:

$$ \nabla=\frac{\partial}{\partial{r}}\vec{r}+\frac{1}{r}\frac{\partial}{\partial{\theta}}\vec{\theta}+\frac{1}{r\sin{\theta}}\frac{\partial}{\partial{\phi}}\vec{\phi} $$

I'm just lost as to whether this is right, and if it is right how to put it all together...any nudges in the right direction would be greatly appreciated.

Thank you.

Addendum for Lydia:

$$ \frac{\partial{\vec{r}}}{\partial{R}}=\sin\theta\cos\phi\mathbf{i}+\sin\theta\sin\phi\mathbf{j}+\cos\theta\mathbf{k} $$

But, how would I evaluate:

$$ \left|\left|{\frac{\partial{\vec{r}}}{\partial{R}}}\right|\right| $$


1 Answer 1


Here are some nudges. I will use $R$ for the radius in spherical coordinates to avoid any confusion with $\vec{r}$, the position vector.

  1. You can write $\vec{r}=R\sin\theta\cos\phi\boldsymbol{i}+R\sin\theta\sin\phi\boldsymbol{j}+R\cos\theta\boldsymbol{k}$.

  2. Let $\{\boldsymbol{e}_R,\boldsymbol{e}_\theta, \boldsymbol{e}_\phi\}$ be the orthonormal basis vectors for spherical coordinates. Then $$\boldsymbol{e}_R=\frac{1}{\left\|\frac{\partial\vec{r}}{\partial R}\right\|}\frac{\partial\vec{r}}{\partial R}, \qquad \boldsymbol{e}_\theta=\frac{1}{\left\|\frac{\partial\vec{r}}{\partial \theta}\right\|}\frac{\partial\vec{r}}{\partial \theta}, \qquad \boldsymbol{e}_\phi=\frac{1}{\left\|\frac{\partial\vec{r}}{\partial \phi}\right\|}\frac{\partial\vec{r}}{\partial \phi}. $$
    Using 1, you can calculate the above basis vectors in terms of $\boldsymbol{i},\boldsymbol{j},\boldsymbol{k}$. You should get $$\boldsymbol{e}_\phi =-\sin\phi\boldsymbol{i}+\cos\phi\boldsymbol{j}, \quad \boldsymbol{e}_\theta =\cos\theta\cos\phi\boldsymbol{i} + \cos\theta\sin\phi\boldsymbol{j}-\sin\theta\boldsymbol{k}, \quad \boldsymbol{e}_R =\sin\theta\cos\phi\boldsymbol{i} + \sin\theta\sin\phi\boldsymbol{j}+\cos\theta\boldsymbol{k}.$$ Another hint is that you should have $$\vec{r}=R\boldsymbol{e}_{R}.$$

  3. Since the Cartesian and spherical forms of $\vec{A}$ must be equivalent, $$A_x\boldsymbol{i}+A_y\boldsymbol{j}+A_z\boldsymbol{k}=A_R\boldsymbol{e}_R+A_\theta\boldsymbol{e}_\theta+A_\phi\boldsymbol{e}_\phi.$$ Taking the dot product of both sides with $\boldsymbol{e}_R$ gives that $$A_R = A_x\boldsymbol{i}\cdot\boldsymbol{e}_R+A_y\boldsymbol{j}\cdot\boldsymbol{e}_R+A_z\boldsymbol{k}\cdot\boldsymbol{e}_R.$$ Use part 2 to evaluate $\boldsymbol{i}\cdot\boldsymbol{e}_R$, $\boldsymbol{j}\cdot\boldsymbol{e}_R$, and $\boldsymbol{k}\cdot\boldsymbol{e}_R$. You should find that $$A_R=\sin\theta\cos\phi A_x+\sin\theta\sin\phi A_y+\cos\theta A_z.$$ Repeat the above process for $\boldsymbol{e}_\theta$ and $\boldsymbol{e}_\phi$. You will now have the components of $\vec{A}$ in the spherical basis.

  4. You can now plug everything in. Note that $$\nabla=\frac{\partial}{\partial R}\boldsymbol{e}_R+\frac{1}{R}\frac{\partial}{\partial\theta}\boldsymbol{e}_\theta+\frac{1}{R\sin\theta}\frac{\partial}{\partial\phi}\boldsymbol{e}_\phi.$$

  • $\begingroup$ Just to ask a silly question, and I apologise if it is really stupid. From all that you have derived and stated what is $\vec{A}$...? Is that what you denote as $A_{R}$... $\endgroup$ Commented Oct 10, 2015 at 16:08
  • $\begingroup$ $\vec{A}$ is the same as in your question. $\endgroup$
    – Lythia
    Commented Oct 10, 2015 at 20:46
  • $\begingroup$ It isn't defined, so could I just set it to any vector? $\endgroup$ Commented Oct 10, 2015 at 21:06
  • 1
    $\begingroup$ Yup. That is what is meant by $$\vec{A}=A_x\boldsymbol{i} +A_y\boldsymbol{j} +A_z\boldsymbol{k}$$ or $$\vec{A}=A_R\boldsymbol{e}_R +A_\theta\boldsymbol{e}_\theta +A_\phi\boldsymbol{e}_\phi$$ $\endgroup$
    – Lythia
    Commented Oct 10, 2015 at 21:07
  • $\begingroup$ When you derived $\mathbf{e}_{\theta}$, shouldn't that be $\mathbf{e}_{\phi}$ $\endgroup$ Commented Oct 13, 2015 at 13:56

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