# Vector calculus - Material derivative in spherical coordinates…

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.

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

But, how would I evaluate:

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

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 $$\nabla=\frac{\partial}{\partial R}\boldsymbol{e}_R+\frac{1}{R}\frac{\partial}{\partial\theta}\boldsymbol{e}_\theta+\frac{1}{R\sin\theta}\frac{\partial}{\partial\phi}\boldsymbol{e}_\phi.$$

• 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 $A_{R}$... – Michael Roberts Oct 10 '15 at 16:08
• $\vec{A}$ is the same as in your question. – Lythia Oct 10 '15 at 20:46
• It isn't defined, so could I just set it to any vector? – Michael Roberts Oct 10 '15 at 21:06
• Yup. That is what is meant by $$\vec{A}=A_x\boldsymbol{i} +A_y\boldsymbol{j} +A_z\boldsymbol{k}$$ or $$\vec{A}=A_R\boldsymbol{e}_R +A_\theta\boldsymbol{e}_\theta +A_\phi\boldsymbol{e}_\phi$$ – Lythia Oct 10 '15 at 21:07
• When you derived $\mathbf{e}_{\theta}$, shouldn't that be $\mathbf{e}_{\phi}$ – Michael Roberts Oct 13 '15 at 13:56