Differentiation spherical coordinates

[ebrattr]

Hi ! I'm trying to inverse a mass matrix so I need to do something like this

\dfrac{q}{\partial \mathbf{r}} where \cos q = \dfrac{\mathbf{r}\cdot \hat{\mathbf{k}}}{r}

However, when \mathbf{r} = \hat{\mathbf{k}} \text{ or } -\hat{\mathbf{k}} I have problems.

¿What can I do?

Thanks !

 

[Mark44]

Hi ! I'm trying to inverse a mass matrix so I need to do something like this

\dfrac{q}{\partial \mathbf{r}} where \cos q = \dfrac{\mathbf{r}\cdot \hat{\mathbf{k}}}{r}

What does $\dfrac{q}{\partial \mathbf{r}}$ mean?
Did you mean to write this?
$\dfrac{\partial q}{\partial \mathbf{r}}$

However, when \mathbf{r} = \hat{\mathbf{k}} \text{ or } -\hat{\mathbf{k}} I have problems.

¿What can I do?

Isn't k$\cdot$k = 1, and similarly, isn't -k$\cdot$k = -1? What problems are you having?

 

[ebrattr]

Yes, I mean \dfrac{\partial q}{\partial \mathbf{r}} As you know if you do \dfrac{\partial}{\partial \mathbf{r}} both sides you get \dfrac{\partial q}{\partial \mathbf{r}} -\sin q = \dfrac{\hat{\mathbf{k}}}{L}. Solving for \dfrac{\partial q}{\partial \mathbf{r}} you get \dfrac{\partial q}{\partial \mathbf{r}} = -\dfrac{\hat{\mathbf{k}}}{L\sin q} So, when q=0 I'm having troubles.

 

[vanhees71]

You have to transform the gradient to spherical coordinates first. This you get with
\mathrm{d} f=\mathrm{d} \vec{r} \cdot \vec{\nabla} f = (\mathrm{d} r \vec{e}_r \cdot +\mathrm{d} \vartheta r \vec{e}_{\vartheta} + \mathrm{d} \varphi r \sin \vartheta \vec{e}_{\varphi}) \cdot \nabla{f}.
Since \vec{e}_{r}, \vec{e}_{\vartheta}, and \vec{e}_{\varphi} build a complete orthonormal system you immediately read off
\vec{\nabla} \cdot \vec{r}=\vec{e}_r \frac{\partial f}{\partial r} + \vec{e}_{\vartheta} \frac{1}{r} \frac{\partial f}{\partial \vartheta} + \vec{e}_{\varphi} \frac{1}{r \sin \vartheta} \frac{\partial f}{\partial \vartheta}.
For f=\vartheta you thus get
\vec{\nabla} \vartheta=\frac{1}{r} \vec{e}_{\vartheta}.
You can of course do the whole thing in Cartesian coordinates:
f=\arccos \left (\frac{\vec{k} \cdot \vec{r}}{r} \right ).
Using the chain rule gives
\vec{\nabla} f=-\frac{1}{\sin \vartheta} \vec{\nabla} \left (\frac{\vec{k} \cdot \vec{r}}{r} \right )=-\frac{1}{\sin \vartheta} \left (\frac{\vec{k}}{r}-\frac{(\vec{r} \cdot \vec{k}) \vec{r}}{r^3} \right ).
Now we can simplify
\frac{\vec{k}}{r}-\frac{(\vec{r} \cdot \vec{k}) \vec{r}}{r^3}<br /> =\frac{1}{r} \left (\vec{k} - \frac{(\vec{r} \cdot \vec{k})<br /> \vec{r}}{r^2} \right )=-\frac{\sin \vartheta}{r} \begin{pmatrix}<br /> \cos \varphi \cos \vartheta \\ \sin \varphi \cos \vartheta \\ -\sin \vartheta<br /> \end{pmatrix}=-\frac{\vec{e}_{\vartheta} \sin \vartheta}{r}.
Plugging this into the equation above you again get
\vec{\nabla} \vartheta=\frac{1}{r} \vec{e}_{\vartheta}.
As you see you always must use the correct expression of the covariant differential operators to get the right result!