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Differentiation spherical coordinates

  1. Aug 23, 2013 #1
    Hi ! I'm trying to inverse a mass matrix so I need to do something like this

    [itex] \dfrac{q}{\partial \mathbf{r}} [/itex] where [itex] \cos q = \dfrac{\mathbf{r}\cdot \hat{\mathbf{k}}}{r} [/itex]

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

    ¿What can I do?

    Thanks !
     
  2. jcsd
  3. Aug 23, 2013 #2

    Mark44

    Staff: Mentor

    What does ##\dfrac{q}{\partial \mathbf{r}} ## mean?
    Did you mean to write this?
    ##\dfrac{\partial q}{\partial \mathbf{r}}##
    Isn't k##\cdot##k = 1, and similarly, isn't -k##\cdot##k = -1? What problems are you having?
     
  4. Aug 23, 2013 #3
    Yes, I mean [itex] \dfrac{\partial q}{\partial \mathbf{r}} [/itex] As you know if you do [itex] \dfrac{\partial}{\partial \mathbf{r}} [/itex] both sides you get [itex] \dfrac{\partial q}{\partial \mathbf{r}} -\sin q = \dfrac{\hat{\mathbf{k}}}{L} [/itex]. Solving for [itex] \dfrac{\partial q}{\partial \mathbf{r}} [/itex] you get [itex] \dfrac{\partial q}{\partial \mathbf{r}} = -\dfrac{\hat{\mathbf{k}}}{L\sin q} [/itex] So, when [itex] q=0 [/itex] I'm having troubles.
     
  5. Aug 27, 2013 #4

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

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