Spherical Polar

• Vectors:
  $$\vr = r\ \hr$$

  But Note Well: $\hr$ is now a function of $(\theta,\phi)$ ! Similarly:

  $$\vA = A_r\ \hr$$

  with $\hr$ a function of the angles $(\theta,\phi)$ that define the direction of $\vA$ . Specifically:

• Unit vectors (relative to Cartesian $\hx,\hy,\hz$ :
  $$\hr = \sin\theta\cos\phi\ \hx + \sin\theta\sin\phi\ \hy$$
  $$\htheta = \cos\theta\cos\phi\ \hx + \cos\theta\sin\phi \hy - \sin\theta \hz$$
  $$\hphi = -\sin\phi \hx + \cos\phi \hy$$
  The fact that $\hr(\theta,\phi), \htheta(\theta,\phi), \hphi(\phi)$ complicates all of the spherical coordinate vector differential forms, although we indicate above a different, more direct way of evaluating them than applying derivatives to the unit vectors themselves before completing the tensor construction of the vector differential operators.
• Direct Length
  $$d\vell = dr\ \hr + rd\theta\ \htheta + r\sin\theta\ d\phi\ \hphi$$

• Directed Area
  $$dA = r^2 \sin\theta\ d\theta\ d\phi$$
  $$dA\ \hn = d\va = r^2 \sin\theta\ d\theta\ d\phi\ \hr = dA\ \hr$$
  And again, there are many other possible $d\va$ 's, for example, for the bounding surface for hemispherical volume where one piece of it would be a circular surface with an normal like (for example) $\hz$ . This is precisely the surface needed for certain problems you will tackle this semester.
• Volume Element
  $$dV = d\tau = r^2 \sin\theta\ d\theta\ d\phi dr = d\va \cdot dr\hr$$

• Gradient:
  $$\grad f = \partialdiv{f}{r} \hr + \frac{1}{r} \partialdiv{f}{\theta} \htheta + \frac{1}{r\sin\theta} \partialdiv{f}{\phi}\hphi$$

• Divergence; The divergence is constructed by the same argument that proves the divergence theorem in a general curvilinear coordinate system, or alternatively picks up pieces from $\grad \hr$ , etc, hence its complexity:
  $$\deldot \vB = \frac{1}{r^2}\partialdiv{\left(r^2B_r\right)}{r} + ... ...right)}{\theta} + \frac{1}{r\sin\theta} \partialdiv{\left(B_\phi\right)}{\phi}$$

  Note that this follows from:

  $$\deldot \vB = \frac{1}{(fgh)} \left[\left(\partialdiv{(gh\ B_u)}{... ...rtialdiv{(fh\ B_v)}{v}\right) + \left(\partialdiv{(fg\ B_w)}{w}\right) \right]$$

  with $u = r$ , $v = \theta$ , $w = \phi$ , and $f = 1$ , $g = r$ , $h = r\sin\theta$ . Take the contribution from $r$ :

  $$\frac{1}{r^2\sin\theta} \partialdiv{(r^2\sin\theta B_r)}{r} = \frac{1}{r^2} \partialdiv{(r^2 B_r)}{r}$$

  because $\sin\theta$ does not depend on $r$ , similarly for the other two pieces.

• Curl The curl is evaluated in exactly the same way from the expression above, but it ends up being much more complex:
  $$\curl{B} = \frac{1}{r\sin\theta} \left[\partialdiv{(\sin\theta B_... ...{r} \left[ \partialdiv{(rB_\theta)}{r} - \partialdiv{B_r}{\theta}\right] \hphi$$

• Laplacian The Laplacian follows by applying the divergence rule to the gradient rule and simplifying:
  $$\lapl f = \frac{1}{r^2} \partialdiv{}{r}\left( r^2\partialdiv{f}{... ...2\sin\theta}\partialdiv{}{\theta}\left(\sin\theta\partialdiv{f}{\theta}\right)$$