L'operador nabla i les seves propietats

L'operador nabla, designat per $\vec{\nabla}$, és un operador diferencial vectorial³ que, en coordenades cartesianes, té la forma, a $\ensuremath{\mathbbm{ R}^3}$,

$$\vec{\nabla}$$ = $$\left(\vphantom{\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z}}\right.$$$${\frac{{\partial}}{{\partial x}}}$$,$${\frac{{\partial}}{{\partial y}}}$$,$${\frac{{\partial}}{{\partial z}}}$$$$\left.\vphantom{\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z}}\right)$$.

Cada una de les components és un operador de derivació i, com a tal, obeeix les regles de derivació. Quan $\vec{\nabla}$ actua sobre funcions escalars o vectorials s'obtenen diverses quantitats importants:

• gradient d'una funció escalar f (x, y, z), és un camp vectorial donat per
  

  $$\vec{\nabla} \left(\vphantom{\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z}}\right. {\frac{{\partial}}{{\partial x}}} {\frac{{\partial}}{{\partial y}}} {\frac{{\partial}}{{\partial z}}} \left.\vphantom{\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z}}\right)$$ =

  $$\left(\vphantom{\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}}\right. {\frac{{\partial f}}{{\partial x}}} {\frac{{\partial f}}{{\partial y}}} {\frac{{\partial f}}{{\partial z}}} \left.\vphantom{\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}}\right)$$ =

  $$\equiv$$ grad f.

  
  En un punt (x, y, z) donat, $\vec{\nabla}$f (x, y, z) apunta en la direcció en el que el valor de f aumenta el més ràpidament possible.
• divergència d'una funció vectorial $\vec{F}$(x, y, z), és una funció escalar, producte escalar de $\vec{\nabla}$ i $\vec{F}$,
  

  $$\vec{\nabla} \vec{F} \left(\vphantom{\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z}}\right. {\frac{{\partial}}{{\partial x}}} {\frac{{\partial}}{{\partial y}}} {\frac{{\partial}}{{\partial z}}} \left.\vphantom{\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z}}\right)$$ =

  $${\frac{{\partial F_x}}{{\partial x}}} {\frac{{\partial F_y}}{{\partial y}}} {\frac{{\partial F_z}}{{\partial z}}}$$ =

  $$\equiv \vec{F}$$

  
  
• rotacional d'una funció vectorial $\vec{F}$(x, y, z), és un camp vectorial, producte vectorial de $\vec{\nabla}$ i $\vec{F}$,
  

  $$\vec{\nabla} \vec{F} \left(\vphantom{\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z}}\right. {\frac{{\partial}}{{\partial x}}} {\frac{{\partial}}{{\partial y}}} {\frac{{\partial}}{{\partial z}}} \left.\vphantom{\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z}}\right)$$ =

  $$\left(\vphantom{ \frac{\partial F_z}{\partial y}-\frac{\partial F... ...al x}, \frac{\partial F_y}{\partial x}-\frac{\partial F_x}{\partial y} }\right. {\frac{{\partial F_z}}{{\partial y}}} {\frac{{\partial F_y}}{{\partial z}}} {\frac{{\partial F_x}}{{\partial z}}} {\frac{{\partial F_z}}{{\partial x}}} {\frac{{\partial F_y}}{{\partial x}}} {\frac{{\partial F_x}}{{\partial y}}} \left.\vphantom{ \frac{\partial F_z}{\partial y}-\frac{\partial F... ...al x}, \frac{\partial F_y}{\partial x}-\frac{\partial F_x}{\partial y} }\right)$$ =

  $$\equiv \vec{F}$$

  
  

• Laplacià d'una funció escalar f (x, y, z), és una funció escalar, la divergència del gradient de f,
  

  $$\nabla^{2}_{} \vec{\nabla} \vec{\nabla} \vec{\nabla}$$ =

  $$\left(\vphantom{\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z}}\right. {\frac{{\partial}}{{\partial x}}} {\frac{{\partial}}{{\partial y}}} {\frac{{\partial}}{{\partial z}}} \left.\vphantom{\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z}}\right) \left(\vphantom{\frac{\partial f}{\partial x}, \frac{\partial y}{\partial y}, \frac{\partial f}{\partial z}}\right. {\frac{{\partial f}}{{\partial x}}} {\frac{{\partial y}}{{\partial y}}} {\frac{{\partial f}}{{\partial z}}} \left.\vphantom{\frac{\partial f}{\partial x}, \frac{\partial y}{\partial y}, \frac{\partial f}{\partial z}}\right)$$ =

  $${\frac{{\partial^2 f}}{{\partial x^2}}} {\frac{{\partial^2 f}}{{\partial y^2}}} {\frac{{\partial^2 f}}{{\partial z^2}}}$$ =

  
  
• Laplacià d'un camp vectorial $\vec{F}$, és un camp vectorial format pels Laplacians de les components,
  $$\nabla^{2}_{}$$$$\vec{F}$$ = $$\left(\vphantom{ \nabla^2 F_x,\nabla^2 F_y,\nabla^2 F_z}\right.$$$$\nabla^{2}_{}$$F_x,$$\nabla^{2}_{}$$F_y,$$\nabla^{2}_{}$$F_z$$\left.\vphantom{ \nabla^2 F_x,\nabla^2 F_y,\nabla^2 F_z}\right)$$.

Evitarem les notacions grad f, div $\vec{F}$ i rot $\vec{F}$ ja que utilitzen de forma inconsistent la notació vectorial.

El fet que $\vec{\nabla}$ sigui alhora un vector i un operador de derivació fa que tingui les propietats d'ambdós tipus d'objectes. Les més importants són les següents:

1. $\vec{\nabla}$(f + g) = $\vec{\nabla}$f + $\vec{\nabla}$g,
2. $\vec{\nabla}$ ^. ($\vec{F}$ + $\vec{G}$) = $\vec{\nabla}$ ^. $\vec{F}$ + $\vec{\nabla}$ ^. $\vec{G}$,
3. $\vec{\nabla}$×($\vec{F}$ + $\vec{G}$) = $\vec{\nabla}$×$\vec{F}$ + $\vec{\nabla}$×$\vec{G}$,
4. $\vec{\nabla}$ ^. (f$\vec{F}$) = ($\vec{\nabla}$f ) ^. $\vec{F}$ + f$\vec{\nabla}$ ^. $\vec{F}$,
5. $\vec{\nabla}$×(f$\vec{F}$) = ($\vec{\nabla}$f )×$\vec{F}$ + f$\vec{\nabla}$×$\vec{F}$,
6. $\vec{\nabla}$ ^. ($\vec{F}$×$\vec{G}$) = $\vec{G}$ ^. ($\vec{\nabla}$×$\vec{F}$) - $\vec{F}$ ^. ($\vec{\nabla}$×$\vec{G}$),
7. $\vec{\nabla}$×($\vec{F}$×$\vec{G}$) = ($\vec{G}$ ^. $\vec{\nabla}$)$\vec{F}$ - $\vec{G}$($\vec{\nabla}$ ^. $\vec{F}$) - ($\vec{F}$ ^. $\vec{\nabla}$)$\vec{G}$ + $\vec{F}$($\vec{\nabla}$ ^. $\vec{G}$),
8. $\vec{\nabla}$($\vec{F}$ ^. $\vec{G}$) = ($\vec{G}$ ^. $\vec{\nabla}$)$\vec{F}$ + ($\vec{F}$ ^. $\vec{\nabla}$)$\vec{G}$ + $\vec{G}$×($\vec{\nabla}$×$\vec{F}$) + $\vec{F}$×($\vec{\nabla}$×$\vec{G}$),
9. $\vec{\nabla}$×($\vec{\nabla}$f )= 0,
10. $\vec{\nabla}$ ^. ($\vec{\nabla}$×$\vec{F}$) = 0,
11. $\vec{\nabla}$×($\vec{\nabla}$×$\vec{F}$) = $\vec{\nabla}$($\vec{\nabla}$ ^. $\vec{F}$) - $\nabla^{2}_{}$$\vec{F}$.

La demostració d'aquestes igualtats és un simple càlcul,⁴ que es deixa al lector, el qual podrà, si ho intenta, descubrir el significat d'algunes de les expressions dels membres drets (significat que, d'altra banda, és el més obvi). Aquest càlcul, però, es simplifica, en els casos més complicats, si s'utilitza l'anomenat conveni de sumació d'Einstein, juntament amb una representació intel-l ^. ligent dels vectors, de $\vec{\nabla}$ i del producte vectorial, tal com es mostra a l'apèndix.

Per acabar aquesta secció, parlarem una mica de canvis de coordenades. En concret, volem saber com s'expressa el Laplacià en coordenades no cartesianes, qüestió aquesta força útil, per exemple per a resoldre problemes d'electrostàtica.

Per concretar, imaginem que volem treballar amb coordenades cilındriques a $\ensuremath{\mathbbm{ R}^3}$ (o coordenades polars a $\ensuremath{\mathbbm{ R}^2}$):

$$\left\{\vphantom{\begin{array}{ccl} x&=&r\cos\phi\\ y&=&r\sin\phi\\ z&=&z \end{array} }\right.$$$$\begin{array}{ccl} x&=&r\cos\phi\\ y&=&r\sin\phi\\ z&=&z \end{array}$$$$\begin{array}{ccl} r&\in&[0,+\infty)\\ \phi&\in&[0,2\pi)\\ z&\in&(-\infty,+\infty) \end{array}$$

Sigui ara una funció f (r,$\phi$, z) de la que, a través de la dependència de les coordenades cilındriques de les cartesianes, volem calcular

$${\frac{{\partial f}}{{\partial x}}}$$(r,$$\phi$$, z).

Aplicant la regla de la cadena tenim

$${\frac{{\partial f}}{{\partial x}}}$$(r,$$\phi$$, z) = $${\frac{{\partial r}}{{\partial x}}}$$$${\frac{{\partial f}}{{\partial r}}}$$ + $${\frac{{\partial \phi}}{{\partial x}}}$$$${\frac{{\partial f}}{{\partial \phi}}}$$ + $${\frac{{\partial z}}{{\partial x}}}$$$${\frac{{\partial f}}{{\partial z}}}$$,

i anàlogament si volem calcular ${\frac{{\partial f}}{{\partial y}}}$ i ${\frac{{\partial f}}{{\partial z}}}$. Expressant-ho tot vectorialment tindrem

$$\left(\vphantom{ \begin{array}{c} \frac{\partial f}{\partial x}... ...artial f}{\partial y}\\ \frac{\partial f}{\partial z} \end{array} }\right.$$$$\begin{array}{c} \frac{\partial f}{\partial x}\\ \frac{\partial f}{\partial y}\\ \frac{\partial f}{\partial z} \end{array}$$$$\left.\vphantom{ \begin{array}{c} \frac{\partial f}{\partial x}... ...artial f}{\partial y}\\ \frac{\partial f}{\partial z} \end{array} }\right)$$ = $$\left(\vphantom{ \begin{array}{ccc} \frac{\partial r}{\partial ... ...rtial\phi}{\partial z} & \frac{\partial z}{\partial z} \end{array} }\right.$$$$\begin{array}{ccc} \frac{\partial r}{\partial x} & \frac{\partia... ... \frac{\partial\phi}{\partial z} & \frac{\partial z}{\partial z} \end{array}$$$$\left.\vphantom{ \begin{array}{ccc} \frac{\partial r}{\partial ... ...rtial\phi}{\partial z} & \frac{\partial z}{\partial z} \end{array} }\right)$$$$\left(\vphantom{ \begin{array}{c} \frac{\partial f}{\partial r}... ...ial f}{\partial \phi}\\ \frac{\partial f}{\partial z} \end{array} }\right.$$$$\begin{array}{c} \frac{\partial f}{\partial r}\\ \frac{\partial f}{\partial \phi}\\ \frac{\partial f}{\partial z} \end{array}$$$$\left.\vphantom{ \begin{array}{c} \frac{\partial f}{\partial r}... ...ial f}{\partial \phi}\\ \frac{\partial f}{\partial z} \end{array} }\right)$$

Això es pot escriure, amb notació matricial compacte i autoexplicativa, com

$$\vec{\nabla}_{{(x,y,z)}}^{}$$f = M$$\vec{\nabla}_{{(r,\phi,z)}}^{}$$f.

Pensant una mica, es veu que la matriu M és

M = $$\left(\vphantom{\left(\frac{\partial(x,y,z)}{\partial(r,\phi,z)}\right)^{-1} }\right.$$$$\left(\vphantom{\frac{\partial(x,y,z)}{\partial(r,\phi,z)}}\right.$$$${\frac{{\partial(x,y,z)}}{{\partial(r,\phi,z)}}}$$$$\left.\vphantom{\frac{\partial(x,y,z)}{\partial(r,\phi,z)}}\right)^{{-1}}_{}$$$$\left.\vphantom{\left(\frac{\partial(x,y,z)}{\partial(r,\phi,z)}\right)^{-1} }\right)^{{T}}_{}$$,

on

$${\frac{{\partial(x,y,z)}}{{\partial(r,\phi,z)}}}$$

és la derivada de l'aplicació que expressa les coordenades cartesianes en termes de les cilındriques:

$${\frac{{\partial(x,y,z)}}{{\partial(r,\phi,z)}}}$$ = $$\left(\vphantom{\begin{array}{ccc} \cos\phi & -r\sin\phi & 0 \\ \sin\phi & r\cos\phi & 0 \\ 0 & 0 & 1 \end{array} }\right.$$$$\begin{array}{ccc} \cos\phi & -r\sin\phi & 0 \\ \sin\phi & r\cos\phi & 0 \\ 0 & 0 & 1 \end{array}$$$$\left.\vphantom{\begin{array}{ccc} \cos\phi & -r\sin\phi & 0 \\ \sin\phi & r\cos\phi & 0 \\ 0 & 0 & 1 \end{array} }\right)$$.

Hom obté aixı

M = $$\left(\vphantom{\begin{array}{ccc} \cos\phi & -\frac{1}{r}\sin\... ... \sin\phi & \frac{1}{r}\cos\phi & 0 \\ 0 & 0 & 1 \end{array} }\right.$$$$\begin{array}{ccc} \cos\phi & -\frac{1}{r}\sin\phi & 0 \\ \sin\phi & \frac{1}{r}\cos\phi & 0 \\ 0 & 0 & 1 \end{array}$$$$\left.\vphantom{\begin{array}{ccc} \cos\phi & -\frac{1}{r}\sin\... ... \sin\phi & \frac{1}{r}\cos\phi & 0 \\ 0 & 0 & 1 \end{array} }\right)$$.

La relació $\vec{\nabla}_{{(x,y,z)}}^{}$f = M$\vec{\nabla}_{{(r,\phi,z)}}^{}$f pot llegir-se, prescindint de la funció f, com

$$\vec{\nabla}_{{(x,y,z)}}^{}$$ = M$$\vec{\nabla}_{{(r,\phi,z)}}^{}$$,

que expressa de manera compacta com canvien les derivades parcials al canviar de sistema de coordenades. Per a un sistema de coordenades (u, v, w) qualsevol es té

$$\vec{\nabla}_{{(x,y,z)}}^{}$$ = M$$\vec{\nabla}_{{(u,v,w)}}^{}$$,

amb

M = $$\left(\vphantom{\left(\frac{\partial(x,y,z)}{\partial(u,v,w)}\right)^{-1} }\right.$$$$\left(\vphantom{\frac{\partial(x,y,z)}{\partial(u,v,w)}}\right.$$$${\frac{{\partial(x,y,z)}}{{\partial(u,v,w)}}}$$$$\left.\vphantom{\frac{\partial(x,y,z)}{\partial(u,v,w)}}\right)^{{-1}}_{}$$$$\left.\vphantom{\left(\frac{\partial(x,y,z)}{\partial(u,v,w)}\right)^{-1} }\right)^{{T}}_{}$$.

Tornant al nostre problema, serà,

$$\nabla_{{(x,y,z)}}^{2}$$f = $$\vec{\nabla}_{{(x,y,z)}}^{}$$ ^. $$\vec{\nabla}_{{(x,y,z)}}^{}$$f = (M$$\vec{\nabla}_{{(r,\phi,z)}}^{}$$) ^. (M$$\vec{\nabla}_{{(r,\phi,z)}}^{}$$f ).

S'ha de tenir en compte que les derivades que hi ha en el primer $\vec{\nabla}_{{(r,\phi,z)}}^{}$ actuen tant sobre el segon M com sobre f. Hom té

M$$\vec{\nabla}_{{(r,\phi,z)}}^{}$$ = $$\left(\vphantom{ \cos\phi\frac{\partial}{\partial r}-\frac{1}{r}... ...\cos\phi \frac{\partial}{\partial\phi}, \frac{\partial}{\partial z} }\right.$$cos$$\phi$$$${\frac{{\partial}}{{\partial r}}}$$ - $${\frac{{1}}{{r}}}$$sin$$\phi$$$${\frac{{\partial}}{{\partial\phi}}}$$, sin$$\phi$$$${\frac{{\partial}}{{\partial r}}}$$ + $${\frac{{1}}{{r}}}$$cos$$\phi$$$${\frac{{\partial}}{{\partial\phi}}}$$,$${\frac{{\partial}}{{\partial z}}}$$$$\left.\vphantom{ \cos\phi\frac{\partial}{\partial r}-\frac{1}{r}... ...\cos\phi \frac{\partial}{\partial\phi}, \frac{\partial}{\partial z} }\right)$$,

M$$\vec{\nabla}_{{(r,\phi,z)}}^{}$$f = $$\left(\vphantom{ \cos\phi\frac{\partial f}{\partial r}-\frac{1}{... ...\phi \frac{\partial f}{\partial\phi}, \frac{\partial f}{\partial z} }\right.$$cos$$\phi$$$${\frac{{\partial f}}{{\partial r}}}$$ - $${\frac{{1}}{{r}}}$$sin$$\phi$$$${\frac{{\partial f}}{{\partial \phi}}}$$, sin$$\phi$$$${\frac{{\partial f}}{{\partial r}}}$$ + $${\frac{{1}}{{r}}}$$cos$$\phi$$$${\frac{{\partial f}}{{\partial \phi}}}$$,$${\frac{{\partial f}}{{\partial z}}}$$$$\left.\vphantom{ \cos\phi\frac{\partial f}{\partial r}-\frac{1}{... ...\phi \frac{\partial f}{\partial\phi}, \frac{\partial f}{\partial z} }\right)$$.

Multiplicant escalarment aquests dos vectors s'arriba a que

$$\nabla_{{(x,y,z)}}^{2}$$f = $${\frac{{\partial^2 f}}{{\partial r^2}}}$$ + $${\frac{{1}}{{r^2}}}$$$${\frac{{\partial^2 f}}{{\partial\phi^2}}}$$ + $${\frac{{1}}{{r}}}$$$${\frac{{\partial f}}{{\partial r}}}$$ + $${\frac{{\partial^2 f}}{{\partial z^2}}}$$.

El membre de la dreta és el que s'anomena el Laplacià en coordenades cilındriques. Simbòlicament, prescindint de la funció f,

$$\nabla_{{(r,\phi,z)}}^{2} {\frac{{\partial^2 }}{{\partial r^2}}} {\frac{{1}}{{r^2}}} {\frac{{\partial^2 }}{{\partial\phi^2}}} {\frac{{1}}{{r}}} {\frac{{\partial}}{{\partial r}}} {\frac{{\partial^2 }}{{\partial z^2}}}$$ =

$${\frac{{1}}{{r}}} {\frac{{\partial}}{{\partial r}}} \left(\vphantom{r\frac{\partial}{\partial r}}\right. {\frac{{\partial}}{{\partial r}}} \left.\vphantom{r\frac{\partial}{\partial r}}\right) {\frac{{1}}{{r^2}}} {\frac{{\partial^2 }}{{\partial\phi^2}}} {\frac{{\partial^2 }}{{\partial z^2}}}$$ =

Exemple 16   Demostreu que el Laplacià en coordenades esfèriques

$$\left\{\vphantom{\begin{array}{ccl} x&=&\rho\sin\theta\cos\phi\\ y&=&\rho\sin\theta\sin\phi\\ z&=&\rho\cos\theta \end{array} }\right.$$$$\begin{array}{ccl} x&=&\rho\sin\theta\cos\phi\\ y&=&\rho\sin\theta\sin\phi\\ z&=&\rho\cos\theta \end{array}$$$$\begin{array}{ccl} \rho&\in&[0,+\infty)\\ \theta&\in&[0,\pi/2]\\ \phi&\in&[0,2\pi) \end{array}$$

val

$$\nabla_{{(\rho,\theta,\phi)}}^{2}$$ = $${\frac{{1}}{{\rho^2}}}$$$${\frac{{\partial}}{{\partial\rho}}}$$$$\left(\vphantom{ \rho^2\frac{\partial}{\partial\rho}}\right.$$$$\rho^{2}_{}$$$${\frac{{\partial}}{{\partial\rho}}}$$$$\left.\vphantom{ \rho^2\frac{\partial}{\partial\rho}}\right)$$ + $${\frac{{1}}{{\rho^2\sin\theta}}}$$$${\frac{{\partial}}{{\partial\theta}}}$$$$\left(\vphantom{\sin\theta\frac{\partial}{\partial\theta}}\right.$$sin$$\theta$$$${\frac{{\partial}}{{\partial\theta}}}$$$$\left.\vphantom{\sin\theta\frac{\partial}{\partial\theta}}\right)$$ + $${\frac{{1}}{{\rho^2 \sin^2\theta}}}$$$${\frac{{\partial^2 }}{{\partial\phi^2}}}$$.