Operatore nabla: differenze tra le versioni

m
a capo in eccesso
m (Annullata la modifica 81097277 di 93.34.229.195 (discussione) vandalismo)
m (a capo in eccesso)
x=r\,\sin\theta\,\cos\phi
</math>
 
 
:<math>
y=r\,\sin\theta\,\sin\phi
</math>
 
 
:<math>
z=r\,\cos\theta
</math>
 
 
Sfruttando la regola di derivazione a catena si può scrivere:
\frac{\partial}{\partial r}=\frac{\partial x}{\partial r}\frac{\partial}{\partial x}+\frac{\partial y}{\partial r}\frac{\partial}{\partial y}+\frac{\partial z}{\partial r}\frac{\partial}{\partial z}
</math>
 
 
:<math>
\frac{\partial}{\partial\theta}=\frac{\partial x}{\partial\theta}\frac{\partial}{\partial x}+\frac{\partial y}{\partial\theta}\frac{\partial}{\partial y}+\frac{\partial z}{\partial\theta}\frac{\partial}{\partial z}
</math>
 
 
:<math>
\frac{\partial}{\partial\phi}=\frac{\partial x}{\partial\phi}\frac{\partial}{\partial x}+\frac{\partial y}{\partial\phi}\frac{\partial}{\partial y}+\frac{\partial z}{\partial\phi}\frac{\partial}{\partial z}
</math>
 
 
la stessa cosa, usando la notazione con matrici e vettori, si scrive:
\end{pmatrix}
</math>
 
 
o anche in forma più compatta:
\nabla_{r}=AB\nabla
</math>
 
 
avendo definito:
\end{pmatrix}
</math>
 
 
:<math>
B^{-1}=B^{T}=b_{1}+b_{2}+b_{3}
</math>
 
 
con
\end{pmatrix}
</math>
 
 
:<math>
b_{2}=\frac{\partial b_{1}}{\partial\theta}
</math>
 
 
:<math>
b_{3}=\frac{1}{\sin\theta}\frac{\partial b_{1}}{\partial\phi}=\frac{1}{\cos\theta}\frac{\partial b_{2}}{\partial\phi}
</math>
 
 
:<math>
b_{i}\cdot b_{j}=\delta_{ij}
</math>
 
 
:<math> b_{i}\times b_{j}=\epsilon_{ijk}b_{k} </math>
 
 
Con quanto suddetto l'operatore gradiente in coordinate polari si esprime:
 
:<math> \nabla=B^{-1}A^{-1}\nabla_{r}= \left(b_{1}\frac{\partial}{\partial r} , b_{2}\frac{1}{r}\frac{\partial}{\partial\theta}, b_{3}\frac{1}{r\,\sin\theta}\frac{\partial}{\partial\phi}\right) </math>
 
 
Si ha:
 
:<math> b_{1}\frac{\partial}{\partial r}\cdot b_{1}\frac{\partial}{\partial r}=b_{1}\cdot\left(\frac{\partial b_{1}}{\partial r}\right)\frac{\partial}{\partial r}+b_{1}\cdot b_{1}\frac{\partial}{\partial r}\frac{\partial}{\partial r}=\frac{\partial^{2}}{\partial r^{2}} </math>
 
 
:<math> b_{1}\frac{\partial}{\partial r}\cdot b_{2}\frac{1}{r}\frac{\partial}{\partial\theta}=b_{1}\cdot b_{2}\frac{\partial}{\partial r}\frac{1}{r}\frac{\partial}{\partial\theta}=0 </math>
 
 
:<math> b_{1}\frac{\partial}{\partial r}\cdot b_{3}\frac{1}{r\,\sin\theta}\frac{\partial}{\partial\phi}=b_{1}\cdot b_{3}\frac{\partial}{\partial r}\frac{1}{r\,\sin\theta}\frac{\partial}{\partial\phi}=0 </math>
b_{2}\frac{1}{r}\frac{\partial}{\partial\theta}\cdot b_{1}\frac{\partial}{\partial r}=b_{2}\cdot\left(\frac{\partial b_{1}}{\partial\theta}\right)\frac{1}{r}\frac{\partial}{\partial r}+b_{2}\cdot b_{1}\frac{1}{r}\frac{\partial}{\partial\theta}\frac{\partial}{\partial r}=\frac{1}{r}\frac{\partial}{\partial r}
</math>
 
 
:<math>
b_{2}\frac{1}{r}\frac{\partial}{\partial\theta}\cdot b_{2}\frac{1}{r}\frac{\partial}{\partial\theta}=b_{2}\cdot\left(\frac{\partial b_{2}}{\partial\theta}\right)\frac{1}{r}\frac{1}{r}\frac{\partial}{\partial\theta}+b_{2}\cdot b_{2}\frac{1}{r}\frac{\partial}{\partial\theta}\frac{1}{r}\frac{\partial}{\partial\theta}=\frac{1}{r^{2}}\frac{\partial^{2}}{\partial\theta^{2}}
</math>
 
 
:<math>
b_{2}\frac{1}{r}\frac{\partial}{\partial\theta}\cdot b_{3}\frac{1}{r\,\sin\theta}\frac{\partial}{\partial\phi}=b_{2}\cdot b_{3}\frac{1}{r}\frac{\partial}{\partial\theta}\frac{1}{r\,\sin\theta}\frac{\partial}{\partial\phi}=0
</math>
 
 
:<math>
b_{3}\frac{1}{r\,\sin\theta}\frac{\partial}{\partial\phi}\cdot b_{1}\frac{\partial}{\partial r}=b_{3}\cdot\left(\frac{1}{\sin\theta}\frac{\partial b_{1}}{\partial\phi}\right)\frac{1}{r}\frac{\partial}{\partial r}+b_{3}\cdot b_{1}\frac{1}{r\,\sin\theta}\frac{\partial}{\partial\phi}\frac{\partial}{\partial r}=\frac{1}{r}\frac{\partial}{\partial r}
</math>
 
 
:<math>
=\frac{\cot\theta}{r^{2}}\frac{\partial}{\partial\theta}
</math>
 
 
:<math>
=\frac{1}{r^{2}\,\sin^{2}\theta}\frac{\partial^{2}}{\partial\phi^{2}}
</math>
 
 
da cui si ricava l'espressione del laplaciano in coordinate polari:
=\frac{1}{r^{2}}\left(\frac{\partial}{\partial r}r^{2}\frac{\partial}{\partial r}+\frac{\partial^{2}}{\partial\theta^{2}}+\frac{1}{\sin^{2}\theta}\frac{\partial^{2}}{\partial\phi^{2}}+\cot\theta\frac{\partial}{\partial\theta}\right)
</math>
 
 
Si trovano facilmente anche gli operatori <math>x\times\nabla</math> (il legendriano)
x\times\nabla=r\, b_{1}\times\left(b_{1}\frac{\partial}{\partial r}+b_{2}\frac{1}{r}\frac{\partial}{\partial\theta}+b_{3}\frac{1}{r\,\sin\theta}\frac{\partial}{\partial\phi}\right)=b_{3}\frac{\partial}{\partial\theta}-b_{2}\frac{1}{\sin\theta}\frac{\partial}{\partial\phi}
</math>
 
 
e calcolando:
b_{2}\frac{1}{\sin\theta}\frac{\partial}{\partial\phi}\cdot b_{2}\frac{1}{\sin\theta}\frac{\partial}{\partial\phi}=b_{2}\cdot\left(\frac{\cos\theta}{\sin\theta}\frac{1}{\cos\theta}\frac{\partial b_{2}}{\partial\phi}\right)\frac{1}{\sin\theta}\frac{\partial}{\partial\phi}+b_{2}\cdot b_{2}\frac{1}{\sin\theta}\frac{\partial}{\partial\phi}\frac{1}{\sin\theta}\frac{\partial}{\partial\phi}=\frac{1}{\sin^{2}\theta}\frac{\partial^{2}}{\partial\phi^{2}}
</math>
 
 
:<math>
b_{2}\frac{1}{\sin\theta}\frac{\partial}{\partial\phi}\cdot b_{3}\frac{\partial}{\partial\theta}=b_{2}\cdot\left(\frac{1}{\sin\theta}\frac{\partial b_{3}}{\partial\phi}\right)\frac{\partial}{\partial\theta}+b_{2}\cdot b_{3}\frac{1}{\sin\theta}\frac{\partial}{\partial\phi}\frac{\partial}{\partial\theta}=-\cot\theta\frac{\partial}{\partial\theta}
</math>
 
 
:<math>
b_{3}\frac{\partial}{\partial\theta}\cdot b_{2}\frac{1}{\sin\theta}\frac{\partial}{\partial\phi}=b_{3}\cdot\left(\frac{\partial b_{2}}{\partial\theta}\right)\frac{1}{\sin\theta}\frac{\partial}{\partial\phi}+b_{3}\cdot b_{2}\frac{\partial}{\partial\theta}\frac{1}{\sin\theta}\frac{\partial}{\partial\phi}=0
</math>
 
 
:<math>
b_{3}\frac{\partial}{\partial\theta}\cdot b_{3}\frac{\partial}{\partial\theta}=b_{3}\cdot\left(\frac{\partial b_{3}}{\partial\theta}\right)\frac{\partial}{\partial\theta}+b_{3}\cdot b_{3}\frac{\partial}{\partial\theta}\frac{\partial}{\partial\theta}=\frac{\partial^{2}}{\partial\theta^{2}}
</math>
 
 
si ottiene:
\left(x\times\nabla\right)^{2}=\frac{\partial^{2}}{\partial\theta^{2}}+\frac{1}{\sin^{2}\theta}\frac{\partial^{2}}{\partial\phi^{2}}+\cot\theta\frac{\partial}{\partial\theta}
</math>
 
 
L'operatore <math>\left(x\times\nabla\right)^{2}</math> rappresenta la parte
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