4.3 The Operator \(\nabla \)
Recall that if a scalar function \(\Phi (x,y,z)\) is continuously differentiable with respect to \(x,y\) and \(z\), then the gradient of \(\Phi \) written \(\grad \Phi \) is defined as the vector \begin {align*} \grad \Phi & = \frac {\partial \Phi }{\partial x} \textbf {i} + \frac {\partial \Phi }{\partial y}\textbf {j} + \frac {\partial \Phi }{\partial z}\textbf {k}\\\\ & = \Bigg (\textbf {i}\frac {\partial }{\partial x} + \textbf {j} \frac {\partial }{\partial y} + \textbf {k} \frac {\partial }{\partial z}\Bigg ) \Phi \end {align*}
where \(\Bigg (\textbf {i}\dfrac {\partial }{\partial x} + \textbf {j} \dfrac {\partial }{\partial y} + \textbf {k} \dfrac {\partial }{\partial z}\Bigg )\) is called a vector differential operator and denoted by \((\nabla )\) called del or nabla. That is \[\nabla = \Bigg (\textbf {i} \frac {\partial }{\partial x} + \textbf {j}\frac {\partial }{\partial y} + \textbf {k} \frac {\partial }{\partial z}\Bigg )\] Note that \(\nabla \) is an operator and therefore must operate on a stated function \(\Phi (x,y,z)\).
Properties
Let \(f, g\) be differentiable functions and \(\alpha \) a scalar then
- 1.
- \(\nabla \alpha f = \alpha \nabla f\)
- 2.
- \(\nabla (f + g) = \nabla f + \nabla g\)
- 3.
- \( \nabla (f/g) = \dfrac {g\nabla f - f \nabla g}{g^2}\hspace {0.4cm}, \hspace {0.4cm} g\neq 0\)
- 4.
- \(\nabla (fg) = f\nabla g + g \nabla f\)
Proof. \begin {align*} \nabla \bigg (\frac {f}{g}\bigg ) & = \nabla \big (fg^{-1}\big )\\ & = \Bigg (\textbf {i} \frac {\partial }{\partial x} + \textbf {j} \frac {\partial }{\partial y} + \textbf {k}\frac {\partial }{\partial z}\Bigg ) \big (fg^{-1}\big )\\\\ & = \textbf {i}\frac {\partial }{\partial x}\big (fg^{-1}\big ) + \textbf {j}\frac {\partial }{\partial y}\big (fg^{-1}\big ) + \textbf {k} \frac {\partial }{\partial z}\big (fg^{-1}\big )\\\\ & = \textbf {i}\Bigg (\frac {\partial f}{\partial x}g^{-1} + f\bigg \{-g^{-2}\frac {\partial g}{\partial x}\bigg \}\Bigg ) + \textbf {j}\Bigg (\frac {\partial f}{\partial y}g^{-1} + \bigg \{-g^{-2}\frac {\partial g}{\partial y}\bigg \}f\Bigg ) + \textbf {k} \Bigg (\frac {\partial f}{\partial z}g^{-1} + f\bigg \{-g^{-2}\frac {\partial g}{\partial z}\bigg \}\Bigg )\\\\ & = \textbf {i}\Bigg (\frac {1}{g}\frac {\partial f}{\partial x}-\frac {f}{g^2}\frac {\partial g}{\partial x}\Bigg ) + \textbf {j}\Bigg (\frac {1}{g}\frac {\partial f}{\partial y}-\frac {f}{g^2}\frac {\partial g}{\partial y}\Bigg ) + \textbf {k}\Bigg (\frac {1}{g}\frac {\partial f}{\partial z}- \frac {f}{g^2}\frac {\partial g}{\partial z}\Bigg )\\\\ & = \frac {1}{g}\Bigg ( \frac {\partial f}{\partial x}\textbf {i} + \frac {\partial f}{\partial y}\textbf {j} + \frac {\partial f}{\partial z}\textbf {k}\Bigg ) - \frac {f}{g^2}\Bigg (\frac {\partial g}{\partial x}\textbf {i} + \frac {\partial g}{\partial y}\textbf {j} + \frac {\partial g}{\partial z}\textbf {k}\Bigg )\\\\ & = \frac {1}{g}\nabla f - \frac {f}{g^2}\nabla g\\\\ & = \frac {g \nabla f - f \nabla g }{g^2}\hspace {0.6cm} \text {as required}\\\\\\ \end {align*} □
Prove that \(\nabla r^{n} = n r^{n-2}\overline {r}\), where \(\overline {r} = x\textbf {i} + y\textbf {j} + z\textbf {k}\) and \(r = \left |\overline {r}\\\right | = \sqrt {x^2 + y^2 + z^2}\).
Solution. By definition \begin {align*} \nabla r^n & = \Bigg ( \textbf {i} \frac {\partial }{\partial x} + \textbf {j} \frac {\partial }{\partial y} + \textbf {k}\frac {\partial }{\partial z}\Bigg ) r^n = \textbf {i} \frac {\partial }{\partial x}r^n + \textbf {j} \frac {\partial }{\partial y}r^n + \textbf {k}\frac {\partial }{\partial z}r^n\\ \end {align*}
\[ \frac {\partial }{\partial x}r^n = n r^{n-1} \frac {\partial r}{\partial x}\hspace {0.5cm}\text {But}\]
\begin {align*} \frac {\partial r}{\partial x} & = \frac {\partial }{\partial x} (x^2 + y^2 + z^2)^{1/2}\\ & = \frac {1}{2}\cdot 2x\cdot (x^2 + y^2 + z^2)^{-1/2}\\ \implies \hspace {0.5cm}\frac {\partial r}{\partial x} & = \frac {x}{r}\\ \end {align*}
\[\text {Similary}\hspace {0.5cm} \frac {\partial r}{\partial y} = \frac {y}{r}\hspace {0.5cm} \text {and} \hspace {0.5cm} \frac {\partial r}{\partial z} = \frac {z}{r}\]
\[\frac {\partial r^n}{\partial x} = nr^{n-1} \frac {x}{r} = nr^{n-2}x\hspace {0.4cm},\hspace {1cm} \frac {\partial r^n}{\partial y} = nr^{n-2}y\hspace {0.5cm},\hspace {1cm} \frac {\partial r^n}{\partial z}= nr^{n-2}z\]
\begin {align*} \text {But} \hspace {0.5cm} \nabla r^n & = \textbf {i}\frac {\partial r^n}{\partial x} + \textbf {j} \frac {\partial r^n}{\partial y} + \textbf {k} \frac {\partial r^n}{\partial z}\\\\ & = nr^{n-2}x \textbf {i} + nr^{n-2} y \textbf {j} + nr^{n-2} z\textbf {k}\\\\ & = nr^{n-2} (x\textbf {i} + y \textbf {j} + z\textbf {k})\\ \implies \hspace {0.5cm} \nabla r^n & = nr^{n-2}\overline {r}\\\\ \end {align*}
\[\frac {\partial }{\partial x} r^n = \frac {\partial }{\partial x} (x^2 + y^2 + z^2 )^{1/2}\]
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