Vector Calculus
Below, \(\mathbf{A}\), \(\mathbf{B}\) \(\mathbf{C}\), and \(\mathbf{D}\) are vector fields (or constant vectors) and \(\phi\) and \(\psi\)
are scalar functions. Then:
(473)\[\begin{align*}
\mathbf{A}\cdot (\mathbf{B}\times\mathbf{C}) = \mathbf{B}\cdot (\mathbf{C}\times \mathbf{A}) =
\mathbf{C} \cdot ( \mathbf{A}\times \mathbf{B}).
\end{align*}\]
(474)\[\begin{align*}
\mathbf{A}\times (\mathbf{B}\times \mathbf{C}) = (\mathbf{A}\cdot \mathbf{C}) \mathbf{B}
- (\mathbf{A}\cdot \mathbf{B}) \mathbf{C}.
\end{align*}\]
(475)\[\begin{align*}
(\mathbf{A}\times \mathbf{B})\cdot (\mathbf{C} \times \mathbf{D})
= (\mathbf{A}\cdot \mathbf{C}) (\mathbf{B}\cdot \mathbf{D})
- (\mathbf{A}\cdot \mathbf{D}) (\mathbf{B}\cdot \mathbf{C})
\end{align*}\]
(476)\[\begin{align*}
\mathbf{\nabla} \cdot (\phi \mathbf{A}) = \phi \mathbf{\nabla}\cdot \mathbf{A} + \nabla \phi \cdot \mathbf{A}.
\end{align*}\]
(477)\[\begin{align*}
\mathbf{\nabla} \times (\phi \mathbf{A}) = \phi \mathbf{\nabla} \times \mathbf{A} + \nabla \phi \times \mathbf{A}.
\end{align*}\]
(478)\[\begin{align*}
\mathbf{\nabla} \cdot (\mathbf{A}\times \mathbf{B}) = - \mathbf{A}\cdot \mathbf{\nabla} \times \mathbf{B} +
\mathbf{B} \cdot \mathbf{\nabla} \times \mathbf{A}.
\end{align*}\]
(479)\[\begin{align*}
\mathbf{\nabla} \times (\mathbf{A}\times \mathbf{B}) = - (\mathbf{A}\cdot \mathbf{\nabla}) \mathbf{B}
+ \mathbf{A} \mathbf{\nabla} \cdot \mathbf{B}
+ ( \mathbf{B} \cdot \mathbf{\nabla}) \mathbf{A}
- \mathbf{B} \mathbf{\nabla} \cdot \mathbf{A}.
\end{align*}\]
(480)\[\begin{align*}
\mathbf{\nabla} (\mathbf{A}\cdot \mathbf{B}) = (\mathbf{A}\cdot \mathbf{\nabla}) \mathbf{B}
+ \mathbf{A} \times \mathbf{\nabla} \times \mathbf{B}
+ ( \mathbf{B} \cdot \mathbf{\nabla}) \mathbf{A}
+ \mathbf{B} \times \mathbf{\nabla} \times \mathbf{A}.
\end{align*}\]
(481)\[\begin{align*}
\mathbf{\nabla} \cdot \mathbf{\nabla} \phi = \Delta \phi,
\end{align*}\]
where \(\Delta = \partial^2/\partial x^2 + \partial^2/\partial y^2
+
\partial^2/\partial z^2\) provided that \((x,y,z)\) is an orthonormal basis.
(482)\[\begin{align*}
\mathbf{\nabla} \times \mathbf{\nabla} \times \mathbf{A}= - \Delta \mathbf{A}
+ \mathbf{\nabla} \mathbf{\nabla} \cdot \mathbf{A}.
\end{align*}\]
Remark: The last formula is only valid in a Cartesian coordinate system. This means that the vector field \(\mathbf{A}\)
must be decomposed on the Cartesian basis and the derivatives must be computed with respect to the corresponding Cartesian coordinates and then \(\Delta \mathbf{A}\) must be interpreted component-by-component: \(\Delta \mathbf{A}\)= (\(\Delta A_x\), \(\Delta
A_y\), \(\Delta A_z)^{T}\), where \(A_x, A_y, A_z\) are components with respect to the Cartesian basis. The formula does not hold in cylindrical or spherical coordinates!.
(483)\[\begin{align*}
\mathbf{\nabla} \times \mathbf{\nabla} \phi =0.
\end{align*}\]
(484)\[\begin{align*}
\mathbf{\nabla} \cdot (\mathbf{\nabla} \times \mathbf{A}) =0.
\end{align*}\]
In addition, the following integral theorems apply
(\(V\) is a volume with surface area \(S\) and outward unit normal
\(\mathbf{n}\)).
Gauss’s Theorem (or divergence theorem):
(485)\[\begin{align*}
\int\!\!\int\!\!\int_{V} \, \mathbf{\nabla} \cdot \mathbf{A} \, dV =
\int\!\!\int_{S} \, \mathbf{A}\cdot \mathbf{n} \, dS.
\end{align*}\]
Apply this to the vector field \(\mathbf{A} = \phi
\mathbf{\nabla} \psi\). Because of (476) and (481) holds
\(\mathbf{\nabla}\cdot \mathbf{A}= \phi \Delta \psi + \mathbf{\nabla} \cdot \mathbf{\nabla}
\psi\) and thus (Green’s Theorem):
(486)\[\begin{align*}
\int\!\!\int\!\!\int_{V}\, \phi \Delta \psi + \mathbf{\nabla} \phi \cdot \mathbf{\nabla} \psi \, dV=
\int\!\!\int_{S} \, \phi \frac{\partial \psi }{\partial n} \, dS.
\end{align*}\]
By subtracting the analogous relation from (486) with
\(\phi\) and \(\psi\) interchanged, one gets:
(487)\[\begin{align*}
\int\!\!\int\!\!\int_{V} \, \phi \Delta \psi - \psi \Delta \phi \, dV=
\int\!\!\int_{S} \, \phi \frac{\partial \psi }{\partial n} -
\psi \frac{\partial \phi}{\partial n} \, dS.
\end{align*}\]
By using (478) and Gauss’s theorem it follows furthermore that
(488)\[\begin{split}\begin{align*}
\int\!\!\int\!\!\int_{V} \mathbf{B}\cdot \mathbf{\nabla} \times \mathbf{A} dV-
\int\!\!\int\!\!\int_{V} \mathbf{A}\cdot \mathbf{\nabla} \times \mathbf{B} dV&=
\int\!\!\int_{S} \, \mathbf{n}\dot (\mathbf{A}\times \mathbf{B}) \, dS \\
&=
\int\!\!\int_{S} \, (\mathbf{n} \times \mathbf{A})\cdot \mathbf{B}) \, dS \\
&=
\int\!\!\int_{S} \,\mathbf{A}\cdot (\mathbf{n} \times \mathbf{B}) \, dS,
\end{align*}\end{split}\]
where in the right-hand side we used
(473).
Stokes’ Theorem (\(S\) is a possibly curved surface with contour \(C\)):
(489)\[\begin{align*}
\int\!\!\int_{S}\, \mathbf{\nabla} \times \mathbf{A} \cdot \mathbf{n} \, dS =
\int_C \, \mathbf{A} \cdot d \mathbf{s},
\end{align*}\]
where \(\mathbf{n}\) is the unit vector field that is perpendicular to
\(S\), which is in the direction to which a right-handed corkscrew points if it is rotated in the positive direction of the line integral along \(C\).
There is also an analogue of Green’s Theorem for the curl operator:
(490)\[\begin{align*}
\int\!\int\!\int_V \mathbf{\nabla} \times \mathbf{A} \cdot \mathbf{B} \,
dS - \int\!\int\!\int_V \mathbf{A} \cdot \mathbf{\nabla} \times
\mathbf{B} \, dS = \int\!\int_S \left(\mathbf{n}\times
\mathbf{A} \right) \cdot \mathbf{B}\, dS = -\int\!\int_S
\mathbf{A} \cdot \left(\mathbf{n}\times \mathbf{B} \right) \,
dS.
\end{align*}\]