We already defined the nabla operator in connection with the gradient in section 39
| \[\vec{\nabla} =\vect{\frac{\partial }{\partial x_1}\\\vdots\\\frac{\partial}{\partial x_N}} \quad \vec{x} \in\mathbb{R}^N\] |
Applying this operator to a function with one component \(f(\vec{x})\) we get the gradient
| \[\mbox{grad} f = \vec{\nabla}f =\vect{\frac{\partial f}{\partial x_1}\\\vdots\\\frac{\partial f}{\partial x_N}}.\] |
Note that the gradient is applied to a scalar and the result is a vector.
One essential aspect of the gradient is the solution of path integrals
| \[\int\limits_{\vec{x}_b}^{\vec{x}_e} \vec{\nabla}f \; d\vec{x} = f(\vec{x}_e) - f(\vec{x}_b).\] |
Calculating the scalar product of the nabla operator and a function with several components \(\vec{f}(\vec{x})\) we get the divergency
| \[\mbox{div} \vec{f} = \vec{\nabla} \vec{f} = \sum\limits_{i} \frac{\partial f_i}{\partial x_i}.\] |
Note that the divergency is applied to a vector and the result is a scalar.
One essential aspect of the divergency is the solution of volume integrals
| \[\iiint\limits_{V} \vec{\nabla}\vec{f} \; dx\;dy\;dz = \oiint\limits_{\partial V} \vec{f}\; d\vec{A}.\] |
Here \(\partial V\) denotes the closed surface of the volume \(V\)
the integration is calculated over.
Calculating the vector product of the nabla operator and a function
with several components \(\vec{f}(\vec{x})\) we get the curl
| \[\mbox{curl} \vec{f} = \mbox{rot} \vec{f} = \vec{\nabla} \times \vec{f}.\] |
Note that the curl is applied to a vector and the result is a vector.
One essential aspect of the curl is the solution of area integrals (Stokes integral equation)
| \[\iint\limits_{A} \vec{\nabla} \times \vec{f} d\vec{A} = \oint\limits_{\partial A} \vec{f} \;d\vec{x}.\] |
Here \(\partial A\) denotes the closed path around the area \(A\)
the integration is calculated over.
Examples using the Maxwell equations:
As an example for a vector function
we already discussed in section 4.2
the electric field of a point source with positive charge \(q\), i.e. the charge density \(\rho(\vec{r})
= q \delta(\vec{r})\). The electrical field strength is calculated from the 1. Maxwell equation
| \[\frac{\rho}{\epsilon_0} = \vec{\nabla} \vec{E}(\vec{r}).\] |
Integrating the Maxwell equation over a sphere with radius \(r\) centered around the point charge we find
| \[\frac{q}{\epsilon_0}=\iiint\limits_{\mathrm{sphere}} \frac{q \delta(\vec{r})}{\epsilon_0}\;dx\;dy\;dz = \iiint\limits_{\mathrm{sphere}} \vec{\nabla} \vec{E}(\vec{r})\;dx\;dy\;dz =\oiint\limits_{\mathrm{surface \; sphere}} \vec{E}(\vec{r})\; d\vec{A}.\] |
Obviously the electrical field strength \(\vec{E}(\vec{r})=E(r)\frac{\vec{r}}{r}\) has a radial symmetry; thus the right hand integral can be simplified
| \[\oiint\limits_{\mathrm{surface \; sphere}} \vec{E}(\vec{r})\; d\vec{A} = \oiint\limits_{\mathrm{surface \; sphere}} E(r)\; dA,\] |
i.e. the scalar product reduces just to the product of the length of vectors, finally leading to
| \[\frac{q}{\epsilon_0} = E(r) 4\pi r^2,\] |
where \(4\pi r^2\) is the surface area of a sphere with radius \(r\), leading to
| \[ \vec{E}(\vec{r})= \frac{q}{4\pi \epsilon_0 r^2} \frac{\vec{r}}{r}.\] |
Since \(\vec{E}=-\vec{\nabla} U\) and taking the potential \(U(\infty)=0\) we find the potential of a point charge as
| \[ U(\vec{r})=\int\limits_{r}^{\infty} (-\vec{\nabla} U) \; d\vec{r} = \int\limits_{r}^{\infty} \vec{E} \; d\vec{r} = \int\limits_{r}^{\infty} E \; dr = \frac{q}{4\pi \epsilon_0 r}.\] |
For this final result the path along the \(\vec{r}\) direction has been chosen.
The 2. Maxwell equation reads
| \[\vec{\nabla} \times \vec{B}(\vec{r}) = \mu_0\, \vec{j}(\vec{r}).\] |
Having a straight wire with a current density \(\vec{j}\) we find according to the Stokes integral equation
| \[ \mu_0\,I = \iint\limits_{A} \mu_0 \vec{j}(\vec{r}) d\vec{A} = \iint\limits_{A} \vec{\nabla} \times \vec{B} d\vec{A} = \oint\limits_{\partial A} \vec{B} \;d\vec{x}.\] |
Choosing for the path in the right hand side integral a circle perpendicular to the wire centered around the center of the wire, which due to symmetry implies \(\vec{B} \;d\vec{x} = const.\) along the path, we finally get
| \[ \mu_0\,I = |\vec{B}| 2 \pi r .\] |
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© J. Carstensen (Math for MS)