Editing Magnetic vector potential (section)

==== Frequency domain ====

The preceding time domain equations can be expressed in the frequency domain.<ref name="Balanis_Ant">{{Citation |last=Balanis |first= Constantine A. |year= 2005 |title= Antenna Theory |edition=third |publisher= John Wiley |isbn= 047166782X |author-link=Constantine A. Balanis}}</ref>{{rp|139}}
* Lorenz gauge <math>\nabla \cdot \mathbf{A} + \frac{j \omega}{c^2} \phi= 0 \qquad </math> or <math>\qquad  \phi = \frac {j \omega}{k^2} \nabla \cdot \mathbf{A} </math>
* Solutions <math> \mathbf{A}\!\left(\mathbf{r}, \omega \right) = \frac{\mu_0}{\ 4\pi\ } \int_\Omega \frac{ \mathbf{J}\left(\mathbf{r}' , \omega \right) }R\  e^{-jkR} d^3 \mathbf{r}' \qquad
\phi\!\left(\mathbf{r}, \omega \right) = \frac{1}{4\pi\epsilon_0} \int_\Omega \frac{ \rho \left(\mathbf{r}', \omega \right) } R \ e^{-jkR} d^3 \mathbf{r}'</math>
* Wave equations <math> \nabla^2 \phi + k^2 \phi = - \frac{\rho}{\epsilon_0}  \qquad  \nabla^2 \mathbf{A} + k^2 \mathbf{A} = - \mu_0\ \mathbf{J} . </math>
* Electromagnetic field equations <math>\mathbf{B} = \nabla \times \mathbf{A}\ \qquad \mathbf{E} = -\nabla \phi - j \omega \mathbf{A}  = - j \omega \mathbf{A} -j \frac {\omega}{k^2} \nabla ( \nabla \cdot \mathbf{A} )</math>
where
: <math> \phi  </math> and <math> \rho </math> are scalar [[phasors]].
: <math> \mathbf{A},  \mathbf{B},  \mathbf{E}, </math> and <math> \mathbf{J} </math> are vector [[phasors]].
: <math> k = \frac \omega c </math>

There are a few notable things about <math>\mathbf{A}</math> and <math>\phi</math> calculated in this way:
* The [[Lorenz gauge condition]] is satisfied: <math>\textstyle \phi = -\frac {c^2}{j \omega}\nabla \cdot \mathbf{A}. </math>  This implies that the frequency domain electric potential, <math>\phi</math>, can be computed entirely from the current density distribution, <math>\mathbf{J}</math>.
* The position of <math>\mathbf{r},</math> the point at which values for <math>\phi</math> and <math>\mathbf{A}</math> are found, only enters the equation as part of the scalar distance from <math>\mathbf{r}'</math> to <math>\ \mathbf{r}.</math> The direction from <math>\mathbf{r}'</math> to <math>\mathbf{r}</math> does not enter into the equation. The only thing that matters about a source point is how far away it is.
* The integrand uses the [[phase shift]] term <math> e^{-jkR} </math> which plays a role equivalent to ''[[retarded time]]''. This reflects the fact that changes in the sources propagate at the speed of light; propagation delay in the time domain is equivalent to a phase shift in the frequency domain.  
* The equation for <math>\mathbf{A}</math> is a vector equation. In Cartesian coordinates, the equation separates into three scalar equations:<ref name=KrausE>{{harvp|Kraus|1984|p=189}}</ref> <math display="block">\begin{align}
        \mathbf{A}_x\!\left(\mathbf{r}, \omega \right) &= \frac{\mu_0}{4\pi} \int_\Omega \frac{ \mathbf{J}_x \left(\mathbf{r}' , \omega \right) }R\  e^{-jkR} \ d^3\mathbf{r}', 
\qquad  \mathbf{A}_y\!\left(\mathbf{r}, \omega \right) &= \frac{\mu_0}{4\pi} \int_\Omega \frac{ \mathbf{J}_y \left(\mathbf{r}' , \omega \right) }R\  e^{-jkR} \ d^3\mathbf{r}',   
\qquad  \mathbf{A}_z\!\left(\mathbf{r}, \omega \right) &= \frac{\mu_0}{4\pi} \int_\Omega \frac{ \mathbf{J}_z \left(\mathbf{r}' , \omega \right) }R\  e^{-jkR} \ d^3\mathbf{r}' 
\end{align}</math> In this form it is apparent that the component of <math>\mathbf{A}</math> in a given direction depends only on the components of <math>\ \mathbf{J}\ </math> that are in the same direction. If the current is carried in a straight wire, <math>\mathbf{A}</math> points in the same direction as the wire.