After hunting in vain* for a documented closed form solution (ie: no integrals or differentiation in the formula) for $\phi$ in terms of A under the Lorenz condition, of an electrically short dipole (Hertzian dipole) transmitter antenna (including the 1/${r}^{2}$ (near) field) I came up with the following derivation. Is it correct?

Starting with the widely documented definition of A for this case:

$$A(x,y,z,t) = [0,0, \,I0\,\mu\,l\,\frac{\mathrm{cos}\left( \omega\,\left( t−\frac{\sqrt{{z}^{2}+{y}^{2}+{x}^{2}}}{c}\right) \right)}{4\,\pi \,\sqrt{{z}^{2}+{y}^{2}+{x}^{2}}}]$$


$I0$ = the magnitude of the dipole current

$l$ = the length of the dipole

Then applying the Lorenz gauge $\nabla$$\cdot$A = -$\mu\,\epsilon\frac{\partial\phi}{{\partial}t}$, first deriving $\nabla$$\cdot$A:

$$\nabla\cdot A = \frac{d}{d\,z}\,\frac{\mu\,l\,\mathrm{cos}\left( \omega\,\left( t−\frac{\sqrt{{z}^{2}+{y}^{2}+{x}^{2}}}{c}\right) \right) \,I0}{4\,\pi \,\sqrt{{z}^{2}+{y}^{2}+{x}^{2}}}$$

$$\nabla\cdot A = \frac{\mu\,\omega\,l\,z\,\mathrm{sin}\left( \omega\,\left( t−\frac{\sqrt{{z}^{2}+{y}^{2}+{x}^{2}}}{c}\right) \right) \,I0}{4\,\pi \,c\,\left( {z}^{2}+{y}^{2}+{x}^{2}\right) }−\frac{\mu\,l\,z\,\mathrm{cos}\left( \omega\,\left( t−\frac{\sqrt{{z}^{2}+{y}^{2}+{x}^{2}}}{c}\right) \right) \,I0}{4\,\pi \,{\left( {z}^{2}+{y}^{2}+{x}^{2}\right) }^{\frac{3}{2}}}$$

$$\nabla\cdot A = −\frac{\mu\,\omega\,l\,z\,\sqrt{{z}^{2}+{y}^{2}+{x}^{2}}\,\mathrm{sin}\left( \frac{\omega\,\sqrt{{z}^{2}+{y}^{2}+{x}^{2}}−\omega\,c\,t}{c}\right) \,I0+\mu\,c\,l\,z\,\mathrm{cos}\left( \frac{\omega\,\sqrt{{z}^{2}+{y}^{2}+{x}^{2}}−\omega\,c\,t}{c}\right) \,I0}{{4\,\pi \,c\,\sqrt{{z}^{2}+{y}^{2}+{x}^{2}}\,\left( {z}^{2}+{y}^{2}+{x}^{2}\right) }}$$

and solving for $\phi$:

$$-\epsilon\,\mu\,\left( \frac{d}{d\,t}\,\Phi\right) =−\mu\,l\,z\,I0\frac{\omega\,\sqrt{{z}^{2}+{y}^{2}+{x}^{2}}\,\mathrm{sin}\left( \frac{\omega\,\sqrt{{z}^{2}+{y}^{2}+{x}^{2}}−\omega\,c\,t}{c}\right) \,+c\,\mathrm{cos}\left( \frac{\omega\,\sqrt{{z}^{2}+{y}^{2}+{x}^{2}}−\omega\,c\,t}{c}\right)}{4\,\pi \,c\,\sqrt{{z}^{2}+{y}^{2}+{x}^{2}}\,\left( {z}^{2}+{y}^{2}+{x}^{2}\right) }$$

$$\mathrm{\phi}\left( x,y,z,t\right) = I0\,l\,z\,\frac{\sqrt{{z}^{2}+{y}^{2}+{x}^{2}}\,\mathrm{cos}\left(\frac{\omega\,\sqrt{{z}^{2}+{y}^{2}+{x}^{2}}−\omega\,c\,t}{c} \right) −\frac{c\,\mathrm{sin}\left(\frac{\omega\,\sqrt{{z}^{2}+{y}^{2}+{x}^{2}}−\omega\,c\,t}{c} \right) }{\omega}}{4\,\pi \,\epsilon\,c\,\sqrt{{z}^{2}+{y}^{2}+{x}^{2}}\,\left( {z}^{2}+{y}^{2}+{x}^{2}\right) }$$

*All of the authoritative sources (including texts such as Van Bladel and Balanis) I access seem to fall short of providing a solution for Φ commensurable, under the Lorenz gauge, with the definition of A they give for the "Hertzian dipole" aka "infinitesimal dipole" aka "electrically short dipole". This shortfall goes beyond the authoritative texts. My online searches yielded exactly one web page with a commensurable solution but was suspect for apparent inconsistency in units and/or dimensions. I also found a physics.stackexchange.com "answer" that provided such a formula for Φ, however neither of these online resources produced a null field for the summation of the terms:

E = - ∇Φ - ∂A/∂t

Many of them provide such a definition for A but then leave the solution for Φ out in their haste to get to a far-field formula for E that does not involve the potentials.

  • $\begingroup$ James, I think (you could have formatted it better) you've got it right. In my lecture notes: $\phi(r,t) = (I_0 l \cos \theta/4\pi\epsilon_0) [ \cos[(\omega(t-r/c)]/rc + \sin[\omega (t-r/c)]/\omega r^2 ]$, where $\cos \theta = z/r$. $\endgroup$ – Rob Jeffries Nov 11 '15 at 19:12
  • $\begingroup$ What more needs to be modified in the question to unblock it? I could re-do the derivation using spherical coordinates instead of Cartesian but Cartesian seems a better choice for what I'm trying to do, which is not only rotate the Tx dipole for an antenna pattern, but rotate the detector, which is basically a test charge with a velocity vector so that the motional term(s) can be added. $\endgroup$ – James Bowery Nov 12 '15 at 15:49