2

The vector potential of a Hertzian dipole falls off spherically as $1/r$. The polar axis of the dipole is a "Null" field -- meaning no electric and magnetic field. The absence of magnetic field is clear enough since there is no curl in the vector potential there. However, the potentials description of the E field involves the equation:

$$\mathbf E = -∇Φ - ∂\mathbf A/∂t$$

which seems to indicate there will be an oscillating electric field in the far field of the Hertzian dipole -- even in the "Null" of the dipole, where the where there should be no electric field.

Clearly, as there is no poynting vector in the "Null", an electron placed there can not oscillate in response to an oscillating electric field as there is no energy locally available, otherwise we'd encounter a violation of conservation of energy or a violation of locality.

How is this conundrum resolved?

Emilio Pisanty
  • 132,859
  • 33
  • 351
  • 666
James Bowery
  • 1,327

1 Answers1

3

The vector potential of an oscillating dipole (using the usual electric dipole approximation) can be written as $$ \vec{A} =\frac{\mu_0 I_0 l}{4\pi r} \cos \omega(t-r/c)\ \hat{z},$$ where the dipole is of length $l$, with a current $I_0 \cos \omega t$ and $\hat{z}$ is a unit vector along the z-axis of the dipole.

Using the Lorenz gauge one can then calculate a corresponding scalar potential $$\phi = \frac{\mu_0 I_0 l c^2\cos \theta}{4\pi} \left( \frac{\cos \omega(t-r/c)}{cr} + \frac{\sin \omega(t-r/c)}{\omega r^2} \right),$$ where $\theta$ is the usual polar angle to the z-axis.

So now the electric field is given by $$\vec{E} = -\frac{\partial \vec{A}}{\partial t} - \nabla \phi$$

If you discard all the terms that have a denominator of $r^2$ or higher from the right hand term (i.e. just consider the far-field), you find the gradient of the scalar potential only has a radial component $$\vec{E} \simeq \frac{\mu_0 I_0 l \omega}{4\pi r} \sin \omega(t-r/c) \hat{z} - \frac{\mu_0 I_0 l \omega \cos \theta}{4\pi r} \sin \omega(t-r/c) \hat{r} $$

But when $\theta =0$ (or $\pi$) along the z-axis, then $\hat{r} \equiv \hat{z}$ and the two terms cancel leaving no far-field.

I am still pondering whether there is a more qualitative way of explaining this.

ProfRob
  • 130,455
  • intuition: to make a dipole you need two equal size opposite charges, along the axis then there cannot be nonzero electric field component; this is true even for moving/oscillating charges. – hyportnex Sep 12 '15 at 21:56
  • @user51748 The question is why is there a ~0 E(t) along the axis of the dipole. A static dipole E-field goes as $r^{-3}$ and is non-zero along the axis. – ProfRob Sep 12 '15 at 23:28
  • i think you have just made my point, $1/r^3$ does dot radiate. – hyportnex Sep 12 '15 at 23:42
  • @user31748 The static dipole field goes as $r^{-3}$ in all directions. What is your point? – ProfRob Sep 13 '15 at 08:36
  • I found a web page with an equivalent formulation that was amenable to implementation in wxMaxima, and verified what you claim to be true is true at least in the limit of the far field. I'm accepting your answer even though I haven't done the algebra to verify your formulation is exactly equivalent. (See equations 1084 and 1087.) http://farside.ph.utexas.edu/teaching/em/lectures/node94.html – James Bowery Sep 15 '15 at 15:03
  • @JamesBowery It is only true in the far field. The near E-field is non-zero along the axis of the dipole. – ProfRob Sep 15 '15 at 15:32
  • What I meant to point out was that even ignoring "the terms that have a denominator of $r^2$ or higher", I can set the frequency to 1Ghz and at 10m (not usually considered the "near field"), and there is still substantial E field oscillation. I'll probably post another question regarding this since the absence of ∇ × A means there should be no E × H poynting vector thence no local energy so an electron moving in response to the changing E would violate conservation of energy or locality. – James Bowery Sep 15 '15 at 15:45
  • @JamesBowery That is the same question. The definition of "near field" is any distance remotely comparable to the wavelength of the radiation. 10m is not really $\gg 0.3$ m. It is not a cut-off; the "far field approximation" is exactly that. I am not sure what you mean by quoting my answer back at me. If you do as the quote suggests, then the E-field is exactly zero, as I showed above. Are you referring to some experiment? – ProfRob Sep 15 '15 at 17:09
  • I'm referring to my wxMaxima model that verified your answer's far field cancellation of ∂A/∂t by ∇Φ along the dipole axis (null field).

    What I need to do is post another question about the near field poynting vector of the dipole axis since, as I understand it, there is no ∇ × A on that axis even in the near field, hence no E × H. You can answer that question when I post it.

     – James Bowery Sep 16 '15 at 16:03
  • 1
    Here is the follow up question about the near field energy. http://physics.stackexchange.com/questions/207399/near-energy-in-the-null-of-a-hertzian-dipole – James Bowery Sep 16 '15 at 23:13