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What is the mechanism by which the near-field non-traveling electromagnetic fields in the vicinity of an antenna are changed into the far-field electromagnetic waves traveling at the speed of light? Clearly, the electric charges in the antenna that generate the near-field electromagnetic fields are not moving at the speed of light. Rather these electrons move at their speed in the antenna material, and the changes in their speed should dictate the rate-of-change of the near-field electromagnetic fields.

https://en.wikipedia.org/wiki/Near_and_far_field

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    $\begingroup$ Your question is based on a totally false assumption: "...non-traveling electromagnetic fields in the vicinity of an antenna are changed into far-field electromagnetic waves". Instead, the generator accelerates the charges in the antenna and the accelerated charges radiate. $\endgroup$ – hyportnex Jul 7 '18 at 22:17
  • $\begingroup$ The speed of the charge is not an indicator if the speed of light emitted from the charge. As @hyportnex pointed out, it's the acceleration that produced the radiation. $\endgroup$ – ggcg Jul 7 '18 at 22:39
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    $\begingroup$ Does no one have a decent quantitative answer to my question? $\endgroup$ – John Petrovic Jul 7 '18 at 23:39
  • $\begingroup$ @hyportnex accelerating a charge results in a change of the EM field aka near field, therefore I dont see what in your comment makes the assumption that near field generates far field incorrect. Please elaborate. $\endgroup$ – Manu de Hanoi Jan 8 '19 at 7:13
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    $\begingroup$ @Manu_de_Hanoi the source of electromagnetic radiation both near and far field is moving (accelerating/decelarating) charges and time-varying currents. Near field does not become far field, E-field does not generate B-field nor vice versa. Those ideas are at best visualization aides, like saying that F=ma means that force creates acceleration and in turn acceleration creates force. To understand it see the very erroneously called "Jefimenko" equations en.wikipedia.org/wiki/Jefimenko%27s_equations , best book to study is Panofsky-Phillips. $\endgroup$ – hyportnex Jan 8 '19 at 12:44
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This question is not a simple one, and the answer is even less so. However, it's not impossible to find an answer, so I will attempt to provide one. Note, however, that antenna analysis is not really my area. (I study microwave engineering).

Background

First, we derive the wave equation for electromagnetic fields. Beginning with Maxwell's equations in a source-free space, it is readily found that all free-space electromagnetic fields obey $$ \tag{1.a}\nabla^2 \mathbf{E} = \frac{1}{c^2} \frac{\partial^2 \mathbf{E}}{\partial t^2} $$ $$ \tag{1.b}\nabla^2 \mathbf{H} = \frac{1}{c^2} \frac{\partial^2 \mathbf{H}}{\partial t^2} $$ Where $c = \frac{1}{\omega\sqrt{\mu \epsilon}}$ is the speed of light in a vacuum. Thus we see that any free-space electromagnetic field must satisfy the 3D wave equation, and evidently all electromagnetic fields in free space propagate with finite velocity $c$. By assuming the direction of propagation $\hat{k}$ is well defined at any given point, we can model electromagnetic waves as changing sinusoidally in time and space. It becomes clear, then, that any such fields which exist in free space will propagate indefinitely.

Our goal is thus to prove how such a field can originate from a charge density $\rho$ or current density $\mathbf{J}$ residing in a conductor.

Expressing the full form of Maxwell's equations (differential form), in phasor form, $$\tag{2.a}\nabla \times \mathbf{E} = -j\omega \mu \mathbf{H}$$ $$\tag{2.b}\nabla \times \mathbf{H} = j\omega \epsilon \mathbf{E} + \mathbf{J}$$ $$\tag{2.c}\nabla \cdot \mathbf{E} = -\frac{\rho}{\epsilon}$$ $$\tag{2.d}\nabla \cdot \mathbf{H} = 0$$ The sources appear in $\text{(2.b)}$ and $\text{(2.c)}$. They are related by the continuity equation, $$\nabla \cdot \mathbf{J} + \frac{\partial \rho}{\partial t} = 0$$

At the interface between a conductor and free space (as in the case of an antenna), we have the following equations which must hold at the boundary. $$ \hat{n} \cdot \mathbf{E} = \frac{\rho_s}{\epsilon} $$ $$ \hat{n} \cdot \mathbf{B} = 0 $$ $$ \hat{n} \times \mathbf{E} = \mathbf{M}_s $$ $$ \hat{n} \times \mathbf{H} = \mathbf{J}_s $$

Where $\mathbf{J}_s$ is the surface current density, $\rho_s$ is the surface charge density, $\hat{n}$ is normal to the surface, and the term $\mathbf{M}_s$ is introduced as the imaginary magnetic current, whose sources include current loops and magnetic dipole currents. For the case where the conductor is sufficiently thick beyond the boundary, $\mathbf{M}_s \approx 0$.

Simple Example of Radiation

Now the stage is set, and we can consider sources of radiation. The first is the simple case of an infinite sheet of current density, with time dependence $e^{j\omega t}$, $$\mathbf{J}_s = J_0 \hat{x}$$ Which is a sinusoidal current, and we define it to exist on an infinite plane surface in the $x,y$ plane. (Note that $\mathbf{J}$ is a phasor, and the time variation is implied).

From the interface conditions, $$\hat{n} \times \mathbf{H} = J_0 \hat{x}$$ And because of the way we defined the surface, $\hat{n} = \hat{z}$, and $$\hat{z} \times \mathbf{H} = J_0 \hat{x}$$ Implies that $$\mathbf{H}_s = -J_0 \hat{y}$$ Also from the interface conditions, $$\mathbf{E}_s = 0$$ And we can readily confirm that solving $\text{(1.a)}$ and $\text{(1.b)}$ in free space produces plane wave solutions, $$\mathbf{E} = \frac{-J_0}{2} \eta_0 e^{-j\beta z}\hat{x} $$ $$\mathbf{H} = \frac{J_0}{2} e^{-j\beta z}\hat{y} $$ Where $\eta_0$ is the impedance of free space. Similar solutions in the opposite direction also exist.

So we can show that near-field and far-field labels are unnecessary here, and radiation occurs from the source.

More Generalized Radiation

Attempting to derive the radiation pattern of an antenna with these equations would prove quite difficult. One method of simplification is to use the scalar and vector potentials, defined such that $$\nabla \times \mathbf{A} = \mathbf{B}$$ $$\mathbf{E} = \nabla\Phi - \frac{\partial \mathbf{A}}{\partial t}$$ Which has the useful feature of reducing two vector fields to one vector field and one scalar field.

Using a fair amount of analysis, we can show that the solution for the vector potential due to a current density $\mathbf{J}$ with $e^{j\omega t}$ time dependence is $$ \mathbf{A}(\mathbf{r}) = \frac{\mu_0}{4\pi} \int_V \mathbf{J}(\mathbf{r}') \frac{e^{j\beta r}}{r} d^3 r' $$ Where $\mathbf{r}$ is the position of observation, $\mathbf{r'}$ is the integration variable, $r = |\mathbf{r} - \mathbf{r}'|$, and $d^3 r'$ means volume integration over $r'$.

This integral equation is complicated by the $r$ term, but we can approximate depending on the distances involved. The three regions commonly investigated for a source of size $d$, point of investigation $r$, and wavelength in free space $\lambda$, are the near field ($d << r << \lambda$), the induction zone ($d << r \approx \lambda$) and the far field ($d << \lambda << r$). In the near field, we can show that $$\mathbf{A}(\mathbf{r}) \approx \frac{\mu_o}{4\pi} \sum_{l,m} \frac{4\pi}{2l + 1} \frac{Y_{lm}(\theta, \phi)}{r^{l+1}} \int \mathbf{J}(\mathbf{r}') r'^l Y*_{lm}(\theta', \phi') d^3 r' $$ Where we have used spherical coordinates, and spherical harmonics. In the far field, our approximation is $$ \mathbf{A}(\mathbf{r}) \approx \frac{\mu_0}{4\pi} \frac{e^{j\beta r}}{r} \int \mathbf{J}(\mathbf{r'}) e^{-j\beta \mathbf{r}'\cdot\hat{n}} d^3r' $$ One can calculate the induction zone approximation, but I will not do so here.

From here, we can calculate the electromagnetic field with $$\mathbf{H} = \frac{1}{\mu_0}\nabla \times \mathbf{A}$$ $$\mathbf{E} = \frac{j\eta_0}{\beta}\nabla \times \mathbf{H}$$ Where again $\eta_0$ is the impedance of free space.

Using the above, we can calculate the electromagnetic field due to an electric dipole, a magnetic dipole, etc. Note that the above approximations are valid for an antenna which is much shorter than the wavelength. Dipole approximations specifically are also common, and are much more efficient. In this case, the electric and magnetic fields in the far field become $$ \mathbf{H} = \frac{c \beta^2}{4 \pi}(\hat{n}\times \mathbf{p}) \frac{e^{j\beta r}}{r} $$ $$ \mathbf{E} = \eta_0 \mathbf{H} \times \hat{n} $$ Where $\mathbf{p}$ is the dipole moment. And thus we see that for a far-field term to exist, we simply need a non-zero dipole moment, $$ \mathbf{p} = \int \mathbf{r}' \rho (\mathbf{r}') d^3 r'$$ Which is the case whenever we have a change charge distribution. This still, however, assumes the length of the dipole is much less than the wavelength.

Center-fed Linear Antennas

Finally, we come to the center-fed linear antenna. For this case, by integration, we find $$ \mathbf{A}(\mathbf{r}) \approx \frac{\mu_0}{4\pi} \frac{2Ie^{j\beta r}}{\beta r} \left[ \frac{ \cos\left(\frac{\beta d}{2} \cos\theta\right) - \cos \left(\frac{\beta d}{2}\right)} {\sin^2\theta} \right]$$ In the far zone, where $I$ is the peak current value. So we see that, for the field to "reach" the far field, the current on the antenna is the only requirement. This holds for all such dipoles. For the case where $\beta d = \pi$ (half-wave dipole), this still applies.

Conclusion

The transition from near-field to far-field is an analytic one, a distinction made between two approximations. In one case, we can make some simplifying assumptions, in another, we use different assumptions. The resulting fields behave quite differently in these limits, so it appears that there is an actual near-field component and far-field component, when in reality they are part of the same solution. Both always exist, as long as our definitions are properly given.


Sources

Pozar, David M. Microwave Engineering, 4ed. 2012.

Jackson, J. D. Classical Electrodynamics, 3ed. 1999.

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The antenna generator accelerates the electrons in the rod.

The laws of physics are invariant under Lorentz transformations, so you can speed up to the speed of the electron, and so you would be in the rest frame of the electron. This constant velocity in this frame of the electron does not produce photons, because in this frame the electron is moving at constant velocity (or is at rest).

Now accelerations is a change in the state of the motion of the electron. By accelerating it, you change its momentum and kinetic energy.

Momentum and energy are conserved quantities, and if the electron's momentum is changed, that change has to be represented by another particle with the exact same momentum that you accelerated the electron with.

When the electron is bound to a nucleus, it can emit a photon by moving to a lower energy level as per QM.

But in conductors, like a metal antenna these electrons are loosely bound, can move mostly freely, like in vacuum.

A free electron, when its momentum is changed because of an electric field, will give off its energy and momentum by emitting a real photon, that is the radio signal.

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  • $\begingroup$ This answer is vaguely response to the original title but completely ignores the much more sophisticated question in the text and the edited title. $\endgroup$ – dmckee --- ex-moderator kitten Jul 9 '18 at 2:03
  • $\begingroup$ @ dmckee can you please tell me the sophisticated question? $\endgroup$ – Árpád Szendrei Jul 9 '18 at 3:30
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When an electric dipole p is radiating, the electromagnetic field takes the form below(the equations are in spherical format):

$$E=\frac{p_0c^2}{r10^7} \langle \frac{2\cos \theta}{r}\left(\frac{1}{r}\cos\left(\omega\left(t-\frac{r}{c}\right)\right)-\frac{\omega}{c}\sin\left(\omega\left(t-\frac{r}{c}\right)\right)\right)|\sin\theta\left(\frac{1}{r^2}\cos\left(\omega\left(t-\frac{r}{c}\right)\right)-\frac{\omega}{cr}\sin\left(\omega\left(t-\frac{r}{c}\right)\right)-\frac{\omega^2}{c^2}\cos\left(\omega\left(t-\frac{r}{c}\right)\right)\right)|0\rangle$$ $$B=-\frac{p_0\omega\sin\theta}{r10^7}\langle0|0|\frac{1}{r}\sin\left(\omega\left(t-\frac{r}{c}\right)\right)+\frac{\omega}{c}\cos\left(\omega\left(t-\frac{r}{c}\right)\right)\rangle$$ At high values of r the equations become: $$E=-\frac{p_0\omega^2}{r10^7} \langle 0|\sin\theta\cos\left(\omega\left(t-\frac{r}{c}\right)\right)|0\rangle$$ $$B=-\frac{p_0\omega^2\sin\theta}{cr10^7}\langle0|0|\cos\left(\omega\left(t-\frac{r}{c}\right)\right)\rangle$$ Which is far field radiation. I hope all of this has increased your understanding and has answered your question.

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    $\begingroup$ Consider using different grouping symbols $()$, $[]$, and $\{\}$ as well as resizing angle brackets \left< and \right> for improved readability. $\endgroup$ – dmckee --- ex-moderator kitten Jul 9 '18 at 2:05

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