Origin of Hertzian dipole radiation

[Txema]

TL;DR Summary

Does Hertzian dipole radiation originate from charges or current?

In various books, blogs, etc., the Hertz dipole radiation is presented and visualized as field lines that lose their link to the pole charges due to the retarded value of the fields at distant points, lines that close on themselves and jump into space as waves. From this it can be deduced that the (far) radiant field of the dipole has a scalar origin, i.e., it is a field generated by the electric charges.

On the other hand, the expressions that determine the Hertz dipole fields are usually obtained from their electric and magnetic potentials.
E=−∇ϕ−∂A∂t
The gradient of the electric potential defines the electric field of scalar origin (charges), while the derivative with respect to time of the magnetic potential defines the electric field with vector origin (current density). Equations are obtained in spherical coordinates that contain terms that decrease with distance according to 1/r3 (cube), 1/r2 (square) and 1/r. The latter is the preponderant term in the far field, represents the dipole radiation, and comes not from the gradient of the electric potential but from the derivative of the magnetic potential, so it can be thought that it is a field generated by the current, not by the charges.
These are contradictory deductions, is one of them correct, and if so, why not the other?
Any help would be appreciated.

 

[Dale]

You are presenting a false dichotomy. To get electromagnetic radiation you need both an electric and a magnetic field. You can get that from oscillating charges or oscillating current loops. It doesn’t matter

 

[Baluncore]

Does Hertzian dipole radiation originate from charges or current?

It originates from both the current and the charge distribution, but antenna design and modelling software always assumes it is the current that generates the magnetic field of EM radiation.

The orthogonal electric field is inferred from the radiation into free space of the magnetic field.

The voltage differences that form the electric field, are much more difficult to compute than is the magnetic field in the vicinity of a conductive antenna.

 

[Txema]

You are presenting a false dichotomy. To get electromagnetic radiation you need both an electric and a magnetic field. You can get that from oscillating charges or oscillating current loops. It doesn’t matter

I am sorry, I have expressed myself badly. I am only asking about the electric field of radiation. It is clear that the magnetic field comes from the current, my doubt is the origin of the electrical part.

 

[Paul Colby]

but antenna design and modeling software always assumes it is the current that generates the magnetic field of EM radiation

The NEC code is a moment method for time harmonic fields which uses the full hertz dipole expression as a green function for the system. Dealing with the $1/r^3$ is a complication in these types of modeling programs.

 

[Dale]

I am sorry, I have expressed myself badly. I am only asking about the electric field of radiation. It is clear that the magnetic field comes from the current, my doubt is the origin of the electrical part.

As I said: it doesn’t matter.

 

[Vanadium 50]

Is this really an A level thread? I don't want to start launching into vector calculus if that won't answer your question. Does $\nabla \times \nabla \times E$ mean anything to you?

 

[sophiecentaur]

It is clear that the magnetic field comes from the current,

It's not clear to me. Where are you going to get an antenna in which the only thing that varies is the current?

 

[Txema]

Is this really an A level thread? I don't want to start launching into vector calculus if that won't answer your question. Does $\nabla \times \nabla \times E$ mean anything to you?

I have read a lot before asking but I could not extract an answer, this has led me to think that it is a subtle question, so I chose level “A”, but possibly I have been wrong, I apologize.
I am not asking for a mathematical development although I know that mathematics is the language of physics. Thanks anyway.

 

[Txema]

The fact that I get more questions than answers makes me think that I have not asked a well-thought-out question. I think I'd better ask a new question with a different approach.
Thank you all.

 

[Baluncore]

The fact that I get more questions than answers makes me think that I have not asked a well-thought-out question.

The electric field is the vector sum of all potentials on the antenna, in the same way that the magnetic field is the vector sum of all the currents on the antenna. Close to the antenna, in the near field, those evanescent fields are really confusing, but the EM radiation then propagates away, into the far field. There, in the direction of any Poynting vector, the magnetic and electric fields are constrained by the intrinsic impedance of free space, to be orthogonal, in-phase, and with a relative magnitude of E/H = 376.73 ohms. If you know either the electric or magnetic field, you can compute the other.

I think I'd better ask a new question with a different approach.

Try out your new question here in this thread. If you knew what question to ask, you could answer it yourself. We will try to find the right question with you.

 

[sophiecentaur]

Close to the antenna, in the near field, those evanescent fields are really confusing, but the EM radiation then propagates away, into the far field

It's like that old college cartoon with a lecturer in front of a huge blackboard, full of equations. At the bottom, just above the 'answer' is written "and here a miracle occurs".

 

[Txema]

The electric field is the vector sum of all potentials on the antenna, in the same way that the magnetic field is the vector sum of all the currents on the antenna. Close to the antenna, in the near field, those evanescent fields are really confusing, but the EM radiation then propagates away, into the far field. There, in the direction of any Poynting vector, the magnetic and electric fields are constrained by the intrinsic impedance of free space, to be orthogonal, in-phase, and with a relative magnitude of E/H = 376.73 ohms. If you know either the electric or magnetic field, you can compute the other.


Try out your new question here in this thread. If you knew what question to ask, you could answer it yourself. We will try to find the right question with you.

Thank you very much.

 

[Txema]

The most common way to determine the fields generated by a distribution of charge and current is through their electric and magnetic potentials. It is interesting, or so it seems to me, to interpret the results from the point of view of Maxwell's equations.
For example, from a separated electric charge we can deduce that:
-if it is stationary, it will have a stationary radial electric field around it (Gauss-Coulomb)
-if it is in uniform motion, it will have a stationary and rotational and solenoidal magnetic field (Biot-Savart-Ampere)(Gauss). Its electric field will be irrotational (Faraday), that is, it moves with it.
-if it is in accelerated motion, it will have a variable magnetic field (Ampere). The electric field lines are deformed and the field can be considered decomposed in radial and transverse directions, with which we would have the quasi-static electric field (irrotational), and a rotational component (Faraday). In addition the variable E field contributes to the rotational magnetic field.
Applying curl to the Faraday and Ampere-Maxwell equations for points in free space leads to wave equations. The time-varying E and B fields are necessarily waves.
We could say that the accelerated charge has an electric field that is composed of a radial quasi-static field and a transverse radiant field, and it has a quasi-static magnetic field and a radiant magnetic field, both with the same rotational geometry.
The Heaviside-Feynmann equation for the electric field of a moving charge has three terms:
-The first is Coulomb's law for a retarded position.
-The second is the forward projection extrapolated from previous data to compensate for the retardation, which makes the electric field lines remain straight despite the movement.
-The third term is proportional to the acceleration of the charge, and constitutes the far field since the other two decrease rapidly.
For a charge in uniform motion, the sum of the first two terms gives us the quasi-static field; for an accelerated charge we also have the third term, which is a radiant field.
There is a correspondence between this equation and the qualitative conclusions of Maxwell's equations.
If instead of a separate charge it is a free charge of a conductor through which a variable current flows, the radial electric field will be cancelled by the electric field of the positive nuclei, but the radiant electric field will remain, and of course the radiant magnetic field as well, the conductor radiates.
I apologize for the long exposition and for possible inaccuracies, I have put it to show my intentions with the dipole.
With similar reasoning, in the Hertz dipole we will have:
-Charges: The electric field of the charge of the poles, which will be variable since the charge varies, therefore it will be a radiant field, but in the vicinity of the dipole it is "very little radiant" because the delay is insignificant and we consider it reactive at a short distance.
-Current: The quasi-static and radiant magnetic fields of the variable current and the radiant electric field of the accelerated free charges.
The spherical coordinate expressions of the dipole fields obtained from the electric and magnetic potentials contain terms that decrease according to: 1/r³, 1/r², 1/r.
Er[1/r³, 1/r²], Eθ[1/r³, 1/r², 1/r]: the terms with 1/r³ and 1/r² come from the gradient of the electric potential, while the term at 1/r comes from the derivative of the magnetic potential.
Bφ[1/r², 1/r]: both terms come from the curl of the magnetic potential.
Focusing on the electric field, since the static dipole has a field that decreases according to 1/r³, it is possible to think that the 1/r³ terms of the Hertz dipole correspond to the electric field of the charges that we have called reactive. The rest is difficult, one possibility would be that the term with 1/r² corresponds to the non-reactive field of charges, to the field lines that are disconnected from said charges, and that the term 1/r corresponds to the radiant electric field of the charges that make up the current, the latter would justify the vectorial origin of said term.
This is the object of my question, since I sense that something escapes my understanding.
I understand that mathematical terms do not have to have a physical meaning, but I do not know if this is the case, in some texts there is a brief interpretation that clarifies little.

 

[sophiecentaur]

The electric field is the vector sum of all potentials on the antenna, in the same way that the magnetic field is the vector sum of all the currents on the antenna.

in some texts there is a brief interpretation that clarifies little.

It's almost a matter of luck that you can usually 'get away' with just considering currents in an antenna. Whilst it has to be true that the current flowing in a part of an antenna is determined by all the nearby fields, you can get a good idea of the radiated field from a dipole by assuming a sinusoidal current distribution. But the pattern of a receiving antenna is virtually the same as for a transmitting antenna whilst the only currents flowing are due to the received signal; that's disturbing.

 

[tech99]

I am sorry, I have expressed myself badly. I am only asking about the electric field of radiation. It is clear that the magnetic field comes from the current, my doubt is the origin of the electrical part.

To create radiation, by which I mean the transfer of energy from matter to EM waves, we need to accelerate charges, and to cause acceleration of the charges in an antenna, we apply an alternating electric field. It is this accelerating field which can be observed as the electric induction near field. The two ends of a dipole antenna create this accelerating field because they are are at opposite potential, but at some distance, the two accelerating fields begin to cancel and this is why you then see a r^-3 variation. It is just geometry.

We can notice that the radiated electric field is in-phase with the accelerating electric field, because acceleration is proportional to force (Newton). The "reaction" (described by Newton) is the opposition of a charge to acceleration, caused by its normally radial field lines being bent as a result of acceleration. The work done against this reaction is lost as radiated energy and is seen as a radiation resistance. Once the electron has finished its acceleration it possesses velocity, and so it constitutes a current and has energy stored in a magnetic field, which we can observe. Current is a measure of radiation, not a cause. Where we have resonance, for instance when the length is half a wavelength, the induction fields, electric and magnetic, constitute an energy store and contain sufficient energy for several waves when excitation is removed.

Having once created an electric wave, if it passes an observer they will also detect a magnetic field due to Relativity.

At a large distance from a radiator we know that the power density falls with the inverse square law, so that the electric and magnetic fields fall with a 1/r law. On the other hand, when we measure the electric induction, or accelerating, field very close to one end of a dipole antenna it falls with r^-2 in the same way as from an isolated charge. And as I have mentioned, at greater distance, the two ends of the dipole have opposing electric fields so tend to cancel, giving the r^-3 relationship for the induction field. (This is an analogous effect to the shape of the magnetic field of a bar magnet, where the poles oppose beyond a certain distance).

The following diagram shows my measurements of the electric fields near a dipole antenna (if it comes out, as I have no IT skills). The circles are lambda/4 radius. The distant field on axis falls with 1/r, and notice that the fields closely surrounding the dipole ends are circular - just those of an isolated charge. The electric field in the central portion at a spacing of about lambda/6 is small, due to cancellation from the ends, and a small spurious patch of electric field is observed surrounding the feed point due to the voltage across the feeder.

Figure showing measurements of electric fields near a dipole antenna

 

[nsaspook]

https://ia801502.us.archive.org/25/...-Lines-Antennas-And-Wave-Guides-1945_text.pdf

A classic book on practical qualitative EM theory.
PDF page 90 for antennas.
PDF page 195 for a discussion of fields from a center driven dipole.

 

[tech99]

I am unsure, from a cursory read of the pages, that the author describes how radiation occurs, in distinction from, say, coupling between nearby coils of wire.

 

[Txema]

-Separated charges have incoming or outgoing radial field lines.
-Free charges in a conductor do not have them due to the cancellation with the field of the positive nuclei to which they are bound.
-Both radiate when accelerated.

To create radiation, by which I mean the transfer of energy from matter to EM waves, we need to accelerate charges

The two ends of a dipole antenna create this accelerating field because they are are at opposite potential, but at some distance, the two accelerating fields begin to cancel and this is why you then see a r^-3 variation. It is just geometry.

It means that the charge of the poles generates the induction field that accelerates the charges of the current, and these free charges radiate electromagnetic waves. So the far field are waves coming from the charges that compose the current, since the field of the charges of the dipole decreases rapidly.
OK?
(This is congruent with the far field in the Eθ coordinate coming mathematically from the time derivative of the magnetic potential).

 

[Txema]

A classic book on practical qualitative EM theory.

Thank you for the reference.

 

[tech99]

-Separated charges have incoming or outgoing radial field lines.
-Free charges in a conductor do not have them due to the cancellation with the field of the positive nuclei to which they are bound.
-Both radiate when accelerated.



It means that the charge of the poles generates the induction field that accelerates the charges of the current, and these free charges radiate electromagnetic waves. So the far field are waves coming from the charges that compose the current, since the field of the charges of the dipole decreases rapidly.
OK?
(This is congruent with the far field in the Eθ coordinate coming mathematically from the time derivative of the magnetic potential).

Notice that although an atom has balanced positive and negative charges, the mass of the positive charges is very great so they do not respond significantly to an accelerating potential. On the other hand the electrons are very light and do respond. Notice that the electron charge on a piece of metal such as a dipole is enormous, so the slightest acceleration creates a large effect.