I Electromagnetic waves from a dipole antenna

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Hello, I've been reading up on this topic and have a couple of questions. This videos shows what I am referring to:
1. What makes the electromagnetic waves from a dipol antenna spread out? There would have to be some type of force that pushed them outwards, right?

2. When the electromagnetic waves do spread out, is there a sweet spot for maximum intake of the waves or is it pretty much homogenous disregarding if someone gets the signal from the middle, below or near the top (talking about far distances from the dipole)?

3. Does the illustration from the video prove that dipole antennas are electromagnetic waves or does it explain that dipole antennas are electromagnetic waves?

4: The B field: is the B-field strongest in the middle or homogenous throughout the antenna? This question refers a bit to question 2. So it would essentially look like:
366?cb=20170107183359.png


but without the sharp edges and in 3D it would be a circular shape.
 

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What makes the electromagnetic waves from a dipol antenna spread out? There would have to be some type of force that pushed them outwards, right?
They move at constant velocity so no force is needed, but the mechanism of how they spread out is described by Maxwell’s equations.

is there a sweet spot for maximum intake of the waves
You get the maximum power from the EM wave in the plane normal to the dipole. You get the least on the line of the dipole. And of course the closer the better.

The B field: is the B-field strongest in the middle or homogenous throughout the antenna?
The B field is strongest where the current is highest. That is in the middle. However, it is a smoother shape than what you have drawn.
 

sophiecentaur

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1. What makes the electromagnetic waves from a dipol antenna spread out? There would have to be some type of force that pushed them outwards, right?
No. You don't need any Force to make the wave spread out.
If you ignore the Energy involved for a start and if you change the current in the wire, the field right next to the wire will change 'more or less' instantly. But it will take time for this change in field to make itself known at a distant point (it can't be instantaneous). If the current change was a simple step then the field would settle down at all distances. But, if you keep varying the dipole current, the effect at a distance will be that the field varies in step with the current variations but delayed. That's the property of a wave; waves Propagate.
There is an additional concept and that is that the changing fields (Electric and Magnetic) actually carry Power away from the dipole. This process presents itself as a Resistance, called the Radiation Resistance which is the real part of the product of the V and I oscillations at the point where the dipole is fed. It's where the transmitter power goes. Another way of looking at the effect is that Space can be looked on as having an Impedance which is near enough 377Ω. This impedance is a description of the fact that power is being lost from the dipole as it radiates away.
 
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They move at constant velocity so no force is needed, but the mechanism of how they spread out is described by Maxwell’s equations.
How do the Maxwell equations explain the how/why E and B spread out? Maxwell used the equations to eventually find that they moved at the velocity c but I'm stuck on something else. Why do they even move away from the first place? When we look at an E-Field from a capacitor, the E-Field stays within the capacitor. Why with a dipole do they move outward?
 

sophiecentaur

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the E-Field stays within the capacitor.
Not all of it!!! There is some field variation at the edges of the gap and that will radiate some power but the Capacitance between the two plates is so much greater that the energy is largely contained between them. Otoh, a dipole that's a significant fraction of a wavelength will have a greater effect on the fields outside it and so it radiates well.
 

jtbell

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If the charge on the capacitor is constant (with respect to time), then the E field is also constant and no power is radiated. The E field does extend beyond the edges of the capacitor, and falls off with distance.

If the charge on the capacitor is made to vary, e.g. by putting it in an AC circuit, then the E field varies with time, and there is also a time-varying B field. In this case, electromagnetic waves do radiate from the capacitor. Far enough away from the capacitor, the radiated waves should look like the ones from an ideal oscillating dipole.
 
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Not all of it!!! There is some field variation at the edges of the gap and that will radiate some power but the Capacitance between the two plates is so much greater that the energy is largely contained between them. Otoh, a dipole that's a significant fraction of a wavelength will have a greater effect on the fields outside it and so it radiates well.
Of course the edges do but we don't consider the scatter field on the edges and only look within the capacitor. But I'm still not understanding why, the B and E field radiate in the first place. In the lecture we've always had a coil, a capacitor, a resistor etc. and though there are B and E fields, they are essentially within a "locked" place. But now these fields move and move forever and at the speed of light. It's not clicking why.
 
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Of course the edges do but we don't consider the scatter field on the edges and only look within the capacitor. But I'm still not understanding why, the B and E field radiate in the first place. In the lecture we've always had a coil, a capacitor, a resistor etc. and though there are B and E fields, they are essentially within a "locked" place. But now these fields move and move forever and at the speed of light. It's not clicking why.
You have to remember what field is. It is the force that would be felt by a test charge due to the electric charge of a source. The field changes only if the source charge changes or moves. In your capacitor example the field doesn’t change because the charges don’t change or move. If they did move, for example like when you charging the capacitor then the field would change. Any change or movement of the source charge can’t instantly be felt at a distance. Instead the change in the field must propagate. For example if you rapidly changed the charge on your capacitor from Q1 to Q2 the field configuration would change from static condition 1 to static condition 2. However, the change from 1 to 2 couldn’t happen everywhere instantly. Instead the change in the field would ripple outward at the speed of light.
 

phyzguy

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As jtbell explained, it's the changing nature of the charges and currents that gives rise to the radiation. In a capacitor with a DC voltage, there will be no radiation. If you have your dipole antenna with a DC current flowing, there will be no radiation. But you are feeding a changing current into your dipole. Maxwell's equations say that the changing current gives rise to a changing B-field. The changing B-field gives rise to a changing E-filed, which gives rise to a changing B-field, and so on... As to why EM fields behave this way, there is really no answer other than, "this is the way the universe behaves, and Maxwell's equation give a quantitative description of the phenomena".
 
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How do the Maxwell equations explain the how/why E and B spread out?
I think that the easiest way to see that is from a result that you can derive from Maxwell’s equations. This derived result is called Jefimenko’s equations. It describes the relationship between charge, current, and the fields, but in a way that explicitly shows the dependence on time and space.

https://en.m.wikipedia.org/wiki/Jefimenko's_equations

When we look at an E-Field from a capacitor, the E-Field stays within the capacitor. Why with a dipole do they move outward?
The rules are the same for a capacitor and a dipole. In the right circumstance both can produce radiation and in the right circumstance both can store energy in the E field. The capacitor is a less effective shape for radiation and an antenna is a less effective shape for storage, but both can do either.
 
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Mister T

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In the lecture we've always had a coil, a capacitor, a resistor etc. and though there are B and E fields, they are essentially within a "locked" place.
When you rub a balloon on your hair it will attract bits of paper that are outside of the balloon.

When you run a current through a wire you find that compass needles deflect even though they are outside of the wire.

When you move a bar magnet you find that compass needles deflect even though they are outside the bar magnet.

When you move a bar magnet near a wire you find that a current flows in the wire, even though it is outside of the magnet.

When you drop a ball it falls towards planet Earth even though it is outside of planet Earth.
 

sophiecentaur

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As jtbell explained, it's the changing nature of the charges and currents that gives rise to the radiation. In a capacitor with a DC voltage, there will be no radiation.
It is very much easier to detect and measure the 'unwanted' alternating fields radiated from a capacitor or an inductor than it is to measure the stray stationary fields. This, of course, is basically because there is no actual ENERGY transported by the static fields. Receivers for all but the lowest frequencies can be made incredibly sensitive.
 

bob012345

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As jtbell explained, it's the changing nature of the charges and currents that gives rise to the radiation. In a capacitor with a DC voltage, there will be no radiation. If you have your dipole antenna with a DC current flowing, there will be no radiation. But you are feeding a changing current into your dipole. Maxwell's equations say that the changing current gives rise to a changing B-field. The changing B-field gives rise to a changing E-filed, which gives rise to a changing B-field, and so on... As to why EM fields behave this way, there is really no answer other than, "this is the way the universe behaves, and Maxwell's equation give a quantitative description of the phenomena".
Isn't it possible to have a very slowly varying current in a wire such that no radiation is actually radiated? Consider a long wire with a DC current that generates a 1/r magnetic field around the wire. The field strength is always 1/r but the signal takes time to get to r so one sees it. Suppose the current switches on, stays on for a while then switches off with the switching time about 1/10th the on time. Won't one just see a decaying pulse like a stone thrown in a lake and not an Maxwellian EM wave? Or am I splitting hairs with definitions between pulses and waves? Should I think of the pulse as a form of EM wave? Thanks.
 

sophiecentaur

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Or am I splitting hairs with definitions between pulses and waves?
I would say they are two distinct things. A Pulse is a time variation and a wave has both spatial and temporal variation. Not all waves are 'sinusoids'. A wave can have the short duration and carry the energy within a short distance interval.
Fourier tells us that a (temporal) pulse has an equivalent spectrum of a range of infinitely long sinusoids. The narrower the pulse in the time domain, the wider the spectrum of sinusoids in the frequency domain. Both descriptions are equally valid.
 

phyzguy

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Isn't it possible to have a very slowly varying current in a wire such that no radiation is actually radiated? Consider a long wire with a DC current that generates a 1/r magnetic field around the wire. The field strength is always 1/r but the signal takes time to get to r so one sees it. Suppose the current switches on, stays on for a while then switches off with the switching time about 1/10th the on time. Won't one just see a decaying pulse like a stone thrown in a lake and not an Maxwellian EM wave? Or am I splitting hairs with definitions between pulses and waves? Should I think of the pulse as a form of EM wave? Thanks.
In your scenario, the current is changing, so there will be EM waves. As sophiecentaur said, waves do not need to be pure sinusoids. Using Fourier analysis, any waveform (including a "pulse") can be decomposed into a superposition of pure sinusoidal waves with different frequencies.

The fact that changes in the field do not propagate at infinite speed (as you said, "the signal takes time to get to r ...") is the reason there must be EM waves whenever the sources (charges and currents) are changing. That is the whole point.
 

bob012345

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In your scenario, the current is changing, so there will be EM waves. As sophiecentaur said, waves do not need to be pure sinusoids. Using Fourier analysis, any waveform (including a "pulse") can be decomposed into a superposition of pure sinusoidal waves with different frequencies.

The fact that changes in the field do not propagate at infinite speed (as you said, "the signal takes time to get to r ...") is the reason there must be EM waves whenever the sources (charges and currents) are changing. That is the whole point.
I was assuming the change is adiabatic. I was also thinking that EM field can be radiated away under certain conditions and be constrained to bound to the antenna under other conditions like soap bubbles either remaining attached or breaking free (forgive my simplistic analogies). Static fields still "propagate" or I should say reach further distances in space at the speed of light yet they are not waves at all. Anytime anything is turned on, it takes forever to completely set up its field. I can see calling it a wave it the beginning or when the pulse is turned off.
 

phyzguy

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Static fields still "propagate" or I should say reach further distances in space at the speed of light yet they are not waves at all.
This where your misconception lies. The propagation of any change to the EM field to "further distances in space at the speed of light" is an EM wave.

I was assuming the change is adiabatic.
Since any change to the charges and currents causes radiation, it is not possible for a change to be adiabatic.
 

sophiecentaur

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Static fields still "propagate" or I should say reach further distances in space at the speed of light yet they are not waves at all.
I think you are suggesting a 'step function' causing the field to change. What spreads out qualifies as a Wave same as all the others.

PS The switch would need to have been 'off' for all time until it is 'on' and remains that way for all subsequent time. Still a wave in operation.
 

bob012345

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This where your misconception lies. The propagation of any change to the EM field to "further distances in space at the speed of light" is an EM wave.



Since any change to the charges and currents causes radiation, it is not possible for a change to be adiabatic.
Thanks. I agree.
 

bob012345

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I think you are suggesting a 'step function' causing the field to change. What spreads out qualifies as a Wave same as all the others.

PS The switch would need to have been 'off' for all time until it is 'on' and remains that way for all subsequent time. Still a wave in operation.
Thanks. Then static fields are idealizations or approximations in time and space.
 

sophiecentaur

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Thanks. Then static fields are idealizations or approximations in time and space.
If you like. But the same thing applies to everything. When we talk in terms of limits and treat variables as continuous etc etc it delivers pretty good answers in most cases. But we also find that Energy occurs in quanta so you have to take your pick
 

Cryo

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Cryo

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Just to clarify. With non-radiating confirguration the current does not have oscillate slowly. In principle it can be arbitrarily fast (even with finite-size source), but it is only possible for special kinds of current distributions.
 

sophiecentaur

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Yep, it is possible. Such special cases are called non-radiating configurations. There is a lot of literature on this topic. The best easy-reading intro I know of is Greg Gbur's thesis ("Nonradiating Sources and the Inversen Source Problem", the intro chapter) https://pages.uncc.edu/greg-gbur/wp-content/uploads/sites/158/2012/11/thesis.pdf
There is a relatively common example of this sort of thing in the design of long lengths of multi-conductor cables (telecoms application) in which the cross talk is reduced by splitting the cable into equal length sections and switching signals systematically from pair to pair to make the cross talk signal cancel out over a long distance. It works for many twisted pairs in one cable.
 

bob012345

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Yep, it is possible. Such special cases are called non-radiating configurations. There is a lot of literature on this topic. The best easy-reading intro I know of is Greg Gbur's thesis ("Nonradiating Sources and the Inversen Source Problem", the intro chapter) https://pages.uncc.edu/greg-gbur/wp-content/uploads/sites/158/2012/11/thesis.pdf
Thanks. I'll look at it with interest. I should have known as I've seen work on non-radiating but accelerating charge configurations before.
http://www.slac.stanford.edu/cgi-wrap/getdoc/slac-pub-3514.pdf

https://en.m.wikipedia.org/wiki/Nonradiation_condition


https://journals.aps.org/pr/abstract/10.1103/PhysRev.135.B281

https://aapt.scitation.org/doi/pdf/10.1119/1.14729?class=pdf
 

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