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Erland

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How can this be explained?

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Erland

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How can this be explained?

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Andrew Mason

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Two wave pulses will simply pass through each other and continue on afterward. When a pulse encounters another pulse there will be some interference as they pass through each other. But they will keep going afterward. There will not be perfect interference, however, because the pulses cannot be monochromatic. They are made up of a range of frequencies.

How can this be explained?

I am not the person to ask, but I expect that a better answer involves principles of quantum electrodynamics.

While electromagnetic waves appear in the macroscopic world, at the quantum level the energy is in the form of photons. So I think the question should be: what happens when two otherwise identical photons collide head-on? You might ask that question in the Quantum Physics forum.

AM

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The same thing which makes the antenna a good emitter of RF also makes it a good absorber of RF. So the net energy out of each antenna will be less with the other antenna since it will be emitting the same energy as before but also absorbing energy from the other antenna for less net energy.

How can this be explained?

Note, the reduction in total energy is NOT due to interference, it is due to absorption. Interference does not reduce energy at all, but your scenario involves more than just interference.

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Erland

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Is this really correct?Interference does not reduce energy at all

Suppose that we have two monochromatic spherical waves with the same wavelengths and amplitudes, but opposite phases, propagating from two points with distance

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Yes, but if you look at WHERE the energy is lost you will find that it is all lost at the points of origin. I.e. It is lost through absorption by matter at the sources, not through interference in free space.Is this really correct?

Suppose that we have two monochromatic spherical waves with the same wavelengths and amplitudes, but opposite phases, propagating from two points with distancehbetween them. Then, it is not difficult to prove that the magnitude of the field (or whatever it is), due to interference is ##O(h/r)## where ##r## is the distance from the field point to one of the points of origin for the waves. Since the magnitude for one of the waves is ##O(1/r)##, and the energy is proportional to the square of the magnitude, the energy is substantially reduced and in fact tends to ##0## at every point outside the originating points (this holds also for the surface integrals of the energy over spheres surrounding these origins) when ##h\to 0##.

In the absence of matter the energy in the wave is conserved, regardless of interference.

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Erland

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1. In the case with the two antennas, what exactly will happen? If we assume that we are running both antennas at the same power, what happens with the energy they get from the electric power supply (which we assume is the source of energy here)? Is it lost as heat or does radiation from the other antenna create impedance so that the first antenna cannot be run at the intended power?

2. The energy density of the electromagnetic field is ##\frac 12(\epsilon |\bf {E}|^2 +\frac 1\mu |\bf {B}|^2)##. From what you say, it seems that this has nothing to do with where energy can be absorbed or emitted locally, but just is a function which gives the total energy if it is integrated over the entire space. Does this energy density have no other significance than this?

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It depends on their mutual inductance and how well their impedances are matched. It could (most likely IMO) wind up as heating in the antenna structure, or it could potentially be seen at the power source in which case the power source could either store the energy or lose it to heat there.1. In the case with the two antennas, what exactly will happen? If we assume that we are running both antennas at the same power, what happens with the energy they get from the electric power supply (which we assume is the source of energy here)? Is it lost as heat or does radiation from the other antenna create impedance so that the first antenna cannot be run at the intended power?

I don't really see what you are trying to say. Obviously, you cannot absorb energy at some location if the density is zero there, and if you emit energy locally then the local density becomes non-zero. So I wouldn't say that it has nothing to do with it. But you are correct that integrating it over the whole space gives the total EM energy.2. The energy density of the electromagnetic field is ##\frac 12(\epsilon |\bf {E}|^2 +\frac 1\mu |\bf {B}|^2)##. From what you say, it seems that this has nothing to do with where energy can be absorbed or emitted locally, but just is a function which gives the total energy if it is integrated over the entire space. Does this energy density have no other significance than this?

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Claude Bile

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Claude.

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