How an electric field is absorbed by a dielectric

[Soumava]

when a em waves strikes a dielectric the atoms vibrate in response to the electric field and if the frequency matches the resonant frequency the Lorentz oscillator the electric field is absorbed how a field can be absorbed we know the em field contain energy how is the em field destroyed is some kind of opposite field created what's the physical picture of fields when a em wave strike a free electron

 

[jfizzix]

It would be very nice to have a play-by-play description of how a classical EM wave is absorbed by atoms. Quantum-mechanically, we talk about the absorption and emission of photons, which are quanta of energy of the EM field.

The best answer that I can give without photons is that atoms are described as elastic electric dipoles, with the negative electron cloud surrounding the positive nucleus.

When the EM wave reaches the atom, the electrons get pulled by the electric field a little bit away from the nucleus inducing an electric dipole in the atom. This electric dipole would oscillate at its natural frequency, but is being driven by the frequency of the EM wave. The oscillating electric dipole radiates its own electric field (as all accelerating charges do) at the same frequency as the driving field, but slightly out of phase due to the inertia of the atoms (among other factors). When the frequency of the EM wave is equal to the natural frequency of the dipole, the dipole oscillates with maximum amplitude, and more energy from the EM wave is converted to kinetic energy of the dipole.

Depending on various properties of these atoms, this dipole radiation can be released quickly or slowly, but is released all the same, and in all directions. Because of multiple atomic energy levels (and therefore multiple natural frequencies), the emitted light need not be the same frequency as the light of the EM wave, and could even be in the infrared while the initial EM wave was visible.

This, of course is a hand-wavey description, and specific situations will vary in the details.

 

[ZapperZ]

when a em waves strikes a dielectric the atoms vibrate in response to the electric field and if the frequency matches the resonant frequency the Lorentz oscillator the electric field is absorbed how a field can be absorbed we know the em field contain energy how is the em field destroyed is some kind of opposite field created what's the physical picture of fields when a em wave strike a free electron

First of all, you need to learn how to use punctuations! Your post above does not contain even a single one, and it makes for a very confusing, difficult reading.

Secondly, in a solid, the property of a solid does NOT depend only on the atoms/molecules that make up the solid, but also the collective property of a large group of atoms/molecules attached to each other. There are not "conduction electrons" or "band gap" in individual atoms, but there are such characteristics in a solid.

One important feature relevant here is the collective vibrational mode of the solid, often called phonons. Optical phonons are "active" to certain range of frequencies of EM wave. So they can affect whether a particular frequency transmits, transmits partially, or gets completely absorbed. This behavior is exploited in many experimental methods, such as UV-VIS and Raman spectroscopy, and why we can study the phonon spectrum using EM radiation.

If the EM wave is absorbed, the energy is typically transformed into vibrational energy of the solid, i.e. heat. If the EM wave has a "high" enough energy, then other phenomena can occur that transform the EM energy into a number of other channels.

Zz.

 

[Soumava]

It would be very nice to have a play-by-play description of how a classical EM wave is absorbed by atoms. Quantum-mechanically, we talk about the absorption and emission of photons, which are quanta of energy of the EM field.

The best answer that I can give without photons is that atoms are described as elastic electric dipoles, with the negative electron cloud surrounding the positive nucleus.

When the EM wave reaches the atom, the electrons get pulled by the electric field a little bit away from the nucleus inducing an electric dipole in the atom. This electric dipole would oscillate at its natural frequency, but is being driven by the frequency of the EM wave. The oscillating electric dipole radiates its own electric field (as all accelerating charges do) at the same frequency as the driving field, but slightly out of phase due to the inertia of the atoms (among other factors). When the frequency of the EM wave is equal to the natural frequency of the dipole, the dipole oscillates with maximum amplitude, and more energy from the EM wave is converted to kinetic energy of the dipole.

Depending on various properties of these atoms, this dipole radiation can be released quickly or slowly, but is released all the same, and in all directions. Because of multiple atomic energy levels (and therefore multiple natural frequencies), the emitted light need not be the same frequency as the light of the EM wave, and could even be in the infrared while the initial EM wave was visible.

This, of course is a hand-wavey description, and specific situations will vary in the details.

how energy of wave is converted into kinetic energy of the particle

 

[jfizzix]

how energy of wave is converted into kinetic energy of the particle

I don't know if there is a good answer to this question.

It may help to imagine the process in reverse.
If you have an oscillating dipole, it will produce electromagnetic radiation because the charges are accelerating. The conservation of energy requires that the energy of the radiation released be equal to the kinetic energy lost.
A fuller description of this mechanism can be found by looking up how the Larmor formula for the power radiated by accelerating charges works.

 

[Soumava]

I don't know if there is a good answer to this question.

It may help to imagine the process in reverse.
If you have an oscillating dipole, it will produce electromagnetic radiation because the charges are accelerating. The conservation of energy requires that the energy of the radiation released be equal to the kinetic energy lost.
A fuller description of this mechanism can be found by looking up how the Larmor formula for the power radiated by accelerating charges works.

what happens to the original electromagnetic wave

 

[Vanadium 50]

Soumava, I've got to admire your fortitude - sticking to your guns and refusing to use punctuation and writing complete sentences. That'll teach us! Yesiree! Couple that with providing us clarifying information one tiny piece at a time and you can make a question that could be answered in five minutes stretch out for days.

Or, you could think about the question you are asking, write it clearly, concisely, and completely - using proper English - and get your answer quickly. Up to you.

 

[jfizzix]

what happens to the original electromagnetic wave

This is a good question to ask, but I can only give a partial answer without talking about photons.

To get some intuition for what happens to an EM wave as it gets absorbed, you will want to look up how EM waves behave in a medium that can conduct electricity, since here, we have free charges interacting with the field.

Looking at Maxwell's equations, you can argue that for an EM wave in a vacuum, the rate of change of the electric field drives the magnetic field, while the rate of change of the magnetic field in turn drives the electric field.

In a conductor, this simple reciprocal relationship is no longer true, as the magnetic field is driven both by the rate of change of the electric field, and the electric current, which in turn is approximately proportional to the electric field (from Ohm's law).

[[]]
This is a hand-wavey way of looking at it, but...
Because the electric field is causing the charges to move, and because moving charges generate magnetic fields, and because accelerating charges generate changing magnetic fields, accelerating charges produce electric fields in the opposite direction of their acceleration, subtracting away from the original electric field, which decreases it.
[[]]

Because of the extra dependence on electric current, the EM wave equation you get by taking the curl of both sides of Faraday's law contains an extra term that depends on the rate of change of the current (i.e., the overall acceleration of these moving free charges). This acceleration term acts as a damping term in the wave equation. Indeed, when you solve Maxwell's equations for a plane-wave incident on a conductor, you see the EM wave decrease in intensity exponentially with distance through the medium.

I know this answer is very mathematical, but I hope it helps a little.

 

[Soumava]

This is a good question to ask, but I can only give a partial answer without talking about photons.

To get some intuition for what happens to an EM wave as it gets absorbed, you will want to look up how EM waves behave in a medium that can conduct electricity, since here, we have free charges interacting with the field.

Looking at Maxwell's equations, you can argue that for an EM wave in a vacuum, the rate of change of the electric field drives the magnetic field, while the rate of change of the magnetic field in turn drives the electric field.

In a conductor, this simple reciprocal relationship is no longer true, as the magnetic field is driven both by the rate of change of the electric field, and the electric current, which in turn is approximately proportional to the electric field (from Ohm's law).

[[]]
This is a hand-wavey way of looking at it, but...
Because the electric field is causing the charges to move, and because moving charges generate magnetic fields, and because accelerating charges generate changing magnetic fields, accelerating charges produce electric fields in the opposite direction of their acceleration, subtracting away from the original electric field, which decreases it.
[[]]

Because of the extra dependence on electric current, the EM wave equation you get by taking the curl of both sides of Faraday's law contains an extra term that depends on the rate of change of the current (i.e., the overall acceleration of these moving free charges). This acceleration term acts as a damping term in the wave equation. Indeed, when you solve Maxwell's equations for a plane-wave incident on a conductor, you see the EM wave decrease in intensity exponentially with distance through the medium.

I know this answer is very mathematical, but I hope it helps a little.

I understand but please provide me a more layman expanation
is any kind of opposite field created as you mentioned