How does electromagnetic radiation produce heating effect in a material?

A laser beam (a form of electromagnetic radiation) has various applications in laser cutting, drilling, welding etc. which is possible by melting or vaporization of target material by heat produced by laser.
My question:

How does electromagnetic radiation produce heating effect in a material?

• How does the sun’s electromagnetic radiation heat the earth? Jul 26, 2020 at 8:55

To understand how EM radiation can cause the heating up of a certain material, it is very important to understand what we mean by the heat energy of the material and how it is stored. It is stored in the degrees of freedom of atoms and molecules.

Heat energy, at a microscopic level, is stored in the degrees of freedom of atoms and molecules. These degrees of freedom are translational, rotational and vibrational. They all store different amounts of energy, depending on the geometry of the atom. Translational degrees of freedom are the atom or molecule moving around in space, and there are always 3 for the 3 dimensions of space. The rotational and vibrational modes come from the geometry of the atom/molecule.

How is heat represented on a quantum level?

There are mainly three types of freedoms in connection with heat capacity:

1. translational

Translational degrees of freedom arise from a gas molecule's ability to move freely in space.

1. rotational

A molecule's rotational degrees of freedom represent the number of unique ways the molecule may rotate in space about its center of mass which a change in the molecule's orientation.

1. vibrational

The number of vibrational degrees of freedom (or vibrational modes) of a molecule is determined by examining the number of unique ways the atoms within the molecule may move relative to one another, such as in bond stretches or bends.

https://en.wikibooks.org/wiki/Statistical_Thermodynamics_and_Rate_Theories/Degrees_of_freedom

Now when a photon interacts with the material's atoms and molecules, it might be absorbed (transfers all its energy and ceases to exist) or inelastically scattered (transfers part of its energy and changes angle).

As the photon transfers its energy to the atom or molecule, then its translational, vibrational or rotational energies might rise, and the material heats up.

Let's set the laser part of your question aside for a moment. When a wave of any type (electromagnetic, sound, etc) meets a boundary separating two different media the following can happen (typically in various combinations):

Reflection: Some portion, usually not all, of the wave will be reflected from the surface. Metals are highly reflective in the visible portion of the spectrum. Some metals, like gold or copper, have some absorption (see below) in the blue region so that is why they have a reddish/yellowish hue to them.

Transmission: Some portion, but maybe none, will be transmitted through the medium. Hearing sound standing outside a closed room is an example. Visible light transmission through glass is another.

Absorption: Some portion, again usually not all, of the wave will be absorbed by the surface. This will occur through some depth of the material. This is what will produce heating in a material.

E&M waves striking surfaces around us obviously have a broad range of wavelengths - some in radio wave, some in infrared, some in visible. If you are outside, some in UV.

There is much that can be said about all of these, but since this is a limited space, I'll take the liberty of generalizing. Hopefully folks will not think I am leaving out something critical.

Radio waves mainly transmit through most non-metallic materials around us (thus you can talk on your cell phone or listen to an old antenna style radio inside your home). Most visible light is reflected within a very short distance of the surface of materials (we see things almost entirely through reflected light).

It is thus mostly infrared radiation that will induce warming of materials. (As stated, this is a generalization. Microwaves can obviously heat food to the point of cooking.) This is because, in some cases, it can directly excite molecular vibrations in materials. The molecules absorb infrared photons and are excited to higher vibration/rotation quantum levels. These molecules then de-excite through collisions which heat the material.

But mainly it is because the molecules, or ions that make up the structure of the solid are slightly distorted by the radiation. Because they are charged or may have dipole moments, they can interact with the E&M field. If you make a simple model of an atom as an electron bound through a spring-like constant to the nucleus and write out a simple equation of motion, it will include a damping factor which is responsible for the absorption. E&M waves interacting with a model system like this represent an example of forced oscillations.

In the book Waves by Frank Crawford, appendix 9 is devoted to E&M radiation in material bodies. There he writes:

The damping force represents transfer of energy from the oscillating charge to the medium. This energy is no longer in either the electromagnetic field components of frequency $$\omega$$ or is it in the oscillation energy of M but is instead in the form of translational and rotational energy of the atoms, and also of "random" vibrations at other frequencies. It is called HEAT.

Much, much more could be written about this subject. At a higher level, Ziman's Principles of the Theory of Solids, discusses much about E&M waves and solids in chapter 8. But this is in essence how E&M waves heat materials: Absorption of the wave, which will occur over different depths depending upon the wavelength of the radiation, will induce motion in the material constituents.

Also, here is a great Physics Today article on atmospheric heating by IR radiation.

There is also a nice general discussion of light and matter by Victor Weisskopf found here.

With specification to lasers and ablation, see this PSE answer. Hope this helps!

Basically because all materials are made from charges (protons and electrons). Their kinetic energy increases as a consequence of the EM wave.

It can be shown that for a plane wave, the work done on the charges, per unit of volume is:$$W = \mathbf E .\mathbf j$$, where $$\mathbf E$$ is the electric field of the wave, and $$\mathbf j$$ is the locally density of current.

I would for that problem a classical dipole oscillator model, which is called the Drude-Lorentz-Model.

At first we need a complex refractive index, which we will denote as $$n = n´ + in´´$$ (refractive index + i extinction coefficient) and will provide a way to model a absorption (due to the imaginary part). The refractive index is directly linked to the complex electric constant / relative permittivity $$\epsilon_r = \epsilon´ + \epsilon´´$$. So lets see how this is related to each other:

At first I´d like to show you why the imaginary part of the complex refractive index is considered to account for absorption:

Lets assume a monochromatic plane wave (laser) moving along x as the electric field:

$$E = E_0 * exp[i(kx-\omega t)]$$

with the wavevector $$k = n\frac{\omega}{c} = n*k_0 = n * \frac{2\pi}{\lambda}$$ we can write:

$$E = E_0* exp[-n´´k_0x]*exp[i(n´k_0x-\omega t)]$$

Here you can see the exponential attenuation of the electric field amplitude. For thoroughness, the Intensity I is proportional to the square of the electric field amplitude $$I \propto \epsilon^*\epsilon$$ and therefore we can connect the extinction coefficient and the absorption coefficient via Beer´s law $$I = I_0 exp(-\alpha x)$$, which will give us $$\alpha = 2n´´k_0 = 2n´´\frac{\omega}{c} = n´´\frac{4\pi}{\lambda}$$

Ok that´s the first step to understand absoption. Now we need to connect our dipole-oscillator-model (wich will provide the relative permittivity $$\epsilon$$) and the complex refractive index, which is given by:

$$\epsilon_r = n^2 = \epsilon´ + i\epsilon´´ = n´^2 - n´´^2 + i2n´n´´$$

for integrity i´d like to point out, that the complex (optical) conductivity $$\sigma$$ is also related to that via $$\epsilon = 1+\frac{\sigma}{\epsilon_0 \omega}$$, which results from the maxwell-equations in matter.

Now we will look at the oscillator model:

$$\frac{\partial^2 x}{\partial t^2} + \gamma\frac{\partial x}{\partial t} + \omega_0^2x = -\frac{e}{m}E$$

where the term with $$\omega_0$$ (our resonance frequency) accounts for the restoring force, the term with $$\gamma$$ is a frictional force and the plane wave will drive the damped oscillation.

Our Ansatz for a planewave, which looks like $$E(x,t)=E(x)*exp[i\omega t)$$, is $$x= x_0 exp(i\omega t)$$. This inserted into the differential equation above will lead us to:

$$x_0 = \frac{-eE_0/m}{\omega_0^2-\omega^2-i\gamma\omega}$$

With the polarization $$P$$, which is $$P= Np$$ (N beeing the number of atoms per volume and p being a dipole moment with $$p= -ex$$, we´ll get:

$$P = \frac{Ne^2}{m}\cdot\frac{1}{\omega_0^2-\omega^2-i\gamma\omega}$$

Now we want to get the relation of the polarization and the relative permittivity. This we can obtain from the well known relation of the displacement the electric field and the polarization:

$$D = \epsilon_0 E + P = \epsilon_0 E + \epsilon_0 \chi E = \epsilon_0 \epsilon E$$

leading us to $$P = \epsilon_0(\epsilon - 1)E$$, where $$\epsilon_0$$ is the dielectric constant in vaccum (compare Coulombs-law) and hence:

$$\epsilon = 1 + \chi + \frac{Ne^2}{\epsilon_0 m}\cdot\frac{1}{\omega_0^2-\omega^2-i\gamma\omega}$$

with the plasma frequency $$\omega_p = \frac{Ne^2}{\epsilon_0 m}$$

If one separate this in its real and imaginary part and makes a plot, one would see a lorentzian peak in the imaginary part at the frequency $$\omega_0$$

This is our absorption for a resonant interaction of a electromagnetic wave with a oscillatory eigenfrequency of a atom or crystal. With this model we can simulate phonons, discrete electronic transitions and even free electrons, if we set the resonance frequency $$\omega_0 = 0$$ (no restoring force).

Now imagine a whole range of resonance frequencies (continuous bands can´t be modelled with this classical approach) and you will see, that the beam will be absorbed (also reflected) and this excites the material, which will result in non-radiative relaxation processes, which end up as heat.

So the link is: absorption coefficient -> extinction coefficient (imaginary part of complex refractive index) -> permittivity -> polarization -> dipole moment -> driven and damped harmonic oscillator -> driving force is the electromagnetic wave

Have a nice time

(If you want to learn more about Stuff like this, I would recommend: Mark Fox, Optical Properties of Solids)

In simple terms what happens is that electromagnetic radiation has energy and when the radiation is absorbed by some atom, the kinetic energy of the atom increases. Heat can be loosely defined as the kinetic energy of atoms, and thus an increase in the kinetic enrgy leads to heating of the material.