# Propagation in an isotropic medium

This part continues from the Polarization of Light tutorial.

The propagation of an optical wave is governed by the wave equation. It depends on the optical property and physical structure of the medium. It also depends on the makeup of the optical wave, such as its frequency contents and its temporal characteristics.

In this part, we consider the basic characteristics of the propagation of a monochromatic plane optical wave in an infinite homogeneous medium. For such a monochromatic wave, there is only one value of k and one value of ω. Its complex electric field is that given by (58) in the Polarization of Light tutorial, in which the field amplitude $$\pmb{\mathcal{E}}$$ is independent of r and t. Thus,

$\tag{76}\mathbf{P}(\mathbf{r},t)=\epsilon_0\pmb{\chi}(\mathbf{k},\omega)\cdot\mathbf{E}(\mathbf{r},t)$

and

$\tag{77}\mathbf{D}(\mathbf{r},t)=\pmb{\epsilon}(\mathbf{k},\omega)\cdot\mathbf{E}(\mathbf{r},t)$

Also, in this part, we shall assume no spatial nonlocality in the media thus neglecting the k dependence of χ and ε. Then,

$\tag{78}\mathbf{P}(\mathbf{r},t)=\epsilon_0\pmb{\chi}(\omega)\cdot\mathbf{E}(\mathbf{r},t)$

and

$\tag{79}\mathbf{D}(\mathbf{r},t)=\pmb{\epsilon}(\omega)\cdot\mathbf{E}(\mathbf{r},t)$

For a monochromatic wave of a frequency ω, the wave equation is simply

$\tag{80}\boldsymbol{\nabla}\times\boldsymbol{\nabla}\times\mathbf{E}+\mu_0\pmb{\epsilon}(\omega)\cdot\frac{\partial^2\mathbf{E}}{\partial t^2}=0$

For an isotropic medium, ε(ω) is reduced to a scalar ε(ω) and

$\tag{81}\boldsymbol{\nabla}\cdot\mathbf{E}=\frac{1}{\epsilon(\omega)}\boldsymbol{\nabla}\cdot\mathbf{D}=0$

Then, by using the vector identity $$\boldsymbol{\nabla}\times\boldsymbol{\nabla}\times=\boldsymbol{\nabla}\boldsymbol{\nabla}\cdot-\nabla^2$$, the wave equation in (80) is reduced to the following simple form:

$\tag{82}\nabla^2\mathbf{E}-\mu_0\epsilon(\omega)\frac{\partial^2\mathbf{E}}{\partial t^2}=0$

For an anisotropic medium, (82) is generally not valid because (81) does not hold.

Note that with $$\pmb{\mathcal{E}}$$ in (58) [in the polarization of light tutorial] being independent of r and t, we can make the following replacement for the operators when operating on E of the form in (58) or H of the same form:

$\tag{83}\boldsymbol{\nabla}\rightarrow\text{i}\mathbf{k},\qquad\frac{\partial}{\partial t}\rightarrow-\text{i}\omega$

#### Free Space

In free space, P = 0 and ε is reduced to the scalar ε0. Substitution of (58) in (82) then yields

$\tag{84}k^2=\omega^2\mu_0\epsilon_0$

The propagation constant in free space is

$\tag{85}k=\frac{\omega}{c}=\frac{2\pi\nu}{c}=\frac{2\pi}{\lambda}$

where ν is the frequency of the optical wave and λ is its wavelength. Because k is proportional to 1/λ, it is also called the wavenumber.

Using (83) and noting that B = μ0H and D = ε0E, Maxwell's equations in (40) - (43) [in the harmonic fields tutorial] become

$\tag{86}\mathbf{k}\times\mathbf{E}=\omega\mu_0\mathbf{H}$$\tag{87}\mathbf{k}\times\mathbf{H}=-\omega\epsilon_0\mathbf{E}$$\tag{88}\mathbf{k}\cdot\mathbf{E}=0$$\tag{89}\mathbf{k}\cdot\mathbf{H}=0$

From (86) and (87), we also have

$\tag{90}\mathbf{E}\cdot\mathbf{H}=0$

Therefore, the three vectors E, H, and k are orthogonal. These relationships also imply that

$\tag{91}\mathbf{S}\parallel\mathbf{k}$

The relationships among the directions of these vectors are shown in figure (7) below.

Using (85), we can also write (86) and (87) in the following form:

$\tag{92}\mathbf{H}=\frac{1}{Z_0}\hat{k}\times\mathbf{E},\qquad\mathbf{E}=Z_0\mathbf{H}\times\hat{k}$

where $$\hat{k}=\mathbf{k}/k$$ is the unit vector in the k direction and

$\tag{93}Z_0=\sqrt{\frac{\mu_0}{\epsilon_0}}\approx120\pi\;\Omega\approx377\;\Omega$

is the free-space impedance. The concept of this impedance is not that of the impedance of a resistor but is analogous to the concept of the impedance of a transmission line.

Because $$\mathbf{S}\parallel\mathbf{k}$$, the light intensity in free space can be expressed as

$\tag{94}I=\hat{k}\cdot\overline{\pmb{S}}=2\frac{|\mathbf{E}|^2}{Z_0}=2Z_0|\mathbf{H}|^2$

#### Lossless Medium

In this case, ε(ω) is reduced to a positive real scalar ε(ω), which is different from ε0. All of the results obtained for free space remain valid, except that ε0 is replaced by ε(ω). This change of the electric permittivity from a vacuum to a material is measured by the relative electric permittivity, ε/ε0, which is a dimensionless quantity also known as the dielectric constant of the material. Therefore, the propagation constant in the medium is

$\tag{95}k=\omega\sqrt{\mu_0\epsilon}=\frac{n\omega}{c}=\frac{2\pi n\nu}{c}=\frac{2\pi n}{\lambda}$

where

$\tag{96}n=\sqrt{\frac{\epsilon}{\epsilon_0}}=(\text{dielectric constant})^{1/2}$

is the index of refraction, or refractive index, of the medium.

In a medium that has an index of refraction n, the optical frequency is still ν, but the optical wavelength is λ/n, and the speed of light is v = c/n. Because n(ω) in a medium is generally frequency dependent, the speed of light in a medium is also frequency dependent. This results in various dispersive phenomena such as the separation of different colors by a prism and the broadening or shortening of an optical pulse traveling through a medium. We also find that

$\tag{97}Z=\frac{Z_0}{n}$

in a medium. The light intensity is than

$\tag{98}I=2\frac{|\mathbf{E}|^2}{Z}=2Z|\mathbf{H}|^2=\frac{2k}{\omega\mu_0}|\mathbf{E}|^2=\frac{2k}{\omega\epsilon}|\mathbf{H}|^2$

#### Medium with Loss or Gain

As discussed in the previous tutorials,  χ and ε become complex when a medium has an optical loss or gain. Therefore,

$\tag{99}k^2=\omega^2\mu_0\epsilon=\omega^2\mu_0(\epsilon'+\text{i}\epsilon'')$

and the propagation constant k becomes complex:

$\tag{100}k=k'+\text{i}k''=\beta+\text{i}\frac{\alpha}{2}$

The index of refraction also becomes complex:

$\tag{101}n=\sqrt{\frac{\epsilon'+\text{i}\epsilon''}{\epsilon_0}}=n'+\text{i}n''$

The relation between k and n in (95) is still valid. Meanwhile, the impedance Z of the medium also becomes complex. Therefore, E and H are no longer in phase, as can be seen from (92) by replacing Z0 with a complex Z, and I is not simply given by (98) but is given by the real part of it.

It can be shown that if we choose β to be positive, the sign of α is the same as that of ε''. In this case, k' and n' are also positive and k'' and n'' also have the same sign as ε''. If we consider as an example an optical wave propagating in the z direction, then $$\hat{k}=\hat{z}$$ and, from (58) and (100), the complex electric field is

$\tag{102}\mathbf{E}(\mathbf{r},t)=\pmb{\mathcal{E}}e^{-\alpha z/2}\exp(\text{i}\beta z-\text{i}\omega t)$

It can be seen that the wave has a phase that varies sinusoidally with a period of 1/β along z. In addition, its amplitude is not constant but varies exponentially with z. Thus, light intensity is also a function of z:

$\tag{103}I\propto{e^{-\alpha z}}$

Clearly,  β is the wavenumber in this case, and the sign of α determines the attenuation or amplification of the optical wave:

1. If χ'' > 0, then ε'' > 0 and α > 0. As the optical wave propagates, its field amplitude and intensity decay exponentially along the direction of propagation. Therefore, α is called the absorption coefficient or attenuation coefficient.
2. If χ'' < 0, then ε'' < 0 and α < 0. The field amplitude and intensity of the optical wave grow exponentially. Then, we define g = - α as the gain coefficient or amplification coefficient.

The unit of both α and g is per meter, often also quoted per centimeter.

Example

The complex susceptibility of GaAs at an optical wavelength of λ = 850 nm is χ = 12.17 + i0.49. Therefore, at this wavelength, GaAs has a complex refractive index of

$n=(\epsilon/\epsilon_0)^{1/2}=(1+\chi)^{1/2}=(13.17+\text{i}0.49)^{1/2}=3.63+\text{i}0.0676$

and an absorption coefficient of

$\alpha=2k''=\frac{4\pi n''}{\lambda}=\frac{4\pi\times0.0676}{850\times10^{-9}}\text{m}^{-1}=10^6\,\text{m}^{-1}$

An optical beam at 850 nm wavelength can travel in GaAs only for a distance of l = - ln(1 - 0.99)/α = 4.6 μm before losing 99% of its energy to absorption, which is obtained by solving

$1-e^{-\alpha l}=0.99\;\text{ with }\alpha=10^6\text{ m}^{-1}$

The next part continues with the Propagation in an Anisotropic Medium tutorial.