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# The dielectric constant and the refractive index

The refractive index of a material, n, is defined as the ratio of the speed of light in a vacuum to the speed of light in that material.

$$n = {c \over {{c_{\rm{m}}}}}$$ where c is the speed of light in a vacuum and cm the speed of light in the material.

It is possible to derive another equation for the speed of light, this time in terms of the electric permittivity (ε) and magnetic permeability (μ) of the material. For this, we need Maxwell’s equations.

$$\nabla \times {\bf{E}} = - {{\partial {\bf{B}}} \over {\partial t}}$$        (1)
$$\nabla \times {\bf{B}} = \mu \varepsilon {{\partial {\bf{E}}} \over {\partial t}}$$      (2)

Taking the curl of both sides of (1) allows us to combine (1) and (2):

\eqalign{ \nabla \times (\nabla \times {\bf{E}}) = & - {{\partial (\nabla \times {\bf{B}})} \over {\partial t}} \cr = & - \mu \varepsilon {{{\partial ^2}{\bf{E}}} \over {\partial {t^2}}} \cr}

In general for any vector a:

$$\nabla \times (\nabla \times {\bf{a}}) = - {\nabla ^2}{\bf{a}} + \nabla \cdot (\nabla \cdot {\bf{a}})$$

Now in a vacuum, $$\nabla \cdot {\bf{E}} = 0$$. In this case the above equation becomes:

$${\nabla ^2}{\bf{E}} = {\mu _0}{\varepsilon _0}{{{\partial ^2}{\bf{E}}} \over {\partial {t^2}}}$$

Which is the wave equation in three dimensions. Let us consider the 1D equivalent, as this is easier to solve.

$${{{\partial ^2}E} \over {\partial {x^2}}} = {\mu _0}{\varepsilon _0}{{{\partial ^2}E} \over {\partial {t^2}}}$$

A possible solution to this equation is a sinusoidal wave of wavelength λ and speed c:

$$E = {E_0}\sin \left( {2\pi {{x - ct} \over \lambda }} \right)$$

Differentiating with respect to x and t:

\eqalign{ & {{{\partial ^2}E} \over {\partial {x^2}}} = - {E_0}{\left( {{{2\pi } \over \lambda }} \right)^2}\sin \left( {2\pi {{x - ct} \over \lambda }} \right) \cr & {{{\partial ^2}E} \over {\partial {t^2}}} = - {E_0}{\left( {{{2\pi c} \over \lambda }} \right)^2}\sin \left( {2\pi {{x - ct} \over \lambda }} \right) \cr}

And substituting back into the 1D wave equation above:

$$- {E_0}{\left( {{{2\pi } \over \lambda }} \right)^2}\sin \left( {2\pi {{x - ct} \over \lambda }} \right) = - {\mu _0}{\varepsilon _0}{E_0}{\left( {{{2\pi c} \over \lambda }} \right)^2}\sin \left( {2\pi {{x - ct} \over \lambda }} \right)$$
Which can be simplified and rearranged to give an expression for c, the speed of light in a vacuum:

$$c = {({\mu _0}{\varepsilon _0})^{{\raise0.7ex\hbox{{ - 1}} \!\mathord{\left/ {\vphantom {{ - 1} 2}}\right.\kern-0em} \!\lower0.7ex\hbox{2}}}}$$

It turns out that a similar equation is applicable to the speed of light in any material, cm:

$${c_{\rm{m}}} = {(\mu \varepsilon )^{{\raise0.7ex\hbox{{ - 1}} \!\mathord{\left/ {\vphantom {{ - 1} 2}}\right.\kern-0em} \!\lower0.7ex\hbox{2}}}}$$

For a material that is not magnetic the permeability is μ0, so that:

$${c_{\rm{m}}} = {({\mu _0}\varepsilon )^{{\raise0.7ex\hbox{{ - 1}} \!\mathord{\left/ {\vphantom {{ - 1} 2}}\right.\kern-0em} \!\lower0.7ex\hbox{2}}}}$$ for any non-magnetic material.

Using the expressions for c and cm, the refractive index of the material can be expressed in terms of ε and μ.

\eqalign{ & n = {\left( {{{{\mu _0}{\varepsilon _0}} \over {{\mu _0}\varepsilon }}} \right)^{{\raise0.7ex\hbox{{ - 1}} \!\mathord{\left/ {\vphantom {{ - 1} 2}}\right.\kern-0em} \!\lower0.7ex\hbox{2}}}} \cr & n = {\left( {{\varepsilon \over {{\varepsilon _0}}}} \right)^{{\raise0.7ex\hbox{1} \!\mathord{\left/ {\vphantom {1 2}}\right.\kern-0em} \!\lower0.7ex\hbox{2}}}} \cr}

Finally, recall the earlier definition of the dielectric constant in terms of permittivity:

$$\kappa = {\varepsilon \over {{\varepsilon _0}}}$$

Therefore κ = n2 .