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---
title: "Dielectric function"
firstLetter: "D"
publishDate: 2022-01-24
categories:
- Physics
- Electromagnetism
- Quantum mechanics
date: 2022-01-20T22:04:13+01:00
draft: false
markup: pandoc
---
# Dielectric function
The **dielectric function** or **relative permittivity** $\varepsilon_r$
is a measure of how strongly a given medium counteracts
[electric fields](/know/concept/electric-field/) compared to a vacuum.
Let $\vb{D}$ be the applied external field,
and $\vb{E}$ the effective field inside the material:
$$\begin{aligned}
\boxed{
\vb{D} = \varepsilon_0 \varepsilon_r \vb{E}
}
\end{aligned}$$
If $\varepsilon_r$ is large, then $\vb{D}$ is strongly suppressed,
because the material's electrons and nuclei move to create an opposing field.
In order for $\varepsilon_r$ to be well defined, we only consider linear media,
where the induced polarization $\vb{P}$ is proportional to $\vb{E}$.
We would like to find an alternative definition of $\varepsilon_r$.
Consider that the usual electric fields $\vb{E}$, $\vb{D}$, and $\vb{P}$
can each be written as the gradient of an electrostatic potential like so,
where $\Phi_\mathrm{tot}$, $\Phi_\mathrm{ext}$ and $\Phi_\mathrm{ind}$
are the total, external and induced potentials, respectively:
$$\begin{aligned}
\vb{E}
= -\nabla \Phi_\mathrm{tot}
\qquad \qquad
\vb{D}
= - \varepsilon_0 \nabla \Phi_\mathrm{ext}
\qquad \qquad
\vb{P}
= \varepsilon_0 \nabla \Phi_\mathrm{ind}
\end{aligned}$$
Such that $\Phi_\mathrm{tot} = \Phi_\mathrm{ext} + \Phi_\mathrm{ind}$.
Inserting this into $\vb{D} = \varepsilon_0 \varepsilon_r \vb{E}$
then suggests defining:
$$\begin{aligned}
\boxed{
\varepsilon_r
\equiv \frac{\Phi_\mathrm{ext}}{\Phi_\mathrm{tot}}
}
\end{aligned}$$
## From induced charge density
A common way to calculate $\varepsilon_r$ is from
the induced charge density $\rho_\mathrm{ind}$,
i.e. the offset caused by the material's particles responding to the field.
We start from [Gauss' law](/know/concept/maxwells-equations/) for $\vb{P}$:
$$\begin{aligned}
\nabla \cdot \vb{P}
= \varepsilon_0 \nabla^2 \Phi_\mathrm{ind}(\vb{r})
= - \rho_\mathrm{ind}(\vb{r})
\end{aligned}$$
This is Poisson's equation, which has the following well-known
[Fourier transform](/know/concept/fourier-transform/):
$$\begin{aligned}
\Phi_\mathrm{ind}(\vb{q})
= \frac{\rho_\mathrm{ind}(\vb{q})}{\varepsilon_0 |\vb{q}|^2}
= V(\vb{q}) \: \rho_\mathrm{ind}(\vb{q})
\end{aligned}$$
Where $V(\vb{q})$ represents Coulomb interactions,
and $V(0) = 0$ to ensure overall neutrality:
$$\begin{aligned}
V(\vb{q})
= \frac{1}{\varepsilon_0 |\vb{q}|^2}
\qquad \implies \qquad
V(\vb{r} - \vb{r}')
= \frac{1}{4 \pi \varepsilon_0 |\vb{r} - \vb{r}'|}
\end{aligned}$$
The [convolution theorem](/know/concept/convolution-theorem/)
then gives us the solution $\Phi_\mathrm{ind}$ in the $\vb{r}$-domain:
$$\begin{aligned}
\Phi_\mathrm{ind}(\vb{r})
= (V * \rho_\mathrm{ind})(\vb{r})
= \int_{-\infty}^\infty V(\vb{r} - \vb{r}') \: \rho_\mathrm{ind}(\vb{r}') \dd{\vb{r}'}
\end{aligned}$$
To proceed, we need to find an expression for $\rho_\mathrm{ind}$
that is proportional to $\Phi_\mathrm{tot}$ or $\Phi_\mathrm{ext}$,
or some linear combination thereof.
Such an expression must exist for a linear material.
Suppose we can show that $\rho_\mathrm{ind} = C_\mathrm{ext} \Phi_\mathrm{ext}$,
for some $C_\mathrm{ext}$, which may depend on $\vb{q}$. Then:
$$\begin{aligned}
\Phi_\mathrm{tot}
= (1 + C_\mathrm{ext} V) \Phi_\mathrm{ext}
\quad \implies \quad
\boxed{
\varepsilon_r(\vb{q})
= \frac{1}{1 + C_\mathrm{ext}(\vb{q}) V(\vb{q})}
}
\end{aligned}$$
Similarly, suppose we can show that $\rho_\mathrm{ind} = C_\mathrm{tot} \Phi_\mathrm{tot}$,
for some quantity $C_\mathrm{tot}$, then:
$$\begin{aligned}
\Phi_\mathrm{ext}
= (1 - C_\mathrm{tot} V) \Phi_\mathrm{tot}
\quad \implies \quad
\boxed{
\varepsilon_r(\vb{q})
= 1 - C_\mathrm{tot}(\vb{q}) V(\vb{q})
}
\end{aligned}$$
## References
1. H. Bruus, K. Flensberg,
*Many-body quantum theory in condensed matter physics*,
2016, Oxford.
2. M. Fox,
*Optical properties of solids*, 2nd edition,
Oxford.
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