Stochastic Wave Propagation in Maxwell’s Equations

Abstract

We use the spectral representation to describe the random electromagnetic fields, which are coupled by Maxwell’s equations with a random source term. The covariance matrix functions of the random electromagnetic fields are expressed in terms of the spectral matrix function of the random source term and the Green function of Maxwell’s equations. When the random source term is Gaussian, the electromagnetic fields exhibit diffusive scaling limits in the sense of distributions.

Appendix: Proof of Proposition 1

In the following, we use the notations $\mathbf {X}(\mathbf {k},t):=\int _{\mathbb {R}^{3}} e^{i\mathbf {x}\cdot \mathbf {k}} \mathbf {X}(\mathbf {x},t)d\mathbf {x}$ and $\mathbf {X}(\mathbf {x},\omega ):=$ $\int _{\mathbb {R}}e^{it\omega } \mathbf {X}(\mathbf {x},t)dt$, $\mathbf {X}\in \{\mathbf {E},\mathbf {H},\mathbf {J}^{r},\varepsilon ,\mu ,\sigma \}$, to represent the spatial and temporal Fourier transform, respectively. By taking the temporal Fourier transform on the first equation in (1) and using the convolution theorem,

$$\begin{aligned} \nabla \times \mathbf {E}(\mathbf {x},\omega ) = i\omega \int _{\mathbb {R}^{3}}\mu (\mathbf {x}-\mathbf {x}^{'},\omega )\mathbf {H}(\mathbf {x}^{'},\omega )d\mathbf {x}^{'}. \end{aligned}$$

(50)

Similarly, from (3) and (4),

$$\begin{aligned} \nabla \times \mathbf {H}(\mathbf {x},\omega ) = \int _{\mathbb {R}^{3}} \left( -i\omega \varepsilon (\mathbf {x}-\mathbf {x}^{'},\omega ) +\sigma (\mathbf {x}-\mathbf {x}^{'},\omega ) \right) \mathbf {E}(\mathbf {x}^{'},\omega )d\mathbf {x}^{'} + \mathbf {J}^{r}(\mathbf {x},\omega ). \end{aligned}$$

(51)

Because $\nabla \cdot \mathbf {B}=\nabla \cdot \mathbf {H}=0$, there exists a vector field $\mathbf {A}(\mathbf {x},t)$ such that

$$\begin{aligned} \mathbf {H}(\mathbf {x},t)=\nabla \times \mathbf {A}(\mathbf {x},t). \end{aligned}$$

(52)

By (52), (50) can be rewritten as

$$\begin{aligned} \nabla \times \big (\mathbf {E}(\mathbf {x},\omega ) -i\omega \int _{\mathbb {R}^{3}}\mu (\mathbf {x}-\mathbf {x}^{'},\omega )\mathbf {A}(\mathbf {x}^{'},\omega )d\mathbf {x}^{'} \big )=0. \end{aligned}$$

(53)

In view of (53), there exists a scalar function $g(\mathbf {x},\omega ):\mathbb {R}^{3}\times \mathbb {R}\rightarrow \mathbb {R}$ such that

$$\begin{aligned} \mathbf {E}(\mathbf {x},\omega ) -i\omega \int _{\mathbb {R}^{3}}\mu (\mathbf {x}-\mathbf {x}^{'},\omega )\mathbf {A}(\mathbf {x}^{'},\omega )d\mathbf {x}^{'} =-\nabla g(\mathbf {x},\omega ). \end{aligned}$$

(54)

By substituting the representations (52) for $\mathbf {H}$ and (54) for $\mathbf {E}$ into (51), (51) can be rewritten as

$$\begin{aligned}&\nabla \times \nabla \times \mathbf {A}(\mathbf {x},\omega ) = \int _{\mathbb {R}^{3}} \left( -i\omega \varepsilon (\mathbf {x}-\mathbf {x}^{'},\omega )+\sigma (\mathbf {x}-\mathbf {x}^{'},\omega ) \right) \nonumber \\&\left\{ i\omega \int _{\mathbb {R}^{3}}\mu (\mathbf {x}^{'}-\mathbf {x}^{''},\omega )\mathbf {A}(\mathbf {x}^{''},\omega )d\mathbf {x}^{''} -\nabla g(\mathbf {x}^{'},\omega ) \big \}d\mathbf {x}^{'} -\nabla g(\mathbf {x}^{'},\omega ) \right\} d\mathbf {x}^{'} + \mathbf {J}^{r}(\mathbf {x},\omega ). \end{aligned}$$

(55)

By using the vector identity $\nabla \times \nabla \times \mathbf {A}=-\nabla ^{2}\mathbf {A}+\nabla \nabla \cdot \mathbf {A}$, (55) can be rearranged into the following:

$$\begin{aligned}&-\nabla ^{2}\mathbf {A} + \int _{\mathbb {R}^{3}} \left( -\omega ^{2}\varepsilon (\mathbf {x}-\mathbf {x}^{'},\omega )-i\omega \sigma (\mathbf {x}-\mathbf {x}^{'},\omega ) \right) \nonumber \\&\quad \left\{ \int _{\mathbb {R}^{3}}\mu (\mathbf {x}^{'}-\mathbf {x}^{''},\omega )\mathbf {A}(\mathbf {x}^{''},\omega )d\mathbf {x}^{''} \right\} d\mathbf {x}^{'} \nonumber \\&\quad =\mathbf {J}^{r}(\mathbf {x},\omega )+ \int _{\mathbb {R}^{3}} \Big ( i\omega \varepsilon (\mathbf {x}-\mathbf {x}^{'},\omega ) -\sigma (\mathbf {x}-\mathbf {x}^{'},\omega ) \Big ) \nabla g(\mathbf {x}^{'},\omega )d\mathbf {x}^{'} -\nabla \nabla \cdot \mathbf {A}(\mathbf {x},\omega ). \end{aligned}$$

(56)

By the generalized Lorentz gauge [11]:

$$\begin{aligned} \int _{\mathbb {R}^{3}} \Big ( i\omega \varepsilon (\mathbf {x}-\mathbf {x}^{'},\omega ) -\sigma (\mathbf {x}-\mathbf {x}^{'},\omega ) \Big ) g(\mathbf {x}^{'},\omega )d\mathbf {x}^{'} -\nabla \cdot \mathbf {A}(\mathbf {x},\omega ) =0, \end{aligned}$$

(57)

(56) can be reduced to

$$\begin{aligned}&-\nabla ^{2}\mathbf {A} + \int _{\mathbb {R}^{3}} \left( -\omega ^{2}\varepsilon (\mathbf {x}-\mathbf {x}^{'},\omega )-i\omega \sigma (\mathbf {x}-\mathbf {x}^{'},\omega ) \right) \nonumber \\&\quad \left\{ \int _{\mathbb {R}^{3}}\mu (\mathbf {x}^{'}-\mathbf {x}^{''},\omega )\mathbf {A}(\mathbf {x}^{''},\omega )d\mathbf {x}^{''} \right\} d\mathbf {x}^{'} =\mathbf {J}^{r}(\mathbf {x},\omega ). \end{aligned}$$

(58)

By the convolution theorem, the spatial Fourier transform of (58) is given by

$$\begin{aligned} \Big ( |\mathbf {k}|^{2}-\omega ^{2}\varepsilon (\mathbf {k},\omega )\mu (\mathbf {k},\omega ) -i\omega \sigma (\mathbf {k},\omega )\mu (\mathbf {k},\omega ) \Big )\mathbf {A}(\mathbf {k},\omega ) =\mathbf {J}^{r}(\mathbf {k},\omega ), \end{aligned}$$

that is,

$$\begin{aligned} \mathbf {A}(\mathbf {k},\omega ) =\frac{\mathbf {J}^{r}(\mathbf {k},\omega )}{ |\mathbf {k}|^{2}-\omega ^{2}\varepsilon (\mathbf {k},\omega )\mu (\mathbf {k},\omega ) -i\omega \sigma (\mathbf {k},\omega )\mu (\mathbf {k},\omega ) }. \end{aligned}$$

(59)

The representation (6) of $\mathbf {H}$ follows by taking the inverse Fourier transform on (59) to get $\mathbf {A}(\mathbf {x},t)$ and then using the relation $\mathbf {H}(\mathbf {x},t)=\nabla \times \mathbf {A}(\mathbf {x},t)$ in (52).

For the electric field $\mathbf {E}$, an explicit form for $\nabla g(\mathbf {x},\omega )$ is needed before using the relation (54) to get $\mathbf {E}$. By taking the spatial Fourier transform of the generalized Lorentz gauge (57), we have

$$\begin{aligned} g(\mathbf {k},\omega )= \frac{-i\ \mathbf {k}^{T}\mathbf {A}(\mathbf {k},\omega )}{i\omega \varepsilon (\mathbf {k},\omega )-\sigma (\mathbf {k},\omega )}. \end{aligned}$$

(60)

By substituting (60) into (54),

$$\begin{aligned} \mathbf {E}(\mathbf {x},\omega )&= i\omega \int _{\mathbb {R}^{3}}\mu (\mathbf {x}-\mathbf {x}^{'},\omega )\mathbf {A}(\mathbf {x}^{'},\omega )d\mathbf {x}^{'} -\frac{1}{(2\pi )^{3}}\nabla _{\mathbf {x}}\int _{\mathbb {R}^{3}}e^{-i\mathbf {x}\cdot \mathbf {k}}g(\mathbf {k},\omega )d\mathbf {k} \nonumber \\&= \frac{i\omega }{(2\pi )^{3}} \int _{\mathbb {R}^{3}}e^{-i\mathbf {x}\cdot \mathbf {k}}\mu (\mathbf {k},\omega )\mathbf {A}(\mathbf {k},\omega )d\mathbf {k}\nonumber \\&\quad -\frac{1}{(2\pi )^{3}}\int _{\mathbb {R}^{3}} (-i\mathbf {k})e^{-i\mathbf {x}\cdot \mathbf {k}} \frac{-i\mathbf {k}^{T}\mathbf {A}(\mathbf {k},\omega )}{i\omega \varepsilon (\mathbf {k},\omega )-\sigma (\mathbf {k},\omega )} d\mathbf {k}. \end{aligned}$$

(61)

Finally, the representation (5) of $\mathbf {E}$ follows by taking the inverse (temporal) Fourier transform of (61).