Fourier expansion electromagnetic field

The electromagnetic (EM) field is of importance as a carrier of solar energy and electronic signals (radio, TV, etc.). As its name suggests, it consists of two tightly coupled vector fields, the electric field E and the magnetic field B. The Fourier expansion of the electromagnetic field is used in the quantization of the field that leads to photons, light particles of well-defined energy and momentum. Further the Fourier transform plays a role in theory of wave propagation through different media and light scattering.

In the absence of charges and electric currents, both E and B can be derived from a third vector field, the vector potential A. In this article the Fourier transform of the fields E, B, and A will be discussed. It will be seen that the expansion of the vector potential A yields the expansions of the fields E and B. Further the energy and momentum of the EM field will be expressed in the Fourier components of A.

Fourier expansion of a vector field

For an arbitrary real vector field F its Fourier expansion is the following:

$\mathbf {F} (\mathbf {r} ,t)=\sum _{\mathbf {k} }\left(\mathbf {f} _{k}(t)e^{i\mathbf {k} \cdot \mathbf {r} }+{\bar {\mathbf {f} }}_{k}(t)e^{-i\mathbf {k} \cdot \mathbf {r} }\right)$

where the bar indicates complex conjugation. Such an expansion, labeled by a discrete (countable) set of vectors k, is always possible when F satisfies periodic boundary conditions, i.e., F(r + p,t) = F(r,t) for some finite vector p. To impose such boundary conditions, it is common to consider EM waves as if they are in a virtual cubic box of finite volume V = L³. Waves on opposite walls of the box are enforced to have the same value (usually zero). Note that the waves are not restricted to the box: the box is replicated an infinite number of times in x, y, and z direction. The vectors k are,

$\mathbf {k} ={\frac {2\pi }{L}}(n_{x},\;n_{y},\;n_{z})\quad {\hbox{with}}\quad n_{x},\,n_{y},\,n_{z}=0,\;\pm 1\;\pm 2,\ldots$

Vector potential

The magnetic field B satisfies the following Maxwell equation:

${\boldsymbol {\nabla }}\cdot \mathbf {B} (\mathbf {r} ,t)=0,$

that is, the divergence of B is zero. This equation expresses the fact that magnetic monopoles (charges) do not exist (or, rather, have never been found in nature). A divergence-free field, such as B, is a also referred to as a transverse field. By the Helmholtz decomposition, B can be written as

$\mathbf {B} (\mathbf {r} ,t)={\boldsymbol {\nabla }}\times \mathbf {A} (\mathbf {r} ,t),$

in which the vector potential A is introduced though the curl ∇×A.

The electric field obeys one of the Maxwell equations, in electromagnetic SI units it reads,

${\boldsymbol {\nabla }}\cdot \mathbf {E} (\mathbf {r} ,t)={\frac {\rho }{\epsilon _{0}}}=0$

because it is assumed that charge distributions ρ are zero. The quantity ε₀ is the electric constant. Hence, also the electric field E is transverse. Since there are no charges, the electric potential is zero and the electric field follows from A by,

$\mathbf {E} (\mathbf {r} ,t)=-{\frac {\partial \mathbf {A} (\mathbf {r} ,t)}{\partial t}}.$

The fact that E can be written this way is due to the choice of Coulomb gauge for A:

${\boldsymbol {\nabla }}\cdot \mathbf {A} (\mathbf {r} ,t)=0.$

By definition, a choice of gauge does not affect any measurable properties (the best known example of a choice of gauge is the fixing of the zero of an electric potential, for instance at infinity). The Coulomb gauge makes A transverse as well, and clearly A is in the same plane as E. (The time differentiation does not affect direction.) So, the vector fields A, B, and E are all in the same plane.

The three fields can be written as a linear combination of two orthonormal vectors, e_x and e_y. It is more convenient to choose complex unit vectors obtained by a unitary transformation,

$\mathbf {e} ^{(1)}\equiv {\frac {-1}{\sqrt {2}}}(\mathbf {e} _{x}+i\mathbf {e} _{y})\quad {\hbox{and}}\quad \mathbf {e} ^{(-1)}\equiv {\frac {1}{\sqrt {2}}}(\mathbf {e} _{x}-i\mathbf {e} _{y})$

which are orthonormal,

$\mathbf {e} ^{(\mu )}\cdot {\bar {\mathbf {e} }}^{(\mu ')}=\delta _{\mu ,\mu '}\quad {\hbox{with}}\quad \mu ,\mu '=1,\,-1.$

Expansions

The Fourier expansion of the vector potential reads

$\mathbf {A} (\mathbf {r} ,t)=\sum _{\mathbf {k} }\sum _{\mu =-1,1}\left(\mathbf {e} ^{(\mu )}(\mathbf {k} )a_{\mathbf {k} }^{(\mu )}(t)\,e^{i\mathbf {k} \cdot \mathbf {r} }+{\bar {\mathbf {e} }}^{(\mu )}(\mathbf {k} ){\bar {a}}_{\mathbf {k} }^{(\mu )}(t)\,e^{-i\mathbf {k} \cdot \mathbf {r} }\right).$

The vector potential obeys the wave equation,

$\nabla ^{2}\mathbf {A} (\mathbf {r} ,t)={\frac {1}{c^{2}}}{\frac {\partial ^{2}\mathbf {A} (\mathbf {r} ,t)}{\partial t^{2}}}$

The substitution of the Fourier series of A into the wave equation yields for the individual terms,

$-k^{2}a_{\mathbf {k} }^{(\mu )}(t)={\frac {1}{c^{2}}}{\frac {\partial ^{2}a_{\mathbf {k} }^{(\mu )}(t)}{\partial t^{2}}}\quad \Longrightarrow \quad a_{\mathbf {k} }^{(\mu )}(t)\propto e^{-i\omega t}\quad {\hbox{with}}\quad \omega =kc\quad {\hbox{and}}\quad k\equiv |\mathbf {k} |.$

It is now an easy matter to construct the corresponding Fourier expansions for E and B from the expansion of the vector potential A. The expansion for E follows from differentiation with respect to time,

$\mathbf {E} (\mathbf {r} ,t)=i\sum _{\mathbf {k} }\sum _{\mu =-1,1}\omega \left(\mathbf {e} ^{(\mu )}(\mathbf {k} )a_{\mathbf {k} }^{(\mu )}(t)\,e^{i\mathbf {k} \cdot \mathbf {r} }-{\bar {\mathbf {e} }}^{(\mu )}(\mathbf {k} ){\bar {a}}_{\mathbf {k} }^{(\mu )}(t)\,e^{-i\mathbf {k} \cdot \mathbf {r} }\right).$

The expansion for B follows by taking the curl,

$\mathbf {B} (\mathbf {r} ,t)=i\sum _{\mathbf {k} }\sum _{\mu =-1,1}\left({\big [}\mathbf {k} \times \mathbf {e} ^{(\mu )}(\mathbf {k} ){\big ]}\;a_{\mathbf {k} }^{(\mu )}(t)\,e^{i\mathbf {k} \cdot \mathbf {r} }-{\big [}\mathbf {k} \times {\bar {\mathbf {e} }}^{(\mu )}(\mathbf {k} ){\big ]}\;{\bar {a}}_{\mathbf {k} }^{(\mu )}(t)\,e^{-i\mathbf {k} \cdot \mathbf {r} }\right).$

Fourier-expanded energy

The electromagnetic energy density is

${\mathcal {E}}_{\mathrm {Field} }={\frac {1}{2}}(\epsilon _{0}\mathbf {E} \cdot \mathbf {E} +{\frac {1}{\mu _{0}}}\mathbf {B} \cdot \mathbf {B} ),$

where μ₀ is the magnetic constant. The total energy (classical Hamiltonian) of a finite volume V is defined by

$H=\iiint _{V}{\mathcal {E}}_{\mathrm {Field} }(\mathbf {r} ,t)\mathrm {d} ^{3}\mathbf {r} .$

Use

${\begin{aligned}\iiint _{V}&\left(\mathbf {e} ^{(\mu ')}(\mathbf {k'} )a_{\mathbf {k'} }^{(\mu ')}(t)\,e^{i\mathbf {k'} \cdot \mathbf {r} }-{\bar {\mathbf {e} }}^{(\mu ')}(\mathbf {k'} ){\bar {a}}_{\mathbf {k'} }^{(\mu ')}(t)\,e^{-i\mathbf {k'} \cdot \mathbf {r} }\right)\cdot \left(\mathbf {e} ^{(\mu )}(\mathbf {k} )a_{\mathbf {k} }^{(\mu )}(t)\,e^{i\mathbf {k} \cdot \mathbf {r} }-{\bar {\mathbf {e} }}^{(\mu )}(\mathbf {k} ){\bar {a}}_{\mathbf {k} }^{(\mu )}(t)\,e^{-i\mathbf {k} \cdot \mathbf {r} }\right)\mathrm {d} ^{3}\mathbf {r} \\&=-2V\,\delta _{\mu ',\mu }\,\delta _{\mathbf {k'} ,\mathbf {k} }\,a_{\mathbf {k} }^{(\mu )}(t){\bar {a}}_{\mathbf {k} }^{(\mu )}(t)\end{aligned}}$

and

${\frac {k^{2}}{\mu _{0}}}={\frac {\omega ^{2}}{c^{2}\mu _{0}}}={\frac {\epsilon _{0}\mu _{0}\omega ^{2}}{\mu _{0}}}=\epsilon _{0}\omega ^{2}.$

Then the classical Hamiltonian in terms of Fourier coefficients takes the form

$H=2V\epsilon _{0}\sum _{\mathbf {k} }\sum _{\mu =1,-1}\omega ^{2}a_{\mathbf {k} }^{(\mu )}(t){\bar {a}}_{\mathbf {k} }^{(\mu )}(t).$

Fourier-expanded momentum

The electromagnetic momentum, P_EM, of EM radiation enclosed by a volume V is proportional to an integral of the Poynting vector (see above). In SI units:

$\mathbf {P} _{\textrm {EM}}\equiv {\frac {1}{c^{2}}}\iiint _{V}\mathbf {S} \,{\textrm {d}}^{3}\mathbf {r} =\epsilon _{0}\iiint _{V}\mathbf {E} (\mathbf {r} ,t)\times \mathbf {B} (\mathbf {r} ,t)\,{\textrm {d}}^{3}\mathbf {r} .$