The density of flux momentum of a plane wave

A plane electromagnetic wave exerts a Lorentz force on a charged particle. The direction of such force depends on the speed of such particle. In general, the movement of the particle will be a combination of circular motion and linear drift, i.e., similar to an helical trajectory. The linear drift suggests that there is a net flux of linear momentum from the plane wave to the particle. Therefore, a plane wave propagating along one direction carries a linear momentum directed along the same direction, which can be transfered to material objects. The Maxwell's Stress tensor quantifies the flux of such linear momentum and the calculation is quite straightforward in this case. A plane wave only has contributions in the direction of wave propagation. The electric field of a polarized plane wave in the $ y$ axis propagating in vacuum along the $ z$ direction can be written as

$\displaystyle \vec{E}(x,y,z,t)=\hat{\jmath}E_y=\hat{\jmath}E_0e^{i(kz-\omega t)}$ (B.1)

where $ E_0$ is the amplitude of the electric field, $ k=2\pi/\lambda$ is the wave number, $ \omega=kc$ is the angular frequency and $ c$ is the speed of light. The corresponding magnetic field is given by

$\displaystyle \vec{B}(x,y,z)=\hat{\imath}B_x=\hat{\imath}B_0e^{i(kz-\omega t)}$ (B.2)

where $ B_0=E_0/c$ according to Maxwell equations. The calculation of the Maxwell's Stress tensor can be done using Eq. A.10
$\displaystyle \overleftrightarrow{T}_{ij}$ $\displaystyle =$ $\displaystyle \epsilon_0\left(E_iE_j-\frac{1}{2}\delta_{ij}E^2\right)+\frac{1}{\mu_0}\left(B_iB_j-\frac{1}{2}\delta_{ij}B^2\right)$  
       
  $\displaystyle =$ $\displaystyle \epsilon_0\left\{\begin{pmatrix}0\\ E_y\\ 0\end{pmatrix}\begin{pm...
...rix}-\frac{1}{2}E_y^2\begin{pmatrix}1&0&0\\ 0&1&0\\ 0&0&1\end{pmatrix}\right\}+$  
    $\displaystyle +\frac{1}{\mu_0}\left\{\begin{pmatrix}B_x\\ 0\\ 0\end{pmatrix}\be...
...rix}-\frac{1}{2}B_x^2\begin{pmatrix}1&0&0\\ 0&1&0\\ 0&0&1\end{pmatrix} \right\}$  
  $\displaystyle =$ $\displaystyle \epsilon_0\left\{\begin{pmatrix}0&0&0\\ 0&E_y^2&0\\ 0&0&0\end{pmatrix}-\begin{pmatrix}E_y^2/2&0&0\\ 0&E_y^2/2&0\\ 0&0&E_y^2/2\end{pmatrix}\right\}+$  
    $\displaystyle +\frac{1}{\mu_0}\left\{\begin{pmatrix}B_x^2&0&0\\ 0&0&0\\ 0&0&0\e...
...ix}-\begin{pmatrix}B_x^2/2&0&0\\ 0&B_x^2/2&0\\ 0&0&B_x^2/2\end{pmatrix}\right\}$  
  $\displaystyle =$ $\displaystyle \epsilon_0\begin{pmatrix}-E_y^2/2&0&0\\ 0&E_y^2/2&0\\ 0&0&-E_y^2/2\end{pmatrix}+$  
    $\displaystyle +\frac{1}{\mu_0}\begin{pmatrix}B_x^2/2&0&0\\ 0&-B_x^2/2&0\\ 0&0&-B_x^2/2\end{pmatrix}$  
  $\displaystyle =$ $\displaystyle \begin{pmatrix}-\frac{\epsilon_0}{2}E_y^2+\frac{1}{2\mu_0}B_x^2&0...
...0}B_x^2&0\\ 0&0&-\frac{\epsilon_0}{2}E_y^2-\frac{1}{2\mu_0}B_x^2\end{pmatrix}~.$ (B.3)

Now, by substituting $ E_x$ and $ B_y$ according to Eqs. B.1 and B.2 and knowing that $ E_0=B_0c$ and $ c^{-2}=\epsilon_0\mu_0$ the Maxwell's Stress tensor can be written as
$\displaystyle \overleftrightarrow{T}_{ij}$ $\displaystyle =$ $\displaystyle \begin{pmatrix}0&0&0\\ 0&0&0\\ 0&0&-\epsilon_0E_x^2\end{pmatrix}~.$ (B.4)

Note that all the components of the stress tensor vanish but one. It is a diagonal component that represents the density flux of momentum along the direction of propagation of the plane wave. If we perform the product of the stress tensor with a differential area element that is perpendicular to that tensor $ d\vec{a}=da\,\hat{k}$, then we obtain the flux of electromagnetic moment (i.e., the force) across this surface
$\displaystyle d\vec{F}=d\vec{\Phi}_\mathrm{field}$ $\displaystyle =$ $\displaystyle \overleftrightarrow{T}_{ij}\cdot d\vec{a}$  
  $\displaystyle =$ $\displaystyle \begin{pmatrix}0&0&0\\ 0&0&0\\ 0&0&-\epsilon_0E_x^2\end{pmatrix}\cdot\begin{pmatrix}0\\ 0\\ da\end{pmatrix}$  
  $\displaystyle =$ $\displaystyle -\epsilon_0E_x^2\cdot da\,\hat{k}~.$ (B.5)

If we divide both sides of the previous equation by the area, we can define a density of flux momentum

$\displaystyle \phi_\mathrm{field}=\frac{d\Phi_\mathrm{field}}{da}=-\epsilon_0E_x^2\cdot\hat{k}~.$ (B.6)

Now we can recall the definition of the Poynting vector $ \vec{S}=\vec{E}\times\vec{H}$ that in the case of a plane wave has the following expression:
$\displaystyle \vec{S}$ $\displaystyle =$ $\displaystyle \frac{1}{\mu_0}\vec{E}\times\vec{B} = \frac{1}{\mu_0}E_x\,B_y(\hat{\imath}\times\hat{\jmath})$  
  $\displaystyle =$ $\displaystyle \frac{1}{\mu_0}E_x\,B_y\,\hat{k} = \frac{1}{\mu_0}E_x\,\frac{E_x}{c}\,\hat{k}$  
  $\displaystyle =$ $\displaystyle \epsilon_0c E_x^2\,\hat{k}$ (B.7)

and rewrite Eq. B.6 as

$\displaystyle \vec{\phi}_\mathrm{field}=\frac{\vec{S}}{c}$ (B.8)

which is equivalent to the definition of the density of flux momentum given in Eq. 2.16.

To sum up, the density of flux momentum across a surface perpendicular to the direction of propagation of light can be written in terms of the Poynting vector. This result can be used to calculate the Maxwell's Stress tensor of electromagnetic fields in the ray optics regime, in which the wave can be seen as a bundle of rays, characterized by their Poynting vector.

JM Huguet 2014-02-12