Electromagnetic Momentum Conservation

Let $g^{(i)}$ be the density of electromagnetic momentum directed parallel to the $i$ th Cartesian axis. (Here, $i=1$ corresponds to the $x$ -axis, $i=2$ to the $y$ -axis, and $i=3$ to the $z$ -axis.) Furthermore, let ${\bf G}^{(i)}$ be the flux of such momentum. We would expect the conservation equation for electromagnetic momentum directed parallel to the $i$ th Cartesian axis to take the form

$$\frac{\partial g^{(i)}}{\partial t} + \nabla\cdot{\bf G}^{(i)} = -(\rho\,{\bf E}+{\bf j}\times {\bf B})_i,$$ (111)

where the subscript $i$ denotes a component of a vector parallel to the $i$ th Cartesian axis. The term on the right-hand side is the rate per unit volume at which electromagnetic fields gain momentum parallel to the $i$ th Cartesian axis via interaction with matter. Thus, the term is minus the rate at which matter gains momentum parallel to the $i$ th Cartesian axis via interaction with electromagnetic fields. In other words, the term is minus the $i$ th component of the force per unit volume exerted on matter by electromagnetic fields. [See Equation (10).] Equation (111) can be generalized to give

$$\frac{\partial {\bf g}}{\partial t} + \nabla\cdot{\bf G} = - (\rho\,{\bf E} + {\bf j}\times {\bf B}),$$ (112)

where ${\bf g}$ is the electromagnetic momentum density (the $i$ th Cartesian component of ${\bf g}$ is thus $g^{(i)}$ ), and ${\bf G}$ is a tensor (see Section 12.5) whose Cartesian components $G_{ij}= {\bf G}^{(i)}\cdot{\bf e}_j$ , where ${\bf e}_j$ is a unit vector parallel to the $j$ th Cartesian axis, specify the flux of electromagnetic momentum parallel to the $i$ th Cartesian axis across a plane surface whose normal is parallel to the $j$ th Cartesian axis. Let us attempt to derive an expression of the form (112) from Maxwell's equations.

Maxwell's equations are as follows:

$$\nabla\cdot {\bf E} =\frac{\rho}{\epsilon_0},$$ (113)

$$\nabla\cdot{\bf B} =0,$$ (114)

$$\nabla\times {\bf E} = -\frac{\partial{\bf B}}{\partial t},$$ (115)

$$\nabla\times {\bf B} = \mu_0\,{\bf j} + \epsilon_0\,\mu_0\, \frac{\partial {\bf E}}{\partial t}.$$ (116)

We can take the vector product of Equation (116) divided by $\mu_0$ with ${\bf B}$ , and rearrange, to give

$$-\epsilon_0\,\frac{\partial {\bf E}}{\partial t}\times {\bf B} = \frac{{\bf B}\times(\nabla\times {\bf B})}{\mu_0} + {\bf j}\times {\bf B}.$$ (117)

Next, we can take the vector product of ${\bf E}$ with Equation (115) times $\epsilon_0$ , rearrange, and add the result to the previous equation. We obtain

$$-\epsilon_0\,\frac{\partial {\bf E}}{\partial t}\times {\bf B}-\e... ... E})+\frac{{\bf B}\times(\nabla\times {\bf B})}{\mu_0} + {\bf j}\times {\bf B}.$$ (118)

Making use of Equations (113) and (114), we get

$$-\frac{\partial}{\partial t} \left(\epsilon_0\,{\bf E}\times{\bf B}\right) = \epsilon_0\,{\bf E}\times (\nabla\times{\bf E})+\frac{{\bf B}\times(\nabla\times {\bf B})}{\mu_0}$$

$$\phantom{=} - \epsilon_0\,(\nabla\cdot {\bf E}) \,{\bf E} - \frac{1}{\mu_0}\,(\nabla\cdot {\bf B}) \,{\bf B}+ \rho\,{\bf E}+{\bf j}\times {\bf B}.$$ (119)

Now,

$$\nabla (E^{\,2}/2) \equiv {\bf E}\times(\nabla\times {\bf E}) + ({\bf E}\cdot{\nabla})\,{\bf E},$$ (120)

with a similar equation for ${\bf B}$ . Hence, Equation (119) can be written

$$-\frac{\partial}{\partial t} \left(\epsilon_0\,{\bf E}\times{\bf B}\right) = \epsilon_0\left[\nabla(E^{\,2}/2) - (\nabla\cdot {\bf E})\,{\bf E} - ({\bf E}\cdot\nabla)\,{\bf E}\right]$$

$$\phantom{=} +\frac{1}{\mu_0}\left[\nabla(B^{\,2}/2) - (\nabla\cdot {\bf B})\,{\bf B} - ({\bf B}\cdot\nabla)\,{\bf B}\right]$$

$$\phantom{=}+ \rho\,{\bf E}+{\bf j}\times {\bf B}.$$ (121)

Finally, when written in terms of components, the above equation becomes

$$-\frac{\partial}{\partial t} \left(\epsilon_0\,{\bf E}\times{\bf B}\right)_i = \frac{\partial}{\partial x_j}\!\left(\epsilon_0\,E^{\,2}\,\delt... ...2 - \epsilon_0\,E_i\,E_j+ B^{\,2}\,\delta_{ij}/2\,\mu_0 - B_i\,B_j/\mu_0\right)$$

$$\phantom{=}+ \left(\rho\,{\bf E}+{\bf j}\times {\bf B}\right)_i,$$ (122)

because $[(\nabla\cdot{\bf E})\,{\bf E}]_i \equiv (\partial E_j/\partial x_j)\,E_i$ , and $[({\bf E}\cdot\nabla)\,{\bf E}]_i \equiv E_j\,(\partial E_i/\partial x_j)$ . Here, $x_1$ corresponds to $x$ , $x_2$ to $y$ , and $x_3$ to $z$ . Furthermore, $\delta_{ij}$ is a Kronecker delta symbol (i.e., $\delta_{ij}=1$ if $i=j$ , and $\delta_{ij}=0$ otherwise). Finally, we are making use of the Einstein summation convention (that repeated indices are summed from 1 to 3). Comparing the previous expression with Equation (112), we conclude that the momentum density of electromagnetic fields takes the form

$${\bf g} = \epsilon_0\,({\bf E}\times {\bf B}),$$ (123)

whereas the corresponding momentum flux tensor has the Cartesian components

$$G_{ij} = \epsilon_0\,(E^{\,2}\,\delta_{ij}/2-E_i\,E_j) + (B^{\,2}\,\delta_{ij}/2-B_i\,B_j)/ \mu_0.$$ (124)

The momentum conservation equation, (112), is sometimes written

$$\rho\,{\bf E} + {\bf j}\times{\bf B} = \nabla\cdot {\bf T} -\frac{\partial}{\partial t}\left(\epsilon_0\,{\bf E}\times{\bf B}\right),$$ (125)

where

$$T_{ij} = - G_{ij} = \epsilon_0\,(E_i\,E_j-E^{\,2}\,\delta_{ij}/2) + (B_i\,B_j-B^{\,2}\,\delta_{ij}/2)/ \mu_0$$ (126)

is called the Maxwell stress tensor.