section_path: Electromagnetic/Conservation/Vacuum
section_name: Vacuum
paragraph_text: Vacuum
  
  The Lorentz force imparts a momentum to the particle, so by Newton's second law the particle must impart a momentum to the electromagnetic fields.
  
  In a vacuum, the momentum per unit volume is
  
  g = 1 μ 0 c 2 E × B, {\displaystyle \mathbf {g} ={\frac {1}{\mu _{0}c^{2}}}\mathbf {E} \times \mathbf {B} \,,}
  
  where μ is the vacuum permeability and c is the speed of light. The momentum density is proportional to the Poynting vector S which gives the directional rate of energy transfer per unit area:
  
  g = S c 2. {\displaystyle \mathbf {g} ={\frac {\mathbf {S} }{c^{2}}}\,.}
  
  If momentum is to be conserved over the volume V over a region Q, changes in the momentum of matter through the Lorentz force must be balanced by changes in the momentum of the electromagnetic field and outflow of momentum. If P is the momentum of all the particles in Q, and the particles are treated as a continuum, then Newton's second law gives
  
  d P mech d t = ∭ Q (ρ E + J × B) d V. {\displaystyle {\frac {{\text{d}}\mathbf {P} _{\text{mech}}}{{\text{d}}t}}=\iiint \limits _{Q}\left(\rho \mathbf {E} +\mathbf {J} \times \mathbf {B} \right){\text{d}}V\,.}
  
  The electromagnetic momentum is
  
  P field = 1 μ 0 c 2 ∭ Q E × B d V, {\displaystyle \mathbf {P} _{\text{field}}={\frac {1}{\mu _{0}c^{2}}}\iiint \limits _{Q}\mathbf {E} \times \mathbf {B} \,dV\,,}
  
  and the equation for conservation of each component i of the momentum is
  
  d d t (P mech + P field) i = ∬ σ (∑ j T i j n j) d Σ. {\displaystyle {\frac {\text{d}}{{\text{d}}t}}\left(\mathbf {P} _{\text{mech}}+\mathbf {P} _{\text{field}}\right)_{i}=\iint \limits _{\sigma }\left(\sum \limits _{j}T_{ij}n_{j}\right){\text{d}}\Sigma \,.}
  
  The term on the right is an integral over the surface area Σ of the surface σ representing momentum flow into and out of the volume, and n is a component of the surface normal of S. The quantity T is called the Maxwell stress tensor, defined as
  
  T i j ≡ ϵ 0 (E i E j − 1 2 δ i j E 2) + 1 μ 0 (B i B j − 1 2 δ i j B 2). {\displaystyle T_{ij}\equiv \epsilon _{0}\left(E_{i}E_{j}-{\frac {1}{2}}\delta _{ij}E^{2}\right)+{\frac {1}{\mu _{0}}}\left(B_{i}B_{j}-{\frac {1}{2}}\delta _{ij}B^{2}\right)\,.}