Covariant formulation of classical electromagnetism

Lorentz tensors of the following kinds may be used in this article to describe bodies or particles:

• four-displacement:
  $${\displaystyle x^{\alpha }=(ct,\mathbf {x} )=(ct,x,y,z)\,.}$$

  
• Four-velocity:
  $${\displaystyle u^{\alpha }=\gamma (c,\mathbf {u} ),}$$

  
  where γ(u) is the Lorentz factor at the 3-velocity u.
• Four-momentum:
  $${\displaystyle p^{\alpha }=(E/c,\mathbf {p} )=m_{0}u^{\alpha }}$$

  
  where ${\displaystyle \mathbf {p} }$ is 3-momentum, ${\displaystyle E}$ is the total energy, and ${\displaystyle m_{0}}$ is rest mass.
• Four-gradient:
  $${\displaystyle \partial ^{\nu }={\frac {\partial }{\partial x_{\nu }}}=\left({\frac {1}{c}}{\frac {\partial }{\partial t}},-\mathbf {\nabla } \right)\,,}$$

  
• The d'Alembertian operator is denoted ${\displaystyle {\partial }^{2}}$,
  $${\displaystyle \partial ^{2}={\frac {1}{c^{2}}}{\partial \over \partial t}{\partial \over \partial t}-\nabla ^{2}.}$$

  

The signs in the following tensor analysis depend on the convention used for the metric tensor. The convention used here is (+ − − −), corresponding to the Minkowski metric tensor:

$${\displaystyle \eta ^{\mu \nu }={\begin{pmatrix}1&0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&-1\end{pmatrix}}}$$

The electromagnetic tensor is the combination of the electric and magnetic fields into a covariant antisymmetric tensor whose entries are B-field quantities.[1]

$${\displaystyle F_{\alpha \beta }={\begin{pmatrix}0&E_{x}/c&E_{y}/c&E_{z}/c\\-E_{x}/c&0&-B_{z}&B_{y}\\-E_{y}/c&B_{z}&0&-B_{x}\\-E_{z}/c&-B_{y}&B_{x}&0\end{pmatrix}}}$$

and the result of raising its indices is

$${\displaystyle F^{\mu \nu }\mathrel {\stackrel {\mathrm {def} }{=}} \eta ^{\mu \alpha }\,F_{\alpha \beta }\,\eta ^{\beta \nu }={\begin{pmatrix}0&-E_{x}/c&-E_{y}/c&-E_{z}/c\\E_{x}/c&0&-B_{z}&B_{y}\\E_{y}/c&B_{z}&0&-B_{x}\\E_{z}/c&-B_{y}&B_{x}&0\end{pmatrix}}\,.}$$

where E is the electric field, B the magnetic field, and c the speed of light.

The four-current is the contravariant four-vector which combines electric charge density ρ and electric current density j:

$${\displaystyle J^{\alpha }=(c\rho ,\mathbf {j} )\,.}$$

The electromagnetic four-potential is a covariant four-vector containing the electric potential (also called the scalar potential) ϕ and magnetic vector potential (or vector potential) A, as follows:

$${\displaystyle A^{\alpha }=\left(\phi /c,\mathbf {A} \right)\,.}$$

The differential of the electromagnetic potential is

$${\displaystyle F_{\alpha \beta }=\partial _{\alpha }A_{\beta }-\partial _{\beta }A_{\alpha }\,.}$$

In the language of differential forms, which provides the generalisation to curved spacetimes, these are the components of a 1-form ${\displaystyle A=A_{\alpha }dx^{\alpha }}$ and a 2-form ${\textstyle F=dA={\frac {1}{2}}F_{\alpha \beta }dx^{\alpha }\wedge dx^{\beta }}$ respectively. Here, ${\displaystyle d}$ is the exterior derivative and ${\displaystyle \wedge }$ the wedge product.

The electromagnetic stress–energy tensor can be interpreted as the flux density of the momentum four-vector, and is a contravariant symmetric tensor that is the contribution of the electromagnetic fields to the overall stress–energy tensor:

$${\displaystyle T^{\alpha \beta }={\begin{pmatrix}\varepsilon _{0}E^{2}/2+B^{2}/2\mu _{0}&S_{x}/c&S_{y}/c&S_{z}/c\\S_{x}/c&-\sigma _{xx}&-\sigma _{xy}&-\sigma _{xz}\\S_{y}/c&-\sigma _{yx}&-\sigma _{yy}&-\sigma _{yz}\\S_{z}/c&-\sigma _{zx}&-\sigma _{zy}&-\sigma _{zz}\end{pmatrix}}\,,}$$

where ${\displaystyle \varepsilon _{0}}$ is the electric permittivity of vacuum, μ₀ is the magnetic permeability of vacuum, the Poynting vector is

$${\displaystyle \mathbf {S} ={\frac {1}{\mu _{0}}}\mathbf {E} \times \mathbf {B} }$$

and the Maxwell stress tensor is given by

$${\displaystyle \sigma _{ij}=\varepsilon _{0}E_{i}E_{j}+{\frac {1}{\mu _{0}}}B_{i}B_{j}-\left({\frac {1}{2}}\varepsilon _{0}E^{2}+{\frac {1}{2\mu _{0}}}B^{2}\right)\delta _{ij}\,.}$$

The electromagnetic field tensor F constructs the electromagnetic stress–energy tensor T by the equation:[2]

$${\displaystyle T^{\alpha \beta }={\frac {1}{\mu _{0}}}\left(\eta ^{\alpha \nu }F_{\nu \gamma }F^{\gamma \beta }+{\frac {1}{4}}\eta ^{\alpha \beta }F_{\gamma \nu }F^{\gamma \nu }\right)}$$

where η is the Minkowski metric tensor (with signature (+ − − −)). Notice that we use the fact that

$${\displaystyle \varepsilon _{0}\mu _{0}c^{2}=1\,,}$$

which is predicted by Maxwell's equations.

The Lorenz gauge condition is a Lorentz-invariant gauge condition. (This can be contrasted with other gauge conditions such as the Coulomb gauge, which if it holds in one inertial frame will generally not hold in any other.) It is expressed in terms of the four-potential as follows:

$${\displaystyle \partial _{\alpha }A^{\alpha }=\partial ^{\alpha }A_{\alpha }=0\,.}$$

In the Lorenz gauge, the microscopic Maxwell's equations can be written as:

$${\displaystyle {\partial }^{2}A^{\sigma }=\mu _{0}\,J^{\sigma }\,.}$$

Lorentz force f on a charged particle (of charge q) in motion (instantaneous velocity v). The E field and B field vary in space and time.

Electromagnetic (EM) fields affect the motion of electrically charged matter: due to the Lorentz force. In this way, EM fields can be detected (with applications in particle physics, and natural occurrences such as in aurorae). In relativistic form, the Lorentz force uses the field strength tensor as follows.[4]

Expressed in terms of coordinate time t, it is:

$${\displaystyle {dp_{\alpha } \over {dt}}=q\,F_{\alpha \beta }\,{\frac {dx^{\beta }}{dt}},}$$

where p_α is the four-momentum, q is the charge, and x^β is the position.

Expressed in frame-independent form, we have the four-force

$${\displaystyle {\frac {dp_{\alpha }}{d\tau }}\,=q\,F_{\alpha \beta }\,u^{\beta },}$$

where u^β is the four-velocity, and τ is the particle's proper time, which is related to coordinate time by dt = γdτ.

Lorentz force per spatial volume f on a continuous charge distribution (charge density ρ) in motion.

The density of force due to electromagnetism, whose spatial part is the Lorentz force, is given by

$${\displaystyle f_{\alpha }=F_{\alpha \beta }J^{\beta }.}$$

and is related to the electromagnetic stress–energy tensor by

$${\displaystyle f^{\alpha }=-{T^{\alpha \beta }}_{,\beta }\equiv -{\frac {\partial T^{\alpha \beta }}{\partial x^{\beta }}}.}$$

The continuity equation:

$${\displaystyle {J^{\beta }}_{,\beta }\mathrel {\overset {\text{def}}{\mathop {=} }} \partial _{\beta }J^{\beta }=\partial _{\beta }\partial _{\alpha }F^{\alpha \beta }/\mu _{0}=0.}$$

expresses charge conservation.

Using the Maxwell equations, one can see that the electromagnetic stress–energy tensor (defined above) satisfies the following differential equation, relating it to the electromagnetic tensor and the current four-vector

$${\displaystyle {T^{\alpha \beta }}_{,\beta }+F^{\alpha \beta }J_{\beta }=0}$$

or

$${\displaystyle \eta _{\alpha \nu }{T^{\nu \beta }}_{,\beta }+F_{\alpha \beta }J^{\beta }=0,}$$

which expresses the conservation of linear momentum and energy by electromagnetic interactions.

In vacuum, the constitutive relations between the field tensor and displacement tensor are:

$${\displaystyle \mu _{0}{\mathcal {D}}^{\mu \nu }=\eta ^{\mu \alpha }F_{\alpha \beta }\eta ^{\beta \nu }\,.}$$

Antisymmetry reduces these 16 equations to just six independent equations. Because it is usual to define F^μν by

$${\displaystyle F^{\mu \nu }=\eta ^{\mu \alpha }F_{\alpha \beta }\eta ^{\beta \nu },}$$

the constitutive equations may, in vacuum, be combined with the Gauss–Ampère law to get:

$${\displaystyle \partial _{\beta }F^{\alpha \beta }=\mu _{0}J^{\alpha }.}$$

The electromagnetic stress–energy tensor in terms of the displacement is:

$${\displaystyle T_{\alpha }{}^{\pi }=F_{\alpha \beta }{\mathcal {D}}^{\pi \beta }-{\frac {1}{4}}\delta _{\alpha }^{\pi }F_{\mu \nu }{\mathcal {D}}^{\mu \nu },}$$

where δ_α^π is the Kronecker delta. When the upper index is lowered with η, it becomes symmetric and is part of the source of the gravitational field.

The Lagrangian density for classical electrodynamics is composed by two components: a field component and a source component:

$${\displaystyle {\mathcal {L}}\,=\,{\mathcal {L}}_{\text{field}}+{\mathcal {L}}_{\text{int}}=-{\frac {1}{4\mu _{0}}}F^{\alpha \beta }F_{\alpha \beta }-A_{\alpha }J^{\alpha }\,.}$$

In the interaction term, the four-current should be understood as an abbreviation of many terms expressing the electric currents of other charged fields in terms of their variables; the four-current is not itself a fundamental field.

The Lagrange equations for the electromagnetic lagrangian density ${\displaystyle {\mathcal {L}}{\mathord {\left(A_{\alpha },\partial _{\beta }A_{\alpha }\right)}}}$ can be stated as follows:

$${\displaystyle \partial _{\beta }\left[{\frac {\partial {\mathcal {L}}}{\partial (\partial _{\beta }A_{\alpha })}}\right]-{\frac {\partial {\mathcal {L}}}{\partial A_{\alpha }}}=0\,.}$$

Noting

$${\displaystyle {\begin{aligned}F^{\lambda \sigma }&=F_{\mu \nu }\eta ^{\mu \lambda }\eta ^{\nu \sigma },\\F_{\mu \nu }&=\partial _{\mu }A_{\nu }-\partial _{\nu }A_{\mu }\,\\{\partial \left(\partial _{\mu }A_{\nu }\right) \over \partial \left(\partial _{\rho }A_{\sigma }\right)}&=\delta _{\mu }^{\rho }\delta _{\nu }^{\sigma }\end{aligned}}}$$

the expression inside the square bracket is

$${\displaystyle {\begin{aligned}{\frac {\partial {\mathcal {L}}}{\partial (\partial _{\beta }A_{\alpha })}}&=-\ {\frac {1}{4\mu _{0}}}\ {\frac {\partial \left(F_{\mu \nu }\eta ^{\mu \lambda }\eta ^{\nu \sigma }F_{\lambda \sigma }\right)}{\partial \left(\partial _{\beta }A_{\alpha }\right)}}\\&=-\ {\frac {1}{4\mu _{0}}}\ \eta ^{\mu \lambda }\eta ^{\nu \sigma }\left(F_{\lambda \sigma }\left(\delta _{\mu }^{\beta }\delta _{\nu }^{\alpha }-\delta _{\nu }^{\beta }\delta _{\mu }^{\alpha }\right)+F_{\mu \nu }\left(\delta _{\lambda }^{\beta }\delta _{\sigma }^{\alpha }-\delta _{\sigma }^{\beta }\delta _{\lambda }^{\alpha }\right)\right)\\&=-\ {\frac {F^{\beta \alpha }}{\mu _{0}}}\,.\end{aligned}}}$$

The second term is

$${\displaystyle {\frac {\partial {\mathcal {L}}}{\partial A_{\alpha }}}=-J^{\alpha }\,.}$$

Therefore, the electromagnetic field's equations of motion are

$${\displaystyle {\frac {\partial F^{\beta \alpha }}{\partial x^{\beta }}}=\mu _{0}J^{\alpha }\,.}$$

which is the Gauss–Ampère equation above.