User:Ignirtoq/Electromagnetic tensor

The electromagnetic tensor or electromagnetic field tensor (sometimes called the field strength tensor, Faraday tensor or Maxwell bivector) is a mathematical object that describes the electromagnetic field of a physical system in Maxwell's theory of electromagnetism. The field tensor was first used after the 4-dimensional tensor formulation of special relativity was introduced by Hermann Minkowski. The tensor allows some physical laws to be written in a very concise form.

Details

 * Mathematical note: In this article, the abstract index notation will be used.

The electromagnetic tensor $$F^{\mu\nu}$$ is commonly written as a matrix:


 * $$F^{\mu\nu} = \begin{bmatrix}

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{bmatrix} $$

or


 * $$F_{\mu\nu} = \begin{bmatrix}

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{bmatrix} $$
 * where
 * E is the electric field,
 * B the magnetic field, and
 * c the speed of light.
 * Caution: The signs in the tensor above depend on the convention used for the metric tensor. The convention used here is +---, corresponding to the metric tensor:
 * $$\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix} $$

Properties
From the matrix form of the field tensor, it becomes clear that the electromagnetic tensor satisfies the following properties:


 * antisymmetry: $$F^{\mu\nu} \, = - F^{\nu\mu}$$ (hence the name bivector).
 * six independent components.

If one forms an inner product of the field strength tensor a Lorentz invariant is formed:
 * $$F_{\mu\nu} F^{\mu\nu} = \ 2 \left( B^2 - \frac{E^2}{c^2} \right) = \mathrm{invariant} $$

The product of the tensor $$F^{\mu\nu} \,$$ with its dual tensor gives the pseudoscalar invariant:
 * $$ \frac{1}{2}\epsilon_{\alpha\beta\gamma\delta}F^{\alpha\beta} F^{\gamma\delta} = \frac{4}{c} \left( \vec B \cdot \vec E \right) = \mathrm{invariant} \,$$

where $$\ \epsilon_{\alpha\beta\gamma\delta} \,$$ is the completely antisymmetric unit pseudotensor of the fourth rank or Levi-Civita symbol. Caution: the sign for the above invariant depends on the convention used for the Levi-Civita symbol. The convention used here is $$\ \epsilon_{0123} \,$$ = +1.

Notice that:
 * $$ \det \left( F \right) = \frac{1}{c^2} \left( \vec B \cdot \vec E \right) ^{2} $$

More formally, the electromagnetic tensor may be written in terms of the 4-vector potential $$A^{\alpha} \,$$:



F_{ \alpha\beta } \ \stackrel{\mathrm{def}}{=}\ \frac{ \partial A_{\beta} }{ \partial x^{\alpha} } - \frac{ \partial A_{\alpha} }{ \partial x^{\beta} } \ \stackrel{\mathrm{def}}{=}\ \partial_{\alpha} A_{\beta} - \partial_{\beta} A_{\alpha}$$ Where the 4-vector potential is:
 * $$A^{\mu} = \left( \frac{\phi}{c}, \vec A \right)$$ and its covariant form is found by multiplying by the Minkowski metric $$\eta \,$$:
 * $$A_{\mu} \, = \eta_{\mu\nu} A^{\nu} = \left( \frac{\phi}{c}, -\vec A \right) $$

Derivation
To derive all the elements in the electromagnetic tensor we need to define the derivative operator:
 * $$\partial_{\alpha} = \left(\frac{1}{c} \frac{\partial}{\partial t}, \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z} \right) = \left(\frac{1}{c} \frac{\partial}{\partial t}, \vec{\nabla} \right) \,$$

and the 4-vector potential:
 * $$A_{\alpha} = \left(\frac{\phi}{c}, -A_x, -A_y, -A_z \right) \,$$

where
 * $$\vec A \,$$ is the vector potential and $$ \left(A_x, A_y, A_z \right) $$ are its components
 * $$\phi \,$$ is the scalar potential and
 * $$c \,$$ is the speed of light.

Electric and magnetic fields are derived from the vector potentials and the scalar potential with two formulas:
 * $$\vec{E} = -\frac{\partial \vec{A}}{\partial t} - \vec{\nabla} \phi \,$$
 * $$\vec{B} = \vec{\nabla} \times \vec{A} \,$$

As an example, the x components are just
 * $$E_x = -\frac{\partial A_x}{\partial t} - \frac{\partial \phi}{\partial x} \,$$
 * $$B_x = \frac{\partial A_z}{\partial y} - \frac{\partial A_y}{\partial z} \,$$

Using the definitions we began with, we can rewrite these two equations to look like:
 * $$E_x = -c \left(\partial_0 A_1 - \partial_1 A_0 \right) \,$$
 * $$B_x = \partial_3 A_2 - \partial_2 A_3 \,$$

Evaluating all the components results in a second-rank, antisymmetric and covariant tensor:
 * $$F_{\alpha\beta} = \partial_{\alpha} A_{\beta} - \partial_{\beta} A_{\alpha} \,$$

Thus, for example,
 * $$F_{12}=\partial_1A_2-\partial_2A_1=\partial_xA_y-\partial_yA_x=B_z,$$

and
 * $$F_{13}=\partial_1A_3-\partial_3A_1=\partial_xA_z-\partial_zA_x=-B_y.$$

Compare with the matrix above.

Significance
Hidden beneath the surface of this complex mathematical equation is an ingenious unification of Maxwell's equations for electromagnetism. Consider the electrostatic equation


 * $$\vec{\nabla} \cdot \vec{E} = \frac{\rho}{\epsilon_0}$$

which tells us that the divergence of the electric field vector is equal to the charge density, and the electrodynamic equation



\vec{\nabla} \times \vec{B} - \frac{1}{c^2} \frac{ \partial \vec{E}}{\partial t} = \mu_0 \vec{J} $$

that is the change of the electric field with respect to time, minus the curl of the magnetic field vector, is equal to negative 4π times the current density.

These two equations for electricity reduce to


 * $$\partial_{\beta} F^{\alpha\beta} = \mu_0 J^{\alpha} \,$$

where
 * $$J^{\alpha} = ( c \, \rho, \vec{J} ) \,$$ is the 4-current.

The same holds for magnetism. If we take the magnetostatic equation



\vec{\nabla} \cdot \vec{B} = 0 $$

which tells us that there are no "true" magnetic charges, and the magnetodynamics equation



\frac{ \partial \vec{B}}{ \partial t } + \vec{\nabla} \times \vec{E} = 0 $$

which tells us the change of the magnetic field with respect to time plus the curl of the electric field is equal to zero (or, alternatively, the curl of the electric field is equal to the negative change of the magnetic field with respect to time). With the electromagnetic tensor, the equations for magnetism reduce to


 * $$F_{ \alpha \beta, \gamma } + F_{ \beta \gamma , \alpha } + F_{ \gamma \alpha , \beta } = 0 \,$$

where the comma indicates a partial derivative


 * $${\partial f \over \partial x^\gamma}\equiv \partial_\gamma f \equiv {f}_{,\gamma}$$

The field tensor and relativity
The field tensor derives its name from the fact that the electromagnetic field is found to obey the tensor transformation law, this general property of (non-gravitational) physical laws being recognised after the advent of special relativity. This theory stipulated that all the (non-gravitational) laws of physics should take the same form in all coordinate systems - this led to the introduction of tensors. The tensor formalism also leads to a mathematically elegant presentation of physical laws. For example, Maxwell's equations of electromagnetism may be written using the field tensor as:


 * $$F_{[\alpha\beta,\gamma]} \, = 0$$ and $$F^{\alpha\beta}{}_{,\beta} \, = \mu_0 J^{\alpha}$$

The second equation implies conservation of charge:


 * $$J^\alpha{}_{,\alpha} \, = 0$$

In general relativity, these laws can be generalised in (what many physicists consider to be) an appealing way:


 * $$F_{[\alpha\beta;\gamma]} \, = 0$$ and $$F^{\alpha\beta}{}_{;\beta} \, = \mu_0 J^{\alpha}$$

where the semi-colon represents a covariant derivative, as opposed to a partial derivative. The elegance of these equations stems from the simple replacing of partial with covariant derivatives. These equations are sometimes referred to as the curved space Maxwell equations. Again, the second equation implies charge conservation (in curved spacetime):


 * $$J^\alpha{}_{;\alpha} \, = 0$$

Lagrangian formulation of classical electromagnetism
Classical electromagnetism and Maxwell's equations can be derived from the action defined:
 * $$\mathcal{S} = \int \left( -\begin{matrix} \frac{1}{4 \mu_0} \end{matrix} F_{\mu\nu} F^{\mu\nu} \right) \mathrm{d}^4 x \,$$

where
 * $$\mathrm{d}^4 x \;$$  is over space and time.

This means the Lagrangian density is

The far left and far right term are the same, because $$\mu$$ and $$\nu$$ are just variables of integration after all. The two middle terms are also the same, so the Lagrangian density is
 * $$\mathcal{L} \,$$
 * $$ = -\begin{matrix} \frac{1}{4\mu_0} \end{matrix} F_{\mu\nu} F^{\mu\nu} \,$$
 * $$ = -\begin{matrix} \frac{1}{4\mu_0} \end{matrix} \left( \partial_\mu A_\nu - \partial_\nu A_\mu \right) \left( \partial^\mu A^\nu - \partial^\nu A^\mu \right). \,$$
 * $$ = -\begin{matrix} \frac{1}{4\mu_0} \end{matrix} \left( \partial_\mu A_\nu \partial^\mu A^\nu - \partial_\nu A_\mu \partial^\mu A^\nu - \partial_\mu A_\nu \partial^\nu A^\mu + \partial_\nu A_\mu \partial^\nu A^\mu \right).$$
 * }
 * $$ = -\begin{matrix} \frac{1}{4\mu_0} \end{matrix} \left( \partial_\mu A_\nu \partial^\mu A^\nu - \partial_\nu A_\mu \partial^\mu A^\nu - \partial_\mu A_\nu \partial^\nu A^\mu + \partial_\nu A_\mu \partial^\nu A^\mu \right).$$
 * }
 * $$ = -\begin{matrix} \frac{1}{4\mu_0} \end{matrix} \left( \partial_\mu A_\nu \partial^\mu A^\nu - \partial_\nu A_\mu \partial^\mu A^\nu - \partial_\mu A_\nu \partial^\nu A^\mu + \partial_\nu A_\mu \partial^\nu A^\mu \right).$$
 * }


 * $$\mathcal{L} \,$$
 * $$ = -\begin{matrix} \frac{1}{2\mu_0} \end{matrix} \left( \partial_\mu A_\nu \partial^\mu A^\nu - \partial_\nu A_\mu \partial^\mu A^\nu \right).$$
 * }

We can then plug this into the Euler-Lagrange equation of motion for a field:
 * $$ \partial_\nu \left( \frac{\partial \mathcal{L}}{\partial ( \partial_\nu A_\mu )} \right) - \frac{\partial \mathcal{L}}{\partial A_\mu} = 0 . \,$$

The second term is zero, because the Lagrangian in this case only contains derivatives. So the Euler-Lagrange equation becomes:
 * $$ \partial_\nu \left( \partial^\mu A^\nu - \partial^\nu A^\mu \right) = 0. \,$$

That term in the parenthesis is just the field tensor, so this finally simplifies to
 * {|cellpadding="2" style="border:2px solid #ccccff"


 * $$ \partial_\nu F^{\mu \nu} = 0. \,$$
 * }

That equation is just another way of writing the two homogeneous Maxwell's equations as long as you make the substitutions:
 * $$~E^i /c \ \ = -F^{0 i} \,$$
 * $$\epsilon^{ijk} B^k = -F^{ij} \,$$

where $$i \,$$ and $$j \,$$ take on the values of 1, 2, and 3.

Role in quantum electrodynamics and field theory
The Lagrangian of quantum electrodynamics extends beyond the classical Lagrangian established in relativity, from $$\mathcal{L}=\bar\psi(i\hbar c \, \gamma^\alpha D_\alpha - mc^2)\psi -\frac{1}{4 \mu_0}F_{\alpha\beta}F^{\alpha\beta},$$ &ensp;to incorporate the creation and annihilation of photons (and electrons).

In quantum field theory, it is used for the template of the gauge field strength tensor. That is used in addition to the local interaction Lagrangian, nearly identical to its role in QED.