Electromagnetic Four-potential
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Electromagnetic Four-potential

An electromagnetic four-potential is a relativistic vector function from which the electromagnetic field can be derived. It combines both an electric scalar potential and a magnetic vector potential into a single four-vector.[1]

As measured in a given frame of reference, and for a given gauge, the first component of the electromagnetic four-potential is conventionally taken to be the electric scalar potential, and the other three components make up the magnetic vector potential. While both the scalar and vector potential depend upon the frame, the electromagnetic four-potential is Lorentz covariant.

Like other potentials, many different electromagnetic four-potentials correspond to the same electromagnetic field, depending upon the choice of gauge.

This article uses tensor index notation and the Minkowski metric sign convention . See also covariance and contravariance of vectors and raising and lowering indices for more details on notation. Formulae are given in SI units and Gaussian-cgs units.

## Definition

The electromagnetic four-potential can be defined as:[2]

SI units Gaussian units
${\displaystyle A^{\alpha }=\left({\frac {1}{c}}\phi ,\mathbf {A} \right)\,\!}$ ${\displaystyle A^{\alpha }=(\phi ,\mathbf {A} )}$

in which ? is the electric potential, and A is the magnetic potential (a vector potential). The units of A? are V·s·m-1 in SI, and Mx·cm-1 in Gaussian-cgs.

The electric and magnetic fields associated with these four-potentials are:[3]

SI units Gaussian units
${\displaystyle \mathbf {E} =-\mathbf {\nabla } \phi -{\frac {\partial \mathbf {A} }{\partial t}}}$ ${\displaystyle \mathbf {E} =-\mathbf {\nabla } \phi -{\frac {1}{c}}{\frac {\partial \mathbf {A} }{\partial t}}}$
${\displaystyle \mathbf {B} =\mathbf {\nabla } \times \mathbf {A} }$ ${\displaystyle \mathbf {B} =\mathbf {\nabla } \times \mathbf {A} }$

In special relativity, the electric and magnetic fields transform under Lorentz transformations. This can be written in the form of a tensor - the electromagnetic tensor. This is written in terms of the electromagnetic four-potential and the four-gradient as:

${\displaystyle F^{\mu \nu }=\partial ^{\mu }A^{\nu }-\partial ^{\nu }A^{\mu }={\begin{bmatrix}0&-E_{x}/c&-E_{y}/c&-E_{z}/c\\E_{x}/c&0&-B_{z}&B_{y}\\E_{y}/c&B_{z}&0&-B_{x}\\E_{z}/c&-B_{y}&B_{x}&0\end{bmatrix}}}$

assuming that the signature of the Minkowski metric is (+ - - -). If the said signature is instead (- + + +) then:

${\displaystyle F'\,^{\mu \nu }=\partial '\,^{\mu }A^{\nu }-\partial '\,^{\nu }A^{\mu }={\begin{bmatrix}0&E_{x}/c&E_{y}/c&E_{z}/c\\-E_{x}/c&0&B_{z}&-B_{y}\\-E_{y}/c&-B_{z}&0&B_{x}\\-E_{z}/c&B_{y}&-B_{x}&0\end{bmatrix}}}$

This essentially defines the four-potential in terms of physically observable quantities, as well as reducing to the above definition.

## In the Lorenz gauge

Often, the Lorenz gauge condition ${\displaystyle \partial _{\alpha }A^{\alpha }=0}$ in an inertial frame of reference is employed to simplify Maxwell's equations as:[2]

SI units Gaussian units
${\displaystyle \Box A^{\alpha }=\mu _{0}J^{\alpha }}$ ${\displaystyle \Box A^{\alpha }={\frac {4\pi }{c}}J^{\alpha }}$

where J? are the components of the four-current, and

${\displaystyle \Box ={\frac {1}{c^{2}}}{\frac {\partial ^{2}}{\partial t^{2}}}-\nabla ^{2}=\partial ^{\alpha }\partial _{\alpha }}$

is the d'Alembertian operator. In terms of the scalar and vector potentials, this last equation becomes:

SI units Gaussian units
${\displaystyle \Box \phi ={\frac {\rho }{\epsilon _{0}}}}$ ${\displaystyle \Box \phi =4\pi \rho }$
${\displaystyle \Box \mathbf {A} =\mu _{0}\mathbf {j} }$ ${\displaystyle \Box \mathbf {A} ={\frac {4\pi }{c}}\mathbf {j} }$

For a given charge and current distribution, and , the solutions to these equations in SI units are:[3]

{\displaystyle {\begin{aligned}\phi (\mathbf {r} ,t)&={\frac {1}{4\pi \epsilon _{0}}}\int \mathrm {d} ^{3}x^{\prime }{\frac {\rho \left(\mathbf {r} ^{\prime },t_{r}\right)}{\left|\mathbf {r} -\mathbf {r} ^{\prime }\right|}}\\\mathbf {A} (\mathbf {r} ,t)&={\frac {\mu _{0}}{4\pi }}\int \mathrm {d} ^{3}x^{\prime }{\frac {\mathbf {j} \left(\mathbf {r} ^{\prime },t_{r}\right)}{\left|\mathbf {r} -\mathbf {r} ^{\prime }\right|}},\end{aligned}}}

where

${\displaystyle t_{r}=t-{\frac {\left|\mathbf {r} -\mathbf {r} '\right|}{c}}}$

is the retarded time. This is sometimes also expressed with

${\displaystyle \rho \left(\mathbf {r} ',t_{r}\right)=\left[\rho \left(\mathbf {r} ',t\right)\right],}$

where the square brackets are meant to indicate that the time should be evaluated at the retarded time. Of course, since the above equations are simply the solution to an inhomogeneous differential equation, any solution to the homogeneous equation can be added to these to satisfy the boundary conditions. These homogeneous solutions in general represent waves propagating from sources outside the boundary.

When the integrals above are evaluated for typical cases, e.g. of an oscillating current (or charge), they are found to give both a magnetic field component varying according to r-2 (the induction field) and a component decreasing as r-1 (the radiation field).[clarification needed]

## Gauge freedom

When flattened to a one-form, A can be decomposed via the Hodge decomposition theorem as the sum of an exact, a coexact, and a harmonic form,

${\displaystyle A=d\alpha +\delta \beta +\gamma }$.

There is gauge freedom in A in that of the three forms in this decomposition, only the coexact form has any effect on the electromagnetic tensor

${\displaystyle F=dA}$.

Exact forms are closed, as are harmonic forms over an appropriate domain, so ${\displaystyle dd\alpha =0}$ and ${\displaystyle d\gamma =0}$, always. So regardless of what ${\displaystyle \alpha }$ and ${\displaystyle \gamma }$ are, we are left with simply

${\displaystyle F=d\delta \beta }$.