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Friday, May 29, 2015

Relativistic electromagnetism


From Wikipedia, the free encyclopedia

This article is about a simplified presentation of electromagnetism, incorporating special relativity. For a more general article on the relationship between special relativity and electromagnetism, see Classical electromagnetism and special relativity. For a more rigorous discussion, see Covariant formulation of classical electromagnetism.
Relativistic electromagnetism is a modern teaching strategy for developing electromagnetic field theory from Coulomb's law and Lorentz transformations. Though Coulomb's law expresses action at a distance, it is an easily understood electric force principle. The more sophisticated view of electromagnetism expressed by electromagnetic fields in spacetime can be approached by applying spacetime symmetries. In certain special configurations it is possible to exhibit magnetic effects due to relative charge density in various simultaneous hyperplanes. This approach to physics education and the education and training of electrical and electronics engineers can be seen in the Encyclopædia Britannica (1956), The Feynman Lectures on Physics (1964), Edward M. Purcell (1965), Jack R. Tessman (1966), W.G.V. Rosser (1968), Anthony French (1968), and Dale R. Corson & Paul Lorrain (1970). This approach provides some preparation for understanding of magnetic forces involved in the Biot–Savart law, Ampère's law, and Maxwell's equations.

In 1912 Leigh Page expressed the aspiration of relativistic electromagnetism:[1]
If the principle of relativity had been enunciated before the date of Oersted’s discovery, the fundamental relations of electrodynamics could have been predicted on theoretical grounds as a direct consequence of the fundamental laws of electrostatics, extended so as to apply to charges relatively in motion as well as charges relatively at rest.

Einstein's motivation

In 1953 Albert Einstein wrote to the Cleveland Physics Society on the occasion of a commemoration of the Michelson–Morley experiment. In that letter he wrote:[2]
What led me more or less directly to the special theory of relativity was the conviction that the electromotive force acting on a body in motion in a magnetic field was nothing else but an electric field.
This statement by Einstein reveals that he investigated spacetime symmetries to determine the complementarity of electric and magnetic forces.

Introduction

Purcell argued that the question of an electric field in one inertial frame of reference, and how it looks from a different reference frame moving with respect to the first, is crucial to understand fields created by moving sources. In the special case, the sources that create the field are at rest with respect to one of the reference frames. Given the electric field in the frame where the sources are at rest, Purcell asked: what is the electric field in some other frame?
He stated that the fundamental assumption is that, knowing the electric field at some point (in space and time) in the rest frame of the sources, and knowing the relative velocity of the two frames provided all the information needed to calculate the electric field at the same point in the other frame. In other words, the electric field in the other frame does not depend on the particular distribution of the source charges, only on the local value of the electric field in the first frame at that point. He assumed that the electric field is a complete representation of the influence of the far-away charges.

Alternatively, introductory treatments of magnetism introduce the Biot–Savart law, which describes the magnetic field associated with an electric current. An observer at rest with respect to a system of static, free charges will see no magnetic field. However, a moving observer looking at the same set of charges does perceive a current, and thus a magnetic field.

Uniform electric field — simple analysis


Figure 1: Two oppositely charged plates produce uniform electric field even when moving. The electric field is shown as 'flowing' from top to bottom plate. The Gaussian pill box (at rest) can be used to find the strength of the field.

Consider the very simple situation of a charged parallel-plate capacitor, whose electric field (in its rest frame) is uniform (neglecting edge effects) between the plates and zero outside.

To calculate the electric field of this charge distribution in a reference frame where it is in motion, suppose that the motion is in a direction parallel to the plates as shown in figure 1. The plates will then be shorter by a factor of:
\sqrt{1 - v^2/c^2}
than they are in their rest frame, but the distance between them will be the same. Since charge is independent of the frame in which it is measured, the total charge on each plate is also the same. So the charge per unit area on the plates is therefore larger than in the rest frame by a factor of:
1\over\sqrt{1 - v^2/c^2}
The field between the plates is therefore stronger by this factor.

More rigorous analysis


Figure 2a: The electric field lines are shown flowing outward from the positive plate

Figure 2b: The electric field lines flow inward toward the negative plate

Consider the electric field of a single, infinite plate of positive charge, moving parallel to itself. The field must be uniform both above and below the plate, since it is uniform in its rest frame. We also assume that knowing the field in one frame is sufficient for calculating it in the other frame.

The plate however could have a non zero component of electric field in the direction of motion as in Fig 2a. Even in this case, the field of the infinite plane of negative charge must be equal and opposite to that of the positive plate (as in Fig 2b), since the combination of plates is neutral and cannot therefore produce any net fields. When the plates are separated, the horizontal components still cancel, and the resultant is a uniform vertical field as shown in Fig 1.

If Gauss's law is applied to pillbox as shown in Fig 1, it can be shown that the magnitude of the electric field between the plates is given by:
|E'|= {\sigma' \over\epsilon_0}\
where the prime (') indicates the value measured in the frame in which the plates are moving. \sigma represents the surface charge density of the positive plate. Since the plates are contracted in length by the factor
\sqrt{1 - v^2/c^2}
then the surface charge density in the primed frame is related to the value in the rest frame of the plates by:
\sigma'\ = {\sigma\over\sqrt{1 - v^2/c^2}}
But the electric field in the rest frame has value σ / ε0 and the field points in the same direction on both of the frames, so
E' = {E\over\sqrt{1 - v^2/c^2}}\
The E field in the primed frame is therefore stronger than in the unprimed frame. If the direction of motion is perpendicular to the plates, length contraction of the plates does not occur, but the distance between them is reduced. This closer spacing however does not affect the strength of the electric field. So for motion parallel to the electric field E,
E' = E \
In the general case where motion is in a diagonal direction relative to the field the field is merely a superposition of the perpendicular and parallel fields., each generated by a set of plates at right angles to each other as shown in Fig 3. Since both sets of plates are length contracted, the two components of the E field are
E'_y = {E_y\over\sqrt{1 - v^2/c^2}}
and
E'_x = E_x \
where the y subscript denotes perpendicular, and the x subscript, parallel.
These transformation equations only apply if the source of the field is at rest in the unprimed frame.

The field of a moving point charge[edit]


Figure 3: A point charge at rest, surrounded by an imaginary sphere.

Figure 4: A view of the electric field of a point charge moving at constant velocity.

A very important application of the electric field transformation equations is to the field of a single point charge moving with constant velocity. In its rest frame, the electric field of a positive point charge has the same strength in all directions and points directly away from the charge. In some other reference frame the field will appear differently.

In applying the transformation equations to a nonuniform electric field, it is important to record not only the value of the field, but also at what point in space it has this value.

In the rest frame of the particle, the point charge can be imagined to be surrounded by a spherical shell which is also at rest. In our reference frame, however, both the particle and its sphere are moving. Length contraction therefore states that the sphere is deformed into an oblate spheroid, as shown in cross section in Fig 4.

Consider the value of the electric field at any point on the surface of the sphere. Let x and y be the components of the displacement (in the rest frame of the charge), from the charge to a point on the sphere, measured parallel and perpendicular to the direction of motion as shown in the figure. Because the field in the rest frame of the charge points directly away from the charge, its components are in the same ratio as the components of the displacement:
{E_y \over E_x} = {y \over x}
In our reference frame, where the charge is moving, the displacement x' in the direction of motion is length-contracted:
x' = x\sqrt{1 - v^2/c^2}
The electric field at any point on the sphere points directly away from the charge. (b) In a reference frame where the charge and the sphere are moving to the right, the sphere is length-contracted but the vertical component of the field is stronger. These two effects combine to make the field again point directly away from the current location of the charge. (While the y component of the displacement is the same in both frames).

However, according to the above results, the y component of the field is enhanced by a similar factor:
E'_y = {E_y\over\sqrt{1 - v^2/c^2}}
whilst the x component of the field is the same in both frames. The ratio of the field components is therefore
{E'_y \over E'_x} = {E_y \over E_x\sqrt{1 - v^2/c^2}} = {y' \over x'}
So, the field in the primed frame points directly away from the charge, just as in the unprimed frame. A view of the electric field of a point charge moving at constant velocity is shown in figure 4. The faster the charge is moving, the more noticeable the enhancement of the perpendicular component of the field becomes. If the speed of the charge is much less than the speed of light, this enhancement is often negligible. But under certain circumstances, it is crucially important even at low velocities.

The origin of magnetic forces


Figure 5, lab frame: A horizontal wire carrying a current, represented by evenly spaced positive charges moving to the right whilst an equal number of negative charges remain at rest, with a positively charged particle outside the wire and traveling in a direction parallel to the current.

In the simple model of events in a wire stretched out horizontally, a current can be represented by the evenly spaced positive charges, moving to the right, whilst an equal number of negative charges remain at rest. If the wire is electrostatically neutral, the distance between adjacent positive charges must be the same as the distance between adjacent negative charges.

Assume that in our 'lab frame' (Figure 5), we have a positive test charge, Q, outside the wire, traveling parallel to the current, at the speed, v, which is equal to the speed of the moving charges in the wire. It should experience a magnetic force, which can be easily confirmed by experiment.

Figure 6, test charge frame: The same situation as in fig. 5, but viewed from the reference frame in which positive charges are at rest. The negative charges flow to the left. The distance between the negative charges is length-contracted relative to the lab frame, while the distance between the positive charges is expanded, so the wire carries a net negative charge.

Inside 'test charge frame'(Fig. 6), the only possible force is the electrostatic force Fe = Q * E because, although the magnetic field is the same, the test charge is at rest and, therefore, cannot feel it. In this frame, the negative charge density has Lorentz-contracted with respect to what we had in lab frame because of the increased speed. This means that spacing between charges has reduced by the Lorentz factor with respect to the lab frame spacing, l:
l_- = {l\sqrt{1-v^2/c^2}}
Thus, positive charges have Lorentz-expanded (because their speed has dropped):
l_+ = l / \sqrt{1-v^2/c^2}
Both of these effects combine to give the wire a net negative charge in the test charge frame. Since the negatively charged wire exerts an attractive force on a positively charged particle, the test charge will therefore be attracted and will move toward the wire.

For v << c, we can concretely compute both,[3] the magnetic force sensed in the lab frame
F_m = {Q v I \over 2 \pi \epsilon _0 c^2 R}
and electrostatic force, sensed in the test charge frame, where we first compute the charge density with respect to the lab frame length, l:
\lambda = {q \over l}_+ - {q \over l}_- = {q \over l}(\sqrt{1-v^2/c^2} - 1/\sqrt{1-v^2/c^2}) \approx {q \over l}(1-0.5({v^2\over c^2}) - 1 - 0.5({v^2\over c^2})) = -{q \over l}{v^2\over c^2}
and, keeping in mind that current I = {q\over t} = q{v\over l}, resulting electrostatic force
F_e = Q E = Q {\lambda \over 2 \pi \epsilon _0 R} = {Q q v^2 \over 2 \pi \epsilon _0 c^2 R l} = {Q v I\over 2 \pi \epsilon _0 c^2 R }
which comes out exactly equal to the magnetic force sensed in the lab frame, F_e = F_m.

The lesson is that observers in different frames of reference see the same phenomena but disagree on their reasons.
If the currents are in opposite directions, consider the charge moving to the left. No charges are now at rest in the reference frame of the test charge. The negative charges are moving with speed v in the test charge frame so their spacing is again:
l_{(-)} = {l\sqrt{1-v^2/c^2}}
The distance between positive charges is more difficult to calculate. The relative velocity should be less than 2v due to special relativity. For simplicity, assume it is 2v. The positive charge spacing contraction is then:
{\sqrt{1-(2v/c)^2}}
relative to its value in their rest frame. Now its value in their rest frame was found to be
l_{(+)} = {l\over\sqrt{1-v^2/c^2}}
So the final spacing of positive charges is:
l_{(+)} = {l\over\sqrt{1-v^2/c^2}}{\sqrt{1-(2v/c)^2}}
To determine whether l(+) or l(-) is larger we assume that v << c and use the binomial approximation that
(1+x)^p \approxeq 1+px\mbox{ when }x<<1
After some algebraic calculation it is found that l(+) < l(-), and so the wire is positively charged in the frame of the test charge.[4]

One may think that the picture, presented here, is artificial because electrons, which accelerated in fact, must condense in the lab frame, making the wire charged. Naturally, however, all electrons feel the same accelerating force and, therefore, identically to the Bell's spaceships, the distance between them does not change in the lab frame (i.e. expands in their proper moving frame). Rigid bodies, like trains, don't expand however in their proper frame, and, therefore, really contract, when observed from the stationary frame.

Calculating the magnetic field

The Lorentz force law

A moving test charge near a wire carrying current will experience a magnetic force dependent on the velocity of the moving charges in the wire. If the current is flowing to the right, and a positive test charge is moving below the wire, then there is a force in a direction 90° counterclockwise from the direction of motion.

The magnetic field of a wire

Calculation of the magnitude of the force exerted by a current-carrying wire on a moving charge is equivalent to calculating the magnetic field produced by the wire. Consider again the situation shown in figures. The latter figure, showing the situation in the reference frame of the test charge, is reproduced in the figure. The positive charges in the wire, each with charge q, are at rest in this frame, while the negative charges, each with charge −q, are moving to the left with speed v. The average distance between the negative charges in this frame is length-contracted to:
\sqrt{1 - v^2/c^2}
where is the distance between them in the lab frame. Similarly, the distance between the positive charges is not length-contracted:
\sqrt{1 - v^2/c^2}
Both of these effects give the wire a net negative charge in the test charge frame, so that it exerts an attractive force on the test charge.
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Thursday, May 28, 2015

Classical electromagnetism and special relativity

From Wikipedia, the free encyclopedia

The theory of special relativity plays an important role in the modern theory of classical electromagnetism. First of all, it gives formulas for how electromagnetic objects, in particular the electric and magnetic fields, are altered under a Lorentz transformation from one inertial frame of reference to another. Secondly, it sheds light on the relationship between electricity and magnetism, showing that frame of reference determines if an observation follows electrostatic or magnetic laws. Third, it motivates a compact and convenient notation for the laws of electromagnetism, namely the "manifestly covariant" tensor form.

Maxwell's equations, when they were first stated in their complete form in 1865, would turn out to be compatible with special relativity.[1] Moreover, the apparent coincidences in which the same effect was observed due to different physical phenomena by two different observers would be shown to be not coincidental in the least by special relativity. In fact, half of Einstein's 1905 first paper on special relativity, "On the Electrodynamics of Moving Bodies," explains how to transform Maxwell's equations.

Transformation of the fields between inertial frames

The E and B fields

This equation, also called the Joules-Bernoulli equation, considers two inertial frames. As notation, the field variables in one frame are unprimed, and in a frame moving relative to the unprimed frame at velocity v, the fields are denoted with primes. In addition, the fields parallel to the velocity v are denoted by \stackrel{\mathbf {E}_{\parallel}}{} while the fields perpendicular to v are denoted as \stackrel{\mathbf {E}_{\bot}}{}. In these two frames moving at relative velocity v, the E-fields and B-fields are related by:[2]
\begin{align} & \mathbf {{E}_{\parallel}}' = \mathbf {{E}_{\parallel}}\\ & \mathbf {{B}_{\parallel}}' = \mathbf {{B}_{\parallel}}\\ & \mathbf {{E}_{\bot}}'= \gamma \left( \mathbf {E}_{\bot} + \mathbf{ v} \times \mathbf {B} \right) \\ & \mathbf {{B}_{\bot}}'= \gamma \left( \mathbf {B}_{\bot} -\frac{1}{c^2} \mathbf{ v} \times \mathbf {E} \right) \end{align}
where
\gamma \ \overset{\underset{\mathrm{def}}{}}{=} \ \frac{1}{\sqrt{1 - v^2/{c}^2}}
is called the Lorentz factor and c is the speed of light in free space. The inverse transformations are the same except v → −v.

An equivalent, alternative expression is:[3]
\begin{align} & \mathbf{E}' = \gamma \left( \mathbf{E} + \mathbf{v} \times \mathbf{B} \right ) - \left ({\gamma-1} \right ) ( \mathbf{E} \cdot \mathbf{\hat{v}} ) \mathbf{\hat{v}} \\ & \mathbf{B}' = \gamma \left( \mathbf{B} - \frac {\mathbf{v} \times \mathbf{E}}{c^2} \right ) - \left ({\gamma-1} \right ) ( \mathbf{B} \cdot \mathbf{\hat{v}} ) \mathbf{\hat{v}}\\ \end{align}
where v̂ is the velocity unit vector.

If one of the fields is zero in one frame of reference, that doesn't necessarily mean it is zero in all other frames of reference. This can be seen by, for instance, making the unprimed electric field zero in the transformation to the primed electric field. In this case, depending on the orientation of the magnetic field, the primed system could see an electric field, even though there is none in the unprimed system.

This does not mean two completely different sets of events are seen in the two frames, but that the same sequence of events is described in two different ways (see Moving magnet and conductor problem below).

If a particle of charge q moves with velocity u with respect to frame S, then the Lorentz force in frame S is:
\mathbf{F}=q\mathbf{E}+q \mathbf{u} \times \mathbf{B}
In frame S', the Lorentz force is:
\mathbf{F'}=q\mathbf{E'}+q \mathbf{u'} \times \mathbf{B'}
If S and S' have aligned axes then:[4]
\begin{align} & u_x'=\frac{u_x+v}{1 + (v \ u_x)/c^2}\\ & u_y'=\frac{u_y/\gamma}{1 + (v \ u_x)/c^2}\\ & u_z'=\frac{u_z/\gamma}{1 + (v \ u_x)/c^2} \end{align}
A derivation for the transformation of the Lorentz force for the particular case u = 0 is given here.[5] A more general one can be seen here.[6]

Component by component, for relative motion along the x-axis, this works out to be the following:
\begin{align} & E'_x = E_x & \qquad & B'_x = B_x \\ & E'_y = \gamma \left( E_y - v B_z \right) & & B'_y = \gamma \left( B_y + \frac{v}{c^2} E_z \right) \\ & E'_z = \gamma \left( E_z + v B_y \right) & & B'_z = \gamma \left( B_z - \frac{v}{c^2} E_y \right). \\ \end{align}
The transformations in this form can be made more compact by introducing the electromagnetic tensor (defined below), which is a covariant tensor.

The D and H fields

For the electric displacement D and magnetic intensity H, using the constitutive relations and the result for c2:
\mathbf{D}=\epsilon_0\mathbf{E}\,, \quad \mathbf{B}=\mu_0\mathbf{H}\,,\quad c^2=\frac{1}{\epsilon_0\mu_0}\,,
gives
\begin{align} \mathbf{D}' & =\gamma \left( \mathbf{D}+\frac{1}{c^2}\mathbf{v}\times \mathbf{H} \right)+(1-\gamma )(\mathbf{D}\cdot \mathbf{\hat{v}})\mathbf{\hat{v}} \\ \mathbf{H}' & =\gamma \left( \mathbf{H}-\mathbf{v}\times \mathbf{D} \right)+(1-\gamma )(\mathbf{H}\cdot \mathbf{\hat{v}})\mathbf{\hat{v}} \\ \end{align}
Analogously for E and B, the D and H form the electromagnetic displacement tensor.

The φ and A fields

An alternative simpler transformation of the EM field uses the electromagnetic potentials - the electric potential φ and magnetic potential A:[7]
\begin{align} & \varphi' = \gamma (\varphi - v A_\parallel)\\ & A_\parallel' = \gamma (A_\parallel - v \varphi /c^2)\\ & A_\bot' = A_\bot \end{align}
where \scriptstyle A_\parallel is the parallel component of A to the direction of relative velocity between frames v, and \scriptstyle A_\bot is the perpendicular component. These transparently resemble the characteristic form of other Lorentz transformations (like time-position and energy-momentum), while the transformations of E and B above are slightly more complicated. The components can be collected together as:
\begin{align} \mathbf{A}' & = \mathbf{A} - \dfrac{\gamma \varphi}{c^2}\mathbf{v} + (\gamma-1) (\mathbf{A}\cdot\mathbf{\hat{v}})\mathbf{\hat{v}} \\ {\varphi}' & =\gamma \left( \varphi - \mathbf{A}\cdot \mathbf{v} \right) \end{align}

The ρ and J fields

Analogously for the charge density ρ and current density J,[7]
\begin{align} & J_\parallel' = \gamma ( J_\parallel - v\rho)\\ & \rho' = \gamma (\rho - v J_\parallel /c^2)\\ & J_\bot' = J_\bot \end{align}
Collecting components together:
\begin{align} \mathbf{J}' & =\mathbf{J}-\gamma \rho \mathbf{v} +\left( \gamma -1 \right)(\mathbf{J}\cdot \mathbf{\hat{v}})\mathbf{\hat{v}} \\ {\rho }' & =\gamma ( \rho - \mathbf{J}\cdot \mathbf{v}/c^2) \end{align}

Non-relativistic approximations

For speeds v ≪ c, the relativistic factor γ ≈ 1, which yields:
\begin{align} \mathbf{E}' & \approx \mathbf{E}+\mathbf{v}\times \mathbf{B} \\ \mathbf{B}' & \approx \mathbf{B}-\frac{1}{c^2}\mathbf{v}\times \mathbf{E} \\ \mathbf{J}' & \approx \mathbf{J}-\rho \mathbf{v}\\ \rho' & \approx \left( \rho -\frac{1}{c^2}\mathbf{j}\cdot \mathbf{v} \right) \end{align}
so that there is no need to distinguish between the spatial and temporal coordinates in Maxwell's equations.

Relationship between electricity and magnetism

Deriving magnetism from electrostatics

The chosen reference frame determines if an electromagnetic phenomenon is viewed as an effect of electrostatics or magnetism. Authors usually derive magnetism from electrostatics when special relativity and charge invariance are taken into account. The Feynman Lectures on Physics (vol. 2, ch. 13-6) uses this method to derive the "magnetic" force on a moving charge next to a current-carrying wire. See also Haskell[9] and Landau.[10]

Fields intermix in different frames

The above transformation rules show that the electric field in one frame contributes to the magnetic field in another frame, and vice versa.[11] This is often described by saying that the electric field and magnetic field are two interrelated aspects of a single object, called the electromagnetic field. Indeed, the entire electromagnetic field can be encoded in a single rank-2 tensor called the electromagnetic tensor; see below.

Moving magnet and conductor problem

A famous example of the intermixing of electric and magnetic phenomena in different frames of reference is called the "moving magnet and conductor problem", cited by Einstein in his 1905 paper on Special Relativity.
If a conductor moves with a constant velocity through the field of a stationary magnet, eddy currents will be produced due to a magnetic force on the electrons in the conductor. In the rest frame of the conductor, on the other hand, the magnet will be moving and the conductor stationary. Classical electromagnetic theory predicts that precisely the same microscopic eddy currents will be produced, but they will be due to an electric force.[12]

Covariant formulation in vacuum

The laws and mathematical objects in classical electromagnetism can be written in a form which is manifestly covariant. Here, this is only done so for vacuum (or for the microscopic Maxwell equations, not using macroscopic descriptions of materials such as electric permittivity), and uses SI units.

This section uses Einstein notation, including Einstein summation convention. See also Ricci calculus for a summary of tensor index notations, and raising and lowering indices for definition of superscript and subscript indices, and how to switch between them. The Minkowski metric tensor η here has metric signature (+ − − −).

Field tensor and 4-current

The above relativistic transformations suggest the electric and magnetic fields are coupled together, in a mathematical object with 6 components: an antisymmetric second-rank tensor, or a bivector. This is called the electromagnetic field tensor, usually written as Fμν. In matrix form:[13]
F^{\mu \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 c the speed of light - in natural units c = 1.

There is another way of merging the electric and magnetic fields into an antisymmetric tensor, by replacing E/c → B and B → − E/c, to get the dual tensor Gμν.
G^{\mu \nu} = \begin{pmatrix} 0 & -B_x & -B_y & -B_z \\ B_x & 0 & E_z/c & - E_y/c \\ B_y & -E_z/c & 0 & E_x/c \\ B_z & E_y/c & -E_x/c & 0 \end{pmatrix}
In the context of special relativity, both of these transform according to the Lorentz transformation according to
F'^{\alpha \beta} = \Lambda^\alpha{}_\mu \Lambda^\beta{}_\nu F^{\mu \nu},
where Λαν is the Lorentz transformation tensor for a change from one reference frame to another. The same tensor is used twice in the summation.

The charge and current density, the sources of the fields, also combine into the four-vector
J^\alpha = \begin{pmatrix} c \rho & J_x & J_y & J_z \end{pmatrix}
called the four-current.

Maxwell's equations in tensor form

Using these tensors, Maxwell's equations reduce to:[13]
Maxwell's equations (Covariant formulation)
\begin{align} & \frac{\partial F^{\alpha \beta}}{\partial x^\alpha} = \mu_0 J^\beta\\ & \frac{\partial G^{\alpha \beta}}{\partial x^\alpha} = 0 \end{align}
where the partial derivatives may be written in various ways, see 4-gradient. The first equation listed above corresponds to both Gauss's Law (for β = 0) and the Ampère-Maxwell Law (for β = 1, 2, 3). The second equation corresponds to the two remaining equations, Gauss's law for magnetism (for β = 0) and Faraday's Law ( for β = 1, 2, 3).

These tensor equations are manifestly-covariant, meaning the equations can be seen to be covariant by the index positions. This short form of writing Maxwell's equations illustrates an idea shared amongst some physicists, namely that the laws of physics take on a simpler form when written using tensors.

By lowering the indices on Fαβ to obtain Fαβ (see raising and lowering indices):
F_{\alpha\beta} = \eta_{\alpha\lambda} \eta_{\beta\mu} F^{\lambda\mu}
the second equation can be written in terms of Fαβ as:
\epsilon^{\delta\alpha\beta\gamma} \dfrac{\partial F_{\beta\gamma}}{\partial x^\alpha} = \dfrac{\partial F_{\alpha\beta}}{\partial x^\gamma} + \dfrac{\partial F_{\gamma\alpha}}{\partial x^\beta} + \dfrac{\partial F_{\beta\gamma}}{\partial x^\alpha} = 0
where \epsilon^{\alpha\beta\gamma\delta} is the contravariant Levi-Civita symbol. Notice the cyclic permutation of indices in this equation: \begin{array}{rc} & \scriptstyle{\alpha\,\, \longrightarrow \,\, \beta} \\ & \nwarrow_\gamma \swarrow \end{array} .
Another covariant electromagnetic object is the electromagnetic stress-energy tensor, a covariant rank-2 tensor which includes the Poynting vector, Maxwell stress tensor, and electromagnetic energy density.

4-potential[edit]

The EM field tensor can also be written[14]
F^{\alpha \beta} = \frac {\partial A^{\beta}}{\partial x_{\alpha}} - \frac {\partial A^{\alpha}}{\partial x_{\beta}} \, ,
where
A^\alpha = (\varphi/c, A_x,A_y,A_z)\,,
is the four-potential and
x_\alpha = (ct,-x,-y,-z )
is the four-position.

Using the 4-potential in the Lorenz gauge, an alternative manifestly-covariant formulation can be found in a single equation (a generalization of an equation due to Bernhard Riemann by Arnold Sommerfeld, known as the Riemann–Sommerfeld equation,[15] or the covariant form of the Maxwell equations[16]):
Maxwell's equations (Covariant Lorenz gauge formulation)
\Box A^\mu = \mu_0 J^\mu
where \Box is the d'Alembertian operator, or four-Laplacian. For a more comprehensive presentation of these topics, see Covariant formulation of classical electromagnetism.
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