advanced26 min studyLesson 262 in Physics

Electromagnetic field tensor components

Construct the antisymmetric field tensor from potentials, check all index signs, and recover an electric and magnetic force from its components.

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What you’ll understand

  • Build the field tensor using the declared potential and metric conventions.
  • Recover electric and magnetic components without confusing raised and lowered indices.
  • Check the component table against the force on a test charge.

Electric and magnetic fields supply six independent numbers at an event. A four-vector has only four components, so combining the fields requires a different object. An antisymmetric rank-two tensor has exactly six independent components in four dimensions. Counting is suggestive, but the real test is stronger: its components must come from the potentials, transform correctly, and reproduce the measured force with the same signs. We will perform these checks rather than memorize an isolated array.

Derivatives of the four-potential

Keep η=diag⁡(1,−1,−1,−1)\eta=\operatorname{diag}(1,-1,-1,-1) and x0=ctx^0=ct. Define the contravariant potential Aμ=(ϕ/c,A)A^\mu=(\phi/c,\mathbf A), so Aμ=(ϕ/c,−A)A_\mu=(\phi/c,-\mathbf A). Here phi is the electric potential in volts and the spatial vector potential has units of tesla metres. Its time component has the same units. The tensor is Fμν=∂μAν−∂νAμF^{\mu\nu}=\partial^\mu A^\nu-\partial^\nu A^\mu. Remember that the spatial components of the raised derivative are negative; omitting this sign reverses the identification of the electric field.

For example, its time-space component is F0i=c−1∂tAi+c−1∂iϕ=−Ei/cF^{0i}=c^{-1}\partial_t A^i+c^{-1}\partial_i\phi=-E_i/c, where AiA^i is the ith spatial component of the contravariant potential. To avoid ambiguity we use bold symbols for spatial vectors and reserve spacetime subscripts for metric lowering. The purely spatial components involve the curl of the vector potential. Antisymmetry is automatic because exchanging the two derivative indices changes the sign; every diagonal entry therefore vanishes.

Fμν=(0−Ex/c−Ey/c−Ez/cEx/c0−BzByEy/cBz0−BxEz/c−ByBx0),Fij=−ϵijkBkF^{\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},\qquad F^{ij}=-\epsilon^{ijk}B_k
Rows and columns are ordered 0,x,y,z. The three-dimensional alternating symbol has epsilon123 = +1. Every entry has units of tesla because E/c and B have the same SI units.↔ Scroll wide equations horizontally.

Index placement changes an array

Lower both indices by multiplying the tensor by the metric on both slots. A time-space entry picks up one spatial minus sign, whereas a space-space entry picks up two. Consequently F0i=+Ei/cF_{0i}=+E_i/c and Fij=FijF_{ij}=F^{ij}. Raising just one index gives yet another matrix. The component array is therefore meaningless without its index positions, metric signature, coordinate ordering, and factors of c. Two books can print different arrays and describe identical physics if their conventions differ consistently.

The field tensor is unchanged by a smooth gauge transformation Aμ↦Aμ+∂μχA^\mu\mapsto A^\mu+\partial^\mu\chi, since the extra mixed second derivatives commute. With our signs this means ϕ↦ϕ+∂tχ\phi\mapsto\phi+\partial_t\chi and A↦A−∇χ\mathbf A\mapsto\mathbf A-\nabla\chi. The frequently used opposite sign for the gauge function simply renames chi. Gauge changes relate different potentials within one frame; Lorentz transformations relate different observers' components. They are distinct operations even though both can alter a displayed potential.

Fμν=ημαηνβFαβ,F0i=+Ei/c,Fij=−ϵijkBk,E=−∇ϕ−∂tA,B=∇×AF_{\mu\nu}=\eta_{\mu\alpha}\eta_{\nu\beta}F^{\alpha\beta},\qquad F_{0i}=+E_i/c,\quad F_{ij}=-\epsilon^{ijk}B_k,\qquad \mathbf E=-\nabla\phi-\partial_t\mathbf A,\quad\mathbf B=\nabla\times\mathbf A
These identities hold in Cartesian inertial coordinates for differentiable potentials. Local potentials describe fields without magnetic monopoles; global topology can require more than one potential patch.↔ Scroll wide equations horizontally.

Audit the array against a measurable force

A reliable sign audit uses a positive test charge at rest. Its lowered four-velocity is (c,0,0,0)(c,0,0,0), so qFiνuν=qEiqF^{i\nu}u_\nu=qE_i. The force points along the electric field, as it must. For motion along positive x in a magnetic field along positive z, the ordinary magnetic force points along negative y. Our array gives F21=BzF^{21}=B_z and u1=−γvxu_1=-\gamma v_x, reproducing that negative sign. This physical audit catches errors that antisymmetry alone cannot detect.

The tensor represents the field at one event; it does not include the test charge's velocity. Magnetic force appears only when the tensor contracts with a moving particle's four-velocity. This separation explains why a field tensor and a force four-vector are different kinds of object. Maxwell's equations constrain how these local field values vary across spacetime, but are additional dynamical statements. Constructing a valid antisymmetric array at an event does not by itself prove that an arbitrary field assignment solves those equations.

An electric potential and a magnetic curl

Take static potentials ϕ=−6x V m−1\phi=-6x\,\mathrm{V\,m^{-1}} and Ay=2x TA_y=2x\,\mathrm T, with other spatial components zero; x is measured in metres. Find the fields, F01F^{01}, and F12F^{12}.

  1. The electric field is minus the potential gradient, giving E=(6,0,0) V m−1\mathbf E=(6,0,0)\,\mathrm{V\,m^{-1}}. The vector potential is static, so it adds no electric contribution.

  2. The curl gives Bz=∂xAy=2 TB_z=\partial_x A_y=2\,\mathrm T, with the other magnetic components zero. The coefficient of x has units of tesla; differentiating this linear vector potential gives a spatially uniform magnetic field.

  3. Our table yields F01=−(6 V m−1)/cF^{01}=-(6\,\mathrm{V\,m^{-1}})/c and F12=−2 TF^{12}=-2\,\mathrm T. Reversing the two indices changes each sign. Lowering both indices changes the electric entry's sign but leaves the magnetic entry unchanged.

The result

The nonzero fields are Ex=6 V m−1E_x=6\,\mathrm{V\,m^{-1}} and Bz=2 TB_z=2\,\mathrm T; with c=3.00×108 m s−1c=3.00\times10^8\,\mathrm{m\,s^{-1}}, F01=−2.00×10−8 TF^{01}=-2.00\times10^{-8}\,\mathrm T.

Explore the model

Interactive lab

Interactive lab · fields

Separate radial and circulating spatial patterns

Reset and set BOTH Source strength q = 1 and Rotation Ω = 0; predict outward arrows. For a fresh run, Reset and set q = 0 and Ω = 1; predict counterclockwise arrows. Inspect the paused patterns or Step to compare source-like and circulation-like geometry.

Separate radial and circulating spatial patternsThe softened illustration is q·r/(r² + 0.2) + Ω(−y, x). Moving markers indicate local direction, not integrated particle paths or a Maxwell solution. Current readings: Fx at (1, 0): 0.833; Fy at (1, 0): 0.5; Field magnitude there: 0.972. All parameters are adjustable with the labeled controls below.−3.2−1.601.63.2−1.5−0.7500.751.5xyq = 1
Local field directionSource strength
Fx at (1, 0)
0.833
Fy at (1, 0)
0.5
Field magnitude there
0.972
t = 0 s

The softened illustration is q·r/(r² + 0.2) + Ω(−y, x). Moving markers indicate local direction, not integrated particle paths or a Maxwell solution. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forThis softened planar source plus rotation field illustrates spatial patterns only. Its controls do not supply E, B, potentials, or tensor components, and it does not solve Maxwell equations.

Check your understanding

Which lowering operation is correct for our contravariant field tensor?

Try it yourself

A static vector potential has Ax=−3y TA_x=-3y\,\mathrm T, with Ay=Az=ϕ=0A_y=A_z=\phi=0. Find BzB_z, F12F^{12}, and the magnetic force direction for a positive charge moving along positive x.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Use the full curl component, including its minus sign, then test the tensor using the lowered spatial four-velocity.

Compare with the solution

The curl is Bz=∂xAy−∂yAx=0−(−3)=3 TB_z=\partial_x A_y-\partial_y A_x=0-(-3)=3\,\mathrm T. Thus F12=−3 TF^{12}=-3\,\mathrm T and F21=3 TF^{21}=3\,\mathrm T. For positive vxv_x, u1=−γvxu_1=-\gamma v_x, so the y component of the four-force is negative. Equivalently, positive x crossed with positive z points along negative y. There is no electric field because both potentials are time independent and the scalar potential vanishes.

Take it with you

  • The six field components form an antisymmetric tensor with declared index placement.
  • Electric entries change sign when both indices are lowered; magnetic entries do not.
  • Check a tensor convention against the rest-charge and right-hand-rule force tests.

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