advanced29 min studyLesson 268 in Physics

The electromagnetic stress-energy tensor

Organize energy, energy flux, momentum density, and momentum flux with consistent c factors and relate field losses to the Lorentz force on matter.

On this page

What you’ll understand

  • Compute electromagnetic energy and momentum components in SI units.
  • Distinguish momentum flux from the opposite-sign Maxwell stress convention.
  • Use local stress-energy balance to account for field–matter exchange.

Fields store energy, move energy across surfaces, and deliver momentum to matter. These are linked relativistic observations rather than separate accounting rules. The stress-energy tensor collects them into one symmetric object. Its signs deserve particular care because the spatial momentum-flux tensor is the negative of a commonly defined Maxwell stress. We will declare that convention explicitly, compute a plane-wave example, and verify that the temporal conservation equation is the familiar Poynting energy balance.

Build a tensor with positive field energy

Keep signature (+,−,−,−)(+,-,-,-), x0=ctx^0=ct, and our field convention F0i=−Ei/cF^{0i}=-E_i/c, Fij=−ϵijkBkF^{ij}=-\epsilon^{ijk}B_k. The symmetric vacuum electromagnetic stress-energy tensor is quadratic in F. Its minus sign in the first term is required by this metric signature. Evaluating the time-time entry gives a positive sum of electric and magnetic energies. This is a useful audit: a formula copied from the opposite metric convention can otherwise make even a simple static electric field appear to have negative energy.

Every tensor component here has units of energy per volume, equivalently pascals. Let u denote field energy density, S energy flux, and g momentum density. Then T00=uT^{00}=u, T0i=Ti0=Si/cT^{0i}=T^{i0}=S_i/c, and gi=T0i/c=Si/c2g_i=T^{0i}/c=S_i/c^2. Flux and density differ by factors of speed because x0x^0 is a length coordinate. Equating momentum density directly to the Poynting vector would therefore fail both dimensional analysis and the local momentum-balance equation.

Tμν=1μ0(−FμαFνα+14ημνFαβFαβ),u=ϵ0E22+B22μ0,S=E×Bμ0T^{\mu\nu}=\frac{1}{\mu_0}\left(-F^{\mu\alpha}F^\nu{}_{\alpha}+\frac14\eta^{\mu\nu}F^{\alpha\beta}F_{\alpha\beta}\right),\qquad u=\frac{\epsilon_0E^2}{2}+\frac{B^2}{2\mu_0},\quad\mathbf S=\frac{\mathbf E\times\mathbf B}{\mu_0}
These formulas concern the microscopic electromagnetic field with vacuum coefficients. The mixed index in the first product is lowered using the declared metric. Material energy and stress require separate contributions.↔ Scroll wide equations horizontally.

Spatial entries describe momentum crossing surfaces

Define TijT^{ij} as the flux of ith momentum across a surface normal to direction j. With our convention it is uδij−ϵ0EiEj−BiBj/μ0u\delta_{ij}-\epsilon_0E_iE_j-B_iB_j/\mu_0. If the Maxwell stress is defined as σij=ϵ0EiEj+BiBj/μ0−uδij\sigma_{ij}=\epsilon_0E_iE_j+B_iB_j/\mu_0-u\delta_{ij}, then Tij=−σijT^{ij}=-\sigma_{ij}. Both conventions are common, but their divergence signs must travel with their definitions. A negative longitudinal momentum-flux entry in a static electric field represents field tension in this convention, rather than negative energy density.

For a plane wave propagating along x, E can point along y and B along z with E=cBE=cB. The electric and magnetic contributions to u are equal, the Poynting vector points along x, and Txx=uT^{xx}=u while the two transverse diagonal entries vanish. Its momentum flux is consequently energy flux divided by c. An absorbing surface and a reflecting surface transfer different amounts of momentum, so converting this flux into radiation pressure requires the surface interaction, not just the incident wave tensor.

T00=u,T0i=Ti0=Si/c,Tij=uδij−ϵ0EiEj−BiBjμ0=−σij,Tμμ=0T^{00}=u,\quad T^{0i}=T^{i0}=S_i/c,\quad T^{ij}=u\delta_{ij}-\epsilon_0E_iE_j-\frac{B_iB_j}{\mu_0}=-\sigma_{ij},\qquad T^\mu{}_{\mu}=0
All displayed T components have units J/m³. Sigma is explicitly the opposite-sign Maxwell stress. The zero trace is a vacuum electromagnetic algebraic identity in four spacetime dimensions, not a statement that field energy vanishes.↔ Scroll wide equations horizontally.

Field losses are matter gains

Using Maxwell equations gives ∂μTμν=−FνλJλ\partial_\mu T^{\mu\nu}=-F^{\nu\lambda}J_\lambda. The source on the right is minus the four-force density on matter. Its temporal component gives ∂tu+∇⋅S=−j⋅E\partial_tu+\nabla\cdot\mathbf S=-\mathbf j\cdot\mathbf E. Positive electric work on matter reduces field energy unless incoming energy flux compensates. The spatial equation is ∂tg−∇⋅σ=−(ρE+j×B)\partial_t\mathbf g-\nabla\cdot\boldsymbol\sigma=-(\rho\mathbf E+\mathbf j\times\mathbf B). This sign follows directly from our explicit relationship between stress and momentum flux.

In a source-free region the electromagnetic stress-energy tensor has zero divergence; with sources, field stress-energy alone generally is not conserved. Adding a compatible matter tensor restores zero total divergence for a closed system. A static field can store energy and stress even when S vanishes. Conversely, a nonzero Poynting vector can describe steady energy transport without changing the local energy density. Local storage, boundary flow, and work are separate terms of one balance, and none should be inferred from another without the actual field configuration.

Energy and momentum in a vacuum wave

At an event in a plane wave traveling along positive x, E=300 V m−1y^\mathbf E=300\,\mathrm{V\,m^{-1}}\hat{\mathbf y} and B=1.00 μTz^\mathbf B=1.00\,\mathrm{\mu T}\hat{\mathbf z}. Use c=3.00×108 m s−1c=3.00\times10^8\,\mathrm{m\,s^{-1}} and ϵ0=8.85×10−12 F m−1\epsilon_0=8.85\times10^{-12}\,\mathrm{F\,m^{-1}}. Find u, S, g, and the nonzero tensor entries.

  1. The relation E=cBE=cB holds. Each energy contribution equals ϵ0E2/2\epsilon_0E^2/2, so their sum is u=ϵ0E2=7.965×10−7 J m−3u=\epsilon_0E^2=7.965\times10^{-7}\,\mathrm{J\,m^{-3}}.

  2. The energy flux is along positive x and has magnitude Sx=cu=238.95 W m−2S_x=cu=238.95\,\mathrm{W\,m^{-2}}. The momentum density is gx=u/c=2.655×10−15 kg m−2 s−1g_x=u/c=2.655\times10^{-15}\,\mathrm{kg\,m^{-2}\,s^{-1}}.

  3. The nonzero tensor entries are T00=T0x=Tx0=Txx=uT^{00}=T^{0x}=T^{x0}=T^{xx}=u. The transverse diagonal entries cancel because each field subtraction equals u in its own direction. The trace is u−u=0u-u=0.

The result

The wave carries positive energy and positive x momentum; its energy density, normalized flux, and longitudinal momentum flux are equal tensor entries in the chosen coordinates.

Explore the model

Interactive lab

Interactive lab · vectors

Project a flux through different surface normals

For one fresh run, Reset and set BOTH Angle = 0° and Magnitude = 2; for another, Reset and set Angle = 90° and Magnitude = 2. Predict that the x projection, representing flux through an x-normal surface in this geometric analogy, changes from 2 to 0. Inspect paused projections or Step.

Project a flux through different surface normalsPlayback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Current readings: x component: 1.393; y component: 0.975; Magnitude: 1.7. All parameters are adjustable with the labeled controls below.−7.46−3.7303.737.46−3.5−1.7501.753.5xyvvₓ = 1.39
Vectorx / y projectionsTrace marker
x component
1.393
y component
0.975
Magnitude
1.7
t = 0 s

Playback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. 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 forThe vector lab tests projection geometry only. It does not compute Poynting vectors, energy density, stresses, radiation pressure, or matter exchange.

Check your understanding

With our stated momentum-flux convention, how is the spatial tensor related to the Maxwell stress sigma?

Try it yourself

A static uniform electric field E=200 V m−1x^\mathbf E=200\,\mathrm{V\,m^{-1}}\hat{\mathbf x} has B = 0. Calculate u, S, and all diagonal T components. Verify the trace using ϵ0=8.85×10−12 F m−1\epsilon_0=8.85\times10^{-12}\,\mathrm{F\,m^{-1}}.

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

Give me a hint

Only the x electric dyadic is nonzero. Its magnitude is twice the electric energy density.

Compare with the solution

The energy density is u=ϵ0E2/2=1.77×10−7 J m−3u=\epsilon_0E^2/2=1.77\times10^{-7}\,\mathrm{J\,m^{-3}}, and S is zero because B is zero. The diagonal tensor is (T00,Txx,Tyy,Tzz)=(u,−u,u,u)(T^{00},T^{xx},T^{yy},T^{zz})=(u,-u,u,u). Its trace is u−(−u)−u−u=0u-(-u)-u-u=0. The negative x momentum-flux entry reflects the stated tension convention, while the energy density remains positive. The corresponding Maxwell stress has the opposite spatial signs.

Take it with you

  • Energy flux is c times the mixed tensor entry, and momentum density is that entry divided by c.
  • Declare whether spatial components mean momentum flux or Maxwell stress before assigning signs.
  • Matter work and force are balanced by the opposite changes in field stress-energy.

Read further

Original sources and deeper explanations to follow at your own pace.

Optional progress, saved in this browser when available. Revisit any lesson at any time.

Lesson not marked complete.