advanced27 min studyLesson 51 in Physics

Relativistic energy and the stress tensor

Connect four-momentum, local energy measurements, pressure, and the stress tensor that supplies gravity’s matter source.

On this page

What you’ll understand

  • Compute energy and momentum from a normalized four-velocity.
  • Interpret the components of a stress tensor in a specified frame.
  • Explain why pressure belongs in a relativistic description of matter.

A moving object carries momentum as well as energy. A fluid also transports momentum across imaginary surfaces, which is what pressure measures. Relativity organizes these observations into one tensor, allowing different observers to describe the same matter without agreeing on every component. We begin in a local inertial frame and build the source needed for general relativity, keeping mass density and energy density distinct.

Measure energy relative to an observer

Throughout this module use signature (+,−,−,−)(+,-,-,-) and x0=ctx^0=ct. A massive observer has four-velocity uμ=dxμ/dτu^\mu=dx^\mu/d\tau with uμuμ=c2u^\mu u_\mu=c^2. The proper time τ\tau is read by a clock moving along that observer’s trajectory. For a particle of rest mass mm, four-momentum is pμ=muμp^\mu=mu^\mu. Its squared norm is invariant even though its energy and spatial momentum change between inertial frames.

An observer with four-velocity wμw^\mu measures energy E=pμwμE=p_\mu w^\mu. In that observer’s inertial rest frame this becomes E=cp0E=cp^0. Thus “the particle’s energy” always implies an observer, whereas its rest mass does not. For a massless particle no rest frame or timelike four-velocity exists; momentum remains meaningful and has zero norm. These distinctions prevent the misleading claim that all energy is simply rest mass multiplied by a universal constant.

pμ=(E/c,p),E2=c2∣p∣2+m2c4,E=γmc2,p=γmvp^\mu=(E/c,\mathbf p),\qquad E^2=c^2|\mathbf p|^2+m^2c^4,\qquad E=\gamma mc^2,\quad \mathbf p=\gamma m\mathbf v
The final two expressions concern a massive particle, with gamma equal to one over the square root of one minus v squared over c squared.↔ Scroll wide equations horizontally.

Momentum flux is a physical stress

In a local Cartesian inertial frame, T00T^{00} is energy per volume. The components cT0icT^{0i} describe energy flux, while TijT^{ij} describe flux of the ith momentum component across a surface normal to direction j. For a symmetric matter stress tensor, Ti0/cT^{i0}/c is momentum density. The factors of c are necessary because all tensor components here have the units of energy density. Forgetting them can produce dimensionally plausible but physically incorrect conservation equations.

Consider a gas at rest on average. Its total momentum density can vanish while molecules still strike a wall and transfer momentum. Isotropy makes the spatial stress diagonal with pressure p. An ideal perfect fluid assumes no viscosity or heat conduction in its own rest frame. It is a useful macroscopic model, not a universal description of plasma turbulence, a solid under shear, or matter far from local equilibrium.

Tμν=ε+pc2uμuν−pgμν,Trestμν=diag⁡(ε,p,p,p)T^{\mu\nu}=\frac{\varepsilon+p}{c^2}u^\mu u^\nu-pg^{\mu\nu},\qquad T^{\mu\nu}_{\mathrm{rest}}=\operatorname{diag}(\varepsilon,p,p,p)
Epsilon is rest-frame energy density, including rest energy; p is isotropic pressure. Both use joules per cubic metre, equivalently pascals.↔ Scroll wide equations horizontally.

Local conservation does not require global time symmetry

In flat spacetime an isolated matter system obeys ∂μTμν=0\partial_\mu T^{\mu\nu}=0. Integrating its time component over a region relates changing energy inside to energy flowing through the boundary. Curved spacetime replaces the partial derivative by the metric-compatible covariant derivative. This expresses local energy-momentum accounting in freely falling frames; it does not automatically produce one globally conserved energy for an arbitrary expanding universe.

A global conserved energy requires additional structure, often a timelike Killing symmetry and suitable boundary conditions. Gravity also cannot generally be assigned a coordinate-independent local stress tensor on the same footing as matter. Later we use asymptotic or approximate gravitational energy measures when appropriate. For now, the essential lesson is that the matter source contains energy density, momentum transport, and stress together; specifying only a mass density throws away relativistic information.

Boost a pressureless stream

A pressureless fluid has rest energy density ε=9 J m−3\varepsilon=9\,\mathrm{J\,m^{-3}} and moves at v=0.6cv=0.6c along x. Compute T00T^{00}, T0xT^{0x}, and TxxT^{xx} in the laboratory frame.

  1. For this speed, γ=1/1−0.36=1.25\gamma=1/\sqrt{1-0.36}=1.25, hence γ2=1.5625\gamma^2=1.5625. Dust means rest-frame pressure is zero, not that every laboratory stress component vanishes.

  2. Insert uμ=γ(c,v,0,0)u^\mu=\gamma(c,v,0,0) into the perfect-fluid expression. The energy component is T00=εγ2=14.0625 J m−3T^{00}=\varepsilon\gamma^2=14.0625\,\mathrm{J\,m^{-3}}.

  3. The mixed component is T0x=εγ2(v/c)=8.4375 J m−3T^{0x}=\varepsilon\gamma^2(v/c)=8.4375\,\mathrm{J\,m^{-3}}, and Txx=εγ2(v/c)2=5.0625 J m−3T^{xx}=\varepsilon\gamma^2(v/c)^2=5.0625\,\mathrm{J\,m^{-3}}. Multiply the mixed component by c for energy flux.

The result

A stream with zero pressure in its own rest frame has nonzero laboratory momentum flux. A tensor transformation changes components without turning dust into a fluid with intrinsic thermodynamic pressure.

Explore the model

Interactive lab

Interactive lab · spacetime

Changing the local inertial observer

Change relative velocity β and event time ct. The diagram Lorentz-transforms a single event fixed at x = 0.6 in flat spacetime, using c = 1. Compare changing coordinate values with the unchanged causal type of its interval.

Changing the local inertial observerA fixed x = 0.6 event is Lorentz transformed with c = 1. Playback highlights the relation between coordinate descriptions; it does not move the event. Current readings: Lorentz factor γ: 1.12; Event x′: 0.168; Event ct′: 0.817. All parameters are adjustable with the labeled controls below.−4.69−2.3502.354.69−2.2−1.101.12.2x or x′ · c = 1ct or ct′EE′ct² − x² = 0.64
Original eventTransformed eventLight cone
Lorentz factor γ
1.12
Event x′
0.168
Event ct′
0.817
t = 0 s

A fixed x = 0.6 event is Lorentz transformed with c = 1. Playback highlights the relation between coordinate descriptions; it does not move the event. 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 lab illustrates a local Lorentz frame change. It does not transform a full stress tensor, calculate a fluid flow, or solve curved spacetime gravity.

Check your understanding

Why can stationary gas have nonzero spatial components of its stress tensor?

Try it yourself

A perfect fluid has ε=12 J m−3\varepsilon=12\,\mathrm{J\,m^{-3}} and p=4 J m−3p=4\,\mathrm{J\,m^{-3}} in its rest frame. Find its invariant trace and the laboratory T00T^{00} for speed 0.6c0.6c.

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

Give me a hint

Contract with the metric for the trace, and retain the pressure subtraction in the boosted time component.

Compare with the solution

The trace is Tμμ=ε−3p=12−12=0T^\mu{}_{\mu}=\varepsilon-3p=12-12=0. At the stated speed γ2=1.5625\gamma^2=1.5625, so T00=(ε+p)γ2−p=16(1.5625)−4=21 J m−3T^{00}=(\varepsilon+p)\gamma^2-p=16(1.5625)-4=21\,\mathrm{J\,m^{-3}}. The trace stays zero after a boost; the energy density does not. A vanishing trace here follows from the specified equation of state and is not a general property of matter.

Take it with you

  • State the observer before assigning energy or momentum components.
  • Pressure and shear are momentum fluxes, so they belong in the gravitational matter source.
  • Covariant local conservation alone does not guarantee a global conserved energy.

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