Stress-Energy Tensors for Particles, Fields, and Fluids

The algebraic form of a stress-energy tensor encodes what the model permits: a point particle is distributional and rank one along its worldline; dust assumes a single pressureless stream; a perfect fluid assumes isotropic rest-frame stress and no dissipative flux; scalar and electromagnetic fields retain stresses fixed by their actions.

Required background. Classical Fields and Their Actions supplies metric-derived field tensors, while Events, Intervals, and Minkowski Geometry fixes observer and index conventions. Helpful background. Calculus and Linear Algebra for Relativity supplies projectors and distributions; Presymplectic Potentials and Currents develops boundary-current ambiguities.

Particles and classical fields

Use a symmetric Hilbert tensor. In an observer’s inertial orthonormal frame, $T^{00}$ is energy density, $T^{0i}$ is momentum density, $T^{i0}$ is energy flux, and $T^{ij}$ is the flux of $j$-momentum across a surface normal to direction $i$. The equality $T^{0i}=T^{i0}$ uses symmetry and $c=1$; restoring SI units makes momentum density equal to energy flux divided by $c^2$. All components have energy-density units in the site’s units.

For a general future-directed unit observer $v^\mu$, the same measurements are obtained by projection. Define

$$h^\mu{}_\nu=\delta^\mu{}_\nu+v^\mu v_\nu, \quad \varepsilon=T_{\mu\nu}v^\mu v^\nu, \quad q^\mu=-h^\mu{}_\alpha T^{\alpha\beta}v_\beta, \quad \Pi^{\mu\nu}=h^\mu{}_\alpha h^\nu{}_\beta T^{\alpha\beta}.$$

Then

$$T^{\mu\nu} =\varepsilon v^\mu v^\nu +v^\mu q^\nu+q^\mu v^\nu +\Pi^{\mu\nu}.$$

Energy density, energy flux, and spatial stress are therefore observer-dependent projections, not bare coordinate components. The projector also makes $v_\mu q^\mu=0$ and $v_\mu\Pi^{\mu\nu}=0$ manifest.

For noninteracting point particles with worldlines $z_A(\tau)$ and future-directed unit tangents $u_A^\mu=dz_A^\mu/d\tau$,

$$T^{\mu\nu}_{\rm pp}(x) =\sum_A m_A\int u_A^\mu u_A^\nu \frac{\delta^{(4)}(x-z_A(\tau))}{\sqrt{-g}}\,d\tau.$$

This is a distribution supported on worldlines. The coordinate delta is defined with $d^4x$, so $1/\sqrt{-g}$ converts it to an invariant scalar distribution under the volume element $\sqrt{-g}\,d^4x$.

The constant-time normalization can be checked rather than memorized. For a future-directed worldline in flat space,

$$\delta\!\left(t-z_A^0(\tau)\right) =\frac{\delta(\tau-\tau_t)}{u_A^0(\tau_t)}.$$

Therefore

$$T_{\rm pp}^{0\nu}(t,\mathbf x) =\sum_A m_Au_A^\nu \delta^{(3)}\!\left(\mathbf x-\mathbf z_A(t)\right),$$

and integrating over the slice gives $\sum_A m_Au_A^\nu=\sum_Ap_A^\nu$. McGreevy, §§4.2–4.3, printed pp. 45–48 performs this delta-function reduction explicitly before taking the fluid limit.

For a minimally coupled real scalar,

$$T_{\mu\nu}^{\phi} =\nabla_\mu\phi\nabla_\nu\phi -g_{\mu\nu} \left[\frac12(\nabla\phi)^2+V(\phi)\right].$$

A spatial gradient selects a direction and generally produces anisotropic stress. Only special scalar configurations behave as perfect fluids. A constant field instead gives $T_{\mu\nu}=-Vg_{\mu\nu}$ and $p=-\rho$.

For electromagnetism,

$$T_{\mu\nu}^{\rm EM} =F_{\mu\rho}F_\nu{}^\rho -\frac14g_{\mu\nu}F_{\rho\sigma}F^{\rho\sigma}.$$

In four dimensions its trace vanishes, and an inertial observer measures $(E^2+B^2)/2\ge0$. A coherent electromagnetic field is generally anisotropic, so it is not automatically a perfect fluid. Scalar and Maxwell metric variations are derived in Tong, §4.5.3, printed pp. 184–186, Eqs. (4.51)–(4.52).

Dust and perfect fluids

Coarse-graining particles into a single timelike velocity field $u^\mu$ and rest energy density $\rho$ gives dust:

$$T_{\rm dust}^{\mu\nu}=\rho u^\mu u^\nu.$$

The assumptions are zero rest-frame pressure, heat flow, and viscous stress, with one velocity at each event. Collisionless streams after trajectory crossing do not satisfy that single-stream assumption even if each constituent is pressureless.

Rest-frame isotropy permits one additional scalar, the pressure $p$. Since the rest-frame components must be $\operatorname{diag}(\rho,p,p,p)$, covariance uniquely gives

$$\boxed{ T_{\rm pf}^{\mu\nu} =(\rho+p)u^\mu u^\nu+pg^{\mu\nu}.}$$

This short isotropy derivation is presented in Carroll, Chapter 1, printed pp. 28–30, Eqs. (1.104)–(1.110) and Tong, §4.5.4, printed pp. 186–187, Eqs. (4.53)–(4.55). The observer dependence of dust density and flux is developed operationally in Hughes, Lecture 10, printed pp. 79–87. “Perfect” excludes heat conduction, shear stress, and bulk-viscous pressure. Conservation of $T^{\mu\nu}$ alone also does not supply a material equation of state; a relation such as $p=p(\rho,s)$ and, when relevant, a particle-current law complete the model.

Comparison at a glance

Model	Basic data and support	Rest-frame spatial stress	Isotropic?	Main boundary of the model
Point particle	$z^\mu(\tau)$; distribution on one worldline	Not a smooth medium stress	Not applicable	Singular support; many particles or coarse-graining are needed for continuum matter
Real scalar field	$\phi$ and its derivatives throughout spacetime	Set by gradients and $V(\phi)$	Only for special configurations or suitable averages	A spatial gradient selects a direction; nonminimal couplings change metric response
Electromagnetic field	$F_{\mu\nu}$ throughout spacetime	Maxwell tension and transverse pressure	Generally no	One coherent field is not an isotropic photon gas
Dust	$\rho$ and one unit velocity $u^\mu$	Zero	Yes, with $p=0$	Fails after multistream crossing or when pressure and dissipation matter
Perfect fluid	$\rho$, $p$, and one unit velocity $u^\mu$	$p\delta_{ij}$	Yes, by definition	Excludes heat flow and viscosity; still needs thermodynamic closure

The table compares constitutive content, not a hierarchy of approximations. A scalar or electromagnetic field may sometimes admit a fluid description, but only after its configuration or statistical state supplies the required timelike velocity and rest-frame isotropy.

Independent checks and limits

In the fluid rest frame, $u^\mu=(1,0,0,0)$ gives $T^{00}=\rho$, $T^{0i}=0$, and $T^{ij}=p\delta^{ij}$. The limit $p\to0$ recovers dust. For radiation made isotropic by averaging over directions, tracelessness gives $-\rho+3p=0$, hence $p=\rho/3$; a single plane wave is not isotropic and cannot use that inference.

For a conserved isolated point particle, the worldline expression has zero distributional divergence precisely when the worldline is geodesic. For charged matter, its divergence instead equals the Lorentz-force density and cancels the electromagnetic divergence only in the combined system.

Exercise: a dust stream seen in another frame

Dust has rest density $\rho$ and four-velocity $u^\mu$. An observer has four-velocity $v^\mu$ and relative Lorentz factor $\gamma=-u_\mu v^\mu$. Find the observer’s energy density and energy flux.

Solution

Projecting $T^{\mu\nu}=\rho u^\mu u^\nu$ twice along $v^\mu$ gives

$$\varepsilon_{(v)} =T_{\mu\nu}v^\mu v^\nu =\rho(u\cdot v)^2 =\rho\gamma^2.$$

The spatial flux is

$$q^\mu =-h^\mu{}_\alpha T^{\alpha\beta}v_\beta =\rho\gamma\left(u^\mu-\gamma v^\mu\right).$$

Write $u^\mu=\gamma(v^\mu+w^\mu)$ with $w^\mu v_\mu=0$. Then

$$q^\mu=\rho\gamma^2w^\mu.$$

At zero relative velocity, $\gamma=1$ and $w^\mu=0$, so $\varepsilon=\rho$ and $q^\mu=0$. The extra factor $\gamma^2$ combines the boosted particle energy with the boosted number density.

Exercise: two cold streams make a stress

Two noninteracting dust streams have equal proper rest density $\rho_b$ and velocities $+\beta$ and $-\beta$ along $x$ in the laboratory. Their four-velocities are $u_\pm^\mu=\gamma(1,\pm\beta,0,0)$. Add their tensors and determine whether their combined matter is dust or a perfect fluid.

Solution

With $\gamma=(1-\beta^2)^{-1/2}$,

$$T^{\mu\nu}=\rho_bu_+^\mu u_+^\nu+\rho_bu_-^\mu u_-^\nu =2\rho_b\gamma^2\operatorname{diag}(1,\beta^2,0,0).$$

The opposing momentum densities cancel, so $T^{0x}=0$. Their momentum fluxes add, so $T^{xx}=2\rho_b\gamma^2\beta^2\gt0$ for $\beta\ne0$. The combined state is not dust: zero mean velocity does not mean zero velocity dispersion. It is also not a perfect fluid, because the stress in this energy-rest frame is anisotropic. An isotropic distribution of stream directions could instead produce isotropic pressure. This is why kinetic theory needs the second velocity moment as well as the mean velocity.

Model boundaries and onward link

A perfect-fluid form is a constitutive idealization, not a consequence of symmetry alone. Viscosity, heat flow, magnetic anisotropy, kinetic multistreaming, polarization, and quantum expectation values require additional structures. Point-particle singularities need distribution theory, while quantum stress-energy needs renormalization.

Continue to Waves, Optics, and Radiation to study how field equations transport these energy and momentum densities through characteristics, wave packets, and radiation zones.

References

• Sean M. Carroll, “Lecture Notes on General Relativity,” arXiv:gr-qc/9712019 (1997), Open PDF, Chapter 1, printed pp. 28–30, Eqs. (1.104)–(1.113).
• Scott A. Hughes, “Describing Matter in Bulk,” MIT 8.033 Introduction to Relativity and Spacetime Physics, Lecture 10, Fall 2024, official PDF, printed pp. 79–87.
• John McGreevy, Physics 225A: General Relativity, University of California San Diego lecture notes, current PDF last updated 18 December 2024, official PDF, §§4.2–4.3, printed pp. 45–48.
• David Tong, General Relativity, University of Cambridge Part III lecture notes (2019), official PDF, §§4.5.3–4.5.4, printed pp. 184–187, Eqs. (4.51)–(4.55).