In the stress-energy tensor of a perfect fluid, the pressure p contributes to the gravitational field. In what physical situation does this pressure contribution become important?
AWhenever the fluid is in hydrostatic equilibrium
BOnly in fluids moving close to the speed of light
CWhen the pressure is comparable to the energy density ρc², as in neutron star interiors or the early universe
DOnly in the presence of viscosity and shear stress
In the stress-energy tensor T_μν = (ρ + p/c²)u_μ u_ν + p g_μν, the pressure appears alongside the energy density as a source of gravity. For ordinary matter, p << ρc², so the pressure contribution is negligible. But in neutron stars (where degeneracy pressure is enormous), in the early universe (where radiation pressure equals ρc²/3), and near the core of collapsing stars, the pressure contribution is significant and can even accelerate gravitational collapse — a purely relativistic effect with no Newtonian analog.
Question 2 True / False
The equation ∇^μ T_μν = 0 in curved spacetime means that total energy is globally conserved in general relativity.
TTrue
FFalse
Answer: False
∇^μ T_μν = 0 is a local conservation law: it says energy-momentum is locally conserved in the sense that the covariant divergence vanishes. However, in curved spacetime, a covariant divergence cannot generally be integrated into a global conservation law because there is no coordinate-invariant way to add up energy densities at different points (parallel transport is path-dependent). Global energy conservation holds in spacetimes with time-translation symmetry (Killing vectors), but in general — for example, in an expanding universe — total energy is not a well-defined conserved quantity.
Question 3 Short Answer
Explain why the stress-energy tensor must be symmetric (T_μν = T_νμ) and what physical property this symmetry ensures.
Think about your answer, then reveal below.
Model answer: The symmetry T_μν = T_νμ is required for consistency with the Einstein field equations, since the Einstein tensor G_μν is symmetric. Physically, the symmetry of T_μν ensures conservation of angular momentum. The component T_{0i} represents both momentum density and energy flux — their equality is the relativistic statement that the flow of energy carries momentum. The spatial components T_{ij} = T_{ji} mean that shear stresses are symmetric, which is the standard condition for the absence of intrinsic torques (no 'couple stresses') in the continuum.
In special relativity, the symmetry of T_μν can be derived from the Belinfante-Rosenfeld procedure for the canonical energy-momentum tensor. In GR, defining T_μν as the variational derivative of the matter action with respect to g^{μν} automatically produces a symmetric tensor.
Question 4 Short Answer
For electromagnetic radiation, T_{μν} is traceless (g^{μν}T_{μν} = 0). What does this imply about the relationship between energy density and pressure of a photon gas?
Think about your answer, then reveal below.
Model answer: For a photon gas treated as a perfect fluid, the stress-energy tensor T_μν = (ρ + p/c²)u_μ u_ν + p g_μν has trace T = g^{μν}T_μν = -ρc² + 3p (in the -+++ convention). Setting T = 0 gives p = ρc²/3, the radiation equation of state. This means the pressure of a photon gas is exactly one-third of its energy density. This result is also derivable from kinetic theory (massless particles moving isotropically exert pressure equal to one-third of the energy density) and is crucial for cosmology, where it determines how radiation energy density scales with the expansion of the universe (as a^{-4}).
The tracelessness of the electromagnetic stress-energy tensor reflects the conformal invariance of Maxwell's equations in four dimensions — massless fields have no intrinsic length or energy scale. This property extends to any massless field, and the equation of state p = ρc²/3 is universal for radiation.