A constitutive stress readout turns matter primitives into the coupled stress. Its conservation law yields the fluid equations only with the normalization and constitutive inputs that make those operations valid.
11. Fluid constitutive readouts and Euler equations
Part D supplies constitutive information that a generic stress tensor cannot recover by itself. In §19 Einstein introduces pressure p, a density variable ρ, a relation between them and a normalized four-velocity. His footnote defines the comoving energy density as ϵ=ρ−p. Thus his ρ is not the variable often called energy density in modern notation. We preserve this convention.
On the existing realization add smooth p,ρ and a calibrated vector v, initially retaining the discrepancies n−1=η(v,v)−1 and Φ(p,ρ) for the supplied equation-of-state relation. Define
Σabfl=−pηab+ρvavb,va=ηabvb,u=E−1v.(11.1)
The physical constitutive form is inherited. Its forward operation gives
TμνflTμν,fl=−pgμν+ρuμuν,Tflμν=−pqμν+ρuμuν,=−pδνμ+ρuμuν,g(u,u)=η(v,v).(11.2)(11.3)
Indeed ETvlower=gu, and the tensor operations already proved recover the remaining forms. These are Einstein (58), Einstein (58a), Einstein (58b). On n=1,
Σabflvavb=ρ−p=ϵ,trηΣfl=ρ−4p,Σcomovingfl=diag(p,p,p,ρ−p).(11.4)
No extra thermodynamic or particle-number law is inferred from that definition.
Define, for a scalar s,
s˙=vaDas,ϑ=Dava,ab=vaDavb.(11.5)
The product rule and Dη=0 give the full raw divergence
Rbfl:=DaΣab,fl=vb(ρ˙+ρϑ)+ρab−Dbp.(11.6)
Its coordinate readout is
EbμRbfl=∇αTαμ,fl=ν−1∂α(νTαμ,fl)−21∂μgαβTflαβ.(11.7)
Substitution into the recovered Einstein (57a) is therefore exactly the source equation Rfl=0, with the same comparison connection as gravity.
Theorem 11.1 (Reversible energy and Euler decomposition)
On η(v,v)=1, put
Πbc=δbc−vbvc,E=(ρ−p)˙+ρϑ,Pb=ρab−ΠbcDcp.(11.8)
Then
Rbfl=vbE+Pb,vbRbfl=E,vbPb=0.(11.9)
Consequently the four conservation equations are equivalent to
ϵ˙+(ϵ+p)ϑ=0,ρab=(δbc−vbvc)Dcp.(11.10)
The second equation has three independent components in this normalized sector.
Proof.
Differentiating η(v,v)=1 gives vbab=0. Contract (11.6) with vb to obtain E, then subtract vbE to obtain P. Conversely the first equality in (11.9) reconstructs the original divergence from the scalar and orthogonal covector. The inverse map is (e,pb)↦vbe+pb, with vbpb=0. No division by ρ is used.
□
The coordinate Euler equation is
ρuα∇αuμ=(δμν−uμuν)∂νp.(11.11)
This expands Einstein’s substitution, rather than adding a distinct material law.
The algebraic decomposition R=vlowerE+P remains true for unnormalized v under the same definitions, but its projection interpretation does not. The exact omitted terms are
vbRbvbPb=E+(n−1)(ρ˙+ρϑ)+2ρn˙,=2ρn˙+(n−1)p˙.(11.12)(11.13)
This distinction is included in the public controls. One earlier test wrongly expected the algebraic decomposition itself to fail off normalization; the corrected test checks the actual projection and orthogonality assertions, with that development correction documented in the audit.
Closure, degeneracy and verification examples
The primitive reverse v=Eu preserves the normalization, equation-of-state relation and target boundary data. The six listed functions p,ρ,uμ and the four divergence equations recover the original system. Their count is not a general well-posedness proof. On ρ=0 and constant p, the stress is −pη and its divergence vanishes for any normalized v; the velocity is not determined. The gravitational conservation identities also prevent counting the material divergence equations as independent additional metric equations. Einstein’s residual coordinate-function count after volume specialization is retained as bookkeeping, not a global gauge-fixing theorem.
The earlier pressureless coupled fixture (10.22) now has its constitutive data: v=(0,0,0,1), p=0, ρ=ϵd. For an independent radiation-fluid fixture, take
Er=diag(r1/5,r1/5,r1/5,r−3/5),ϵr=25κ3r−4/5,p=3ϵr,ρ=34ϵr.(11.14)
The constitutive operation gives Σ=diag(ϵr/3,ϵr/3,ϵr/3,ϵr), with zero trace. Both projected fluid equations and G=κΣ hold in the same geometry. Substituting ϵr for Einstein’s ρ would change both the stress and the equations.
A separate fixed-background test exercises a nonzero pressure gradient and acceleration: with N=1+x, E=diag(1,1,1,N), v=(0,0,0,1), choose p=2/N−1 and ρ=2/N on 0<x<1. Then ϵ=1, a1=−1/N, and ∂1p=−2/N2, so (11.10) holds. This is a fluid calculation on a prescribed flat accelerated geometry, not a coupled solution with that geometry and nonzero stress. The classification prevents a successful matter example from being joined to a different gravitational preparation.