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Exact Reynolds and Mach nondimensionalization — worked interpretation

Physical scope

Formal periodic smooth-model example with explicitly selected SI parameters and fields. The values are mathematical inputs, not measurements or representative data for a named fluid. No approximation, limit, turbulence, or empirical conclusion is asserted.

Example

Use SI units and select the formal inputs ρ0=103 kg m−3, μ=10−3 Pa s, U=1 m s−1, L=10−1 m, and c=102 m s−1. These are exact illustrative parameters, not measurements. Then ν=μ/ρ0=10−6 m2 s−1, Re⁡=UL/ν=105, Ma⁡=U/c=10−2, and Ma⁡−2=104.

Let J=(−1,1) be a dimensionless open time interval containing Θ=0 and let the physical time interval be I=(L/U)J=(−L/U,L/U) seconds. For a side length ℓ>0, write Tℓ3:=R3/(ℓZ3) and let [x]ℓ denote the class of x∈R3 modulo ℓZ3. Define the dimensionless flat torus T2π3 and the physical flat torus T2πL3. The map DL:T2πL3→T2π3,DL([x]2πL)=[x/L]2π, is a well-defined smooth bijection with inverse [X]2π↦[LX]2πL. The quotient, not the open cube, is the periodic spatial domain; [0,2π)3 is only a representative cell.

Define the periodic lift u~∗:J×R3→R3 and pressure field p~inc∗:J×R3→R by u~∗(Θ,X)=(e−Θ/Re⁡sin⁡X2,0,0),p~inc∗(Θ,X)=0. For every k∈Z3, u~∗(Θ,X+2πk)=u~∗(Θ,X), so these smooth fields descend to u∗:J×T2π3→R3 and pinc∗:J×T2π3→R. Spatial derivatives on this flat torus are defined by differentiating the periodic lifts. They satisfy the scaled incompressible equations ∂Θu∗+(u∗⋅∇X)u∗=−∇Xpinc∗+Re⁡−1ΔXu∗,div⁡Xu∗=0. Under the spatial dilation DL and the time map Θ=Ut/L, define the corresponding physical fields on I×T2πL3 by u(t,[x]2πL)=Uu∗(Ut/L,DL([x]2πL)),p(t,[x]2πL)=pref+ρ0U2pinc∗(Ut/L,DL([x]2πL))=pref. Equivalently, the physical velocity lift is u~(t,x)=Uu~∗(Ut/L,x/L), which is periodic under x↦x+2πLk for every k∈Z3; physical spatial derivatives on the quotient are defined by this lift. A representative physical cell has side length 2πL. This is an exact solution of the selected incompressible equations.

On the same dimensionless spacetime domain J×T2π3, for the separate barotropic branch choose the pressure law P:(0,∞)→R given by P(ρ)=105 Pa+c2(ρ−ρ0) and the constant fields ρ∗:J×T2π3→(0,∞), u∗:J×T2π3→R3, and pbar∗:J×T2π3→R, with values ρ∗=1, u∗=0, and pbar∗=0. For these fields the scaled equations ∂Θρ∗+div⁡X(ρ∗u∗)=0,∂Θu∗+(u∗⋅∇X)u∗=−Ma⁡−2(ρ∗)−1∇Xpbar∗ hold identically. They are smooth, P′(ρ0)=c2>0, and their corresponding physical fields ρ:I×T2πL3→(0,∞) and u:I×T2πL3→R3 have the constant values ρ(t,[x]2πL)=ρ0 and u(t,[x]2πL)=0. This is an exact equilibrium of the unforced barotropic model. The computed coefficients and fields illustrate algebra in the adopted equations only; they report no observation and imply no low-Mach or high-Reynolds approximation.

Facts & Assumptions

Given: The exact parameter selection, periodic cell, smooth fields, and barotropic pressure law in the Example.

[F1]

Exact Reynolds and Mach nondimensionalization — conditional physical prediction: for positive scales, the unforced constant-density incompressible Newtonian system and the unforced positive-density barotropic Euler system have the exact scaled equations stated there. This example is conditional on those adopted model formulations and is not empirical evidence.

Verification

Given: The formal SI inputs and the two selected model fields in the Example. Use Θ for dimensionless time, distinct from any physical temperature.

1.1givenalgebra

The definitions give ν=μ/ρ0=10−3/103=10−6 m2 s−1. Therefore Re⁡=UL/ν=(1)(10−1)/10−6=105 and Ma⁡=U/c=1/102=10−2, so Ma⁡−2=104.

2.1F1givenstep 1.1algebra

For the selected shear field, div⁡Xu∗=0 because its first component is independent of X1, and (u∗⋅∇X)u∗=0 for the same reason. Direct differentiation gives ∂Θu∗=−Re⁡−1u∗ and ΔXu∗=−u∗. With pinc∗=0, the momentum equation is thus satisfied componentwise. Apply [F1] to the smooth periodic lift on the open domain R3; its physical lift satisfies the dimensional incompressible equations and is 2πL-periodic in each spatial coordinate, so it descends to the stated physical torus.

2.2F1givenstep 1.1algebra

For the linear pressure law, P′(ρ0)=c2>0. At ρ∗=1, pbar∗=0; the selected constant density and zero velocity make both the mass equation and the barotropic momentum equation vanish on each side. The coefficient 104 multiplies a zero pressure gradient in this equilibrium and is not an approximation parameter conclusion.

3.1F1step 1.1step 2.1step 2.2∎

Every selected value and field is a formal input or exact solution of the stated model equations; none is a measured result. The calculation supplies no limit or empirical claim.

Depends on

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