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Linear barotropic equation of state and its enthalpy coordinates

Example

Fix a dimensionless constant w with 0<w<1, take J=(0,∞), choose a reference energy density ϵ∗>0, and set

p(ϵ)=wϵ.

Choose N∗>0 with units m−3 as in Strict charged barotropic tensor system. Then ϵ+p=(1+w)ϵ>0 and p′=w∈(0,1) on J, so this is an instance of the strict EOS. The EOS-defined number-density and enthalpy coordinates are

N(ϵ)=N∗(ϵϵ∗)1/(1+w),r(ϵ)=w1+wlog⁡(ϵϵ∗).

Here the ratios and r are dimensionless. The map r:J→R is strictly increasing and onto; in these variables N=N∗er/w. These are mathematical EOS variables, not a microscopic derivation of particle number.

Facts & Assumptions

Given: 0<w<1, ϵ>0, ϵ∗>0, N∗>0, and the EOS-defined functions of Strict charged barotropic tensor system and Strict charged-fluid symmetric reduction.

[F1]

Strict charged barotropic tensor system: N(ϵ)=N∗exp⁡(∫ϵ∗ϵ(z+p(z))−1 dz), with N∗>0.

[F2]

Strict charged-fluid symmetric reduction: r(ϵ)=∫ϵ∗ϵp′(z)/(z+p(z)) dz and r′=p′/(ϵ+p)>0 for the strict EOS.

Proof

technique · direct integral evaluation and differentiation
1.1givenalgebra

The derivative of p(ϵ)=wϵ is p′=w. Since w>0 and ϵ>0, p′=w∈(0,1) and ϵ+p=(1+w)ϵ>0 on J. Thus the strict EOS hypotheses hold.

1.2F1givenalgebra

For every z>0, (z+p(z))−1=((1+w)z)−1. Substituting this in [F1] and integrating gives N=N∗exp⁡((1+w)−1log⁡(ϵ/ϵ∗))=N∗(ϵ/ϵ∗)1/(1+w). The logarithm is dimensionless because both arguments are energy densities.

2.1F2givenstep 1.2algebra∎

Since p′(z)=w, [F2] gives r=(w/(1+w))log⁡(ϵ/ϵ∗). Differentiating yields r′=w/((1+w)ϵ)>0, so r is strictly increasing. As ϵ↓0 its logarithm tends to −∞, and as ϵ→∞ it tends to +∞; the positive prefactor preserves these limits, so its range is all of R. Solving the displayed formula gives ϵ/ϵ∗=exp⁡((1+w)r/w); inserting this in step 1.2 yields N=N∗er/w.

Depends on

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