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Gaussian Maxwellian normalization, stress, and heat flux

Statement

Adopt Countable Choice, denoted ${A C}_{ω}$ , as in the kinetic-density definition. Let $n > 0$ , $u \in R^{3}$ , $T > 0$ , particle mass $m > 0$ , and Boltzmann constant $k_{B} > 0$ . Set $θ = k_{B} T / m$ and define $M_{n, u, T} (v) : = n (2 π θ)^{- 3 / 2} \exp (- ∣ v - u ∣^{2} / (2 θ))$ . Its Lebesgue moments are $\int_{R^{3}} M_{n, u, T} d v = n, \int_{R^{3}} v M_{n, u, T} d v = n u,$ $m \int_{R^{3}} (v - u) \otimes (v - u) M_{n, u, T} d v = n k_{B} T I_{3}, \frac{m}{2} \int_{R^{3}} ∣ v - u ∣^{2} (v - u) M_{n, u, T} d v = 0,$ and $\frac{m}{2} \int_{R^{3}} ∣ v ∣^{2} M_{n, u, T} d v = \frac{m n ∣ u ∣^{2}}{2} + \frac{3 n k_{B} T}{2} .$ Thus its pressure tensor is isotropic and its centered heat flux vanishes; these conclusions concern this Gaussian profile and do not impose isotropy on a general kinetic density.

Statement references

Tong, §2.4.2, full lecture notes

Zamponi, §3.2, full lecture notes

Facts & Assumptions

Given: Countable Choice and $M_{n, u, T} (v) = n (2 π θ)^{- 3 / 2} \exp (- ∣ v - u ∣^{2} / (2 θ))$ , with $θ = k_{B} T / m > 0$ .

[F1]

The Gaussian integral $\int_{- \infty}^{\infty} e^{- x^{2}} d x = \sqrt{π}$ : $\int_{R} e^{- z^{2} / 2} d z = \sqrt{2 π}$ , obtained from the one-dimensional Gaussian integral by rescaling.

[F2]

Change of variable in an improper integral: The affine substitution $v = u + \sqrt{θ} z$ transforms the stated Gaussian integrals to integrals in $z$ .

[F3]

Tonelli's theorem for nonnegative measurable functions on a sigma-finite product applied iteratively to the nonnegative absolute products establishes their integrability from the finite one-dimensional Gaussian-polynomial integrals; Fubini's theorem for L^1 functions on a sigma-finite product then factors the resulting signed moments.

[F4]

If $u, v$ are differentiable on $[a, b]$ with $u^{'}, v^{'}$ integrable, then $\int_{a}^{b} u v^{'} = u (b) v (b) - u (a) v (a) - \int_{a}^{b} u^{'} v$ : Gaussian decay makes the boundary term vanish and yields $\int_{R} z^{2} e^{- z^{2} / 2} d z = \int_{R} e^{- z^{2} / 2} d z$ .

[F5]

Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ : $θ > 0$ has the positive square root used in the substitution.

[F6]

A nonnegative improper Riemann integral on a half-line agrees with the Lebesgue integral: Under Countable Choice, a nonnegative function on $[b, \infty)$ that is Riemann integrable on every compact truncation and has a finite improper Riemann integral is Lebesgue integrable there with the same integral. Apply this also to the reflected function for a negative half-line; the Gaussian-polynomial factors here are continuous on every compact truncation.

[F7]

The exponential dominates every fixed nonnegative integer power at $+ \infty$ : Polynomial factors times $e^{- z^{2} / 2}$ are integrable at infinity, since the exponential dominates every inverse power after setting $x = z^{2}$ .

[A1]

The parameters $n, m, k_{B}, T$ are positive, and Countable Choice is adopted as in Nonnegative kinetic density and moment domain. It is used for the Riemann-to-Lebesgue conversion in [F6]; no stronger choice principle is assumed.

Proof

1.1A1F1F2F3F5F6F7algebra

Apply [F2] coordinatewise with $v = u + \sqrt{θ} z$ . Then $d v = θ^{3 / 2} d z$ , and the prefactor becomes $n (2 π)^{- 3 / 2}$ . By [F1], [F3], and [F6], the one-dimensional Gaussian Lebesgue integrals are finite and equal their improper Riemann values, and Tonelli/Fubini factors the three-dimensional product integral to $n$ . Each centered first moment is zero because $z_{i} e^{- ∣ z ∣^{2} / 2}$ is odd in $z_{i}$ and absolutely integrable. Hence $\int M d v = n$ and $\int v M d v = n u$ .

1.2A1F1F2F3F4F5F6F7algebra

For $i \neq j$ , the integral of $z_{i} z_{j} e^{- ∣ z ∣^{2} / 2}$ is zero by oddness in one coordinate. For each $i$ , [F4] gives $\int_{R^{3}} z_{i}^{2} (2 π)^{- 3 / 2} e^{- ∣ z ∣^{2} / 2} d z = 1,$ using [F1] and [F6] for the other coordinates. Thus $\int (v - u)_{i} (v - u)_{j} M d v = n θ δ_{i j}$ , and multiplication by $m$ gives $m n θ I_{3} = n k_{B} T I_{3}$ .

2.1A1F1step 1.1step 1.2algebra∎

Each component of $∣ z ∣^{2} z e^{- ∣ z ∣^{2} / 2}$ is odd under $z \mapsto - z$ , with absolute integrability, so its integral is zero and the centered heat flux vanishes. Finally write $∣ v ∣^{2} = ∣ u ∣^{2} + 2 u \cdot (v - u) + ∣ v - u ∣^{2}$ . The middle term integrates to zero by step 1.1, and the last term has integral $3 n θ$ by step 1.2. Multiplying by $m / 2$ yields the stated energy.

Gaussian Maxwellian normalization, stress, and heat flux

Statement

Statement references

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources