3.6 A lower bound for the pulse speed of a non-degenerate gas

Since in (47View Equation) the integral ∫ panap αpβfEdp is symmetric and ∫ pαpβfEdp is symmetric and positive definite, it follows from linear algebra3 that
∫ (pan − V m )p p f dp is negative semi-definite. (48 ) a max α β E
Boillat & Ruggeri [5] have used this knowledge to derive an estimate for V max in terms of N, the highest tensorial degree of the moments. The estimate reads
∘ --(-------)- Vmax- ≥ 6- N − 1- . (49 ) c0 5 2
Therefore, indeed, as more and more moments are drawn into the scheme of extended thermodynamics, the pulse speed goes up and, if N tends to infinity, so does Vmax.

The proof of (49View Equation) rests on the realization that – because of symmetry – pα = pi1pi2 ...pil has only 12 (l + 1)(l + 2) independent components and they are simply powers of p1,p2 and p3, so that pα may be written as (p1 )p(p2)q(p3)r with p + q + r = l. Accordingly pβ = pj1pj2 ...pjk may be written as 1 s 2t 3 u (p ) (p )(p ) with s + t + u = k.


Table 1: Pulse speed in extended thermodynamics of moments. n: Number of moments, N: Highest degree of moments, V ∕c max 0: Pulse speed.
n N Vmax∕c0 n N Vmax ∕c0
4 1 0.77459667 2600 23 6.59011627
10 2 1.34164079 2925 24 6.75262213
20 3 1.80822948 3276 25 6.91176615
35 4 2.21299946 3654 26 7.06774631
56 5 2.57495874 4060 27 7.22074198
84 6 2.90507811 4495 28 7.37091629
120 7 3.21035245 4960 29 7.51841807
165 8 3.49555791 5456 30 7.66338362
220 9 3.76412372 5984 31 7.80593804
286 10 4.01860847 6545 32 7.94619654
364 11 4.26098014 7140 33 8.08426549
455 12 4.26098014 7770 34 8.22024331
560 13 4.71528716 8436 35 8.35422129
680 14 4.92949284 9139 36 8.48628432
816 15 5.13625617 9880 37 8.61651144
969 16 5.33629130 10660 38 8.74497644
1140 17 5.53020569 11480 39 8.87174833
1330 18 5.71852112 12341 40 8.99689171
1540 19 5.90168962 13244 41 9.12046722
1771 20 6.08010585 14190 42 9.24253184
2024 21 6.25411673 15180 43 9.36313918
2300 22 6.42402919      

Therefore (48View Equation) assumes the form

∫ (pana − Vmaxm )(p1)p+s(p2)q+t(p3)r+ufEdp – negative semi-definite. (50 )

The elements of a semi-definite matrix aij satisfy the inequalities aiiajj ≥ a2ij and therefore (50View Equation) implies

( ) ∫ (pan − V m )(p1)2p(p2)2q (p3)2r f dp × a max E × ( ∫ (pan − V m )(p1)2s(p2)2t(p3)2uf dp) ≥ a max E (51 ) ( ∫ a 1 2p+s 2 2q+t 3 2r+u )2 ≥ (p na − Vmaxm )(p ) (p ) (p ) fEdp

Since fE is an even function of p we obtain

( ) (V )2m2 ∫ (p1)2p(p2)2q (p3)2r f dp × ( max ) E × ∫ (p1)2s(p2)2t(p3)2uf dp ≥ E (52 ) (∫ a 1 p+s 2 q+t 3 r+u )2 ≥ (p na − Vmaxm )(p ) (p ) (p ) fEdp
This estimate depends on the choice of the exponents p through u and we choose, rather arbitrarily p = N, s = N − 1 and all others zero. Also we set na = (1,0,0). In that case (52View Equation) implies
∫ 12N 1 ( ) V 2 ≥ ---(p-)---fEdp---= 6-⋅ 5-k-T N − 1- (53 ) max (p1)2(N −1)fEdp1 5 3 m 2
which proves (49View Equation).

An easy check will show that for each N the value ∘ ----------- 6 ( 1) 5 N − 2 lies below the corresponding values of Table 1, as they must. It may well be possible to tighten the estimate (49View Equation).


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