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

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

Since in (47

) the integral $∫ panap αpβfEdp$ is symmetric and $∫ pαpβfEdp$ is symmetric and positive definite, it follows from linear algebra³ that

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

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 (49 ) 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 (48 ) assumes the form

The elements of a semi-definite matrix $aij$ satisfy the inequalities $aiiajj ≥ a2ij$ and therefore (50 ) 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 (52

) implies

which proves (49

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 (49 ).