Nikolai V. Brilliantov and Thorsten Pöschel

Print publication date: 2004

Print ISBN-13: 9780198530381

Published to Oxford Scholarship Online: January 2010

DOI: 10.1093/acprof:oso/9780198530381.001.0001

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(p.283) Appendix C Solutions to Problems

Source:
Kinetic Theory of Granular Gases
Publisher:
Oxford University Press

Solution of problem 2.1

Derive the collision law (2.7)!

First, we change from the velocities $υ → 1$ and $υ → 2$ to the centre of mass and relative velocity:

(C.1)
where M = m 1 + m 2. Then the inverse transformation reads
(C.2)

Consider the normal and transversal components of the relative velocity $υ → 12$at the point of contact of the colliding spheres:

(C.3)

We assume that the interaction force has only a normal component, that is, the tangential component is negligible. This assumption is approximatively valid for smooth spheres (but see the discussion on page 35). The collision affects, hence, only the normal component of the relative velocity while the tangential component of does not change. The components of $υ → 12$ the relative velocity after the collision read

(C.4)
so that
(C.5)

Hence

(C.6)

Similarly, we obtain

(C.7)

(p.284) Solution of problem 2.2

Two spheres collide with an angle of their paths α. Express the angle of their traces after the collision as function of the coefficient of restitution!

The motion of two colliding spheres occurs always in a plane as it follows immediately from the conservation of momentum (show that!). Since the dissipation during a collision concerns only the velocity component in normal direction $e → n = | r → 1 - r → 2 | / ( r → 1 - r → 2 ) ,$ the angle after the collision is always smaller than before for all ε < 1. From simple geometry the result tan α′ = ε tan α follows

Solution of problem 2.3

Assume ε = const. A particle falls from height H and rebounces recurrently from the floor. At which time t does it come to rest? Can t be finite?

At the first contact with the floor, the particle's velocity is given by υthe equivalence of kinetic and potential energy 02 /2 = mgH. So it reaches the floor with $υ 0 = 2 g H$ after time $t 0 = 2 H / g$ The time between the first and the second bounce is $t 1 = 2 ɛ 2 H / g$ In general, the delay between the ith and (i + 1)th bounce (i ≥ 1) is $t i = 2 ɛ i 2 H / g$Hence, the total time is

(C.8)

This means that the bouncing sphere comes to rest after finite time.

Solution of problem 3.1

Assume ε depends on the impact velocity as ε(υ) = 1 – 1/5 (This formula is an approximation of (3.22)). A particle is again dropped from height H to a rigid floor and rebounces recurrently. After which time $t ∞ υ$ this sphere will come to rest? Compare the bouncing times for the cases ε = const. and ε = ε(υ)! Assume that at the first contact with the floor the velocity dependent coefficient of restitution and the constant one have the same value!

The velocity before the kth bounce is υk, that is, υ k+1 = ευk. The corresponding energies are E k+1 = ε 2 Ek. The time lag is $t k + 1 - t k = 2 2 E k / m g 2$ Therefore

(C.9)

Assuming small energy differences and time lags, we formulate the according differential equation

(C.10)

For ε = const. we abbreviate $E ˙ = a E$ with the solution $E = a t / 2 + E 0 ,$ where E 0 = mgH is the initial energy. The energy is completely dissipated after the time

(p.285)

(C.11)

Note that $t ∞ c$ corresponds to the starting time t = 0 at the ground. Since the ball is dropped from the height H the total time is by $t 0 = 2 H / g$ larger. The result (C.11) and the solution of problem 2.3, equation (C.8) (with t 0 substracted for the comparison), differ from each other, although they describe the same physical situation. This difference comes from the approximation of continuous time which has been made in (C.10). This approximation is justified if the kinetic energy between successive bounces decays only by a small amount, that is, for ε → 1. In this limit, both solutions, (C.8) and (C.11) coincide. However, (C.8) is rigorous since no assumption about ε has been made. For the velocity dependent coefficient of restitution, the summation in (C.8) cannot be performed.

For the velocity dependent coefficient of restitution we obtain 1—ε 2 = 2 1/5 with $υ = 2 E / m$ Equation (C10) for this case reads

(C.12)
with the solution $E 2 / 5 = 2 b t / 5 + E 0 2 / 5$. The energy is dissipated after
(C.13)
again with E 0 = mgH. We wish to compare the times for the cases ε = const. and ε = 1 − 1/5. To this end we assume that for the first bounce the coefficients of restitution are identical, that is,
(C.14)
and insert this quantity into (C.11):
(C.15)

With (C.13) we obtain

(C.16)
that is, the ball which bounces with ε = const. dissipates its energy in 80% of the time the viscoelastic ball needs.

(p.286) Solution of problem 3.2

Derive (3.36)!

For the inverse collision

(C.17)

According to the definition of the inverse collision, which is just the direct collision in reverse time, the energy at the point of maximal compression (turning point) and at the beginning of the collision reads

(C.18)

Then integrating the LHS of (C.17), we obtain

(C.19)

On the other hand, since

(C.20)
integration of the RHS of (C.17) yields
(C.21)
where we use the definition (3.33). Comparing (C.19) and (C.21) we arrive at (3.36).

Solution of problem 3.3

Derive the propagation rule (3.52) by means of (3.51, 3.49)!

Consider the collision of two spherical particles of radius R with velocities $υ → 1 , υ → 2$ and angular velocities $ω → 1 , ω → 2 .$ Let $r → 12 = e → R .$ The velocity of the surface of the first sphere at the point of contact is $υ → 1 + R ( e → × ω → 1 )$and that of the second sphere is $υ → 2 - R ( e → × ω → 2 )$ Then the relative velociy at the point of contact reads

(C.22)
where we introduce $ω → 12 ≡ ω → 1 + ω → 2 .$ The normal and transverse components of this velocity are given by (3.49). From the first two equations of (3.51) we obtain

(p.287)

(C.23)

Combining (3.48, 3.49), the tangential velocity after the collision reads

(C.24)

With $ω → ′ 12$ given by (C.23) we write

(C.25)
where we use the relation $A → × B → × C → = B → ( A → . C → ) - C → ( A → . B → ) ,$introduce qI/ (mR 2) and take into account the definition of $g → n ,$(3.49) and (3.48), which relates $g → n$and $( g → n ) ′ .$ We rewrite (C.25) in the form
(C.26)
that is,
(C.27)

With the conservation of momentum

(C.28)
we obtain the result
(C.29)
which coincides with the first equation in the system (3.52). If we subtract (C.27) from (C.28) we obtain the second equation in (3.52). From (3.51), the third equation in (3.52) follows:
(C.30)

The last equation in (3.52) may be obtained completely analogously.

(p.288) Solution of problem 3.4

The reduced moment of inertia qI/(mR 2) characterizes the distribution of particle material inside the grain. How does this quantity affect the coupling between the rotational and translational motion? Look at two opposite cases: (i) all mass is distributed in a very thin shell of radius R, and (ii) the mass is concentrated in a very small volume around the centre of the particle. What is the value of q for the grains of a uniform density? Does q depend on mass, density, or radius in the latter case?

For the case when all mass is distributed in a very thin shell I = mR 2 and q = q max = 1; the coupling between the rotational and translational degrees of freedom is maximal, as it follows from (3.52). For the opposite case, when all mass is located at the centre of the sphere, I = 0 and q = q min = 0, so that 1/(1 + 1/q) = 0; (3.52) hence shows that there is no coupling between rotational motion and translational motion. If the grains are of uniform density I = 2/5 mR 2 and q = 2/5, this quantity depends neither on the masses, the density nor on the radii of the particles.

Solution of problem 4.1

Derive (4.34) and (4.36)!

The extremal value of n = n* follows from the equation $d υ ′ n / d n = 0$ or from $d log ( υ ′ n ) / d n = 0$ since the logarithm is a monotonic function. Hence

(C.31)
we introduce x 0 = (m n/m 0)1/n this equation turns into
(C.32)
which is equivalent with (4.35):
(C.33)

From the definition of x 0 follows (4.34) for n*. Using the definition of x 0 we can also write

(p.289)

(C.34)
which proves (4.36).

Solution of problem 5.1

Estimate the ratio of triple and binary collisions for a gas of soft particles of radius R and mass m which interact with a repulsive potential Φ(r) = α, where ξ is the compression of particles (see Chapter 3 )! The gas has temperature T and number density n.

To estimate the ratio of triple and binary collisions we consider a typical collision of particles with velocities close to a thermal velocity $υ T = 2 T / m$ When particles collide, the typical kinetic energy of their relative motion, $m eff υ T 2 / 2 ( m eff = m / 2 )$ transforms partly into the potential energy of the elastic deformation so that the conservation of energy gives

(C.35)

The maximal deformation at the collision ξ max then reads

(C.36)
and the collision duration is
(C.37)
with the constant
(C.38)
where B(x, y) and Γ(x) are, respectively, the Beta and the Gamma functions. Using the definitions of υT and m eff we can also write
(C.39)

(p.290) The average time of the free flight (mean collision time) τc reads (see (5.3))

(C.40)

The ratio of the collision duration and the mean collision time may then be written using (C.36),

(C.41)
where η = (4/3)πR 3 n is the packing fraction, which illustrates that the approximation of instantaneous collisions becomes better for harder particles and for more dilute gases.

The fraction of time which a particle spends in binary collisions may be estimated as a ratio of the collision duration and the time of the mean free flight, that is, by (C.41). Therefore, if the total number of particles is N, then at each time instant about N (τ coll/τc) of them are involved in a binary collisions. Thus, in a unit volume there exists 1/2 n (τ coll/τc) pairs of particles which are in contact, that is, the concentration of pairs n pair is 1/2 n (τ coll/c).

Now it is easy to calculate the collision frequency of a single particles with pairs, that is, the collision frequency of triple collisions, $τ c,trip - 1 .$ This may be done exactly in the same way as before for the case of binary collisions (see Chapter 5):

(C.42)
where D is a numerical factor of the order of unity which accounts for the effective cross-section of the collision between a single particle and a pair and for the effective mass (equal to 2m/3). Hence, we obtain for the ratio of the collision frequencies of triple and binary collisions,
(C.43)
which again shows that the more dilute a gas and the harder the interaction potential, the more accurate is the binary collision approximation.

Solution of problem 5.2

Using the results of the Exercise 5.1 estimate the temperature dependence of the ratio of triple and binary collisions for a gas of viscoelastic particles!

From the Hertz law (3.2) follows the potential energy of compressed viscoelastic particles, $ϕ ( ξ ) = 2 5 ρ ξ 5 / 2$ and thus the maximal compression,

(C.44)

(p.291) where we assume that the dissipation is small so that the value of ξ max may be approximated by the Hertz law. Similarly we approximate the collision duration. Then the ratio of the collision frequencies of triple and binary collisions is given by (C.43), which we write using (C.44)

(C.45)

The last equation shows that this ratio scales with temperature as T 2/5, that is, the binary collision approximation becomes better as temperature decreases, that is, the particles become appearently harder for smaller collision velocities at low temperature.

Solution of problem 8.1

Recast the basic property of the collision integral (6.25) into its dimensionless form (8.4)!

Let ψ in (6.25) be a function of the dimensionless velocity c 1. Using the definition of the dimensionless collision integral (8.3) we write

(C.46)
which equals the RHS of (6.25) (without the omitted factor g 2(σ)):
(C.47)

Comparing (C.46) with (C.47), we obtain the basic property of the dimensionless collision integral (8.4).

(p.292) Solution of problem 8.2

Derive the expression for µ 2, given by (8.23) in terms of Basic Integrals as defined by (8.30)!

Substituting (8.26, 8.28) into (8.23) for p = 2 we obtain

(C.48)

From the definition (8.30) of the Basic Integrals it follows that µ 2 in the last equation may be written in the form (8.36).

Solution of problem 8.3

Derive µ 4 given by (8.38) using Maple!

The solution of this problem is discussed in more detail on page 260. It is easy to check that the result given by Maple may be written in the form (8.38, 8.39).

Solution of problem 10.1

Find the solution of (10.22), that is, find the exponent ν and the prefactor for the Ansatz φ ∝ (1 + t/τ 0)ν !

Substituting the Ansatz φ = A (1 + t/τ 0)ν into equation (10.22) we obtain

(C.49)

Comparing the exponents in the LHS and RHS we conclude that ν − 1 = —5/6, that is, ν = 1/6. Substituting this exponent into (C.49) we obtain

(C.50)
where we use (9.29) for τ 0/τc (0) and (10.24) for b.

(p.293) Solution of problem 13.1

Derive (13.37) using the definition of the collision operator (13.35) and the collision rules (6.5)!

Since the indices i, j are arbitrary we choose i, j = 1, 2. From the definition of the binary collision operator (13.35) and of the operator $b ˆ i j e →$ (13.36) follows

(C.51)
where according to the collision rules (6.5)
(C.52)

Thus,

(C.53)

Similarly, we can write

(C.54)
where we use $υ → 21 = - υ → 12 , r → 21 = - r → 12$ and take into account $b ˆ 21 e → = b ˆ 12 e → ,$ according to the definition (13.36) of this operator. If we change $e → → - e →$ in the integrand and notice that $b ˆ 12 - e → = b ˆ 12 - e → ,$ as it follows from the collision rule (6.5), we obtain
(C.55)

Solution of problem 13.2

Prove (13.39)!

Since

(C.56)
the term with the free-streaming operators does not contribute. According to the definition (13.35) the expression $T ˆ i j υ 1 2$ with i < j contains the factor $( b ˆ i j e → - 1 ) υ 1 2$.

(p.294) From the definition of $b ˆ i j e →$ which describes the collision between a pair (i, j) follows

(C.57)
since if i ≠ 1 the particle 1 is not involved in the collision (recall that j > i). If i = 1, the summation over the index j > 1 in (13.39) gives (N − 1) identical terms, the same for all j. With j = 2 we arrive at the final result (13.39).

Solution of problem 14.1

Prove the property (14.3) of the binary collision operator!

From the identity of the particles and (13.37) follows

(C.58)

Solution of problem 14.2

Derive (14.6)!

We integrate over $d r → 3 ⋯ d r → N d υ → 3 ⋯ d υ → N$ by means of (13.42),

(C.59)
and use the approximation (13.43),
(C.60)
which corresponds to the assumption of molecular chaos. The integration over$d r → 1 d r → 2$ may be performed using the transformation of variables $d r → 1 , d r → 2 → d r → 1 , d r → 12$ with the result
(C.61)

Substituting this result into (14.5) we arrive at (14.6).

(p.295) Solution of problem 14.3

Derive the velocity correlation time for a granular gas of particles that interact with ε = const. in linear approximation with respect to a 2 : first perform the manual calculations, then use Maple (see Appendix A ) to get the expression for τυ in terms of Basic Integrals. Finally, derive (14.22)!

In linear approximation with respect to a 2 one can write (14.9) using (14.11) and expressing the factor $( S 2 ( c 1 2 ) + S 2 ( c 2 2 )$ in terms of the new variables $c → 12$ and $C → :$

(C.62)
where
(C.63)
from the definition of the Basic Integrals (8.30), the expression for τυ follows:
(C.64)

Substituting the values of the Basic Integrals, which may be found from (8.31), we obtain (14.22). The same result may be obtained by Maple:

which gives as expected (14.22).

Solution of problem 14.4

Find the self-diffusion coefficient up to the second order in a 2 !

For the complete description of the self-diffusion on the level of the first non-vanishing Sonine coefficient a 2 (with all high-order coefficients neglected) one has to take into account the term $a 2 2 S 2 ( c 1 2 ) S 2 ( c 2 2 )$ in (14.9) (which was omitted in (p.296) the linear analysis) and use (8.65) for τ 0, which refers to the complete description of the evolution of temperature (on the level of a 2). Using Maple we obtain

(C.65)
and then with (14.17, 14.18, 14.20) we arrive at
(C.66)
where in the last expression τ 0 from (8.65) is to be used. The Maple program for the evaluation of τυ reads in this case

Solution of problem 14.5

Find the velocity correlation time τυ up to the second order in δ and in a 2 by means of Maple!

The result of the Maple program

(C.67)

(p.297) Solution of problem 15.1

Prove (15.22) and (15.23)!

Consider the vectorial integral

(C.68)

(here we will use $g →$ in place of $ω →$ in (15.20, 15.21)). The result of the integration over $e →$ must be a vector. Due to the symmetry of the problem, this vector should be directed along $g →$ since there is no other preferred directions. Thus,

(C.69)
with the function G(g) to be determined. We chose the coordinate system with the Z axis directed along $g →$ and multiply (C.68) by $g →$ With the definition (C.69) of G(g) we find
(C.70)
that is, G(g) = (π/2)g and thus
(C.71)
with (C.71) the and definition of $A →$ (15.20), we obtain (15.22). To prove the other relation (15.23) we consider the tensor with the components {i, j} = {x, y, z}:
(C.72)

The integrand in (C.72) is a tensor with components given by products of the components of the vector $e →$. Since integration over $e →$ does not change the symmetry, we expect that the components of the resultant tensor are functions of the components of the vector $g →$, which is the only involved vector, plus the unit tensor:

(C.73)

The trace (that is the sum of diagonal elements) of the tensor n ij reads

(C.74)
where we take into account that δii = 1 + 1 + 1 = 3, $g i g i = g x 2 + g y 2 + g z 2 = g 2$ and eiei = 1 Multiplaying nij by gigj summing over i and j we obtain

(p.298)

(C.75)
where we use δijgigj = g 2 and $e i g i = e j g j = g → . e → .$ The system
(C.76)
which follows from (C.74, C.75), has the solution
(C.77)

(C.78)

With (C.78) and the definition (15.21) of Nij we obtain the second relations

Solution of problem 15.2

Prove the second line of (15.66)!

Transforming to time variable t and multiplying (15.63) by υT(t) we obtain the force

(C.79)

(C.80)
where we use (15.64) and take into account
(C.81)
since /dt = γ(t)(1 − b). Then from (15.37) we notice that $υ T 2 = γ ¯ ( a γ )$ and obtain
(C.82)
where the relation a = (1 − b)/2 (15.44) was used.

(p.299) Solution of problem 17.1

Prove (17.18) for the multiplication of a vector and a dyad!

The multiplication of the tensor $a → b →$by the vector $c →$yields a vector whose ith component reads

(C.83)
where the summation convention has been used. Similarly
(C.84)

Solution of problem 17.2

Derive the hydrodynamic equation (17.32) from the Boltzmann equation (17.8) by multiplying it with $m υ 1 2 / 2$ and integrating over $υ → 1 !$

For the first term in the LHS of (17.8) after this transformation we obtain (using the local velocity $V → = υ → - u →$ and omitting the subscript ‘1’)

(C.85)
where we use (17.14) for ∂n/∂t and (17.25) for $∂ u → / ∂ t$ and take into account that
(C.86)

Correspondingly for the second term on the LHS we can write

(C.87)

(p.300) Using the definitions for the heat flux (17.33), the pressure tensor (17.22), density (17.4) and temperature (17.6) and (17.23) we obtain for the LHS

(C.88)

Finally, taking into account that

(C.89)
and
(C.90)
and
(C.91)
and summing (C.85) and (C.88) we obtain for the LHS of (17.8)
(C.92)

Evaluation of the RHS as it was already shown in (17.30, 17.31) gives − (3/2)nTζ. Division of all terms of the obtained equation by (3/2)n yields the hydrodynamic equation (17.32).

Solution of problem 18.1

Derive equations (18.3) and (18.4)!

We assume for simplicity that the kinetic coefficients are constants and consider the ith component of the vector $∇ → ⋅ P ˆ$:

(C.93)
where we use (17.36) and where summation over repeating indices is implied. With
(C.94)
we obtain
(C.95)

Using the (formal) coefficient λ at each power of the gradient we obtain the second equation in (18.3). Next we consider

(p.301)

(C.96)
where we use the definition of the tensor product (17.34) and (C.94). With
(C.97)
which follows from (17.36), we obtain
(C.98)

Using in the above equations λk at each kth power of ∇, we arrive at the third equation in (18.3) together with (18.4).

Sometimes the expression for $P ˆ : ∇ → u →$ is written in another form: we use the relation

(C.99)
and the definition of the tensor product (17.34), which allows us to write
(C.100)

Similarly, using

(C.101)
we obtain
(C.102)
and finally
(C.103)

This form is frequently used for $P ˆ : ∇ → u →$.

(p.302) Solution of problem 18.2

Prove (18.14)!

According to the definition of the collision integral I(a, b) (8.1), its dependence on the functions a and b enters only via the factor

(C.104)

For a = b = h 1 + λh 2 + ··· this factor reads

(C.105)

Substituting this factor into the collision integral we obtain

(C.106)
which proves (18.14).

Solution of problem 18.3

Find the derivatives of f (0) with respect to n, $u →$ and T, that is, derive (18.18)!

The first two equations in the system (18.18) follow from the functional form (18.10) of the distribution function in zeroth order f (0). To prove the third equation we write (see also (15.35))

(C.107)

With

(C.108)
we then obtain
(C.109)
which is the third equation in (18.18).

(p.303) Solution of problem 18.4

Derive the coefficients $A → , B →$ and Cij in (18.26)!

To obtain coefficients A and B, we substitute (18.18) into (18.24) and use the property (18.27). Do obtain Cij we start from (18.21) and using $∇ → ⋅ u → = δ i j ∇ j u i$ we write all terms in the RHS of (18.21) which contain ∇jui in the form

(C.110)
where we use (18.18) for T f (0). The expression in square brackets gives the coefficient Cij in (18.26) whose final form again follows from (18.27).

Solution of problem 18.5

Prove that the coefficient ζ (1), defined by (18.16), vanishes! Use the first-order distribution function f (1) as given on page 201.

ζ (1) is proportional to the integral

(C.111)

The integration is to be performed over the absolute values of $V → 1$ and $V → 2$ and over the directions of these vectors. Integration over the direction of $V → 2$ may be performed without restrictions, while performing integration over $V → 1$ we chose the OZ axis along the vector $V → 2$. Denoting the polar and azimuthal angles as ϕ and θ, we can write the integral in the form

(C.112)
where we take into account that according to (C.145) from Exercise 20.9 the integral in square brackets vanishes. This proves ζ (1) = 0.

Solution of problem 18.6

Prove (18.36)!

The first equation in (18.36) reads

(C.113)

(p.304) where the results of Exercise 20.9 have been used. From (18.29, 20.39) and using the definition of h 1(V) from Exercise 20.9 we obtain

(C.114)

Correspondingly,

(C.115)
since
(C.116)
because the integrands in the last two equations are antisymmetric functions.

Solution of problem 18.7

Prove $∫ d V → S i f ( 0 ) = 0$ and $∫ d V → S i γ k l = 0$ !

Using the definition (17.35) of $S → ( V → )$ we write

(C.117)
where (C.116) from Exercise 18.6 is used. Similarly using (20.6) for γkl we obtain
(C.118)
where again (C.116) has been applied.

Solution of problem 19.1

Prove (19.8)!

Using the scaling form of the velocity distribution function

(C.119)
we write for the integral
(C.120)

(p.305) where integration over the angles gives 4π, and where we use $υ T 2 = 2 T / m$ and the integral

(C.121)

Solution of problem 19.2

In general, the coefficient $α →$ is expressed as an expansion,

(C.122)

with unknown scalar coefficients αk. In this series the coefficient α 0 is always trivial, that is, α 0 = 0. Explain why!

Since the RHS of (19.30) does not have terms proportional to $S 0 ( 3 / 2 ) ( c 2 ) = 1$, we conclude that α 0 = 0.

Solution of problem 19.3

Compute the numerical coefficient $Ω k el$ manually and confirm the result using Maple!

With (19.22), which give the pre-collision velocities $c → 1 , c → 2$ as well as after-collision velocities $c → ′ 1 , c → ′ 2$ in terms of the centre of mass $C →$ and the relative $c → 12$ velocity before the collision, we can write

(C.123)

Substituting (C.123) into (19.42, 19.43) and using the definition of the Basic Integrals (8.31–8.33) together with the properties of the Gamma-function Γ(x + 1) = xΓ(x) and $Γ ( 1 / 2 ) = π$ we obtain

(C.124)
where

(p.306)

(C.125)

Thus,

(C.126)

The same result is obtained by Maple:

Solution of problem 20.1

Prove (20.3)!

From the definition of the dimensionless velocity, c = V/υT with $υ T 2 = 2 T / m$ follows

(C.127)

Similarly the definition of the Maxwell distribution (19.33) yields

(C.128)

(p.307) Solution of problem 20.2

Prove that the second term in the RHS of (20.7) equals 10nT!

(C.129)

The second term on the RHS vanishes since Dij is the traceless tensor, that is, due to the summation convention

(C.130)

The first term may be integrated by parts. Then, with

(C.131)
according to
(C.132)
we arrive at
(C.133)
where the definition of temperature has been employed, as well as the relation
(C.134)
which follows form the summation convention.

Solution of problem 20.3

The relation γ 0 = —η/(nT) has been obtained for gases of elastic particles. For which preconditions is it also valid for granular gases with ε = const.?

The relation γ 0 = —η/(nT) follows from (19.16), which uses the approximation (19.5) for the coefficient γij. Therefore, this relation is valid also for dissipative gases with constant coefficient of restitution as well as for gases of viscoelastic particles. Naturally, this relation is not correct if an expansion for γij is used that containes more than one term.

(p.308) Solution of problem 20.4

The viscosity coefficient (20.18) has been obtained in linear approximation with respect to a 2. How does one derive the viscosity coefficient in second order? Is it enough just to use (8.45) for µ 2, which is correct up to $ ( a 2 2 )$?

This may be done if the field gradients are very small, that is, if terms ∝ a 2jui may be neglected with respect to terms $∝ ( a 2 2 )$.

Solution of problem 20.5

Prove (20.20) for $A →$ by means of (18.26) and (20.4)!

Using the definition (18.26) of $A →$ and the relation (20.4) for the velocity derivative of the distribution function we write

(C.135)
where we use
(C.136)
which follow from the definition of the Sonine polynomials (7.14).

Solution of problem 20.6

Prove (20.26)!

With (19.1) for $A →$ and (19.28) for $S →$ one can write

(C.137)

Changing to dimensionless variables, using the definitions of the Sonine polynomials (7.14) and performing angular integration (which yields 4π) we obtain for the RHS of (C.137)

(p.309)

(C.138)
where the Maxwell distribution φ(c) = exp(—c 2)/π 3/2, the definition of υT and the integral
(C.139)
have been used.

Solution of problem 20.7

Derive the complete dependence κ (a 2), neglecting the terms of order $ ( a 2 ∇ T )$ !

The complete dependence of κ (a 2) on a 2 without the terms $ ( a 2 ∇ T )$ may be obtained from (20.32) if one substitutes into this equation (20.29) for $Ω k ( ɛ )$ and (8.45) for µ2, which correspond to the complete dependence of these quantities on a 2.

Solution of problem 20.8

Calculate the coefficient µ, following the same procedure as for the coefficient of thermal conductivity!

The calculation of κ and µ are completely analogous to the derivation of the coefficient of thermal conductivity i.e. we multiply (18.34) by $- S → ( V → 1 ) / T$ and integrate over $V → 1$. The approximation (20.35) implies

(C.140)
thus we can use the results, which have been obtained for the coefficient κ. The only difference is in the term
(C.141)
and in the temperature dependence of µ, which scales as µ ~ T 3/2.

(p.310) Solution of problem 20.9

Check the normalization of the distribution function $f ( V → )$ given by (20.39)!

For the first part of the distribution function we can write:

(C.142)
where the normalization of φ(c) and the properties of the Sonine polynomials (7.6) have been used. The second part of the distribution function can be written in the form
(C.143)
where the form of the functions h 1/2(V) follows from (20.39) and the definitions (17.27) and (17.35) of $D i j ( V → )$ and $S →$ and the unit vector $V → ˆ = V → / V$ has been introduced. Then we obtain
(C.144)
where
(C.145)

The last equation follows from

(C.146)

To prove (C.146) we consider these identities in components, for example

(C.147)

(p.311) Similarly

(C.148)

Hence

(C.149)
and we arrive at (C.146). From (C.142) and (C.145) it follows that the function f(V) for inhomogeneous granular gases has the same normalization as for a homogeneous gas.

Solution of problem 21.1

Calculate the coefficient µ for a granular gas of viscoelastic particles! Follow the same procedure as for the derivation of the coefficient of thermal conductivity!

According to (18.45) the expression for µ reads

(C.150)
where $β →$ is the solution of the equation
(C.151)
with
(C.152)

Therefore, we seek $β →$ in a form similar to (20.22):

(C.153)

Following the same scheme as in the preceding exercise we multiply (C.151) by $S → ( V → 1 ) / T$ and integrate over $V → 1$. Then with the definition (18.45) of κ and µ we obtain

(C.154)
where we evaluate the integral
(C.155)

(p.312) which equals (up to the coefficient a 2) the result (19.34). The integral

(C.156)
is similar to (21.23). We also employ (20.36)
(C.157)
which relates µ and β 1. Using for µ the expansion
(C.158)
and substituting it into (C.154) together with
(C.159)
we find the coefficients $μ → 1$ and $μ → 2$.

(p.313) Note that on line , the root for a 2δ′ is to be chosen.

Solution of problem 22.1

Prove that in (22.30) all terms that contain factors $a ˙ k υ → ⋅ ∂ G → / ∂ a k$ vanish upon integration over $υ →$ for k ≥ 2!

For the definition of the Sonine expansion

(C.160)
follows from (22.21)
(C.161)

Hence we can write

(C.162)
since, according to (7.17) ν k2 = 0 for k > 1.

Solution of problem 24.1

Derive (24.3)–(24.5)!

The derivation of the hydrodynamic equations for the space-independent coefficients has been performed in the Exercise 18.1. Therefore, these may be obtained by taking λ = 1 in the equations (18.3) and (18.4).

(p.314) Solution of problem 25.1

Find expressions for the coefficients η(ε), k*(ε), μ(ε) and ζ*(ε) using the relations for the kinetic coefficients and the cooling rate from Chapter 16 !

We substitute the relations (20.17) for μ 2, (20.12) for $Ω η ( ɛ )$ and (19.26) for η 0 into (20.16) and obtain

(C.163)

Similarly, from (20.32, 20.29, 19.46) we obtain

(C.164)

The quantity ζ*(ε) may be found from (18.11, 20.17, 19.26):

(C.165)

Finally, the value of μ*(ε) may be obtained from the results of Exercise 20.8:

(C.166)

The above results are accurate up to $ ( a 2 2 )$, since we keep only linear terms with respect to a 2 in ζ*. Therefore, they coincide with the direct evaluation of these quantities from the general expressions (23.4, 23.6) for the case d = 3, which are also accurate up to terms $ ( a 2 2 )$

Solution of problem 25.2

Derive the function $k ⊥ * ( ɛ )$ and show that it depends only on the coefficient of restitution! For ε ≲ 1 show that $k ⊥ * ( ɛ ) ∝ 1 - ɛ 2$ !

From (25.24) and the results of Exercise 25.1 for η*(ε) and ζ*(ε) it follows that $k ⊥ * ( ɛ )$ indeed depends only on ε. Moreover, for 1 − ε 2 ≪ 1 we can neglect a 2 as compared to unity and approximate (3 − ε)(1 + ε)/4 ≈ 1. This gives η*(ε) ≈ 1 and

(C.167)

(p.315) Solution of problem 25.3

Find the function $w → k → ⊥ ( t )$, using the relation between the laboratory time t and the time τ measured in the accumulated number of collisions! Find the function $u → k → ⊥ ( t )$ and show that the unscaled velocity field decays always with time!

From (25.10), definition of τc(t) (14.18) and evolution of temperature in the homogeneous cooling state (8.64) follows

(C.168)

Therefore, using (25.22, 25.23) we obtain

(C.169)

Taking into account

(C.170)
according to (8.65) and using (25.23) for $λ ⊥ ( k → )$ together with (C.165) for ζ*, we obtain the exponent in (C.169):
(C.171)
where the last equation is written in linear approximation with respect to a 2 for μ 2 from (8.37). With $u → k → ⊥ ( t ) = 2 T h ( t ) / m w → k → ⊥ ( t )$ and using the temperature dependence Th(t) we obtain
(C.172)

The last expression shows that the exponent of the power law dependence of $u → k → ⊥ ( t )$ is always negative, that is, this quantity always decays.

(p.316) Solution of problem 25.4

Derive the dependence of the marginal wave vector on the coefficient of restitution: $k H * ( ɛ ) ∝ 1 - ɛ 2$ for 1 − ε 2 ≪ 1!

Similarly as in Exercise 25.2 we notice that for 1 − ε 2 ≪ 1, k* ≈ 1, μ* ≪ 1 and a 2 ≪ 1. Therefore, from (25.29) we obtain the estimate

(C.173)

Solution of problem 27.1

Find the dependence of the non-reduced shear mode $u → k → ⊥$ on the non-reduced (laboratory) time t for 1 − ε 2 ≪ 1! Hint: use the dependence of temperature on time in the homogeneous cooling state and the relation between the laboratory time t and the reduced time τ.

It is shown in Exercise 25.3 that $u → k → ⊥$ depends on time as

(C.174)
with the exponent νk. given by (C.172). For 1 − ε 2 ≪ 1 we can approximate a 2 ≈ 0 and η* ≈ 1 (see Exercise 25.2). Hence, the exponent νk reads
(C.175)
which coincides with (27.1).