# (p.283) Appendix C Solutions to Problems

# (p.283) Appendix C Solutions to Problems

# Solution of problem 2.1

*Derive the collision law* (2.7)*!*

First, we change from the velocities ${\overrightarrow{\upsilon}}_{1}$ and ${\overrightarrow{\upsilon}}_{2}$ to the centre of mass and relative velocity:

*M*=

*m*

_{1}+

*m*

_{2}. Then the inverse transformation reads

Consider the normal and transversal components of the relative velocity ${\overrightarrow{\upsilon}}_{12}$at the point of contact of the colliding spheres:

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 ${\overrightarrow{\upsilon}}_{12}$ the relative velocity after the collision read

Hence

Similarly, we obtain

# (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 ${\overrightarrow{e}}_{n}=\left|{\overrightarrow{r}}_{1}-{\overrightarrow{r}}_{2}\right|/({\overrightarrow{r}}_{1}-{\overrightarrow{r}}_{2}),$ 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 *mυ*0^{2} /2 = *mgH*. So it reaches the floor with ${\upsilon}_{0}=\sqrt{2gH}$ after time ${t}_{0}=\sqrt{2H/g}$ The time between the first and the second bounce is ${t}_{1}=2\varepsilon \sqrt{2H/g}$ In general, the delay between the *i*th and (*i* + 1)th bounce (*i* ≥ 1) is ${t}_{i}=2{\varepsilon}^{i}\sqrt{2H/g}$Hence, the total time is

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 – *Cυ* ^{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 ${t}_{\infty}^{\upsilon}$ 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 *k*th bounce is *υ _{k}*, that is,

*υ*

_{k+1}=

*ευ*. The corresponding energies are

_{k}*E*

_{k+1}=

*ε*

^{2}

*E*. The time lag is $t{}_{k+1}-{t}_{k}=2\sqrt{2{E}_{k}/m{g}^{2}}$ Therefore

_{k}Assuming small energy differences and time lags, we formulate the according differential equation

For *ε* = const. we abbreviate $\dot{E}=a\sqrt{E}$ with the solution $\sqrt{E}=at/2+\sqrt{{E}_{0}},$ where *E* _{0} = *mgH* is the initial energy. The energy is completely dissipated after the time

Note that ${t}_{\infty}^{c}$ corresponds to the starting time *t* = 0 at the ground. Since the ball is dropped from the height *H* the total time is by ${t}_{0}=\sqrt{2H/g}$ 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*Cυ* ^{1/5} with $\upsilon =\sqrt{2E/m}$ Equation (C10) for this case reads

*E*

_{0}=

*mgH*. We wish to compare the times for the cases

*ε*= const. and

*ε*= 1 −

*Cυ*

^{1/5}. To this end we assume that for the first bounce the coefficients of restitution are identical, that is,

With (C.13) we obtain

*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

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

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

On the other hand, since

# 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 ${\overrightarrow{\upsilon}}_{1},{\overrightarrow{\upsilon}}_{2}$ and angular velocities ${\overrightarrow{\omega}}_{1},{\overrightarrow{\omega}}_{2}.$ Let ${\overrightarrow{r}}_{12}=\overrightarrow{e}R.$ The velocity of the surface of the first sphere at the point of contact is ${\overrightarrow{\upsilon}}_{1}+R(\overrightarrow{e}\times {\overrightarrow{\omega}}_{1})$and that of the second sphere is ${\overrightarrow{\upsilon}}_{2}-R(\overrightarrow{e}\times {\overrightarrow{\omega}}_{2})$ Then the relative velociy at the point of contact reads

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

With ${{\overrightarrow{\omega}}^{\prime}}_{12}$ given by (C.23) we write

*q*≡

*I*/ (

*mR*

^{2}) and take into account the definition of ${\overrightarrow{g}}^{n},$(3.49) and (3.48), which relates ${\overrightarrow{g}}^{n}$and $({\overrightarrow{g}}^{n}{)}^{\prime}.$ We rewrite (C.25) in the form

With the conservation of momentum

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

# (p.288) Solution of problem 3.4

*The reduced moment of inertia q* ≡ *I*/(*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

The extremal value of *n* = *n** follows from the equation $d{{\upsilon}^{\prime}}_{n}/dn=0$ or from $dlog({{\upsilon}^{\prime}}_{n})/dn=0$ since the logarithm is a monotonic function. Hence

*x*

_{0}= (

*m*

_{n}/

*m*

_{0})

^{1/n}this equation turns into

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

# 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*) = *Aξ ^{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 ${\upsilon}_{T}=\sqrt{2T/m}$ When particles collide, the typical kinetic energy of their relative motion, ${m}^{\text{eff}}{\upsilon}_{T}^{2}/2({m}^{\text{eff}}=m/2)$ transforms partly into the potential energy of the elastic deformation so that the conservation of energy gives

The maximal deformation at the collision *ξ* _{max} then reads

*B*(

*x, y*) and Γ(

*x*) are, respectively, the Beta and the Gamma functions. Using the definitions of

*υ*and

_{T}*m*

^{eff}we can also write

(p.290)
The average time of the free flight (mean collision time) *τ _{c}* reads (see (5.3))

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

*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}/*τ _{c}*) of them are involved in a binary collisions. Thus, in a unit volume there exists 1/2

*n*(

*τ*

_{coll}/

*τ*) pairs of particles which are in contact, that is, the concentration of pairs

_{c}*n*

_{pair}is 1/2

*n*(

*τ*

_{coll}/

*/τ*).

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

*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 2

*m*/3). Hence, we obtain for the ratio of the collision frequencies of triple and binary collisions,

# 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, $\varphi (\xi )=\frac{2}{5}\rho {\xi}^{5/2}$ and thus the maximal compression,

(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)

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

*g*

_{2}(

*σ*)):

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

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 corresponding Maple program reads.

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

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

*τ*

_{0}/

*τ*(0) and (10.24) for

_{c}*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 ${\stackrel{\u02c6}{b}}_{ij}^{\overrightarrow{e}}$ (13.36) follows

Thus,

Similarly, we can write

# Solution of problem 13.2

*Prove* (13.39)*!*

Since

*i*<

*j*contains the factor $({\stackrel{\u02c6}{b}}_{ij}^{\overrightarrow{e}}-1){\upsilon}_{1}^{2}$.

(p.294)
From the definition of ${\stackrel{\u02c6}{b}}_{ij}^{\overrightarrow{e}}$ which describes the collision between a pair (*i, j*) follows

*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

# Solution of problem 14.2

*Derive* (14.6)*!*

We integrate over $d{\overrightarrow{r}}_{3}\cdots d{\overrightarrow{r}}_{N}d{\overrightarrow{\upsilon}}_{3}\cdots d{\overrightarrow{\upsilon}}_{N}$ by means of (13.42),

# (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 τ _{υ} 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 ${\overrightarrow{c}}_{12}$ and $\overrightarrow{C}:$

*τ*follows:

_{υ}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

*τ*

_{0}from (8.65) is to be used. The Maple program for the evaluation of

*τ*reads in this case

_{υ}# Solution of problem 14.5

*Find the velocity correlation time τ _{υ} up to the second order in δ and in a*

_{2}

*by means of Maple!*

The result of the Maple program

reads

# (p.297) Solution of problem 15.1

Consider the vectorial integral

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

*G*(

*g*) to be determined. We chose the coordinate system with the

*Z*axis directed along $\overrightarrow{g}$ and multiply (C.68) by $\overrightarrow{g}$ With the definition (C.69) of

*G*(

*g*) we find

*G*(

*g*) = (

*π*/2)

*g*and thus

*i, j*} = {

*x, y, z*}:

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

The trace (that is the sum of diagonal elements) of the tensor *n* _{ij} reads

*δ*= 1 + 1 + 1 = 3, ${\text{g}}_{i}{\text{g}}_{i}={\text{g}}_{x}^{2}+{\text{g}}_{y}^{2}+{\text{g}}_{z}^{2}={\text{g}}^{\text{2}}$ and

_{ii}*e*= 1 Multiplaying

_{i}e_{i}*n*by

_{ij}*g*summing over

_{i}g_{j}*i*and

*j*we obtain

*δ*=

_{ij}g_{i}g_{j}*g*

^{2}and ${e}_{i}{\text{g}}_{i}={e}_{j}{\text{g}}_{j}=\overrightarrow{\text{g}}.{\overrightarrow{e}}_{.}$ The system

Hence, (C.73) reads

With (C.78) and the definition (15.21) of *N _{ij}* 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 *υ _{T}*(

*t*) we obtain the force

Its second moment reads

*dτ*/

*dt*=

*γ*(

*t*)(1 −

*b*). Then from (15.37) we notice that ${\upsilon}_{T}^{2}=\overline{\gamma}(a\gamma )$ and obtain

*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 $\overrightarrow{a}\overrightarrow{b}$by the vector $\overrightarrow{c}$yields a vector whose *i*th component reads

# Solution of problem 17.2

*Derive the hydrodynamic equation* (17.32) *from the Boltzmann equation* (17.8) *by multiplying it with $m{\upsilon}_{1}^{2}/2$ and integrating over ${\overrightarrow{\upsilon}}_{1}!$*

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

*n*/∂

*t*and (17.25) for $\partial \overrightarrow{u}/\partial t$ and take into account that

Correspondingly for the second term on the LHS we can write

(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

Finally, taking into account that

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 *i*th component of the vector $\overrightarrow{\nabla}\cdot \stackrel{\u02c6}{P}$:

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

Using in the above equations *λ ^{k}* at each

*k*th power of ∇, we arrive at the third equation in (18.3) together with (18.4).

Sometimes the expression for $\stackrel{\u02c6}{P}\text{:}\overrightarrow{\nabla}\overrightarrow{u}$ is written in another form: we use the relation

Similarly, using

This form is frequently used for $\stackrel{\u02c6}{P}\text{:}\overrightarrow{\nabla}\overrightarrow{u}$.

# (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

For *a* = *b* = *h* _{1} + *λh* _{2} + ··· this factor reads

Substituting this factor into the collision integral we obtain

# Solution of problem 18.3

*Find the derivatives of f* ^{(0)} *with respect to n*, $\overrightarrow{u}$ *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))

With

# (p.303) Solution of problem 18.4

*Derive the coefficients* $\overrightarrow{A},\overrightarrow{B}$ *and C _{ij} in* (18.26)

*!*

To obtain coefficients *A* and *B*, we substitute (18.18) into (18.24) and use the property (18.27). Do obtain *C _{ij}* we start from (18.21) and using $\overrightarrow{\nabla}\cdot \overrightarrow{u}\text{=}{\delta}_{ij}{\nabla}_{j}{u}_{i}$ we write all terms in the RHS of (18.21) which contain ∇

*in the form*

_{j}u_{i}*∂*

_{T}f^{(0)}. The expression in square brackets gives the coefficient

*C*in (18.26) whose final form again follows from (18.27).

_{ij}# 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

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

*ζ*

^{(1)}= 0.

# Solution of problem 18.6

*Prove* (18.36)*!*

The first equation in (18.36) reads

(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

Correspondingly,

# Solution of problem 18.7

*Prove* $\int d\overrightarrow{V}{S}_{i}{f}^{(0)}\text{=}\text{0}$ *and* $\int d\overrightarrow{V}{S}_{i}{\gamma}_{kl}\text{=}\text{0}$ *!*

Using the definition (17.35) of $\overrightarrow{S}(\overrightarrow{V})$ we write

*γ*we obtain

_{kl}# Solution of problem 19.1

*Prove* (19.8)*!*

Using the scaling form of the velocity distribution function

(p.305)
where integration over the angles gives 4*π*, and where we use ${\upsilon}_{T}^{2}\text{=}\text{2}T/m$ and the integral

# Solution of problem 19.2

*In general, the coefficient* $\overrightarrow{\alpha}$ *is expressed as an expansion*,

*with unknown scalar coefficients α _{k}. 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 ${S}_{0}^{(3/2)}({c}^{2})\text{=}\text{1}$, we conclude that *α* _{0} = 0.

# Solution of problem 19.3

*Compute the numerical coefficient* ${\Omega}_{k}^{\text{el}}$ *manually and confirm the result using Maple!*

With (19.22), which give the pre-collision velocities ${\overrightarrow{c}}_{1},{\overrightarrow{c}}_{2}$ as well as after-collision velocities ${{\overrightarrow{c}}^{\prime}}_{1},{{\overrightarrow{c}}^{\prime}}_{2}$ in terms of the centre of mass $\overrightarrow{C}$ and the relative ${\overrightarrow{c}}_{12}$ velocity before the collision, we can write

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) = *x*Γ(*x*) and $\Gamma (1/2)\text{=}\sqrt{\pi}$ we obtain

Thus,

The same result is obtained by Maple:

# Solution of problem 20.1

*Prove* (20.3)*!*

From the definition of the dimensionless velocity, *c* = *V*/*υ _{T}* with ${\upsilon}_{T}^{2}\text{=}\text{2}T/m$ follows

Similarly the definition of the Maxwell distribution (19.33) yields

# (p.307) Solution of problem 20.2

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

We start with (18.26) for the coefficient *C _{ij}*:

The second term on the RHS vanishes since *D _{ij}* is the traceless tensor, that is, due to the summation convention

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

# 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 *γ _{ij}*. 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

*γ*is used that containes more than one term.

_{ij}# (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* $\uf10e\text{(}{a}_{2}^{2}\text{)}$?

This may be done if the field gradients are very small, that is, if terms ∝ *a* _{2} ∇ * _{j}u_{i}* may be neglected with respect to terms $\propto \text{(}{a}_{2}^{2}\text{)}$.

# Solution of problem 20.5

*Prove* (20.20) *for* $\overrightarrow{A}$ *by means of* (18.26) *and* (20.4)*!*

Using the definition (18.26) of $\overrightarrow{A}$ and the relation (20.4) for the velocity derivative of the distribution function we write

# Solution of problem 20.6

*Prove* (20.26)*!*

With (19.1) for $\overrightarrow{A}$ and (19.28) for $\overrightarrow{S}$ one can write

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

*φ*(

*c*) = exp(—

*c*

^{2})/

*π*

^{3/2}, the definition of

*υ*and the integral

_{T}# Solution of problem 20.7

*Derive the complete dependence κ* (*a* _{2}), *neglecting the terms of order* $\uf10e\text{(}{a}_{2}\nabla T\text{)}$ *!*

The complete dependence of *κ* (*a* _{2}) on *a* _{2} without the terms $\uf10e\text{(}{a}_{2}\nabla T\text{)}$ may be obtained from (20.32) if one substitutes into this equation (20.29) for ${\Omega}_{k}^{(\varepsilon )}$ and (8.45) for *µ*2, 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 $-\overrightarrow{S}({\overrightarrow{V}}_{1})/T$ and integrate over ${\overrightarrow{V}}_{1}$. The approximation (20.35) implies

*κ*. The only difference is in the term

*µ*, which scales as

*µ*~

*T*

^{3/2}.

# (p.310) Solution of problem 20.9

*Check the normalization of the distribution function* $f(\overrightarrow{V})$ *given by* (20.39)*!*

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

*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

*h*

_{1/2}(

*V*) follows from (20.39) and the definitions (17.27) and (17.35) of ${D}_{ij}(\overrightarrow{V})$ and $\overrightarrow{S}$ and the unit vector $\stackrel{\u02c6}{\overrightarrow{V}}\text{=}\overrightarrow{V}/V$ has been introduced. Then we obtain

The last equation follows from

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

(p.311) Similarly

Hence

*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

Therefore, we seek $\overrightarrow{\beta}$ in a form similar to (20.22):

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

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

*µ*and

*β*

_{1}. Using for

*µ*the expansion

The corresponding Maple program reads

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

# Solution of problem 22.1

*Prove that in* (22.30) *all terms that contain factors* ${\dot{a}}_{k}\overrightarrow{\upsilon}\cdot \partial \overrightarrow{G}/\partial {a}_{k}$ *vanish upon integration over $\overrightarrow{\upsilon}$ for k* ≥ 2*!*

For the definition of the Sonine expansion

Hence we can write

*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 ${\Omega}_{\eta}^{(\varepsilon )}$ and (19.26) for *η* _{0} into (20.16) and obtain

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

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

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

The above results are accurate up to $\uf10e\text{(}{a}_{2}^{2}\text{)}$, 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 $\uf10e\text{(}{a}_{2}^{2}\text{)}$

# Solution of problem 25.2

*Derive the function* ${k}_{\perp}^{*}(\varepsilon )$ *and show that it depends only on the coefficient of restitution! For ε* ≲ 1 *show that* ${k}_{\perp}^{*}(\varepsilon )\propto \sqrt{1-{\varepsilon}^{2}}$ *!*

From (25.24) and the results of Exercise 25.1 for *η**(*ε*) and *ζ**(*ε*) it follows that ${k}_{\perp}^{*}(\varepsilon )$ 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

# (p.315) Solution of problem 25.3

*Find the function ${\overrightarrow{w}}_{\overrightarrow{k}\perp}\text{(}t\text{)}$, using the relation between the laboratory time t and the time τ measured in the accumulated number of collisions! Find the function ${\overrightarrow{u}}_{\overrightarrow{k}\perp}\text{(}t\text{)}$ and show that the unscaled velocity field decays always with time!*

From (25.10), definition of *τ _{c}*(

*t*) (14.18) and evolution of temperature in the homogeneous cooling state (8.64) follows

Therefore, using (25.22, 25.23) we obtain

Taking into account

*ζ**, we obtain the exponent in (C.169):

*a*

_{2}for

*μ*

_{2}from (8.37). With ${\overrightarrow{u}}_{\overrightarrow{k}\perp}\text{(}t\text{)}\text{=}\sqrt{2{T}_{h}(t)/m}{\overrightarrow{w}}_{\overrightarrow{k}\perp}(t)$ and using the temperature dependence

*T*(

_{h}*t*) we obtain

The last expression shows that the exponent of the power law dependence of ${\overrightarrow{u}}_{\overrightarrow{k}\perp}\text{(}t\text{)}$ 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:* ${k}_{H}^{*}(\varepsilon )\propto \sqrt{1-{\varepsilon}^{2}}$ *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

# Solution of problem 27.1

*Find the dependence of the non-reduced shear mode ${\overrightarrow{u}}_{\overrightarrow{k}\perp}$ 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 ${\overrightarrow{u}}_{\overrightarrow{k}\perp}$ depends on time as

*ν*. given by (C.172). For 1 −

_{k}*ε*

^{2}≪ 1 we can approximate

*a*

_{2}≈ 0 and

*η** ≈ 1 (see Exercise 25.2). Hence, the exponent

*ν*reads

_{k}