technical note, 12/13/2000

Coefficient of restitution does not exceed unity

Hal Tasaki^{1}

1 Definitions and main results

We consider a classical “ball” which consists of N point-like particles whose canonical coordinates and momenta are denoted as q_{i} and p_{i}, respectively, for i = 1, 2, …, N. The Hamiltonian which determines the motion of the ball is

H_{tot} = ∑_{i} \frac{|p_{i}|^{2}}{2μ_{i}} + ∑_{i≠j} V_{i,j}(q_{i} − q_{j}) + ∑_{i} V_{i}^{wall}(q_{i}),  (1.1)

where μ_{i} is the mass of the i-th particle, V_{i,j} is the potential of the force between the particles, and V_{i}^{wall} is the potential of the force that the wall exerts on the i-th particle.

As usual we introduce the coordinate q = (q_{x}, q_{y}, q_{z}) and the momentum p = (p_{x}, p_{y}, p_{z}) of the center of mass, and denote by γ the collection of canonical coordinates for the (N −1) internal degrees of freedom (which do not contain potential of the wall). Then the total Hamiltonian (1.1) can be rewritten as

H_{tot} = \frac{|p|^{2}}{2m} + H(γ) + V^{wall}(q, γ),  (1.2)

where m = ∑_{i} μ_{i} is the total mass of the ball, and H(γ) is the Hamiltonian for the internal degrees of freedom. We assume that the potential of the wall V^{wall}(q, γ) is vanishing whenever q_{x} ≥ −L for some constant L > 0.

The initial state of the ball is sampled according to a measure ρ(Γ)dΓ, where dΓ is the Lebesgue measure, and

ρ(Γ) = ρ_{pos}(q)ρ_{mom}(p)ρ_{int}(γ),  (1.3)

where ρ_{pos}(q) is a uniform distribution in a ball defined by |q − o| ≤ a where o = (0, 0, 0) and 0 < a < L. Similarly ρ_{mom}(p) is a uniform distribution in a ball with |p−(−p_{0}, 0, 0)| ≤ p_{th} where p_{0} > 2p_{th}. The thermal momentum is defined as

p_{th} = \sqrt{\frac{m}{β}}.  (1.4)

This is the typical momentum that the ball gets from the thermal fluctuation. Since this thermal momentum must be much less than any observable momentum (if a ball is macroscopic) we expect p_{th} ≪ p_{0}.

Theorem 1 allows us to take the a → 0 limit, where the center of mass always starts from the origin. As for the initial momentum, however, we are not able to replace the above distribution by a delta function from some technical reasons. We of course believe there is nothing physically essential about the distribution of p′, and the same result should hold for the case of the delta function. (We note in passing that in reality it is extremely hard to give the ball a total momentum with fluctuation less than p_{th}.)

1Department of Physics, Gakushuin University, Mejiro, Toshima-ku, Tokyo 171, JAPAN

electronic address: hal.tasaki@gakushuin.ac.jp

The initial distribution of the internal degrees of freedom is given by the Gibbs distribution at inverse temperature β as

ρ_{int}(γ) = \frac{e^{−βH(γ)}}{Z(β)} ,  (1.5)

with

Z(β) = ∫ dγ e^{−βH(γ)}.  (1.6)

We consider the time evolution of the ball according to the Hamiltonian dynamics. We assume^{2} that there is a time T > 0 at which the ball is certainly bounced back from the wall and is entirely in the region q_{x} > 0.

We denote the map which describes the time evolution from t = 0 to t = T as

Γ = (q, p, γ) → Γ′(Γ) = (q′(Γ), p′(Γ), γ′(Γ)).  (1.7)

Since we are interested in the coefficient of restitution, we consider the average

⟨\frac{|p′|}{|p|}⟩ = ∫ dΓρ(Γ) \frac{p′(Γ)|}{|p|}.  (1.8)

Then our main results are as follows.

Theorem 1: Let the wall potential satisfy

V^{wall}(q + (0, y, z), γ) = V^{wall}(q, γ),  (1.9)

for any y and z, i.e., it is invariant under translations in y and z directions. Then for any 0 < a < L, we have

⟨\frac{|p′|}{|p|}⟩ ≤ 1 + C_{1} \frac{p_{th}}{p_{0}} + C_{2} (\frac{p_{0}}{p_{th}} )^{2} exp (−\frac{p_{0}}{p_{th}} ) ≤ 1 + C_{3} \frac{p_{th}}{p_{0}} ,  (1.10)

where C_{i} are numerical constants. The result is valid in the limit a → 0 as well.

In the limit p_{th}/p_{0} → 0, (1.10) reduces to ⟨p′_{x}⟩/p_{0} ≤ 1, which is the statement of the title.

We have much weaker result which does not depend on any special symmetry.

Theorem 2: We do not make any symmetry assumptions on the potential, but we assume that there is a constant A such that, after any collision with the final momentum p′, the center of mass at time t = T is within a region of the volume A(|p′|/m)T. (In short, A denotes the maximum possible cross section in various collisions.) Then we have

⟨\frac{|p′|}{|p|}⟩ ≤ 1 + C_{1} \frac{p_{th}}{p_{0}} + C_{4} \frac{T Ap_{0}}{a^{3}m} (\frac{p_{0}}{p_{th}} )^{2} exp (−\frac{p_{th}}{p_{0}} ) ,  (1.11)

2We are essentially assuming that the wall does not trap the ball for too long a time.

Compared with the previous bound (1.10), it is obvious that (1.11) with an extra factor (T Ap_{0})/(a^{3}m) is much more crude. In particular we do not get any bound in the a → 0 limit. In practice, however, the volume (T Ap_{0})/m of the final position of the ball cannot be greater than, say, 1 km^{3} while it is not quite meaningful to set a to be less than, say, 1 μm. This combination yields the prefactor of (T Ap_{0})/(a^{3}m) ∼ 10^{18}, which may seem like a huge factor. But one should note that p_{0}/p_{th} is usually of order 10^{8}, and the third term in the right-hand side of (1.11) is fairly small.

Although the Theorem 1 can be proved only for the present problem of collision with a translation invariant wall, the method to prove Theorem 2 is rather robust and can be easily extended to other problems including collisions of two macroscopic balls.

We also note that, by a trivial extension of the proof, it can be proved that the probability that the final momentum |p′| is greater than p_{0} + 2p_{th} is very small (less than of the order exp(−p_{0}/p_{th})).

2 Proof of Theorem 1

With a slight abuse of notation, we denote the inverse image of the map (1.7) as

Γ(Γ′) = (q(q′, p′, γ′), p(q′, p′, γ′), γ(q′, p′, γ′)).  (2.1)

The Liouville’s theorem reads

ρ(Γ)dΓ = ρ(Γ(Γ′))dΓ′,  (2.2)

and the energy conservation law implies

H(γ) + \frac{|p|^{2}}{2m} = H(γ′) + \frac{|p′|^{2}}{2m} ,  (2.3)

since the ball does not feel the potential from the wall in the initial and the final moments. By using (2.2) and a trivial decomposition of the integral, we see that

⟨|p′|⟩ = ∫ dΓ′ |p′| ρ(Γ(Γ′))
= ∫_{|p′|≤p_{0}+δ} dΓ′ |p′| ρ(Γ(Γ′)) + ∫_{|p′|≥p_{0}+δ} dΓ′ |p′| ρ(Γ(Γ′))
≤ (p_{0} + δ) + ∫_{|p′|≥p_{0}+δ} dΓ′ |p′| ρ(Γ(Γ′)),  (2.4)

where the constant δ > 0 will be chosen later. In what follows we will show that the quantity

R = ∫_{|p′|≥p_{0}+δ} dΓ′ |p′| ρ(Γ(Γ′))  (2.5)

is small. By recalling (1.3), we have

R = ∫_{|p′|≥p_{0}+δ} dq′dp′dγ′ |p′| ρ_{pos}(q(q′, p′, γ′))ρ_{mom}(p(q′, p′, γ′))ρ_{int}(γ(q′, p′, γ′)). (2.6)

The reader who is interested in quickly seeing the outline of the proof is advised to now jump to the next section where we prove Theorem 2 (which is much less technical).

In order to evaluate the q′ integral, we make further definitions. Let

q̃(p′, γ′) = q(o, p′, γ′), p̃(p′, γ′) = p(o, p′, γ′), γ̃(p′, γ′) = γ(o, p′, γ′),  (2.7)

where o = (0, 0, 0). We define

τ = \frac{mq′_{x}}{p′_{x}} ,  (2.8)

which is the time that the ball with final momentum p′ spends after leaving the x = 0 plane until reaching the point q′. Denote by T_{τ}(γ) the time evolution of γ for the time τ according to the Hamiltonian H(γ).

Note that, for τ as in (2.8), we have

p(q′, p′, T_{τ}(γ′)) = p̃(p′, γ′).  (2.9)

To see this, suppose that we find the ball in the state (q′, p′, T_{τ}(γ′)) at t = T. Then when the (center of mass of the) ball leaved the x = 0 plane at t = T − τ, its momentum and internal coordinates were (p′, γ′) for any q′. This, with the translation invariance of the wall potential, implies that the collision process with the wall does not depend on q′, and we get (2.9). Correspondingly, we have

γ(q′, p′, T_{τ}(γ′)) = T_{τ}(γ̃(p′, γ′)),  (2.10)

and

q(q′, p′, T_{τ}(γ′)) = q̃(p′, γ′) + \frac{p̃(p′, γ′)}{m}τ + (0, q′_{y}, q′_{z}),  (2.11)

since the initial velocity of the center of mass is p̃(p′, γ′)/m.

Since T_{τ}(·) is volume preserving for fixed τ (which means fixed q′ and p′), we can replace γ′ in the integrand of (2.6) by T_{τ}(γ′) as

R = ∫_{|p′|≥p_{0}+δ} dq′dp′dγ′ |p′| ρ_{pos}(q(q′, p′, T_{τ}(γ′)))ρ_{mom}(p(q′, p′, T_{τ}(γ′)))ρ_{int}(γ(q′, p′, T_{τ}(γ′)))
= ∫_{|p′|≥p_{0}+δ} dq′dp′dγ′ |p′| ρ_{pos} (q̃(p′, γ′) + \frac{p̃(p′, γ′)}{p′_{x}}q′_{x} + (0, q′_{y}, q′_{z})) ×
×ρ_{mom}(p̃(p′, γ′))ρ_{int}(T_{τ}(γ̃(p′, γ′))),  (2.12)

where we used (2.8), (2.9), (2.10), and (2.11).

We can change the order of integrations and integrate over q′ first. Since ρ_{pos}(q) is equal to {(4/3)πa^{3}}^{−1} for |q| ≤ a and vanishing otherwise, we need to integrate {(4/3)πa^{3}}^{−1} by q′ over the region such that

|q̃(p′, γ′) + \frac{p̃(p′, γ′)}{p′_{x}}q′_{x} + (0, q′_{y}, q′_{z})| ≤ a,  (2.13)

whose volume is (4/3)πa^{3}(p′_{x}/|p̃_{x}(p′, γ′)|). Thus for any 0 < a < L, we have

R = ∫_{|p′|≥p_{0}+δ} dp′dγ′ |p′|\frac{p′_{x}}{|p̃_{x}(p′, γ′)|} ρ_{mom}(p̃(p′, γ′))ρ_{int}(γ̃(p′, γ′)),  (2.14)

where we also noted that ρ_{int}(·) is invariant under T_{τ}.

Noting that p′_{x} ≤ |p′| and |p̃_{x}(p′, γ′)| ≥ p_{0} − p_{th} ≥ p_{0}/2, and recalling (1.5) for ρ_{int}(·), we see that

R ≤ ∫_{|p′|≥p_{0}+δ} dp′dγ′ \frac{2|p′|^{2}}{p_{0}} ρ_{mom}(p̃(p′, γ′))ρ_{int}(γ̃(p′, γ′))
= \frac{2}{p_{0}} ∫_{|p′|≥p_{0}+δ} dp′dγ′|p′|^{2}ρ_{mom}(p̃(p′, γ′))\frac{exp[−βH(γ̃(p′, γ′))]}{Z(β)}
= \frac{2}{p_{0}Z(β)} ∫_{|p′|≥p_{0}+δ} dp′dγ′|p′|^{2}ρ_{mom}(p̃(p′, γ′)) exp [−β \{H(γ′) + \frac{|p′|^{2}}{2m} − \frac{|p̃(p′, γ′)|^{2}}{2m} \}]
≤ \frac{2}{p_{0}Z(β)} ∫_{|p′|≥p_{0}+δ} dp′dγ′|p′|^{2}ρ_{mom}(p̃(p′, γ′)) exp [−β \{H(γ′) + \frac{|p′|^{2}}{2m} − \frac{(p_{0} + p_{th})^{2}}{2m} \}]
≤ \frac{6}{4π(p_{th})^{3}p_{0}Z(β)} ∫_{|p′|≥p_{0}+δ} dp′dγ′|p′|^{2} exp [−β \{H(γ′) + \frac{|p′|^{2}}{2m} − \frac{(p_{0} + p_{th})^{2}}{2m} \}] (2.15)

where we used the energy conservation law (2.3) and the bound |p̃(p′, γ′)| ≤ p_{0} + p_{th}. To get the final inequality, we have simply replaced ρ_{mom}(p̃(p′, γ′)) by its maximum value {(4/3)π(p_{th})^{3}}^{−1}.

Now the integrations in p′ and γ′ are decoupled, and using (1.6) we get

R ≤ \frac{3}{2π(p_{th})^{3}p_{0}} exp [\frac{β}{2m}(p_{0} + p_{th})^{2}] ∫_{|p′|≥p_{0}+δ} dp′|p′|^{2} exp [−\frac{β}{2m}|p′|^{2}] .  (2.16)

The integral can be easily bounded from above as

∫_{|p′|≥p_{0}+δ} dp′|p′|^{2} exp [−\frac{β}{2m}|p′|^{2}] = 4π ∫_{p_{0}+δ}^{∞} dp p^{4} exp [−\frac{β}{2m}p^{2}]
≤ 4π ∫_{0}^{∞} dr (r + p_{0} + δ)^{4} exp [−\frac{β}{2m}\{r^{2} + (p_{0} + δ)^{2}\}]
≤ C exp [−\frac{β}{2m}(p_{0} + δ)^{2}] (p_{0} + δ)^{4}p_{th},  (2.17)

where C is a numerical constant that one can work out if necessary. By choosing δ = 2p_{th}, we get

R ≤ \frac{C}{2π(p_{th})^{3}p_{0}} exp [\frac{β}{2m}\{(p_{0} + p_{th})^{2} − (p_{0} + 2p_{th})^{2}\}] (p_{0})^{4}p_{th}
≤ C′ exp [−\frac{β}{m}p_{0}p_{th}] p_{0} (\frac{p_{0}}{p_{th}} )^{2}
≤ C′ exp [−\frac{p_{0}}{p_{th}} ] p_{0} (\frac{p_{0}}{p_{th}} )^{2} ,  (2.18)

where C′ is another constant.

Now going back to (2.6), we can bound the desired quantity as

⟨\frac{|p′|}{|p|}⟩ ≤ \frac{⟨|p′|⟩}{p_{0} − p_{th}}
≤ \frac{p_{0} + 2p_{th}}{p_{0} − p_{th}} + C′ \frac{p_{0}}{p_{0} − p_{th}} (\frac{p_{0}}{p_{th}} )^{2} exp [−\frac{p_{0}}{p_{th}} ]
≤ 1 + 6\frac{p_{th}}{p_{0}} + 2C′ (\frac{p_{0}}{p_{th}} )^{2} exp [−\frac{p_{0}}{p_{th}} ] .  (2.19)

3 Proof of Theorem 2

We substitute (1.5) for ρ_{int}(·) into (2.6) and use the energy conservation (2.3) to get

R = ∫_{|p′|≥p_{0}+δ} dq′dp′dγ′ |p′| ρ_{pos}(q(q′, p′, γ′))ρ_{mom}(p(q′, p′, γ′))\frac{exp[−βH(γ(q′, p′, γ′))]}{Z(β)}
= \frac{1}{Z(β)} ∫_{|p′|≥p_{0}+δ} dq′dp′dγ′ |p′| ρ_{pos}(q(q′, p′, γ′))ρ_{mom}(p(q′, p′, γ′)) ×

× exp [−β \{H(γ′) + \frac{|p′|^{2}}{2m} − \frac{|p(q′, p′, γ′)|^{2}}{2m} \}]

≤ \frac{3}{4πp_{th}^{3}Z(β)} ∫_{|p′|≥p_{0}+δ} dq′dp′dγ′ |p′| ρ_{pos}(q(q′, p′, γ′)) ×
× exp [−β \{H(γ′) + \frac{|p′|^{2}}{2m} − \frac{(p_{0} + p_{th})^{2}}{2m} \}] ,  (3.1)

where we replaced ρ_{mom}(p(q′, p′, γ′)) and |p(q′, p′, γ′)|^{2} by their maximum possible values.

Since the only term in the integrand which depends on q′ is ρ_{pos}(q(q′, p′, γ′)), we can perform the q′ integral first. By replacing ρ_{pos}(q(q′, p′, γ′)) by its maximum value 3/(4πa^{3}) and using the assumption about the possible range of q′, we get

R ≤ \frac{9AT}{(4π)^{2}a^{3}p_{th}^{3}Z(β)} ∫_{|p′|≥p_{0}+δ} dp′dγ′ |p′|^{2} exp [−β \{H(γ′) + \frac{|p′|^{2}}{2m} − \frac{(p_{0} + p_{th})^{2}}{2m} \}]
= \frac{9AT}{(4π)^{2}a^{3}p_{th}^{3}} exp [\frac{β}{2m}(p_{0} + p_{th})^{2}] ∫_{|p′|≥p_{0}+δ} dp′ |p′|^{2} exp [−\frac{β}{2m}|p′|^{2}] .  (3.2)

We have thus obtained the estimate essentially the same as (2.16) with an extra factor. The rest of the proof is the same as in the previous section.

It is a pleasure to thank Hisao Hayakawa and Shin-ichi Sasa for valuable discussions.