Correction of momentum transferred in equation of state

m (109)

where the plus sign is for Eq.74 (forward collision) and the minus sign is for Eq.92 (Class 1 backward collision). So the effect of  reaches its maximum when v_j⁰ = v_k⁰.

But why do they cancel exactly? Let’s check what happens for v_j⁰ ≈ v⁰_k. Define ζ to be ζ ≡| v_j⁰ − v_k⁰ |. For v_j⁰ ≈ v_k⁰, ζ → 0, the situation is shown in Fig. 22. In the center-of-mass frame, for both forward collision and backward collision, the situation is the collision between ¹₂ζ and −¹₂ζ. Therefore, the effects of particle-particle interaction in forward collision and backward collision in Fig. 21 are exactly the same. And for the population, f v_j⁰− ζ
≈ f v_j⁰+ ζ
for ζ → 0. Since the magnitudes of the effect of particle-particle interaction and the population of v⁰_kare all the same, the magnitude of the effect of forward collision for v_j⁰ ≈ v⁰_k is the same as the effect of backward collision for v_j⁰ ≈ v_k⁰, and hence the two divergences cancel out.

3.4.5 Correction of momentum transferred in equation of state

Now the last mile to a full description of the correction to the equation of state: the computation of the corrected momentum transferred to the wall. The effect of ∆v⁰_j

appears in the equation of state with the form of P

Plugging Eq.106 into Eq.110, we have

X

Introducing a change of variables by

x ≡ u⁰_j + u⁰_k

Z ∞ 0

du⁰_je^−(u0j)2 Z ∞

u⁰_j

du⁰_ke⁻(^u0k)²u⁰_j − u⁰_k u⁰_j + u⁰_k = −

Z ∞ 0

dxe^−x2 x

Z x 0

dye^−y2y = −ln2

4 . (114) Putting Eq.113 and Eq.114 into Eq.111, we get

X

j

2m∆v⁰_j

T_j⁰ = 0. (115)

In order to obtainP

j 2m∆v_j⁰

T_j⁰ , we built up a mechanical model and went through all the calculations from Eq.52 to Eq.115, and yet the final result is just zero! What’s going on?

It will be unsatisfactory regarding Eq.115 as just a coincidence. It is hard to believe that such simple result doesn’t have any simple explanation. For obtaining such simple result after a long calculation, we had better give a story. So here is the story.

Note that what we have is P

j 2m∆v_j⁰

T_j⁰ = 0 but not ^2m∆v

0 j

T_j⁰ = 0. A summation giving zero often relates to conservation law. This is the first hint for the physics behind Eq.115.

In order to figure out the physics behind P

j 2m∆v_j⁰

T_j⁰ , we should ask, what exactly is this ∆v⁰_j? In our construction, v_j⁰+ ∆v_j⁰ is the collision velocity that a particle hits the wall. But we can take another point of view. What is the role of the wall? A particle reverses its velocity when hitting the wall due to the reflection provided by the wall.

A different point of view is that, the wall provides some momentum to the box of gas every time a particle hits the wall. That is, the wall is a momentum provider. This is the central idea to understanding Eq.115.

With the idea of the wall being the momentum provider, a natural step is checking the momentum of the box of gas. By our very assumption, the total momentum of the system is zero, so that the box of gas just sits there without “drifting” away. Since total momentum gives us nothing, let’s check the positive momentum of the system, that is, the total momentum of particles with positive velocity. When the box of gas

is at thermal equilibrium, it is a steady state. For steady state, the total positive momentum is time-independent. We are going to utilize the basic assumption from steady state.

total positive momentum is time − independent (116) After the system evolves for ∆t, what is the total positive momentum now? A particle with free particle velocity v_j⁰ > 0 may hit the right wall and reverse its velocity, if it starts with a position near the right wall. To be more precise, for a particle with free particle velocity v_j⁰ > 0 having initial position in L − v⁰_j∆t, L
, its velocity will be negative at t = ∆t. When counting the total positive momentum at t = ∆t, such particle has no contribution since it has a negative velocity at t = ∆t and hence no longer belongs to the class of particles having positive momentum. The particle disappears from the class of particles having positive momentum. Due to the basic assumption that position distribution is homogeneous, such particles have the population of

v_j⁰∆t

L = ∆t

1

2T_j⁰. (117)

When counting the total positive momentum at t = ∆t, momentum loss due to the effect that a particle can change its direction is given by

momentum loss =X

j

∆t

1

2T_j⁰ −mv_j⁰
. (118) On the other hand, we also have momentum gain, since a particle with negative velocity at t = 0 may have positive velocity at t = ∆t, due to the reflection provided by the left wall. When regarding the wall as the momentum provider, the momentum gain can be seen as provided by the wall, since momentum gain comes from particles that reverse their velocities basing on the mechanism that the wall gives them the momentum. For each collision, the left wall provides a momentum of m v_j⁰+ ∆v_j⁰
to

v

v

r>r

r<r

v

' v

'

Figure 23: Two particles in the box. The two particles are free particles in the upper situation and are coupled in the lower situation.

the total positive momentum of the system. That is, the momentum gain of the total positive momentum of the system provided by the left wall during time interval of ∆t is given by

momentum gain =X

j

∆t

1

2T_j⁰ m v_j⁰+ ∆v_j⁰

(119)

Since the total positive momentum of the system at t = ∆t should be the same as the one at t = 0, by Eq.118 and Eq.119, we have

X

j

∆t

1

2T_j⁰ −mv_j⁰
+X

j

∆t

1

2T_j⁰ m v_j⁰+ ∆v⁰_j

= 0 (120) and hence we get

X

j

2m∆v_j⁰

T_j⁰ = 0 (121)

But why should we use −mv_j⁰ instead of −mv_j |_average in Eq.118 for the momentum loss? The answer is hidden in what is meant by the total positive momentum of the system, which we did not state clearly in the above argument. Consider the very simple case when there are just two interacting particles in the box, as shown in Fig. 23. What is the total positive momentum of the system, mv₁ or mv₁⁰? In Fig. 23, compared with the upper situation that the two particles are decoupled, the two particles are coupled

and hence their momenta (one positive and one negative) are also coupled in the lower situation. For a box of N particles, if total positive momentum couples to total negative momentum, negative momentum gain from the right wall and negative momentum loss should be taken into account when dealing with total positive momentum. In order to have an idea of total positive momentum that is separate from total negative momentum, the appropriate definition should be constructed in the way that all coupled particle pairs are pull apart as free particles. That is, we take a snapshot for the box of gas and pull apart all particle pairs, and then define total positive momentum and total negative momentum by these N decoupled particles. It is in this sense of total positive momentum that we should use −mv_j⁰ but not −mv_j |_average in Eq.118, and this is also why the negative momentum gain provided by the right wall and the negative momentum loss play no role when dealing with total positive momentum.

Notice that we don’t use any information of particle-particle interaction in the above argument. It is a result of steady state assumption. To be more precise, it is a result of the requirement that the total positive momentum of the system is time-independent.

Since it has nothing to do with the form of particle-particle interaction, the result is general and valid for generic particle-particle interactions.

By requiring that the total positive momentum of the system to be time-independent, we can get the correction of momentum transferred in equation of state (namelyP

j 2m∆v⁰_j

T_j⁰ ), which is the total effect of ∆v_j⁰ for all v_j⁰. However, the total effect exhibited in the equation of state cannot tell apart the behavior of ∆v_j⁰, namely the v⁰_j − dependence of

∆v⁰_j, since we have summed over all v_j⁰ and thus lost such information. Such informa-tion can be figured out only if we study individual ∆v⁰_j instead of P

j 2m∆v⁰_j

T_j⁰ . This is one merit of the mechanical model for calculating ∆v_j⁰ (Eq.105 and Fig. 21).