Physics 47100
Advanced Laboratory II
Computational Lab 1
Equation of state
for hard disks
(Pre-lab: Understanding Collisions)
1)
Read: Phase Transition in Elastic Disks, B. J.
Alder and T. E. Wainwright, Phys. Rev. 127, 359 – Published 15 July 1962: https://gibbs.ccny.cuny.edu/teaching/s2020/labs/Hard_Disk_Simulation/Alders&Wainwright1962.pdf
2)
In this paper,
Alder and Wainwright simulate hard disks moving with constant velocities (i.e.,
no external forces). Hard means that the
disks never overlap, but they can collide when the distance between two disks
is equal to their diameter D.
a. If disk n, has diameter D, initial center position
vector Xn0 and initial
velocity vector Vn0, and
disk k has diameter D, initial center position vector Xk0 and initial velocity vector Vk0, then the position vector of disk n as a function of
time t is Xn(t)
= Xn0 + Vn0 t and Xk(t)
= Xk0 + Vk0 t. Draw a sketch of the two disks labeled n and
k, with positions Xn0 and
Xk0 and velocities Vn0 and Vk0.
b. Show that:
D2nk(t) = Dnk(t)∙Dnk(t) = (Dnk0
+ Vnk0 t)2 = (Dnk0 + Vnk0 t) ∙(Dnk0 + Vnk0
t),
where
D2nk(t)
is the squared distance between the centers of disks n and k, Dnk(t)
is the vector pointing from the center of disk n to the center of disk k at
time t, Dnk0 is the initial vector pointing from the center of
disk k to the center of disk n, and Vnk0
is the relative velocity vector between disk k and disk n.
c. Notice that the equation in 2b above is equivalent to
transforming to the (moving) frame of reference of disk n. Draw a sketch like that of part 2a in this
frame with disk n at the origin with zero velocity.
d. Notice further that without changing the dynamics, we
can rotate the axes in the diagram from 2c so that disk k is moving in the
direction of the x-axis. Redraw the
diagram in this way. Now disk n is not
moving at the origin and the center of disk k is at position (xk,yk)
moving in the + or – x-direction.
e. Expanding the equation from 2b:
D2nk(t) = D2nk0
+2Bnk0 t + V2nk0
t2,
where Dnk0
= (Dnk0∙Dnk0)1/2,
Bnk0 = Dnk0 ∙Vnk0, and
Vnk0 = (Vnk0∙Vnk0.)1/2. Find the values of t where D2nk(tc) = D2. These are the times tc where n and k are colliding or just
touching.
f.
Find the values
of |xk| and |yk|
from 2d in terms of variables from 2e and explain the solutions for two
collision tc. Since the tc equation is quadratic there are two
solution tc1 and tc2. List and explain what happens for different
situations. For example: If tc1=tc2
then yk=+/-D and sign(tc1)
= sign(-Bnk0). What if tc1>0
and tc2<0, or tc1 and tc2 complex, etc?
What are the conditions for a collision at some point in the future?
3)
In a collision
the velocities are changed according to conservation of energy and
momentum. Confirm that for disks of
equal mass the change in velocity for particle n during a collision with
particle k, ΔVn
= - ΔVk
= −Dnk(tc) (Dnk(tc) ∙ Vnk0)/ D2,
where Dnk(tc) is the vector pointing from particle n to
particle k. Note this make sense,
because the change in momentum must point in the direction of the force between
the centers of the disks. For this part
the center-of-mass frame of the collision may be easier to use. In this frame the origin is at the point of
collision, the frame moves with the center of mass velocity: Vcm
= (Vn
+ Vk)/2,
and the velocities of the particles are equal and opposite. Vn – Vcm = Vn – (Vn + Vk)/2 = Vn /2
– Vk/2 = – (Vk –
(Vn
+ Vk)/2)
= –(Vk – Vcm).
4)
We will be
reproducing Figure 1 during this lab.
a. Briefly explain what the two axes represent.
b. The y axis could be written [pA/(NkT) – 1] + 1. The term in square brackets is called the collisional pressure or the excess pressure. In 3D the area
A would be replaced by the volume V. For ideal gases in 2D or 3D the collisional pressure is zero. Look at eq.1 in https://gibbs.ccny.cuny.edu/teaching/s2020/labs/Hard_Disk_Simulation/Alders&Wainwright1960.pdf
and explain briefly how the collisional pressure is
calculated in the simulations.