Physics 450 - Problem Set 1 Solutions - January 23 2001

1. (a) The probability of any one choice of some particular n objects is 1/2n. Since there are N!/[n! (N-n)!] ways to choose n objects from a set of N, the probability of finding n particles on the LH side of the box is just
P(n) =  1 
2N		 N! 
n! (N-n)!
 
(b) Plot of P(n) for N = 10, note n = 0,1,2¼10; also note that the distribution already looks a lot like a Gaussian distribution, which is the exact limiting distribution as N ® ¥.

2. (a) The energy is E = mv2/2 + kx2/2, so the probability of the position and velocity being (x,v) is
P(x,v) µ exp[-E/kB T] = exp[-(mv2/2 + kx2/2)/kB T]
 
The µ indicates proportionality; we won't need to calculate the normalization constant in front of the distribution in this problem.

This joint probability factorizes into probability distributions for x and v:

P(x,v) µ P(x) P(v)
 
where
P(v) µ exp[-mv2/(2kB T)]
 
and
P(x) µ exp[-kx2/(2kB T)]
 
thanks to the important factorization property of the exponential, eA+B = eA eB.

So, the probability distribution for position is just a gaussian distribution in x:

P(x) µ exp[-kx2 / (2kB T)]
 
(b) The typical value of x2 is just that which has an energy of about kB T associated with it, or kx2 » kB T, or
x2 » kB T/k
 
An alternative way to obtain this result is to note that it is the half-width of the gaussian distribution above.

(c) For the stiff cantilever, k = 1 nN/10-9 N/ 10-10 m = 10 N/m. From (b), we therefore have

x2 »  kB T 
k		 =  4×10-21    J 
10    N/m
 
= 4 ×10-22 m2 (recall 1 N = 1 J/m).
Thus the typical value of x is about 4 ×10-11 m = 0.4 Å. For this case, thermal fluctuations are small compared with the size of an atom, and such a cantilever is suitable for studying atomic-scale structure of a surface.

For the flimsy cantilever, k = 1 pN/m = 10-12 N/10-6 m = 10-6 J/m. In this case, x2 » kB T/k = 4×10-15 m2 giving |x| » 6 ×10-8 m = 60 nm. Here the thermal bending flucutations of this cantilever are nearly large enough to observe in the light microscope.

The above is enough is enough for full credit on (a), but at least the physics students should go on to normalize the distribution. The normalization condition is
ó 
õ 		 ¥ 

-¥ 

 dx P(x) = 1
 
and if we suppose that P(x) = 1/Z exp[-kx2/(2kB T)], we have
ó 
õ 		 ¥ 

-¥ 

 dx exp[-kx2/(2kB T)] = Z
 
Now, we know the most important integral in physics,
ó 
õ 		 ¥ 

-¥ 

 dx exp[-ax2 /2 ] =  æ 
ç 
è 		 2p 
a		 ö 
÷ 
ø 		 1/2 

 
 
which tells us
Z =  æ 
ç 
è 		 2pkB T 
k 		 ö 
÷ 
ø 		 1/2 

 
 
and thus the normalized probability distribution for x is
P(x) =  æ 
ç 
è 		 k 
2 pkB T		 ö 
÷ 
ø 		 1/2 

 

 exp[-kx2/(2 kB T)]
 
Also, in (b) and (c), physics students should remember the exact result that for the (classical) harmonic oscillator, equipartition holds, so that there is exactly kB T/2 of energy associated with each harmonic degree of freedom. Therefore k < x2 > /2 = kB T/2, indicating the important exact result
< x2 > = kB T
 
This important result can be verified using P(x), by direct calculation of the average value of x2,
< x2 > =  ó 
õ 		 +¥ 

-¥ 

 dx   x2 P(x)
 
where you will need the second most important integral in physics,
ó 
õ 		 +¥ 

-¥ 

 dx   x2 exp[-ax2/2] =  æ 
ç 
è 		 2p 
a3 		 ö 
÷ 
ø 		 1/2 

 
 
3. We know that particles in a gas move with kinetic energies of m v2 /2 » kB T (the exact result is m v2/2 = 3 kB T/2), and therefore have velocities of magnitude v » (kB T / m)1/2

Consider a side of the container of area A. Particles hit the side of the container with some momentum p and transfer up to 2 times that momentum to the wall. A transfer of momentum per unit time is by Newton's second law of mechanics, a force. So we want to add up all the momentum transfer that occurs to the wall in some time interval t.

In our time interval t, particles can travel from up to a distance v t away from the wall to hit it, so the number of particles that hit the wall during our time interval is n » A v tr where r = N/V is the number density of particles in the gas.

Now we put everything together. The force on the wall is

f »  n p 
t		 »  A v trp 
t
 
Cancelling the t's (indicating that the time interval considered is not crucial to the calculation) and using p = mv we have
f » A (p2 / m) r » A kB T r
 
and therefore the force per area, or the pressure exterted on the wall, is
P =  f 
A		 » kB T r = kB T N / V
 
Thus, up to a numerical constant (which the physics students should be able to show to be equal to 1) we have derived the ideal gas law, PV = N kB T, or PV = (N/NA) R T where NA is Avogadro's number and R = NA kB is the molar gas constant.

4. (a) The terminal velocity of a small particle moving through liquid is given by the balance of drag and driving forces:
fdriving = 6phR v
 
or in our case,

fdriving = 6 p×0.001 kg/(m·sec)×1.5 ×10-6 m ×10-5 m/sec 30 ×10-14 kg·m/sec2 = 0.3 pN

(b) For gravitational sedimentation, fdriving = Dm g = (4p)/3 R3   Drg where Dr is the mass density difference between the molecule and the surrounding fluid, which in our case is 500 kg/m3.

So, as discussed in class and in the notes, the sedimentation velocity is

v =  2 
9		 R2   Dr  g 
h
 
which in this case gives
v =  2 
9		 (1.5 ×10-6    m)2 ×500    kg/m3×10     m/sec2 
10-3    kg·m/sec 
 
or v = 2.5 ×10-6 m/sec = 2.5 microns/sec.

File translated from TEX by TTH, version 2.53.
On 23 Jan 2001, 23:02.