Cornell University, Physics Department
PHYS-7653 Statistical Physics II
Fall 2012
Prof. Veit Elser
Homework Set 1
(Dated: 09/05/2012)
Due Thursday, 09/06/2012
1. Random points on a sphere
N points are placed at random on the surface of a sphere in D dimensions. Calculate the
probability, PD (N ), that all N points lie on the same hemisphere.
(a) Calculate P1 (N )
In 1D, the surface of a sphere essentially contains only 2 points. The problem therefore
reduces to finding the probability that all N points fall on the same side.
P1 (N ) = 2 · 2−N
= 21−N
(b) Show that PD (N ) = 1 for N ≤ D
It may seem evident that all N points lie on the same hyperplane, which tells us which
hemisphere the points are in, but let’s try to be a little more rigorous. Note first that
((D − 1)-dimensional) hyperplanes in <D can be represented by linear equations
c0 +
D
X
ci xi = 0.
i=1
(1)
(1)
(N )
(N )
Given N ≤ D points (x1 , . . . , xD ), . . . , (x1 , . . . , xD ), we have N homogeneous equations for ci . With D + 1 variables and N < D + 1 equations, we are guaranteed to have
non-trivial solutions.
Suppose now that {c0 , . . . , cD } is a solution. If c0 = 0, the points lie on the edge of both
hemispheres with the poles along the (c1 , . . . , cD ) axis. If c0 6= 0, the points all lie on the
hemisphere with its pole in the direction of sgn(c0 ) · (c1 , . . . , cD ).
(c) The general case
Hints:
(1) Instead of sampling N random points directly, sample N random diameters first,
then select one of the two endpoints on each diameter. The probability PD0 (N ) given
the N diameters is independent of the choice of diameters (except in cases with zero
probability measure), thus PD0 (N ) = PD (N ).
1
(2) Consider the hyperplanes perpendicular to the diameters chosen in (1). These hyperplanes divide the sphere into regions (”chambers”), each of which corresponds
to a particular choice of endpoints that will make all N points fall on the same
hemisphere.
Pick a random point X on the sphere, and for each diameter pick the side of the
corresponding hyperplane that X falls on. It is obvious from this construction that
crossing a hyperplane will result in a different choice of endpoints to get all N points
on the same hemisphere. The total number of chambers, CD (N ), is proportional to
PD0 (N ).
(3) There is a simple, linear relation between CD−1 (N ), CD (N ), and CD (N + 1)1 . You
can reason this out by considering what happens when you add 1 extra hyperplane to
a sphere in D dimensions, with N hyperplanes already added. The new hyperplane
will slice some old chambers into two smaller ones.
To see now many chambers are sliced up by the new hyperplane, note that the new
hyperplane is essentially a (D−1)-dimensional sphere. Each of the N old hyperplanes
will intersect the new (D−1)-sphere along (D−2)-dimensional ”hyperlines”, and the
number of chambers marked out on this (D − 1)-sphere is the number of chambers
sliced up by the new hyperplane.
As derived in class (and explained above), we know that
CD (N + 1) = CD (N ) + CD−1 (N )
(1)
Beginning with CD (1) = 0 for D < 1 and CD (1) = 2 for D ≥ 1, we obtain the following
table of values:
D
D
D
D
D
=
=
=
=
=
1
2
3
4
5
N=1
2
2
2
2
2
N=2
2
4
4
4
4
N=3
2
6
8
8
8
N=4
2
8
14
16
16
N= 5
2
10
22
30
32
Observe how the values ”propagate” by adding from left to right in the table, from the
N = 1 column of 0’s for D < 1, and 2’s for D ≥ 1. The allowed steps for propagation
are ”Right” ((N, D) → (N + 1, D)), and ”Right-Down”((N, D) → (N + 1, D + 1)).
Given these, the value at each (N, D) point on the table is the value 2 multiplied by the
total number of ways in which one can step from the 2’s in the N = 1 column to that
(N, D) point.2 Equivalently, CD (N ) is 2 multiplied by the number of ways to step from
1
There are exceptions for certain choices of diameters, e.g., when 3 different hyperplanes coincide on the
same ”hyperline” of dimension D − 2, when D > 2. These cases, however, have zero probability measure.
2
Technically, you should also get contributions from the 0’s in the N = 1 column, but zero multiplied by
the number of possible paths gives zero...
2
(N, D) to any point (1, D) for D ≥ 1, given the moves ”Left” ((N, D) → (N − 1, D))
and ”Left-Up” ((N, D) → (N − 1, D − 1)).
It is easy to see that the number of ways to step from (N, D) to (1, D0 ) is
A ≡ 0 for B > A.
where B
Result: CD (N ) = 2
Pk
i=0
N −1
i
, k = min(D − 1, N − 1)
N −1
D−D0
,
for D ≤ 1, N ≤ 1.
For the more rigorous-minded, note that the above result can be proved quite easily by
induction.
It is obvious that the probability given N diameters,
PD0 (N ) = CD (N )/2N
k X
N −1
1−N
=2
i
i=0
, where k = min(D − 1, N − 1).
Note that for any given D, CD (N ) = 2N for N ≤ D, confirming our previous result
that PD (N ) = 1 for N ≤ D.
(d) Find the value of N/D at which the probability transition takes place, and
the width of the transition.
We first look at how PD (N ) changes
as vary D at fixed N , for D < N . Note that
PD+1 (N ) − PD (N ) = 2−N ND−1 . The greatest increase occurs at D = (N − 1)/2, where
N −1
N
≈
D
N/2
N!
=
(N/2)!2
(N/e)N
≈
(N/2e)N
≈ 2N
At large N , all the changes in PD (N ) is concentrated at the transition point N/D = 2 !
This is why we see that as N (and hence D) get larger, the probability transition gets
sharper, and PD (N ) approaches a step function.
To see how the transition width changes with D, we can approximate the binomial
coefficients near the transition point with a gaussian:
1
2
2
N
−N
2
≈ √ e−(i−µ) /(2σ )
(2)
i
σ 2π
3
where µ = N/2, and σ 2 = N/4.3
We can approximate the sum in PD (N ) to be an integral. Since the binomial coefficients are only significant near i = N/2, where the gaussian approximation is good, we
can safely take
Z D
1
2
2
dx √ e−(x−µ) /(2σ )
PD (N ) = 1 −
(3)
σ 2π
−∞
√
√
which gives an error function with a spread of (∆D)|N = σ ∼ N ∼ D. The desired
transition width is
√
√
(∆N )|D /D ∼ (∆D)|N /D ∼ D/D = 1/ D
which agrees with our observation that as D increases, the transition gets sharper.
One way to see how the approximation works is to note that pN (i) = 2−N Ni is the exact probability
distribution function for i successes in N binomial trials with probability p = 1/2. By the central limit
theorem, pN (i) can be approximated by a normal distribution of the same mean (µ = N p) and variance
(σ 2 = N p(1 − p)) in the large-N limit.
3
4