Download 763620S STATISTICAL PHYSICS Solution Set 8 Autumn

763620S
STATISTICAL PHYSICS
Solution Set 8
Autumn 2012
1. Temperature of the Earth Calculate the surface temperature of the Earth by assuming that the
Earth is a black body with a constant surface temperature in radiation balance with the Sun. The
diameter of the Sun is 1.4 × 106 km, distance from the Earth 1.5 × 108 km, and the surface temperature
5800 K.
Solution First of all, let us calculate the power radiated by a spherical black body. Take a small
surface element dA, let ρ = n(ω)/V be the (frequency dependent) particle density, and c the velocity
of the photons. The photon flux through A is ρc · dA, and since the radiation inside the surface is
isotropic, we just average that over all directions of c, counting only the flux that is leaving the volume:
Z
Z
Z
ρc
ρc 2π π/2
n(ω)c
ρc
cos θ sin θ dθ dφ dA = dA =
ĉ · n̂ dΩ dA =
dA.
4π ĉ·n̂>0
4π 0
4
4V
0
Multiply that by h̄ω to get the emitted power at ω, integrate over the whole surface, and then over all
the photon states, to get the total power:
Z
2
πRSun
c ∞
2
4
h̄ωn(ω)D(ω) dω = 4πRSun
σTSun
.
(1)
PSun =
V
0
|
{z
}
4
=EV = 4c σV TSun
OK, so how much of that reaches Earth? If d is the Sun-Earth distance, then the fraction of radiation
hitting Earth (essentially a disc of radius REarth ) is
2
πREarth
.
4πd2
Supposing now that Earth radiates as a black body with power PEarth , given by Eq. (1) but with Earth
parameters, and that PInput = PEarth , we get
PInput =
2
πREarth
PSun
4πd2
2
π
REarth
2
4
2
4
Sun
4πR
σTSun
4π
REarth
σTEarth
=
2
4πd
r
RSun
TEarth =
TSun .
2d
Neat! Now plugging in the numbers, we get an estimate that is not bad at all:
PEarth =
TEarth ' 280 K = 7◦ C.
2. Thermodynamics of a Black Hole
The energy and entropy for a non-rotating black hole of mass M can be defined as
E = M c2
S=
kB A
,
8π`2P
wherepA = 4πR2 is the surface area of the black hole, with R = 2GM/c2 the Schwarzschild radius, and
lP = h̄G/c3 is the Planck length, and the gravitation constant G = 6.673 × 10−11 Nm2 /(kg)2 .
(a) Calaculate the temperature of the black hole as a function of mass using thermodynamical identities.
(b) Black hole radiates like an (almost) perfect black body (why?). Calculate the radiation power P
and the timescale associated with the loss of mass, τ = E/P .
(c) Calculate the temperature T and lifetime τ in years for a black hole of mass 1012 kg.
(d) Currently the Universe is filled with radiation of T ≈ 2.7 K, as a remnant of the Big Bang. How
does this affect the development of the above black hole (qualitatively).
Solution
(a) We can find the temperature of the black hole by calculating
1
∂S(E, V, N )
=
.
T
∂E
VN
We can easily express S in terms of E by plgging in the area of black hole, then its radius in terms
of its mass, and finally setting M = E/c2 :
2
2
kB
2kB G2 2
kB
2GM
2GE
kB
=
=
4πR2 =
4π
4π
E .
S=
8π`P
8π`P
c2
8π`P
c4
c8 `2P
Temperature turns out to be inversely proportional to E and consequently to M as well:
1
4kB G2
4kB G2
= 8 2 E= 6 2 M
T
c `P
c `P
⇒
T =
c6 `2P 1
c3 h̄ 1
=
.
4kB G2 M
4GkB M
(b) Obviously, a black hole is a perfect absorber (how it can emit particles is a longer story!). The
radiation power can be found using the formula, derived elsewhere:
P = 4πR2 σT 4 .
One merely plugs in R and T as a function of M to get the power and lifetime τ :
4
3
2GM
c6 h̄π 3 1
c h̄ 1
P = 4π
=
,
σ
c2
4GkB M
960G2 M 2
960G2
τ = 3 4 M 3.
π h̄c
More massive the black hole is, the longer it will be around.
(c) Fixing the mass, M = 1012 kg, we get some definite values for T , P , and τ :
T = 7.71 × 1011 K,
P = 5.55 × 1011 W,
τ = 5.13 × 109 a.
(d) Although the cosmic microwave background radiation will slow down the evaporation of this black
hole, it will only get hotter as its mass radiates away. Should it have been more (much more)
massive, it could have become cool enough to reach thermal equilibrium with the CMB and thus
survive much longer.
3. Gregarious Bosons
Consider a Bose-Einstein condensate with all N bosons in the single-particle ground state ψ0
Ψ(r1 , r2 , . . . , rN ) = ψ0 (r1 ) · · · ψ(rN ).
Suppose a new particle is gently injected into the system, into an equal superposition of the M lowest
single-particle states
1
φ(rN +1 ) = √
ψ0 (rN +1 ) + ψ1 (rN +1 ) + · · · + ψM −1 (rN +1 ) .
M
The state Φ(r1 , . . . , rN +1 ) of the N + 1-boson system after the insertion is obtained by symmetrizing
the product function
Ψ(r1 , r2 , . . . , rN )φ(rN +1 ).
Calculate the symmetrized initial state of the system with the injected particle. Show that the ratio
of the probability that the new boson enters the ground state is enhanced over that of its entering an
empty space by a factor N + 1. (Hint: First do it for N = 1.)
Solution To symmetrize the a given wavefunction, we simply sum over all the possible permutations
of the parameters {ri }. Denoting by P (s) all the permutations of a sequence s, the symmetrized Φ
reads:
X
Φ(r1 , . . . , rN +1 ) = N
Ψ(rp1 , rp2 , . . . , rpN )φ(rpN +1 ),
p∈P (1,...,N +1)
where N is a normalization factor (which will magically absorb any constant coefficients), to be determined later. For N = 1 the above becomes:
Φ(r1 , r2 )
= N [Ψ(r1 )φ(r2 ) + Ψ(r2 )φ(r1 )]
#
"
M −1
M −1
1 X
1 X
ψm (r2 ) + ψ0 (r2 ) √
ψm (r1 )
= N ψ0 (r1 ) √
M m=0
M m=0
"
#
M
−1
M
−1
X
X
= N ψ0 (r1 )ψ0 (r2 ) + ψ0 (r1 )
ψm (r2 ) + ψ0 (r2 )ψ0 (r1 ) + ψ0 (r2 )
ψm (r1 )
m=1
"
= N 2ψ0 (r1 )ψ0 (r2 ) + ψ0 (r1 )
M
−1
X
m=1
ψm (r2 ) + ψ0 (r2 )
m=1
M
−1
X
#
ψm (r1 ) .
m=1
Note now that the terms appearing in the last line above are all mutually orthogonal, and that only
the first term (appearing with twice the amplitude of the other ones) has all the particles in the ground
state. Normalization constant is then
N =p
1
22
+ 2(M − 1)
,
and the probability of observing the new particle in the ground state is
P (All at ground state) =
22
.
N2
The asked ratio of ground state to excited state probabilities is
P (All at ground state)
4
2
P (All at ground state)
=
=
=
.
P (Not all at ground state)
1 − P (All at ground state)
4 + 2(M − 1) − 4
M −1
√
P
Since the probability of finding φ(r) = m ψm (r)/ M in the ground state is clearly 1/M , the above
ratio is simply 1/(M − 1). We see that in the symmetrized case, that probability is 2 = N + 1 times
higher.
So does this hold for general N ? One can pretty much now guess the form of the symmetrized Φ from
what we got above: adding one more particle into
PMthe ground state will increase the amplitude of the
ground state by one, and introduce one more m=1 term into Φ, just with different permutation of
r1,2,3 . But let us for completeness go through the above steps for general N :
Φ(r1 , . . . , rN +1 )
X
=N
Ψ(rp1 , rp2 , . . . , rpN )φ(rpN +1 ),
p∈P (1,...,N +1)
=N
N
+1
X

X

k=1
=N

N
+1
X
Ψ(rp1 , rp2 , . . . , rpN ) φ(rk ),
p∈P (1,...,k−1,k+1,...,N +1)
N !Ψ(r1 , . . . , rk−1 , rk+1 , . . . , rN +1 )φ(rk )
k=1
=N
M −1
1 X
Ψ(r1 , . . . , rk−1 , rk+1 , . . . , rN +1 ) √
ψm (rk )
M m=0
k=1
N
+1
X
Single out the pN +1 index; rename
it to say k and sum it separately,
taking care to remove the k term
from the permutation p summation.
Since Ψ(r1 , . . . , rN ) is already symmetric, each term in the permutation summation is just the same;
replace the sum by the factor
N !Ψ(r1 , . . . , rk−1 , rk+1 , . . . , rN +1 ).
Next just remember that Ψ is simply a product of ground state wave functions:
Φ(r1 , . . . , rN +1 )
NX
+1
N
+1
M
−1
X
X
Ψ(r1 , . . . , rk−1 , rk+1 , . . . , rN +1 )ψ0 (rk ) +
Ψ(r1 , . . . , rk−1 , rk+1 , . . . , rN +1 )
ψm (rk )
=N
{z
}
|
m=1
k=1
k=1
Ψ(r1 ,...,rN +1 )
N
+1 M
−1
X
X
= N (N + 1)Ψ(r1 , . . . , rN +1 ) +
ψ0 (r1 ) · · · ψ0 (rk−1 )ψm (rk )ψ0 (rk+1 ) · · · ψ0 (rN +1 )
k=1 m=1
Again, we have the ground state term appearing with the amplitude N + 1, and the rest (N + 1)(M − 1)
terms with amplitudes 1. In each term a single-particle state with a different rk is raised to a different
excited state m. Therefore, all terms are mutually orthogonal and so the normalization is
N =p
1
(N + 1)2 + (N + 1)(M − 1)
=p
1
(N + 1)(N + M )
.
The ratio of ground state to excited state probabilities is then
P (All at ground state)
N +1
(N + 1)2
=
=
.
1 − P (All at ground state)
(N + 1)(N + M ) − (N + 1)2
M −1
Indeed. This is again N + 1 times larger than the value for φ. The preferrability of the ground state
increases by a factor of N + 1 after the new particle is injected into the condensate, compared to it
being in empty space.
4. Bohr-van Leeuwen Theorem
Consider a system of charged particles (not dipoles), obeying classical mechanics and classical statistics.
Show that the magnetic susceptibility of this system is identically zero.
Solution Suppose B = B ẑ, and use the Landau gauge to express the vector potential A, A = Bxŷ.
The Hamiltonian (expressed in canonical momenta that include the vector potential) reads
H=
Px2 + (Py − qBx)2 + Pz2
+ eφ(x, y, z)
2m
where φ is the scalar potential.
Magnetization can again be evaluated using the formula
1 ∂
ln Z,
β ∂B
Z
Z1 = e−βH d3 x d3 P.
M=
Z = [Z1 ]N ,
It is quite easy to now see that Z1 is independent of B:
Z
Z1 = e−βH d3 x d3 P
2
Z
Px + (Py − qBx)2 + Pz2
= exp −β
+ eφ(x, y, z)
d3 x d3 P
2m
(
"
#)
Z
Px2 + Py2 + Pz2
= exp −β
+ eφ(x, y, z)
d3 x d3 P
2m
Change of variable Py → Py + qBx;
integration bounds still −∞ . . . ∞
Notice that B has now disappeard from Z1 ! Since ∂Z1 /∂B = 0, must then the magnetization M , and
consequently, the susceptibility be also zero.
As a quick proof of the applicability of the our equation for the magnetization, note that we can write
the Hamiltonian also in terms of the kinetic momenta p, and a potential associated with the external
magnetic field:
H=
p2
− m · B + V (r)
2m
where m = qr × p/2m is the magnetic moment of a point charge on a circular path. Upon inspection
of ∂ ln Z1 /∂B, the formula for the magnetization M quickly follows.
5. Entropy of an Isolated Quantum System
Show that the von Neumann entropy
S = Tr(ρ̂ ln ρ̂)
of an isolated quantum system is time-independent.
Solution Let us actually prove a very slightly more general statement, as it turns out to be somewhat
less messy operation. We will show that
d
Tr F (ρ̂) = 0,
dt
where F is an arbitrary analytic function, defined on the unit interval. The statement regarding entropy
is obtained with the special case of F (x) = x ln x.
All we need is the time-evolution of ρ̂, and a series expansion for F :
1
∂ ρ̂
= [Ĥ, ρ̂],
∂t
ih̄
F (x) =
∞
X
Fk x k .
k=0
A straight-forward calculation gives:
∞
∞
k
k=0
k=0
`=0
X X
∂ ρ̂
d X
d
Fk
Tr ρ̂k−` ρ̂`
Fk ρ̂k =
Tr F (ρ̂) = Tr
dt
dt
∂t
=
∞
X
k=0
Fk
k
X
Tr
=
∞
X
k=0
Fk k
∂ ρ̂ k
∂t ρ̂
1
{Tr Ĥ ρ̂ρ̂k − Tr ρ̂Ĥ ρ̂k }
ih̄ | {z } | {z }
Tr Ĥ ρ̂k+1
|
= 0.
∞
X
∂ρ k
Fk k Tr
Tr ρ̂
ρ̂ =
ρ̂
∂t
∂t
{z
} k=0
`=0 |
k−` ∂ ρ̂ `
Tr Ĥ ρ̂k+1
{z
0
}