022 - pr 06 - grab bag of problems to use QMTP on (go nuts, buddy): Disclaimers: (1) None of the questions in this problem requires any deep knowledge of the field the problem comes from. (2) Not all parameters which determine the result are mentioned in the formulation of the problem. (3) Some of the mentioned parameters may be irrelevant . Goal: Use your best judgment and intuition to determine the minimum set of relevant parameters. Even if you happen to know the exact solution to the problem, still use qualitative methods to estimate the result . (a) Gravitational collapse (black hole formation) occurs if the radius of a gravitating body of mass M is less than some critical value, r g. Estimate r g. For a hole to be black, the escape velocity of a particle must equal the speed of light. Thus, our parameters should be: mass, radius, the speed of light, and Newton’s gravitational constant. We then put a power-law-ansatz to estimate this critical-radius, 3a + b = 1 c = a = − 12 b Gm 3a 2a b 3a b 1 a b c a b c c a b 2a → R~ R ~ G c m = [ L M T L T M ] = [ L M T (1.1) ] = [L ] → c − a = 0 ↔ 2 c a =1 b + 2a = 0 −
−
−
+
−
− −
The accepted critical radius is R = 2Gm / c 2 , so we missed a pure factor of 2. (b) Under certain conditions, stars can be described as ultra-relativistic Fermi gases, i.e., ensembles of free fermions moving moving at speeds speeds very very close to the speed speed of light, light, c: The dispersio dispersion n relation at such such speeds speeds is the electron electron momentum. momentum. Estimate the dependence dependence of the specific specific heat (per (per unit ε ≈ c p = c ⋅ ℏ k , where p is the 1
volume) of such a gas on its density n . Let an “atom” of a star have mass m. Then, a b c d e b 2c c c d d 2e e e b e c 1 3a 2e [CV ] = EK = n m ℏ c k B = L M L T M L T L T M K = M T −
−
−
−
−
−
b = −c b +1+ c = 1 →C ~ n ↔ ↔ V −c = d = b −2(1) − c − d = −2 2(1) + d + 2c − 3a = 2 a = − 13 d e =1
+
+
2e −c − d
−
2e + d + 2c − 3a
L
K
−e
e =1
−d
/3
m
d
ℏ
−d
d
c kB
d =1
=
(
mc 3
nℏ
)d k B
(1.2)
The missing unknown: There are two physical interpret interpretations ations for the missing missing unknown, for (1) Even so we have five unknowns and four equations, the extra missing piece of information comes from the fact that in one dimension, k F ~ n1 , whereas in 2-dimensions, kF ~
n , and in three, kF ~
3
n . So, while we’re at it, we might
as well write down the dimensional-analysis-order-of-magnitude estimate for an ultra-relativistic system of dimensionality D. Secondly, (2) suppose the ultra-relativistic fermion-ensemble was made of massless fermions (actually, our above analysis is valid for bosons as well). Then: we must find a way for the term mc to make no contribution to the heat capacity, and this is done by setting d = 0 . 3
[
n mc ℏ
d
]
=
V
1
−
=
−
L
D
=
−
L
d ( D / 3)− d
↔
D = d(
D
3
+
1) ↔ d
=
D
1 D + 3
1
;
D = (1, 2, 3) → d = ( 34 , 65 , 32 )
[zero rest mass] → d = 0
(1.3)
(c) An ultra-relativistic free Fermi-gas is placed in a box of side L. The temperature is equal to zero. Estimate the pressure that the gas exerts on the walls . For a free Fermi gas to be ultra-relativistic and have a temperature of zero, it must have a maximum density. Let there be N free electrons. Suppose we have a k-space cell with q free electrons per unit cell2. You then have,
1 2
Despite a star being made of a multitude of elements, one can consider an average atom’s mass. E.g., q can take on various values for BCC, FCC, orthorhombic, HCP, etc., unit cells.
P = EL 3 −
=
L 1T 2 M −
−
ℏ
a b
=
c
cV N
−
N 4/3 ℏc N ) ; → 2(1) + b + 3c = − 1 → P = ( V q V −(1) − b = −2 a =1
c
=
q
a
M L2
a + b + 3c
−
T
a− b
=
n=
1 3π 2
kF 3 ; (1.4)
(d) A hydrogen atom is placed into a strong magnetic field B. The orbital effect of the field is to squeeze the electron orbit. Estimate the field when this effect becomes well-pronounced . There is an energy-density 1
associated with a uniform magnetic field, B = B zˆ → u B
= 2
2
B / µ 0 . Let us also compute an energy density
assicated with the hydrogen atom’s ground state, which is exactly E 0 volume can be approximated by V H ~ 43 π a0 3 in which a0
orbit u B deform → E / V H 0
B 2V H
=
ɶ >
2 E0 µ0
2 E 0 µ0
1 ↔ B >ɶ
=
=
4 πε 0 ℏ2 2
me
= −
2
e 13.6 eV = − 2mℏ2 ( 4πε ) 2 , and whose 0
, the Bohr radius. Thus,
2(13.6 ×1.602 × 10
19
−
)(4π × 10 7 ) −
4π 13 (5.29 × 10
V H
11 3
−
)
T = 3000 T
(1.5)
Further justification: We can regard the energy density of the B-field and the energy density of the -13.6 eV in the (4/3)*Pi*a0^3 volume as being two fluids with opposing pressures. The B-field’s pressure has a directionality such that it radially-squeezes a ring of current3, which can be seen from the Lorentz force law, F = qv × B = q (vφ ˆ) × ( Bzˆ ) = qvBsˆ = [radially-directed force] (1.6)
(e) Free massive bosons condense into a state in which all particles have zero energy (Bose-Einstein condensation). For a given density of bosons, estimate the temperature of Bose-Einstein condensation . 4 Parameters you have running around: consider only intensive parameters. Let the bosons have number-density n = N / V , and have a mass m. Then, d = −1 d = −1 d = −1 a
T ~n m
b
⋅ℏ
c
k B
d
=
1
[K ]
=
3a
−
b
c
c
d
[L M E T E K
−d
] →
c=2 ↔ c − 2c − 2d = 0 4 − 2 − 3a = 0 b+c+d = 0 b + 2 −1 = 0
2c + 2d − 3a = 0
c=2 ↔ a = 23 b = −1
(1.7)
This yields a transition-temperature of TC ~ n 2/3ℏ 2 / ( k B m) . This is off from the correct answer by a factor of 2π ≈ 2.612 2/3
3
3.31 .
Regard the hydrogen atom as a superposition of current loops, whose radius is given by
hydrogen-wavefunction ground state 4
x |100
= ɺ
(π a0 3 )
1/ 2
−
e
−
r / a0
R
. This justifies saying J
=
=
100 | R |100 , where we have the
J φˆ , for φ ˆ a cylindrical unit vector.
We are asserting that the two extensive properties, N and V, must enter as an intensive property, n = N/V. Note N is dimensionless, so this requires the (not so deep!) knowledge that a phase-transition as Bose-condensation is an intensive phenomenon.
