Thermal Physics
• Too many particles… can’t keep track!
• Use pressure (p) and volume (V) instead.
• Temperature (T) measures the tendency of an object to
spontaneously give up/absorb energy to/from its
surroundings. (p and T will turn out to be related to the
too many particles mentioned above)
Pressure, Volume, Temperature
P, V, T
F/A
L3
Something to
do with heat
Equations of state
•An equation of state is a mathematical relation between state
variables, e.g. p, V & T.
•This reduces the number of independent variables to two.
General form: f (p,V,T) = 0
Example:
pV – nRT = 0
(ideal gas law)
•Defines a 2D surface in p-V-T state space.
•Each point on this surface represents an unique state of the system.
f (p,V,T) = 0
Equilibrium state
Ideal gas equation of state
Boyle’s law
p  1/V
Robert Boyle (1627 – 1691)
Charles’ law
pV = NkB T
VT
Jacques Charles (1746 – 1823)
Gay-Lussac’ law
pT
Joseph Louis Gay-Lussac (1778 - 1850)
kB = 1.38  10-23 J/K
Heat is energy in transit
Surroundings
System
Universe
(system +
surroundings)
What is temperature?
Temperature is what you measure with a thermometer
Temperature is the thing that’s the same for two
objects, after they’ve been in contact long enough.
Long enough so that the two objects are in thermal
equilibrium.
Time required to reach thermal equilibrium is the
relaxation time.
Zeroth law of thermodynamics
A
C
If two systems are separately
in thermal equilibrium with a
third system, they are in
thermal equilibrium with each
other.
Diathermal
wall
B
C
C can be considered the
thermometer. If C is at a
certain temperature then A
and B are also at the same
temperature.
How can we define temperature using
the microscopic properties of a
system?
Most likely macrostate the system will
find itself in is the one with the maximum
number of microstates.
Number of Microstates ()
1.2e+029
1e+029
8e+028
6e+028
4e+028
2e+028
0
0
20
40
x
Macrostate
60
80
100
1.Each microstate is equally likely
2.The microstate of a system is
continually changing
3.Given enough time, the system
will explore all possible
microstates and spend equal time
in each of them (ergodic
hypothesis).
Most likely macrostate the system will find
itself in is the one with the maximum
number of microstates.
E1
E2
1(E1)
(E)
2(E2)
Most likely macrostate the system will find
itself in is the one with the maximum
number of microstates.
𝐸 = 𝐸1 + 𝐸2 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
Total microstates = Ω(𝐸1 , 𝐸2 )
Ω 𝐸1 , 𝐸2 = Ω1 (𝐸1 )Ω2 (𝐸2 )
EE
1
1
1(E
1) )
(E
1
1
EE
2
2
2(E2)
2(E2)
To maximize Ω:
𝑑Ω
=0
𝑑𝐸1
Most likely macrostate the system will find
itself in is the one with the maximum
number of microstates.
E1
E2
1(E1)
2(E2)
d ln 1 d ln  2
1


dE1
dE2
k BT
Using this definition of temperature we need to
describe real systems
E
(E)
Microcanonical ensemble:
An ensemble of snapshots
of a system with the same
N, V, and E
Canonical ensemble: An ensemble of snapshots
of a system with the same N, V, and T (red box
with energy  << E.
E-
(E-)

 I(  )
Red box is small only in terms of energy, its volume could still be large
Boltzmann Factor
(canonical ensemble)
P ( )  e

k BT
Canonical ensemble
Reservoir
The red ball is the particle from the canonical ensemble in thermal equilibrium with
the reservoir. It occupies the same volume as the reservoir which in this case are
the rest of particles in an ideal gas.
Spherical
coordinates
𝑑𝑉 = 𝑟 2 sin 𝜃𝑑𝑟𝑑𝜃𝑑𝜑
𝑑𝐴 = 𝑟 2 sin 𝜃𝑑𝜃𝑑𝜑
Monatomic ideal gas
First try to find the probability that the red particle has a certain velocity
𝑓 ′ 𝑣 𝑑𝑣𝑥 𝑑𝑣𝑦 𝑑𝑣𝑧 ∝
𝑓 ′ 𝑣 𝑑𝑣𝑥 𝑑𝑣𝑦 𝑑𝑣𝑧 ∝ 𝑒
𝑓 ′ 𝑣 𝑑𝑣𝑥 𝑑𝑣𝑦 𝑑𝑣𝑧 ∝
𝑚𝑣 2
−
𝑒 2𝑘𝐵 𝑇 𝑑𝑣𝑥 𝑑𝑣𝑦 𝑑𝑣𝑧
𝑚(𝑣𝑥2 +𝑣𝑦2 +𝑣𝑧2 )
−
2𝑘 𝑇
𝐵
𝑑𝑣𝑥 𝑑𝑣𝑦 𝑑𝑣𝑧
𝑚𝑣𝑦2
𝑚𝑣𝑥2
𝑚𝑣𝑧2
−
−
−
2𝑘
𝑇
2𝑘
𝑇
𝐵
𝐵
𝑒
𝑑𝑣𝑥 𝑒
𝑑𝑣𝑦 𝑒 2𝑘𝐵 𝑇 𝑑𝑣𝑧
∝ 𝑔 𝑣𝑥 𝑑𝑣𝑥
𝜃
𝑣
𝜑
𝑓 ′ 𝑣 𝑑𝑣𝑥 𝑑𝑣𝑦 𝑑𝑣𝑧 ∝
𝑚𝑣 2
−
𝑒 2𝑘𝐵 𝑇 𝑑𝑣𝑥 𝑑𝑣𝑦 𝑑𝑣𝑧
𝑓 ′ 𝑣 𝑣 2 sin 𝜃 𝑑𝑣𝑑𝜃𝑑𝜑 ∝
𝑚𝑣 2
−2𝑘 𝑇 2
𝑒 𝐵 𝑣 sin 𝜃 𝑑𝑣𝑑𝜃𝑑𝜑
Integrating over the two angular variables we can get the probability that
the speed of a particle is between 𝑣 and 𝑣 + 𝑑𝑣:
𝑓 ′ 𝑣 𝑣 2 sin 𝜃 𝑑𝑣𝑑𝜃𝑑𝜑 ∝
⇒ 𝑓 𝑣 𝑑𝑣 ∝
𝑚𝑣 2
−
𝑒 2𝑘𝐵 𝑇 𝑣 2 sin 𝜃 𝑑𝑣𝑑𝜃𝑑𝜑
𝑚𝑣 2
−
𝑒 2𝑘𝐵 𝑇 𝑣 2 𝑑𝑣
For 𝑓 𝑣 to be a proper probability distribution/density function:
∞
𝑓 𝑣 𝑑𝑣 = 1
0
4
𝑚
⇒ 𝑓 𝑣 𝑑𝑣 =
𝜋 2𝑘𝐵 𝑇
3 2
𝑚𝑣 2
−
𝑣 2 𝑒 2𝑘𝐵𝑇 𝑑𝑣
Maxwell-Boltzmann speed distribution
0.12
0.1
T = 10
0.08
4
𝑚
⇒ 𝑓 𝑣 𝑑𝑣 =
𝜋 2𝑘𝐵 𝑇
3 2
𝑚𝑣 2
−
𝑣 2 𝑒 2𝑘𝐵𝑇 𝑑𝑣
0.06
0.04
0.02
0
0.04
0
10
20
30
40
50
60
70
80
90
100
0.012
0.035
0.01
0.03
T = 1000
0.008
0.025
T = 100
0.02
0.006
0.015
0.004
0.01
0.002
0.005
0
0
10
20
30
40
50
60
70
80
90
100
0
0
10
20
30
40
50
60
70
80
90
100
∞
𝑣 =
𝑣𝑓 𝑣 𝑑𝑣 =
0
∞
𝑣
2
3𝑘𝐵 𝑇
2
𝑣 𝑓 𝑣 𝑑𝑣 =
= 𝑣𝑟𝑚𝑠
𝑚
2
=
0
8𝑘𝐵 𝑇
𝜋𝑚
Solid angle Ω =
𝐴
𝑟2
In velocity space:
𝑣𝑧
𝑑𝐴
𝑑Ω = 2
𝑟
Or since its velocity space
𝑑𝐴
𝑑Ω = 2
𝑣
𝜃
𝑣
𝑣𝑦
𝜑
𝑣𝑥
This tiny solid angle 𝑑Ω will include all the
particles travelling between angles 𝜃 and
𝜃 + 𝑑𝜃 and 𝜑 and 𝜑 + 𝑑𝜑
Solid angle Ω =
𝐴
𝑟2
In velocity space:
𝑣𝑧
𝑑Ω′
𝑑𝐴′
=
𝑑𝐴
𝑑Ω = 2
𝑟
𝑑𝐴′
= 2
𝑣
Or since its velocity space
𝑑𝐴
𝑑Ω = 2
𝑣
𝜃
2𝜋𝑣 2 sin 𝜃𝑑𝜃
𝑣
𝑣𝑦
𝜑
𝑣𝑥
This shaded solid angle 𝑑Ω′ includes all the
particles travelling between angles 𝜃 and
𝜃 + 𝑑𝜃
Solid angle Ω =
𝐴
𝑟2
In velocity space:
𝑣𝑧
𝑑𝐴′
= 2
𝑣
𝑑𝐴′ = 2𝜋𝑣 2 sin 𝜃𝑑𝜃
𝑑𝐴
𝑑Ω = 2
𝑟
𝑑Ω′
⇒
𝑑Ω′
Or since its velocity space
𝑑𝐴
𝑑Ω = 2
𝑣
𝜃
= 2𝜋 sin 𝜃𝑑𝜃
𝑣
𝑣𝑦
𝜑
𝑣𝑥
Since the total solid angle is 4𝜋 and the ideal
gas is isotropic i.e. no preferred direction
for 𝑣, the fraction of particles moving
between angles 𝜃 and 𝜃 + 𝑑𝜃 is
𝑑Ω′
4𝜋
Once again:
Probability that a particle in a monatomic ideal gas has a speed between 𝑣
and 𝑣 + 𝑑𝑣 is given by:
4
𝑚
⇒ 𝑓 𝑣 𝑑𝑣 =
𝜋 2𝑘𝐵 𝑇
3 2
𝑚𝑣 2
−
𝑣 2 𝑒 2𝑘𝐵𝑇 𝑑𝑣
If the total number of particles is 𝑁 then the number per unit volume is 𝑛 =
Therefore, the number per unit volume in a monatomic ideal which have
speeds between 𝑣 and 𝑣 + 𝑑𝑣 is 𝑛𝑓 𝑣 𝑑𝑣
These particles are travelling in all possible directions i.e. the entire 4𝜋
steradians of solid angle.
Hence the fraction of 𝑛𝑓 𝑣 𝑑𝑣 travelling at polar angles between 𝜃 and
𝜃 + 𝑑𝜃 i.e. into a solid angle of 𝑑Ω′ is 𝑛𝑓 𝑣 𝑑𝑣 ×
𝑑Ω′
4𝜋
𝑁
𝑉
The number per unit volume in a monatomic ideal which have speeds between 𝑣
and 𝑣 + 𝑑𝑣 and travelling at polar angles between 𝜃 and 𝜃 + 𝑑𝜃 is:
′
𝑑Ω
2𝜋 sin 𝜃𝑑𝜃
1
𝑑𝑛′ = 𝑛𝑓 𝑣 𝑑𝑣
= 𝑛𝑓 𝑣 𝑑𝑣
= 𝑛𝑓 𝑣 𝑑𝑣 sin 𝜃𝑑𝜃
4𝜋
4𝜋
2
𝑣𝑧
𝜃
𝑣
𝑣𝑦
𝜑
𝑣𝑥
Remember all this is happening in velocity space
This is what happens in real space
𝑣𝑥

𝜑
𝑣𝑧
𝑣𝑥
𝑣𝑦
𝜃
A
𝑣𝑧
𝑣
𝜑
𝜃
𝑣 𝑣
𝜃
𝑣𝑦
𝑣𝑧
𝜑
𝑣𝑦
𝑣𝑥

A
𝑑𝑉 = 𝐴𝑣𝑑𝑡 cos 𝜃
The number of particles which have speeds between 𝑣 and 𝑣 + 𝑑𝑣 and travelling at
polar angles between 𝜃 and 𝜃 + 𝑑𝜃 and hitting the wall of area 𝐴 in time 𝑑𝑡:
1
𝑑𝑁 = 𝑑𝑛 𝑑𝑉 = 𝑛𝑓 𝑣 𝑑𝑣 sin 𝜃𝑑𝜃𝐴𝑣𝑑𝑡 cos 𝜃
2
′
Change in momentum of
each particle = 2𝑚𝑣 cos 𝜃
𝜃
The total change in momentum of all the number of particles which have speeds
between 𝑣 and 𝑣 + 𝑑𝑣 and travelling at polar angles between 𝜃 and 𝜃 + 𝑑𝜃 and
hitting the wall of area 𝐴 in time 𝑑𝑡 is:
1
𝑑 𝑝 = 𝑑𝑁 × 2𝑚𝑣 cos 𝜃 = 𝑛𝑓 𝑣 𝑑𝑣 sin 𝜃𝑑𝜃𝐴𝑣𝑑𝑡 cos 𝜃 × 2𝑚𝑣 cos 𝜃
2
The total force on the wall due to all the particles which have speeds between 𝑣 and
𝑣 + 𝑑𝑣 and travelling at polar angles between 𝜃 and 𝜃 + 𝑑𝜃 and hitting the wall of
area 𝐴 in time 𝑑𝑡 is:
𝑑𝑝
1
𝑑𝐹 =
= 𝑛𝑓 𝑣 𝑑𝑣 sin 𝜃𝑑𝜃𝐴𝑣 cos 𝜃 × 2𝑚𝑣 cos 𝜃
𝑑𝑡
2
The pressure on the wall due to all the particles which have speeds between 𝑣 and
𝑣 + 𝑑𝑣 and travelling at polar angles between 𝜃 and 𝜃 + 𝑑𝜃 and hitting the wall of
area 𝐴 in time 𝑑𝑡 is:
𝑑𝐹
1
𝑑𝑝 =
= 𝑛𝑓 𝑣 𝑑𝑣 sin 𝜃𝑑𝜃𝑣 cos 𝜃 × 2𝑚𝑣 cos 𝜃
𝐴
2
⇒ 𝑑𝑝 = 𝑛𝑚𝑣 2 𝑓 𝑣 𝑑𝑣 sin 𝜃 cos 2 𝜃 𝑑𝜃
The pressure on the wall due to all the particles in the gas is:
𝜋
∞
𝜋/2
2
𝑝 = 𝑛𝑚
0
= 𝑛𝑚
sin 𝜃 cos2 𝜃 𝑑𝜃
𝑣 𝑓(𝑣) 𝑑𝑣
𝑣2
0
1
3
3𝑘𝐵 𝑇 1
= 𝑛𝑚
𝑚 3
𝑁
= 𝑛𝑘𝐵 𝑇 = 𝑘𝐵 𝑇
𝑉
⇒ 𝑝𝑉 = 𝑁𝑘𝐵 𝑇
Only till 2 to include
only those particles
hitting the wall from
the left
𝑐
𝐴′
𝐴1
𝜋
−𝜃
2
𝜃
𝐴
𝑏 𝐴2
𝐴2
𝜃
𝑐
𝑏
𝑎
𝐴′
1
= 𝐴1 + 2𝐴2 = (𝑏 cos 𝜃)(𝑐 − 𝑏 sin 𝜃) + 2 ∙ ∙ 𝑏 sin 𝜃 ∙ 𝑏 cos 𝜃 = 𝑏𝑐 cos 𝜃
2
𝑉 = 𝑎 ∙ 𝐴′ = 𝑎𝑏𝑐 cos 𝜃 = 𝐴𝑐 cos 𝜃
𝑏
𝑎
𝑐
𝑉 =𝑎∙𝑏×𝑐