Number of Modes in a 1D Line
edit
Each mode is a standing wave
For standing waves with nodes at each end, the wavefunction is:
ψ
(
x
)
=
ψ
0
sin
(
k
x
)
=
ψ
0
sin
(
n
π
L
x
)
{\displaystyle \psi (x)=\psi _{0}\sin(kx)=\psi _{0}\sin({\frac {n\pi }{L}}x)}
where n is a positive integer
Periodic Boundary Conditions
edit
What if the line is wrapped round into a circle?
Now the boundary condition is
ψ
(
x
)
=
ψ
(
x
+
L
)
{\textstyle \psi (x)=\psi (x+L)}
. Since the wave is periodic, psi can be written in the form:
ψ
(
x
)
=
e
i
k
x
{\displaystyle \psi (x)=e^{ikx}}
.
ψ
(
x
)
=
ψ
(
x
+
L
)
∴
e
i
k
x
=
e
i
k
(
x
+
L
)
e
i
k
x
=
e
i
k
x
e
i
k
L
∴
1
=
e
i
k
L
e
0
,
e
2
π
i
,
e
4
π
i
,
…
=
e
i
k
L
⇒
0
,
2
π
,
4
π
,
…
=
k
L
⇒
k
=
2
n
π
L
{\displaystyle {\begin{aligned}\psi (x)&=\psi (x+L)\\\therefore e^{ikx}&=e^{ik(x+L)}\\e^{ikx}&=e^{ikx}e^{ikL}\\\therefore 1&=e^{ikL}\\e^{0},e^{2\pi i},e^{4\pi i},\dots &=e^{ikL}\\\Rightarrow 0,2\pi ,4\pi ,\dots &=kL\\\Rightarrow k&={\frac {2n\pi }{L}}\end{aligned}}}
We want to sum up each mode
∴
⟨
E
⟩
T
O
T
=
∑
n
(
⟨
E
n
⟩
)
{\displaystyle \therefore \langle E\rangle _{TOT}=\sum _{n}(\langle E_{n}\rangle )}
since there's a unique value k for each n, we can change this to:
⟨
E
⟩
T
O
T
=
∑
k
(
⟨
E
k
⟩
)
{\displaystyle \langle E\rangle _{TOT}=\sum _{k}(\langle E_{k}\rangle )}
We can change this to an integral for large values of L by considering the no. of modes in a range of k-values.
Each mode takes up
2
π
/
L
{\displaystyle 2\pi /L}
in 1D k-space, therefore no. of modes between
±
k
{\displaystyle \pm k}
is given by:
k
−
−
k
2
π
L
=
2
k
L
2
π
=
k
L
π
=
L
2
π
∫
−
∞
∞
d
k
{\displaystyle {\frac {k--k}{\frac {2\pi }{L}}}={\frac {2kL}{2\pi }}={\frac {kL}{\pi }}={\frac {L}{2\pi }}\int _{-\infty }^{\infty }dk}
therefore, the number of modes between
±
∞
=
L
2
π
∫
−
∞
∞
d
k
{\displaystyle \pm \infty ={\frac {L}{2\pi }}\int _{-\infty }^{\infty }dk}
3D Periodic Boundary Conditions
edit
Imagine a box where if you go a distance L in any direction, you end up at a place that looks identical to where you started. We'll ignore edge effects as we're more interested in the local conditions.
In 3D, the waves can be described in the form of exponentials with vector exponents:
ψ
(
x
)
=
e
i
k
⋅
r
{\displaystyle \psi (x)=e^{i{\boldsymbol {k}}\cdot {\boldsymbol {r}}}}
where
k
=
(
k
x
k
y
k
z
)
=
2
π
L
(
n
x
n
y
n
z
)
{\displaystyle {\boldsymbol {k}}={\begin{pmatrix}k_{x}\\k_{y}\\k_{z}\end{pmatrix}}={\frac {2\pi }{L}}{\begin{pmatrix}n_{x}\\n_{y}\\n_{z}\end{pmatrix}}}
therefore, to sum over all modes we can use:
no. of modes
=
L
2
π
∫
(
L
2
π
∫
(
L
2
π
∫
d
k
z
)
d
k
y
)
d
k
x
∑
k
=
(
L
2
π
)
3
∫
∫
∫
d
k
z
d
k
y
d
k
x
∑
k
=
V
(
2
π
)
3
∫
d
k
{\displaystyle {\begin{aligned}{\text{no. of modes}}&={\frac {L}{2\pi }}\int {\Bigl (}{\frac {L}{2\pi }}\int ({\frac {L}{2\pi }}\int dk_{z})dk_{y}{\Bigr )}dk_{x}\\\sum _{k}&={\Bigl (}{\frac {L}{2\pi }}{\Bigr )}^{3}\int \int \int dk_{z}dk_{y}dk_{x}\\\sum _{k}&={\frac {V}{(2\pi )^{3}}}\int d{\boldsymbol {k}}\\\end{aligned}}}
Since sound waves have 3 polarisations, we need to triple this expression:
∑
m
o
d
e
s
=
3
∑
k
=
3
V
(
2
π
)
3
∫
d
k
{\displaystyle \sum _{modes}=3\sum _{k}={\frac {3V}{(2\pi )^{3}}}\int d{\boldsymbol {k}}}
Converting to Spherical Polar Co-ordinates
edit
Since we're assuming speed of sound is isotropic, then we can rewrite this expression in spherical polar co-ordinates:
3
V
(
2
π
)
3
∫
d
k
=
3
V
(
2
π
)
3
∫
0
∞
4
π
k
2
d
k
{\displaystyle {\frac {3V}{(2\pi )^{3}}}\int d{\boldsymbol {k}}={\frac {3V}{(2\pi )^{3}}}\int _{0}^{\infty }4\pi k^{2}dk}
We can then use the linear velocity dispersion relation to integrate in terms of angular frequency instead of wave number:
k
=
ω
v
∴
d
k
d
ω
=
1
v
∴
d
k
=
d
ω
v
{\displaystyle {\begin{aligned}k&={\frac {\omega }{v}}\\\therefore {\frac {dk}{d\omega }}&={\frac {1}{v}}\\\therefore dk&={\frac {d\omega }{v}}\end{aligned}}}
Substituting in the expressions for k and dk, we get:
∑
m
o
d
e
s
=
3
V
(
2
π
)
3
∫
0
∞
4
π
ω
2
v
3
d
ω
=
12
V
π
(
2
π
)
3
v
3
∫
0
∞
ω
2
d
ω
{\displaystyle {\begin{aligned}\sum _{modes}&={\frac {3V}{(2\pi )^{3}}}\int _{0}^{\infty }4\pi {\frac {\omega ^{2}}{v^{3}}}d\omega \\&={\frac {12V\pi }{(2\pi )^{3}v^{3}}}\int _{0}^{\infty }\omega ^{2}d\omega \end{aligned}}}
The integrand here is called the density of states.
Definition (Density of States) :
The density of states is a function
g
{\displaystyle g}
of angular frequency
ω
{\displaystyle \omega }
, such that:
∑
m
o
d
e
s
=
∫
0
∞
g
(
ω
)
d
ω
{\displaystyle \sum _{modes}=\int _{0}^{\infty }g(\omega )d\omega }
In this case
g
(
ω
)
=
12
π
V
(
2
π
)
3
v
3
ω
2
{\displaystyle g(\omega )={\frac {12\pi V}{(2\pi )^{3}v^{3}}}\omega ^{2}}
.
If N is defined as the number of states across all frequencies in one polarisation:
∴
3
N
=
∫
0
∞
g
(
ω
)
d
ω
{\displaystyle \therefore 3N=\int _{0}^{\infty }g(\omega )d\omega }
We can also write the density of states in terms of a value called the Debye frequency (we'll see where this comes from later):
ω
D
=
6
π
2
(
N
V
)
v
3
3
=
6
π
2
ρ
v
3
3
⇒
g
(
ω
)
=
N
9
ω
2
ω
D
3
{\displaystyle {\begin{aligned}\omega _{D}&={\sqrt[{3}]{6\pi ^{2}{\Bigl (}{\frac {N}{V}}{\Bigr )}v^{3}}}={\sqrt[{3}]{6\pi ^{2}\rho v^{3}}}\\\Rightarrow g(\omega )&=N{\frac {9\omega ^{2}}{\omega _{D}^{3}}}\end{aligned}}}
Getting Specific Heat Capacity
edit
Going back to our initial expression for the total average energy, we can now write it in terms of an integral of
ω
{\displaystyle \omega }
:
⟨
E
⟩
T
O
T
=
∑
m
o
d
e
s
(
ℏ
ω
m
o
d
e
(
n
B
(
β
ℏ
ω
)
+
1
2
)
)
⟨
E
⟩
T
O
T
=
∫
0
∞
(
g
(
ω
)
(
n
B
(
β
ℏ
ω
)
+
1
2
)
ℏ
ω
)
d
ω
{\displaystyle {\begin{aligned}\langle E\rangle _{TOT}&=\sum _{modes}(\hbar \omega _{mode}(n_{B}(\beta \hbar \omega )+{\frac {1}{2}}))\\\langle E\rangle _{TOT}&=\int _{0}^{\infty }{\Bigl (}g(\omega ){\Bigl (}n_{B}(\beta \hbar \omega )+{\frac {1}{2}}{\Bigr )}\hbar \omega {\Bigr )}d\omega \end{aligned}}}
Note that this expression is actually incorrect! The upper limit of infinity suggests that there are an infinite number of modes, and evaluating the integral will give an infinite answer, due to the term of 1/2 - this is known as the zero-point energy problem . However, for the purpose of calculating and expression for heat capacity, it doesn't have much effect as it doesn't affect the derivative. So for now, don't worry about this - we'll come back later and fix this.
Ignoring the zero-point term, we can plug in our expression for the density of states:
⟨
E
⟩
T
O
T
=
∫
0
∞
(
9
N
ω
2
ω
D
3
(
n
B
(
β
ℏ
ω
)
+
1
2
)
ℏ
ω
)
d
ω
=
9
N
ℏ
ω
D
3
∫
0
∞
ω
3
n
B
(
β
ℏ
ω
)
d
ω
=
9
N
ℏ
ω
D
3
∫
0
∞
ω
3
(
1
e
β
ℏ
ω
−
1
)
d
ω
{\displaystyle {\begin{aligned}\langle E\rangle _{TOT}&=\int _{0}^{\infty }{\Bigl (}{\frac {9N\omega ^{2}}{\omega _{D}^{3}}}{\Bigl (}n_{B}(\beta \hbar \omega )+{\frac {1}{2}}{\Bigr )}\hbar \omega {\Bigr )}d\omega \\&={\frac {9N\hbar }{\omega _{D}^{3}}}\int _{0}^{\infty }\omega ^{3}n_{B}(\beta \hbar \omega )d\omega \\&={\frac {9N\hbar }{\omega _{D}^{3}}}\int _{0}^{\infty }\omega ^{3}{\Bigl (}{\frac {1}{e^{\beta \hbar \omega }-1}}{\Bigr )}d\omega \end{aligned}}}
Substituting
x
=
β
ℏ
ω
{\displaystyle x=\beta \hbar \omega }
:
ω
=
x
β
ℏ
d
ω
d
x
=
1
β
ℏ
d
ω
=
d
x
β
ℏ
∴
⟨
E
⟩
TOT
=
9
N
ℏ
ω
D
3
∫
0
∞
(
x
β
ℏ
)
3
(
1
e
x
−
1
)
1
β
ℏ
d
x
=
9
N
ℏ
ω
D
3
β
4
ℏ
4
∫
0
∞
x
3
e
x
−
1
d
x
{\displaystyle {\begin{aligned}\omega &={\frac {x}{\beta \hbar }}\\{\frac {d\omega }{dx}}&={\frac {1}{\beta \hbar }}\\d\omega &={\frac {dx}{\beta \hbar }}\\\therefore \left\langle E\right\rangle _{\text{TOT}}&={\frac {9N\hbar }{\omega _{D}^{3}}}\int _{0}^{\infty }\left({\frac {x}{\beta \hbar }}\right)^{3}\left({\frac {1}{e^{x}-1}}\right){\frac {1}{\beta \hbar }}dx\\&={\frac {9N\hbar }{\omega _{D}^{3}\beta ^{4}\hbar ^{4}}}\int _{0}^{\infty }{\frac {x^{3}}{e^{x}-1}}dx\\\end{aligned}}}
Fixing the Zero-Point Energy Problem
edit