Thermally Induced Agitation of a Simple Pendulum—C.E. Mungan, Spring 2000
The generalized equipartition theorem is
E=
1
kT (2 − δ a 0 − δ b 0 )
n
(1a)
provided the energy per conjugate pair of degrees of freedom q and p can be written in the form
n
n
E ≡ KE + PE = a p + b q .
(1b)
where n > 0. The absolute value bars in Eq. (1b) ensure that the energy is always positive. The
Krönecker delta functions in Eq. (1a) reduce the average energy if some degree of freedom is
either frozen out or otherwise not present. For example, three common applications of Eq. (1)
are:
• A classical free particle for which KE = p2 / 2 m and PE = 0 . In this case n = 2 and b = 0, so
that E = kT / 2 .
• A photon for which KE = pc and PE = 0 . In this case n = 1 and b = 0, so that E = kT .
• A simple harmonic oscillator for which KE = p2 / 2 m and PE = ks q 2 / 2 . In this case n = 2, so
that E = kT .
A derivation of Eq. (1) is in the Appendix.
To apply this to a simple pendulum, whose geometry is sketched below, we can consider
using any of three different choices for our conjugate pair of variables.
θ
l
l
α
α
z
φ
x
y
We see that
2α + θ = 180˚ ⇒ α = 90˚−θ / 2 ⇒ φ = 90˚−α = θ / 2
(2a)
z = 2l sin(θ / 2)
(2b)
and
so that
x = z cos φ ≅ lθ and
y = z sin φ ≅ 12 lθ 2
⇒ y ≅ x 2 / 2l
(2c)
assuming small angle θ.
One choice of conjugate variables is θ and the angular momentum L ≡ Iθ̇ ,
KE = 12 Iθ˙ 2 = L2 / 2 I and
PE = mgy ≅ 12 mglθ 2 ,
(3)
while another is x and px ≡ mx˙ ,
KE = 12 Iθ˙ 2 ≅ px2 / 2 m and
PE = mgy ≅ mgx 2 / 2l .
(4)
In obtaining the kinetic energy in Eq. (4), I differentiated Eq. (2c) for x and used I ≡ ml 2 .
However, since
KE = p2 / 2 m = ( px2 + py2 ) / 2 m ,
this is equivalent to putting py ≡ my˙ ≅ 0 . Hence, choosing y and py as the conjugate variables
implies
KE = py2 / 2 m ≅ 0 and
PE = mgy .
(5)
Now applying Eq. (1) to any of Eqs. (3)–(5) in all cases gives the same answer, namely
E = kT
(6)
in agreement with the third application discussed above.
Appendix—Derivation of the Generalized Equipartition Theorem
The average energy is given by
∑ Ese −βE
E = states s
∑ e −βE
s
s
.
(7)
states s
We can convert the summations into integrals by introducing the density of states g,
∑
+∞
→ ∫ g(q, p)dqdp .
states s
(8)
−∞
But according to Heisenberg’s Uncertainty Principle, each quantum state occupies a volume of h
in q–p phase space, so that g = 1/ h . Using Eq. (1b), Eq. (7) thus becomes
∞∞
E = KE + PE =
n −βap n −βbq n
∫ ∫ ap e
00
e
∞∞
∞∞
dpdq + ∫ ∫ bq ne −βap e −βbq dpdq
∫∫e
n
n
00
−βap
n
e
−βbq n
(9)
dpdq
00
where I canceled a factor of 4/h in front of each integral. This can be further simplified to
∞
n −βap n
∫ ap e
E=
0
∞
∫e
∞
dp
+
−βap n
n −βbq
∫ bq e dq
0
n
∞
∫e
dp
0
−βbq n
dq
= β −1
I1
(2 − δ a 0 − δ b 0 )
I2
(10)
0
where I put
∞
n −xn
I1 ≡∫ x e
∞
dx and
0
I2 ≡∫ e − x dx
n
(11)
0
by defining x ≡ (βa)1 / n p in the first ratio of integrals and x ≡ (βb)1 / n q in the second, with the
delta functions allowing for the possibilities that either a or b is zero. We now simply integrate I1
by parts,
n
n
1
u = x ⇒ du = dx and dv = x n −1e − x dx ⇒ v = − e − x
n
n
x
∴ I1 = − e − x
n
∞
0
1
1
+ I2 = I2
n
n
where I used l’Hôpital’s rule to evaluate the term at ∞ with n > 0:
x
y≡
e
since
xn
⇒ ln y = ln x − x n → −∞ as x → ∞
ln x
1/ x
→ x −n → 0 and thus y → 0.
n →
x
nx n −1
Substituting Eq. (12) into Eq. (10) gives Eq. (1a).
(12)