Poisson Distribution: In Distribution: In the Drude Drude model the probability probability of an electron electron suffering suffering a collision collision in any infinitesimal interval dt is just dt / τ , , P r(collide ∈ [t , t + dt ]) = (1 / τ )dt
[I.1]
(a) Show that an electron picked at random at a given moment had no collision during the preceding t ′ seconds with probability e −t / . Notice the “prime” I put on the “preceding t ′ seconds”. Also, let dt = lim n→∞ ∆nt . Then, τ
e
− t / τ
= lim(1 + n→ ∞
=
− t / τ n
n
n
n
) = lim(1 − [ dt / τ ]) = lim(1 − Pr(collide ∈[ t, t + dt])) = (Pr( not collide ∈[ t, t + dt])) n →∞
n →∞
Pr( not collide ∈ [0, 0 + t ′])
n
[I.2]
Show that it will have no collision during the next t seconds with the same probability. n
Pr( not collide ∈ [t ′, t ′ + t ]) = lim (1 − Pr(collide ∈ [t ′, t ′ + dt ])) = lim(1 + n→ ∞
n →∞
/ τ n − t
) = e
n
− t / τ
[I.3]
b) Show that the probability that the time interval between two successive collisions of an electron falls in the range between t and t + dt is ( dt / τ )e − t / . Because of the independence of velocity from collisions, the act of travelling for some certain time without a collision and the event of collision itself are independent events. Therefore, travel for t seconds collision in / − t [I.4] Pr = Pr P r × = e × ( dt / τ ) ′ w i t h o u t t t t d t a c o l l i s i o n t i m e [ , ] ∈ + τ
τ
(c) Show as a consequence of (a) that at any moment the mean time back to the last collision (or up to the next collision) averaged over all electrons is τ .
The time back to the last collision has the probability distribution [I.2], and the time until the next collision similarly has the probability distribution [I.3]; computing the mean time,
∫ t ≡ ∫
∞
te − t / dt dt τ
0 ∞
e
0
− t /τ
τ
2
∫
∞
0
=
dt
τ
(t / τ )e − t / d (t / τ ) τ
∫
∞
0
e
=
2 τ
0!
τ
−(0 − 1)
− t / τ
d (t / τ )
=
[I.5]
τ
d) Show as a consequence of (b) that the mean time between successive collisions of an electron is
τ
.
Finding the first moment of the probability distribution [I.4], the average, t =
∫
∞
0
t ′ Pr =
∫
∞
0
t ′e
′ τ − t /
d (t ′ / τ ) = τ
∫
∞
0
−x
xe dx = τ ⋅ 0 ! =
τ
[I.6]
(e) Part (c) implies that at any moment the time T between the last and next collision averaged over all electrons is 2τ . Explain why this is not inconsistent with the result in (d). (a thorough explanation should include a derivation of the probability distribution for T). Qualitative reasoning: Notice that the electron’s trajectory, with collisions, appears as,
[I.7] Consider each X in [I.7] to denote a collision. Notice that probabilistic “bins” are formed, then. To pick an electron from all electrons is to pick from one of these bins. But that means you are more likely to pick a wider bin. Thus, your selection of a random electron is biased towards longer-T’s between collisions. Quantitative reasoning: to put this more rigourously, consider the out-journey and in-journey of the electron as independent events. Then, the probability distribution for T is found from the “product rule”,
Pr no collision Pr no collision −2 t / × =e T = T (t ) = in last t seconds in next t seconds
τ
→ t
∫ = ∫
∞
0
∫
0
τ
τ
τ
∫
T
0
Te −T /
τ
exp
− t ′+ ( t ′− T ) τ
dt ′ =
τ
2
→ T
≡
∫
t
∞
0
Crap! The answer is, T p ( t ′) p (T − t ′) 1 P (T ) = dt ′ = 2
te −2 / dt e
2
τ
−2 t /τ
dt
τ
=
4
∞
∫ ∫
∞
0
T ⋅ P (T )dT =
∞
0
∫
t τ
τ
τ
2
e −2 / d ( 2 t )
2t
0
e
τ
− 2 t /τ
d( )
τ
2
[I.8]
τ
Te −T /
τ
∞
0
2 t
=
T
τ
2
dT = 2τ [I.9]
A failure to appreciate this subtlety led Drude to a conductivity only half of Eq. (1.6). He did not make the same mistake in the thermal conductivity, whence the factor of two in his calculating the Lorenz number (see p. 23).
