Download 1 mean velocity of electrons

1
mean velocity of electrons
The question is like this: Consider the classical electrons in solid under a
electrostatic field E, suppose the relaxation time is τ and after the electrons
collide with the irons, the velocity becomes random and the mean is zero.
To be convenient, we can remove the random part and consider the velocity
caused the by electronic field. The model becomes this: the velocity of
electron becomes zero after colliding with irons (we omit the collision between
electrons), and the relaxation time is τ . When the mean velocity won’t
change any more, what’s the mean velocity?
1.1
One unclear method
I saw one method in the lecture of Phys. 551 online and relevant books
looks like this (To be convenient, I’ll omit the negative sign below):
The force is F = eE and after time t, it collides with an iron, so the
increment of the velocity is v = eEt/m. Take the average or replace t with
the relaxation time τ , we get v = −eEτ /m
I was confused here. Even taking the average, this average is just the
average of the terminal velocity. If we calculate the whole average, we could
get v = eEτ /(2m). But, every book and my lecture before says the original
one is correct. Where does the problem lie?
I calculated first using the following stupid model:
1.2
One wrong model
Assume every electron collides with the iron after exactly τ , then no
matter what the initial distribution is, the final state would be that the
velocity is uniformly distributed in [0, eEτ /m]. The average is one half of
the previous one v = eEτ /(2m). I thus got into confusion now.
1.3
One correct method
I found one method in my old lecture:
In an infinitesimal time dt, the probability of collision is dt/τ . Then, the
average momentum p(t) of one electron satisfies p(t + dt) = p(t) + eEdt −
(dt/τ )p(t). If the state is steady, the average momentum won’t change and
thus p(t + dt) = p(t). We have p(t) = eEτ and thus v = eEτ /m
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Even though I agree that this method is correct, how can we solve the
confusion about the first method and what’s the problem of the the second
model?
Using this correct method, I even got a more stupid result:
If the probability of collision with dt is dt/τ (this holds only if dt is
infinitesimal), the density function of collision time is 1/τ . The distribution
is uniform distribution.(*)
The the average collision time is τ /2. Then the average velocity is
v = eEτ /m = 2eEt/m. How can this be?!!–the maximum increment is half
of this!
1.4
Another correct and convincing method
At this point, I consider the problem like this: Fix time t, and consider
what electrons can come to this time point. However, we don’t know the
time point the last collisions happens and t may not be the time point where
the collision happens. This means we don’t know the distribution of the time
from the last collision till now even though we know the distribution of the
time needed for a collision to happen.
Then, I realized that we can use the equation for distribution. Suppose
the distribution is steady and is independent of position. f (v) is the distribution. The acceleration is a = eE/m. f (v) = f (v − adt) − f (v)dt/τ
if v > 0. Here, we made an assumption that the possibility of collision is
independent
of Rv. To conserve the whole quantity of electrons, we
R ∞must have
R adt
∞
1
dt
dv.
Which
is
equivalent
to
af
(0)
=
f
(v)dv
=
f
(v)
τ
0
0
R ∞ 0 f (v) τ dv.
And the definition of density says we should have F (0) + 0 f (v)dv = 1.
Finally, we get f (v) = aτ1 exp(−v/aτ ), F (0) = 0. The mean velocity is
aτ . This is actually the same with the above method. This is actually the
simplest Boltzmann equation
Let’s move on to the distribution of time where the first collision happens. Suppose p(t) is the probability that the electron still exists at time t,
and then, we have p(t + dt) = p(t)(1 − dt/τ ). The solution is p(t) = e−t/τ .
This is the exponential distribution instead of the uniform distribution. (*)
is wrong! The mean time is τ .
Having this result, we can check why the above methods fail.
Firstly, even though the mean increment is eEτ /m, the time used is
different. The initial velocity is also different. We can’t simply take the
average. Consider the distribution of f (v) and the evolution, and we can see
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how the confusion in the first part can be solved. The stupid model is wrong
because the collision can happen any time.
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Debye’s model
We can know the density of state function for 3D phonon is D(ω) =
Rω
In Debye’s model, ω = qv. We want to determine 0 D D(ω)dω.
The right hand side is the total number of modes. It should be equal to N
where N is the number of cells, because theRtotal number of q in the first BZ
ω
is N. We have three acoustic waves. Thus, 0 D 3D(ω)dω = 3N .
My question is that I saw the total modes should be equal to the degree
of freedom, which is 3N 0 where N 0 is the total number of atoms instead of
cells. What if there are more than one atoms in one cell?
In Debye’s model, N should still be the number of cells. What we omit
is the number of optical modes. If we count this in, then the whole q should
have the number 3N 0 . In the limit T → ∞, the Cv should go to 3N 0 kB and
the optical modes have contributed. Usually, in normal distribution, only
the acoustic modes contribute.
V q2
.
2π 2 dω/dq
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