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Solid state physics
lectures
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!72
rework the drude derivation
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rework the drude derivation
drude_better_calc
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Anton Akhmerov
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drude_better_calc
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5 years ago
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maybe that works better
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5 years ago
by
Anton Akhmerov
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@@ -135,20 +135,22 @@ Our goal is then to compute the *average* velocity.
happens with each individual element.
**
Let us compute how the average velocity changes with time.
The equation with the Lorentz force we just average right away:
Consider the effect that scattering has over a small time $dt$.
A fraction $dt/τ$ of the electrons scatters, and that their average velocity becomes zero.
The rest of the electrons $(1 - dt/τ)$ accelerates by the Lorentz force, and after $dt$ their velocity becomes
$$
m
\
f
rac{d⟨
\m
athbf{v}⟩}{dt} = -e
\l
eft
(
\m
athbf{E}
+⟨
\m
athbf{v}
⟩×
\m
athbf{B}
\r
ight
).
m
\
m
athbf{v}(t + dt) - m
\m
athbf{v}(t) = - dt e
(
\m
athbf{E}
+
\m
athbf{v}
×
\m
athbf{B}).
$$
Almost there, but we still need to do something with the change of the average velocity due to scattering.
Consider the effect that scattering has over a small time $dt$.
Most electrons continue with the same velocity, however a fraction $dt/τ$ will scatter, and that their average velocity becomes zero.
Therefore we get
Combining the two groups of particles, we get
$$
⟨
\m
athbf{v}(t+dt)⟩ = ⟨
\m
athbf{v}(t)⟩(1 - dt/τ) + 0⋅(dt/τ) ⇒
\f
rac{d⟨
\m
athbf{v}⟩}{dt} = -
\f
rac{⟨
\m
athbf{v}⟩}{τ}.
\b
egin{align}
m⟨
\m
athbf{v}(t+dt)⟩ &= (m⟨
\m
athbf{v}(t)⟩ - dt e (
\m
athbf{E} +
\m
athbf{v} ×
\m
athbf{B}))(1 - dt/τ) + 0⋅(dt/τ)
\\
&= m⟨
\m
athbf{v}(t)⟩ - dt [e (
\m
athbf{E} +
\m
athbf{v} ×
\m
athbf{B}) - m⟨
\m
athbf{v}(t)⟩/τ]
\\
&+ dt²
[
e (\mathbf{E} + \mathbf{v} × \mathbf{B}) m⟨\mathbf{v}(t)⟩/τ
\end{align}
$$
That's it!
We now combine both contributions into a single equation and get
We now neglect the term proportional to $dt²$ (it vanishes faster when $dt → ∞$).
Finally, we recognize that $(⟨\mathbf{v}(t+dt)⟩ - (⟨\mathbf{v}(t)⟩)/dt = d⟨\mathbf{v}(t)⟩)/dt$, and arrive to
$$
m\frac{d⟨\mathbf{v}⟩}{dt} = -m\frac{⟨\mathbf{v}⟩}{τ} -e\left(\mathbf{E}+⟨\mathbf{v}⟩×\mathbf{B}\right).
$$
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