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Using the Drude Model, we get that $$\vec{J}=\frac{ne^2 \tau}{m_e}\vec{E}$$ where $\tau$ is the mean free time of each electron and $n$ is the free electron density. This is simply ohms law and hence the Drude model provides a microscopic explanation for Ohms Law. But vital in this formula is the fact that $\tau$ is independent of $\vec{E}$. But this assumption seems to tear the logic of the whole model into shreds. Just as a starter, If $\tau$ is independent of $\vec{E}$, then an increase in $\vec{E}$ will not change the amount of collisions that occur per unit time (because the amount of collisions per unit time is exclusively a function of $\tau$ and $n$). The same amount of collisions per unit time means that the amount of energy dissipated in the wire/resistor should be independent of the electric field. This is obviously absurd though.

Secondly, the fact that the Drude model predicts Ohms law means that it also shows that the energy dissipated in the collisions (within a resistor) is proportional to $E^2$ on average (since $P=V\cdot I=V^2/R$). Now if we examine the situation of a single electron in a vacuum in a constant electric field $\vec{E}$ and suppose that at time $t=0$ it has zero kinetic energy($K(0)=0$), then at time $t=t_f$, it will have a kinetic energy equal to $$K(t)=\frac{1}{2m_e}(eEt_f)^2$$ From this we can see that for a given time interval $\Delta t=t_f-0$, the kinetic energy gained by the electron (that is, the work done on the electron) is directly proportional to the square of the electric field. The reason for this quadratic dependence is that if we double $\vec{E}$, then we necessarily also double the distance that the force acts over within a given time interval because the particle moves faster and hence further in that time interval. Thus doubling $\vec{E}$ leads to 4 times the work for a constant time interval. However, if we now go back to the drude model, the mean free time and the mean free distance are assumed to be independent of $\vec{E}$ and only dependent on T. So if we do in fact have a constant $\tau$ and a constant mean free distance that are both independent of $\vec{E}$, then doubling the electric field strength cannot possibly quadruple the power because the electric force is still being exerted over the same distance in any given time interval. The only way that we can explain the $E^2$ power dissipation in a resistor by the Drude model seems to be that we must assume that the mean free distance increases in direct proportion with the electric field. But no reputable sources seem to indicate this. For example, in Ashcroft and Mermin, they assume that that the mean free distance is solely determined by the temperature (the average thermal speed). So how can the Drude model make any sense whatsoever given what I've just described?

Any help on this issue would be most appreciated as this issue has been driving me mad!

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  • $\begingroup$ in your Eq.1, the proportionality coefficient between the current and the field is called conductivity ($\sigma$). Do you mean that conductivity must depend on the electric field? $\endgroup$
    – sleepy
    Commented Mar 29, 2021 at 20:01
  • $\begingroup$ @sleepy yes I am aware that it is called conductivity. In the drude model the conductivity is dependent on $n$, $ m_e$, $e$ and $\tau$. Obviously the first three of these are independent of E. But it is not obvious at all that $\tau$ should be independent of E. For if we increase E , then we increase the velocities of the electrons meaning they should travel their mean free distances in a shorter amount of time. Thus $\tau$ should decrease with increasing E if we are to believe the drude model $\endgroup$ Commented Mar 29, 2021 at 20:25

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Good catch! You've identified the fundamental problem with the Drude model: you get different assumptions if you assume the electrons travel a constant distance before colliding, and if you assume they travel a constant time.

If you assume a constant time $\tau$, then the average velocity of the electrons is $v \propto E \tau \propto E$, which is the expected result. But if you assume a constant distance $\ell$, then the typical speed $v$ of the electrons obeys $mv^2 \sim E \ell$ which means $v \propto \sqrt{E}$, a completely different result! But of course, if we think in terms of what causes collisions in the first place, a constant distance sounds much more reasonable than a constant time, which means the Drude model doesn't seem to make sense.

The reason solid state textbooks don't mention this subtlety is because the Drude model is not remotely true anyway; it's just a stepping stone to learn the more accurate Sommerfeld model. In the Sommerfeld model, the conducting electrons have extremely high speeds because of the Pauli exclusion principle, even when the electric field is off. This means that the electric field has little effect on the typical speed, so a collision per constant distance traveled is the same as a collision per constant time, and there's no need to distinguish between the two.

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  • $\begingroup$ Thanks for the great response! Just a couple questions: First, in the Drude model, we already do make the assumption that the typical speed of an electron is very fast $v_{th}=\sqrt{3 \cdot k_B T/m_e}\approx 10^5 m/s $ and hence the field does not significantly alter the speed or distance traveled per unit time. So even if we double the electric field, $\tau$ and $l$ are roughly constant according to the drude model. A constant $\tau$ implies that doubling $\vec{E}$ should not affect the amount of collisions per sec. A constant $l$ implies that doubling $\vec{E}$ should only double ... $\endgroup$ Commented Mar 30, 2021 at 6:23
  • $\begingroup$ the energy dissipated per collision because $E_{energy}=F\cdot \Delta x =qE_{field}\cdot l \Rightarrow 2E_{energy} = 2E_{field}\cdot l$. Thus, according to the drude model, doubling the field should only double the power dissipated in a resistor. Yet from ohms law (which the drude model predicts), we get that $P_{dissipated} =V^2/R \propto E^2$. This leads me to think that there is a deep inconsistency in drudes model that few people seem to understand ( its more likely I am the one who does not understand). $\endgroup$ Commented Mar 30, 2021 at 6:26
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    $\begingroup$ @SalahTheGoat The problem with your reasoning is evaluating $\Delta x$. Because there is a dot product, this quantity is not the distance traveled between collisions. It is instead the average displacement when you average over all possible directions of the initial velocity. $\endgroup$
    – knzhou
    Commented Mar 30, 2021 at 6:40
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    $\begingroup$ For example, suppose there is no force. An electron can go either $l$ to the right or $l$ to the left before colliding. Now if we turn on a force, an electron can go either $l + \Delta l$ to the right, or $l - \Delta l$ to the left. The average energy dissipated per collision is not $F l$, it's $F \Delta l$. Since $\Delta l \propto E$ you get the quadratic scaling of power. $\endgroup$
    – knzhou
    Commented Mar 30, 2021 at 6:41
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    $\begingroup$ @SalahTheGoat No problem, I'm just passing on wisdom I got from this site when I was in your shoes! $\endgroup$
    – knzhou
    Commented Mar 30, 2021 at 6:46

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