Crude Drude model analysis

Here's the question. Between two points, $a$ and $b$, if an object is accelerated between them, it will cross the distance between them in time $t=\sqrt{2\frac{d}{a}}$. So, if the acceleration is increased, the time decreases with the square root of $a$.

If we take the average $t$ for sinusoidally distributed $d$ values, our $\tau$ should be $\tau = \frac{t}{\sqrt{2}}$. Therefore, increasing the $E$ field should decrease the value of $\tau$ by the square root of the increase. Therefore, the current density equation $j=nq^2\frac{\tau}{m}E$ should be scaling up $j$ by the square root of $E$ since we substitute $\tau=\sqrt{\frac{d}{E}}$. This defeats Ohm's Law. I don't want it to though because its seemingly wrong, but I don't know why.

  • $\begingroup$ Some kind of schematic would be nice. This setup is hard to imagine. $\endgroup$
    – KvdLingen
    Jul 15, 2014 at 18:14
  • $\begingroup$ I have TeXified your post, but I must say that it is still hard to parse for me, mainly due to unexplained notation/context: Apart from the title, there is no indication that we are talking about the Drude model of conduction. Are you using the notation from that article? $\endgroup$
    – ACuriousMind
    Jul 15, 2014 at 18:14
  • $\begingroup$ Thanks a lot @ACuriousMind. Yes the distance d between points a and b is modeled after the distance between two positive ions in the metal lattice. $\endgroup$ Jul 15, 2014 at 18:16
  • $\begingroup$ I may attempt a picture. Be advised I'm no art major $\endgroup$ Jul 15, 2014 at 18:17
  • $\begingroup$ I've seen worse pictures ;) Now, I don't really know the Drude model, but I don't think the sinusoidal distribution of $d$ you use is obvious (sorry if it is). $\endgroup$
    – ACuriousMind
    Jul 15, 2014 at 18:26

2 Answers 2


The electric field (or whatever force) has only a very small effect on the time between collisions. The reason is that electrons are very light, and their thermal velocity exceeds any change in the velocity due to the applied fields.
One thus treats the scattering time (tau) as a constant.
The applied electric field then causes a drift velocity, tau*acceleration. Oh, it's assumed that every collision resets the electron velocity randomly... it doesn't remember it's previous velocity.

  • $\begingroup$ The previous section helped me clear up this important concept where drift velocity =slight compared to electron velocity. Thank you, it's very interesting $\endgroup$ Jul 16, 2014 at 15:02
  • $\begingroup$ So how does current affect temperature so pronouncedly if the time between collisions is always on average~the same? $\endgroup$ Jul 16, 2014 at 17:15

That seems right, but due to your unconventional (most of the time, like below, they do it with velocity, not distance) derivation of the electron kinematics (which are, I should point out, just an approximation), I think the source of your problem is that you still have $d$ in your final Ohm's law. If you replaced that by the more usual terms ($\tau$ and $E$), you'll probably get the right relation.

Here is the quick little derivation from p7 of Ashcroft and Mermin:

Common definition of current: $j = nqv_{avg}$

$t$ is time since electron's last collision. In this time it has force $qE$ acting on it, so accelerates at $a=qE/m$, so in time $t$ will gain velocity $v=at=qEt/m$. Plugging this into the above definition for current, $j=nqv=nq^2tE/m$.

(an important little detail they mention is that we assume the electron left its last collision with a random velocity $v_0$, but those all average out among all the electrons. The velocity from $E$ doesn't, though.)

  • $\begingroup$ True. My final equation has a d and an a, but I'm precisely saying that t, the time between collisions, should decrease with an increasing electric field. I'm deriving this (no doubt slightly incorrectly) $\endgroup$ Jul 15, 2014 at 18:35
  • $\begingroup$ It seems that the above equation doesn't allow t to change based on how hard the electrons are being pushed $\endgroup$ Jul 15, 2014 at 18:38
  • 1
    $\begingroup$ @AndresSalas, I think you've stated the Drude model: $\tau$ is independent of $E$. Again, it's an approximation, but I think it works because (in a semiclassical model) the average electron speed is very high and can be estimated from $.5mv^2=1.5k_BT$ and the distance between atoms is (mostly) fixed, so you have $l=v_0 \tau$. $\endgroup$ Jul 15, 2014 at 18:43
  • $\begingroup$ So you're saying that tau is independent of E? How can this be? If I push air through a tube with a lattice of internal sticks, the average time between air and stick collisions is certainly dependent on the hardness of the push, otherwise in the limit of zero pushing there wouldn't be less collisions than at the time when some pushing occurs. But of course, blowing air causes more collisions. $\endgroup$ Jul 15, 2014 at 18:46
  • $\begingroup$ Sorry, didn't catch your edit, I see I've opened up a whole can of worms. The actual electron speed is very high, so the difference between normal electrons and current is slight. It's hard to say that there were collisions beforehand though for the simple reason that the electrons were in the ground state. How can you impart momentum in the ground state? You can't. Thanks for the added complexity though. $\endgroup$ Jul 15, 2014 at 18:50

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