The derivation of drift velocity in case of electrons is equivalent to the case of an charged ionic gas and therefore all the arguments also apply there. Now for an ideal "ionic" gas which interacts with each others only as perfect spheres would (for arguments sake), when we apply an external electric field, they would just show drift, that means that they "favour" one direction more than the others where drift velocity is given by $v_d=(eE/m)\tau$ where $\tau$ is the average relaxation time. Hence the net velocities of the electrons wont vectorially add to zero, and moreover the speed as given by kinetic theory of matter would not depend only on the temperature but also depend on electric field. Does this not contradict two fundamental postulates of kinetic theory?

Moreover in case of electrons in a circuit with many series resistors, the relaxation time would be different. The resistivity as defined by $E=\rho J$ and $\rho=m/(ne^2\tau)$ where j=current density, n=number of electrons per unit volume and e=electronic charge, should be different for all the series resistors. But then will not the drift velocity be different for all the resistors if the top equation is correct since the resistivities might defer, causing the relaxation time to differ. Doesn't this contradict the postulation of a steady current which requires constant drift velocity. Moreover suppose we use the same material of different cross section, since there is no area dependence in the drift velocity, it shouldn't change but then again we don't get a steady current as J remains a constant and I is simply J times A.

  • $\begingroup$ I learnt the derivation without considering the electrons as an ideal gas. Do you want me to post it. $\endgroup$
    – udiboy1209
    Commented Jul 1, 2013 at 6:55
  • $\begingroup$ Well, post it and let us see if it helps. $\endgroup$ Commented Jul 1, 2013 at 17:05

1 Answer 1


Alternate method to derive drift velocity:

Consider a field $\vec{E}$ inside the conductor. Using equations of motion we can say that for every charge inside the conductor, $$\vec{v_1}=\vec{u_1}+\frac{\vec{E}e}{m}t_1$$ $$\vec{v_2}=\vec{u_2}+\frac{\vec{E}e}{m}t_2$$ $$.$$ $$.$$ $$.$$ $$\vec{v_n}=\vec{u_n}+\frac{\vec{E}e}{m}t_n$$
where $t_1,t_2,...t_n$ are the times until each of them collide with another particle respectively.

Summing them and dividing by the number of charge particles($N$) we get, $$\sum_{i=1}^n\frac{\vec{v_i}}{N}=\sum_{i=1}^n\frac{\vec{u_i}}{N}+\frac{\vec{E}e}{m}\sum_{i=1}^n \frac {t_i}{N}$$

We can say that $$\sum_{i=1}^n\vec{u_i}=0$$ as the charges initially are in perfect random motion. Thus substituting $\sum_{i=1}^n\frac{\vec{v_i}}{N}$ as $\vec{v_d}$, the drift velocity and $\sum_{i=1}^n \frac {t_i}{N}$ as $\tau$, the relaxation time for each particle we get $$\vec{v_d} = \frac{\vec{E}e}{m}\tau$$

  • $\begingroup$ Thanks.This is pretty nice. That resolves the question cited in the first paragraph of my question. But what about the drift velocity in case of different circuit elements in series? $\endgroup$ Commented Jul 4, 2013 at 3:47
  • $\begingroup$ You are saying that every component in the series will have different relaxation times, and thus different drift velocities, leading to different currents through them. But this is all assuming the number of charge carriers($n$) remain the same. Maybe the number of charge carriers in each component changes, leading to a constant current through all the components in series, even if they have different drift velocities. $\endgroup$
    – udiboy1209
    Commented Jul 4, 2013 at 9:05
  • $\begingroup$ I know this is an old post, but wouldn't the drift velocity need to get divided by two? It's supposed to represent average velocity in the direction of motion, right? For example, if I accelerate from 0 to 10 meters per second with a constant acceleration, my average velocity in that time interval won't be 10 meters per second; it'll be 5 meters per second. $\endgroup$ Commented Mar 16, 2019 at 9:53
  • 1
    $\begingroup$ @JoshuaRonis We aren't calculating the average velocity in time for each particle. If that was the case, then there would be a factor of two. We are calculating the final velocity of each particle and averaging for all particles. The average final velocity of all electrons before impact constitutes drift velocity. $\endgroup$
    – udiboy1209
    Commented Mar 17, 2019 at 8:33

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