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Referring to the Drude Model, I've seen a lot of excellent questions on whether $\tau$ should be thought of as the "average time between collisions" or the "average time until the next collision", and whether or not those definitions are the same or should differ by a factor of $2$.

Some examples:

Definition of mean free time in the Drude model

Drift velocity in Drude model


As much as I still don't understand the answer to the above question, this is NOT what I wish to ask here. Here, my question has completely to do with how, hypothetically, this "average time" should be calculated.

Calculation 1:

Say we have a piece of metal, and we can somehow keep track of every single electron within that metal.

For a period of time, we count the number of times each electron collides with an atom. Different electrons will have collided a different number of times.

Then, for each of the electrons, we divide the number of collisions it went through by the time that we waited for. That’ll be the average number of collisions per unit of time for a single electron.

To get the average collisions per unit of time over all the electrons, we would add up all the average collisions per unit of time for each of the electrons, and divide by the number of electrons. Then, to get the average time between collisions, $\tau$ we would take the reciprocal of that.

For example:

Say Bob and Mary are the only two electrons in a material. Yep, the electrons have names - whatever.

We time for $20$ seconds and find that Bob collided $300$ times while Mary collides $260$ times. That means that Bob is colliding at an average of $15$ collisions per second, while Mary is colliding at an average of $13$ collisions per second.

To get the average number of collisions per second, we add 15+13 and divide by 2, the number of electrons. $ \frac{15+13}{2} =14$.

In other words, on average, an electron in this material (the only electrons are Bob and Mary) collide $14$ times per second.

The average time between collisions, $\tau$, would then be $\frac{1}{14}$ of a second.

We could’ve done this same calculation faster like so:

Take the total amount of time each electron traveled for over all electrons and divide by the total number of collisions.

Bob travelled for 20 seconds and collided 300 times, while Mary also travelled for 20 seconds and collided 260 times.

$ \frac{2(20)}{300+260} = \frac{1}{14}$

Calculation 2:

$\tau$ could also be calculated by first calculating the average time between collisions for each of the electrons, and then taking the average over all the electrons:

We take the same situation as above - find that Bob collides an average of $15$ times per second, while Mary collides an average of 13 times per second.

That means the average time between collisions for Bob is $\frac{1}{15}$ of a second while the average time between collisions for Mary is $\frac{1}{13}$ of a second. To get the average time between collisions over all electrons, we would do $\frac{(\frac{1}{15} + \frac{1}{13})}{2}$, the sum of the average times for each electron divided by the number of electrons. That gives us $\frac{14}{195}$ as the average time between collisions.

.....

In summary, I’m confused whether an average time between collisions means we sum the average times for all the electrons and divide by the number of electrons (Calculation 2), or we calculate the average number of collisions per second for each electron and then sum over all electrons and take the reciprocal of that (Calculation 1).

Both calculations seem correct, but they’re giving me slightly different answers.

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You need to find the average not the average of the averages.

You might have met this idea when doing a kinematic problem to find the average speed after having travelled a total of 100 kilometre spending 6 hours for the first half of the trip and then 4 hours for the remaining half.

In your case the “distance” is the time and the “time” is the number of collisions.

So the average time between collisions is the total time divided by the total number of collisions which is what you did for your first calculation.

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  • $\begingroup$ Thanks! Ha, I feel bad that I spent so much time writing such a long question (and you had to spend so long reading) for such a simple answer, but I really do appreciate it. Now on to understanding the first part...!!! $\endgroup$ – Joshua Ronis Mar 16 at 18:13

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