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I have been following Feynman lectures on Physics, lecture 43 — Diffusion.

There he derives the average value of time till next collision over all molecules. He finds it to be equal to the average time between collisions $\tau$. This concept is pretty clear.

Then he goes on to derive the drift velocity of molecules (and in my case of interest, electrons). Here he states that the average time between collisions also to be equal to the average number of collisions per second. My doubt lies here:

I'm not able to mathematically prove how the average value of time since last collision to be equal to the average time between collisions.

I checked up some online resources too, none of them explain this properly.

How can it be proved mathematically?

Edit: Another question cropped up in this same topic. The author derives the probability that the molecule survives t seconds without collision is

$$P(t)=\exp\left(\frac{-t}{\tau}\right)$$

Since all these events are independent shouldn't the integral of this probability function from $t=o$ to infinity should be equal to $1$? But if we evaluate this integral we will actually get $\tau$ as the answer. Isn't this a violation of the axioms of probability?

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  • $\begingroup$ What makes you think that the event of a molecule surviving from one moment to the next is independent of all of the other previous events/moments? You can't "un-collide" the molecule; once the molecule fails to survive without collision during a particular moment, it also automatically fails to survive without collision for all future moments, since it has already done so. $\endgroup$ – probably_someone Jan 22 at 0:14
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Here he states that the average value of time since last collisions also to be equal to the average number of collisions per second.

I cannot find such a statement in the reference that you have given.

If the average time between collisions is 0.1 seconds then the average number of collisions per second is $\dfrac{1}{0.1} = 10$

The author derives the probability that the molecule survives t seconds without collision is P(t)=exp(-t/τ).

Your integration should not equal 1.

This equation gives you the probability of something happening.
In this case no collisions occurring after a time $t$.
At $t$ increases the probability of such an event decreases which is what one would expect.

This is equivalent to the probability of a radioactive nucleus with decay constant $\lambda$ not decaying in a time $t$.

$P(t) = \exp(-\lambda t)$

From the probability function one can find the average time before a collision $<t>$.

$<t> = \dfrac{\int ^\infty_0 t \; \exp(t/\tau)\; dt}{\int ^\infty_0 \; \exp(t/\tau)\; dt} = \tau$

which in the nuclear context was the average lifetime of a radioactive nucleus with $<t> = \dfrac 1 \lambda$.

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  • $\begingroup$ The statement is found in the last lines of the paragraph before equation 43.13 $\endgroup$ – Danny Sep 14 '16 at 9:17
  • $\begingroup$ Feynman write *"Now the average time since the last collision must be the same as the average time until the next collision, . . . " $\endgroup$ – Farcher Sep 14 '16 at 9:30
  • $\begingroup$ Which in turn is equal to $\tau$ . Which is the average number of collisions per second $\endgroup$ – Danny Sep 14 '16 at 11:28
  • $\begingroup$ Where is it stated that tau is the average number of collisions per seconds which would mean that tau would have the units of $\text{time}^{-1}$ whereas tau actually has the units of $\text{time}$? $\endgroup$ – Farcher Sep 14 '16 at 11:46
  • $\begingroup$ Damn , how couldn't I have though about that? I'm really sorry.Its the average time between collisions. Btw how do we prove that the average time since last collision also equals tau . Thanks for your patience! $\endgroup$ – Danny Sep 14 '16 at 12:04

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