Exponential growth or decay of neutrons in fission I am considering the derivation for the number of neutrons as a function of time, in a fission chain reaction.
In the derivation I am following, let $\tau$ be the time interval between fission events, $q$ be the probability that a neutron induces a fission event, and $\nu$ be the number of neutrons released per fission. It is then said that the average number of neutrons released by a newly created neutron in time $\tau$ is $q \nu - 1$. I don't understand why this is. My reasoning is as follows:
Let's 'follow' a particular neutron. In every timestep $\tau$, we have a probability $q$ that $\nu$ new neutrons will be created due to this neutron, and so the net gain in neutrons would be $\nu - 1$, since we have lost the original neutron we were following. The probability that the neutron won't induce a fission is $1 - q$, and so the probability that the net gain in neutrons is zero is $1 - q$. So, the expected value per $\tau$ of neutron gain due to a given neutron is $g = q(\nu - 1) + (1 - q) 0 = q(\nu - 1)$. This is different from the expression in the derivation I have seen.
What am I missing in my reasoning? Thanks!
 A: I think there might be a semantics problem here. If $\tau$ is an average time, and not the time for "step", then the formula makes sense. So start out by ignoring $\tau$:
There is 1 neutron. It has a probability of $q$ of inducing fission. What goes into $q$ is everything: fission cross sections weighted by neutron energy which includes moderation and any other processes that change the energy spectrum. It also includes engineering factors like absorption, escape, and reflection back into the system.
If a neutron does induce fission it releases, on average $\nu$ new neutrons, so the expectation per neutron is $q\nu$...
...but the origin neutron is gone, so we are left with $q\nu-1$ in the end.
Now we can ask: what is the timescale? $\tau$. Fission events will then be (roughly) exponentially distributed in time, with bounds on both ends: at the low end, two U (or Pu, etc) nuclei can only be so close to each other. On the top end, things like energetic disassembly of the system factor in.
A: Your model does not explicitly allow for escape, capture without fission or other losses so any $\nu>1$ should increase the number of neutrons, however slowly.  
Thus take $\nu=2$ but $q=1/1000$.  This should lead to very slow growth, i.e. you have to wait 1000 steps to get, on average, $1$ extra neutron.  
Your expression correctly predicts a small increase in neutrons but the original expression predicts a significant decrease in neutrons for every time step.  Indeed it is easy to see that, as soon as $q\nu<1$ there would be no new neutrons released as per $q\nu-1$ even if $\nu>1$, i.e. even if each reaction - however improbable - results in a net increase in the number of neutrons.
