To be precise, shouldn't that be the expected value for that "rate of change"? This reminds me of something that bothers me when they say Kelly maximizes the "bankroll growth rate" for repeated gambles. More precisely, it maximizes the expected value for the bankroll growth rate.

As for lamda not being exactly the probability of nucleus decay per second. It sounds like what it really is, is the limit of probabilities per unit of time as that time goes to 0. i.e.

lamda = lim[t-->0] ( (Pr(decay by time t) / t )

Is that right?

PairTheBoard

The appropriate unit of time is very small relative to the half-life. This is not any different from any other derivative-like situation. Here's how your argument sounds to me:

The speed of a baseball is 100 miles per hour. This means an hour from now, it's going to be 100 miles away. But we know that it won't be, so we've picked a bad definition for speed.

Sure. You can nit it up all you want here and you're right. Technically, it's a rate because it's defined by a derivative.

You must hate it when theoretical physicists set c = 1.

Right this is why i called it dP=λ*dt ie the probability rate (derivative dP/dt=λ).

Also yes on the fractional rate of change being the avg really because obviously you have random fluctuations for small populations but typically in these cases we are talking N>10^15 radioactive systems (nuclei) since Avogadro's number (typically a few grams) is already ~10^24. In that sense in the decay law the number of nuclei left from the initial sample is also always the avg not an exact number since the effect is random anyway. But the error is so unimportant for standard applications.

Your speed argument is instructive, but again you seem to be defining a quantity by its units. If you define speed as distance traveled per unit time then you're just getting an average speed over whichever time interval you choose. If you define it properly as rate of change of distance then it's clear that it's the instantaneous speed and you would have no problems with imagining the baseball flying far away.

The definition of probability per unit time seems to me like it needs something extra like "provided the time interval is small compared to the half life (and maybe even specifying how small).

It's like stating Boyle's law as pressure is inversely proportional to volume and omitting to mention that you need to keep the mass and the temperature constant for it to hold.

And set c=1 all you want, whatever system of (correct) units is convenient is fine, especially if it simplifies things.

I have no problem imagining the baseball flying far away even without defining the velocity as a derivative. I don't see how that changes anything. That's not where the problem lies.

Isn't that implicit in the mere fact that probability is a dimensionless quantity whose value is between 0 and 1, but the decay constant has units of inverse time?

I guess that's an ok heuristic if the unit of time is small but I agree with you that it's bad as a definition. It's a little like saying the integral of a function f from 0 to 1 is f(0).

If I have this right, the probability distribution for the time at which the decay of a nucleus happens is exponential. So its probability density function (pdf) is

f(t) = lambda * exp[ - lambda * t]

So you could say that lamda is the value at time t=0 of the pdf for when the nucleus decays.

or as I said in my first post, lambda is the limit of probabilities per time as the time goes to 0.

lambda = lim[t-->0] ( [Pr(decay by t)] / t)


PairTheBoard

No I don't think so. The very fact that the definition is so loose that it allows you to choose a unit of time means that you can get probabilities greater than 1 so clearly there is something wrong with the definition. It is not precise enough. I think we have to agree to disagree on this one. If you're happy with that definition as it is, I'll leave that down to your professional judgement.

In one sec., the fraction of no. of atoms that got decayed is decay constant, λ. Its equation is dN/dt = -λN

Read above and you will see why that is flawed...

If I'm remembering correctly, doesn't radioactive decay have to do with the Pauli loan principle? The more potential energy that an atom possesses the more likely it is to decay or something like that?

This should help: