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Exponential decay

  1. Feb 15, 2008 #1
    Somewhere the connection is not being made. I have seen all the analogies (flipping pennies, popcorn, etc) and know all the equations.

    What is the simplest self-contained explanation for why radioactive (ie random) decay is exponential, rather than linear, for example? How do you translate saying that "something has an equal probability of occurring at any given time" into an exponential curve on a graph, and why is that curve exponential rather than linear? I understand that "the amount of decay events will reduce", as will "the amount of material that's left to decay", but how do those phrases translate into a non-linear decay over time.

  2. jcsd
  3. Feb 15, 2008 #2
    Well a non formal answer would be is if you solve the differential equation


    Then you will get a decaying exponential.

    Last edited: Feb 15, 2008
  4. Feb 16, 2008 #3
    My visualization for exponential decay consists of discrete, quasi-continuous steps where the amount of radioactive material determines the amount of decay determines the amount of radioactive material determines the amount of decay, etc..., approximating John's formula, above.
  5. Feb 16, 2008 #4
    The ensemble mean though should be described pretty closely by the above equation.
  6. Feb 16, 2008 #5
    I think I figured out the best way to think about it. If I have two carbon-14 atoms, with a half-life of 5700 years, it doesn't mean that it will take 5700 years for one of the atoms to finally decay. It means that on average it will take that long. Maybe in my specific case, it takes 10 seconds, or 10 years, or 100 years. But on average it would take 5700 years.

    So what does "on average" mean? Suppose that I had 100 jars sitting on a table, and put two carbon-14 atoms into each jar. Now, if I measure how long it took for each jar to lose one of its atoms (reducing the number of atoms in that jar by 50%), I would find a lot of different results. In jar A it took 10 years, in jar B it took 12000 years. If I averaged all of those years together, I would get about 5700.

    Now suppose instead that I took all of those 100 jars, each containing two atoms, and poured all of those jars into a bucket, filling the bucket with 200 carbon-14 atoms. The situation must be the same, correct? So, if I sat there and waited for 100 of those atoms to decay, it would be like averaging the results of the previous example, I would get about 5700 years.

    The same would be true no matter how many pairs of carbon-14 atoms I added to that bucket.
  7. Feb 16, 2008 #6
    How about:

    You have 128 coins.
    Halve them you get 64.
    Have them again you get 32.
    Have them again you get 16.
    Have again you get 8. etc etc

    129 to 64 to 32 to 16 to 8 is not linear. It is decreasing exponential.

  8. Feb 16, 2008 #7
  9. Feb 16, 2008 #8
    Yes, as I said, I've seen that explanation, and it is probably the best, but how do you go from flipping coins (50-50 chance of getting heads/tails) to segments of time each of which have equal probability of resulting in a given decay. I guess you could say that the half-life point is the time when the probability of decay equals the probability of flipping a coin and getting heads, but that adds a level of disconnect for anyone trying to understand it.
  10. Feb 16, 2008 #9
    As I said, I already know all of these equations better than I'd like to say. I've taken several courses on every subject in this forum. It was just never explained with words in a fundamentally-convincing way, and I was never willing to accept equations as an understandable explanation.
  11. Feb 16, 2008 #10
    Well, if you understand how all these things are derived then I'm not sure what you are looking for. My next step would be to take expectations of each side of some stochastic differential equation in order to get something close if not exactly to the differential equation above.
  12. Feb 16, 2008 #11


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    Staff: Mentor

    Chris didn't say anything about flipping the coins, but your explanation actually works better: Mathematically, it's exactly the same thing as decay. In each timeslice you have a 50% chance of decay just like for each coin flip you have a 50% chance of getting heads. You can start with a handful of coins and every time you throw all of them, you separate out the ones that came up heads, cutting the size of the pile in half and keeping the random element in there as well.

    I guess I'm not seeing the problem here either. It sounds like you understand this issue just fine. The thing is, people are explaning the concept of exponential decay to you - are you instead asking why, physically, radioactive elements don't decay linearly? Physically, that wouldn't make any logical sense: each element would need it's own internal clock and they'd all have to be scheduled in such a way as to distribute the decay linearly. You can't have a random element in such a model.
    Last edited: Feb 16, 2008
  13. Feb 16, 2008 #12
    You need to know the definition of Euler's constant to understand this problem:

    e=(lim n->infinity) (1+1/n)^n

    of equivalently 1/e=(lim n->infinity) (1-1/n)^n

    Now the connection of exp function and particle decay is obvious:

    1/n of the existing particles decay in a time dt=lifetime/n, so
    their number multiplies by (1-1/n) on every step dt, causing the number of particles after
    t=lifetime to fall on 1/e of the initial number.
    Last edited: Feb 16, 2008
  14. Feb 16, 2008 #13


    Staff: Mentor

    Do you understand compound interest and how, in the limit of instantaneously compounding interest, it reduces to exponential growth? If you understand that then exponential decay is just like instantaneously compounded (negative) interest.
  15. Feb 16, 2008 #14
    Thanks for all the help. This is definitely much clearer to me.
  16. Feb 16, 2008 #15
    You have 100 atoms with a half-life of 10 years.

    Of course that means that only 50 will decay in 10 years.

    Though don't you think it should be more likely that the closer you get to the ten years the more will decay?

    Do you think you have more of a chance of 4 or 5 atoms decaying in the first day or the 1753 day?

    The closer you get to the endpoint, the more of a chance they will decay.

    That is exponential.
  17. Feb 16, 2008 #16


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    even so, i'll stab at it and twist my knife:

    let's say that you have this parameter, [itex]\tau[/itex], (that we happen to call a "half-life", but we could call it a "farg" or "foo", doesn't matter). in a period of time [itex]\tau[/itex], for every unstable atom, there is a fixed probability (in this case 50% probability, but it could be some other fixed number between 0 and 1) that this atom will fall apart (decay). so, if you have ten zillion of these atoms, in that time [itex]\tau[/itex], you expect the same fixed proportion (in this case 50%) of those 10 zillion atoms to decay. so after [itex]\tau[/itex] seconds (or whatever unit of time you use), you have 5 zillion unstable atoms left. now, in the next [itex]\tau[/itex] seconds, what do you expect to happen to those 5 zillion atoms that have not yet decayed?
    Last edited: Feb 16, 2008
  18. Feb 16, 2008 #17


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    whooops. looks like Riogho made the same point with fewer atoms.

    that's teach me to read before clicking "Quote".
  19. Feb 16, 2008 #18


    Staff: Mentor

    Actually, exponential decay is essentially "history free". With 100 atoms and a half-life of 10 years there is a 50% chance that one decays about every 70 days or so. If 70 days pass with no decay then the probability that we get a decay in the following 70 days is still 50%. It does not go up or down based on the history, only based on the number of atoms.
  20. Feb 16, 2008 #19
    Yeah, that much is obvious, although apparently not to everyone.
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