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Radioactive decay versus compound interest

  1. Sep 17, 2011 #1
    The fact that radioactive decay and continuous compound interest end up with the same formula (with the "rate" being negative in the former and positive in the latter) seems to me to be more a result of the ubiquity of the exponential function in solving differential equations than any common physical principle underlying both processes --the mechanism for the interest requires a "memory" and is dependent on the total, whereas the probability for each particle cannot be dependent on any memory or influence from other particles . But I could easily be wrong. If I am, could someone elucidate this principle?
  2. jcsd
  3. Sep 17, 2011 #2


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    I wouldn't say either requires a memory. Suppose you have something at 10% compound interest. $1 is going to turn into $1.10 after that period regardless if it's the only dollar in the account or if it's part of $1000 in the account. The new money created really doesn't care where it came from. That new $.10 in the $1 account is going to earn 10% interest itself and generate $0.01 the next period. All that matters is that it exists and the same process will happen to it as happened to the full $1 earlier.

    So in neither situation is the actual $.10 or a single particle dependent on the total. That $.10 will earn $.01 more regardless if there is $1.10 in the account or $1,000.10 in the account.

    I think your line of thinking is missing the idea that the particle acts independent of what's going on in the whole, but not realizing that a dollar will act independent of what's going on in the whole as well.
  4. Sep 17, 2011 #3
    dN/dt is proportional to N in both cases, which is why you get exponentials. I don't think there's anything more mystical than this - in both cases the rate of change dN/dt depends on how much you have, N, at that particular time t.
  5. Sep 18, 2011 #4
    Thanks for the answers, Pengwuino and JeffKoch.
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