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I know how to derive the formula for "quantity at time $t$" for some decaying materials. You can see the derivation here. But, what I don't get is that what the number $e$ is doing here? We get the value of $e$ from the equation of compound interest. Compound interest and decay (like, radioactive decay) are two different things. Is there any intuitive way to see what the Euler's number doing here?

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    $\begingroup$ It's not clear what you're asking. You seem to understand the derivation and that this is the solution to the equation. What more of an answer do you want than that? $\endgroup$
    – Brick
    Commented Jul 19, 2021 at 16:09
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    $\begingroup$ Compound interest is exponential growth. Radioactivity is exponential decay. Same math, just backwards. $\endgroup$
    – Jon Custer
    Commented Jul 19, 2021 at 16:10
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    $\begingroup$ This question is about math or possibly the history of math, but not physics $\endgroup$
    – Paul T.
    Commented Jul 19, 2021 at 16:15
  • $\begingroup$ You question is either unclear or it should be asked in Math SE. If you are asking why we have $e^{\lambda t}$ and not some other number say $a^{\lambda t}$ when it belongs to Math SE. $\endgroup$ Commented Jul 19, 2021 at 16:25

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Compound interest and radioactive decay both vary exponentially with time. That simply means that in any set period of time, the value changes by the same fractional amount.

Any exponential function can be written using any base. If the number of remaining atoms in a sample is given by $A * e^{-Bt}$, then it is also given by $A * 2^{-Ct}$, for the appropriate value of $C$.

Regardless of the base you choose, the number $e$ pops out when you calculate the rate of change of the exponential function (which is another exponential function). So, assuming you are interested in, say, both the number of atoms and how fast that number is changing, you will be stuck with an $e$ somewhere or other anyway. Choosing it as the base eliminates a constant that would otherwise appear in the expressions.

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    $\begingroup$ In fact, the $2^{-t/t_{1/2}}$ is quite a natural way to express the decay in terms of half-life. $\endgroup$
    – Ruslan
    Commented Jul 19, 2021 at 17:02
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By definition, the natural log base $\mathrm{e}$ has the special property that

$$ \tfrac{\rm d}{{\rm d}t} \mathrm{e}^t = \mathrm{e}^t $$

This leads us to understand that all solutions to

$$ \tfrac{\rm d}{{\rm d}t} f(t) = f(t) $$ have basis functions $f(t) \propto \mathrm{e}^t$.

A subset of these problems are that of

$$ \tfrac{\rm d}{{\rm d}t} x(t) = -a\, x(t) $$ which have solutions of the form $x(t) = X \mathrm{e}^{-a t}$

This is the basic exponential decay formula, and the interpretation of the differential equation is that the amount decaying is proportional to the amount that exists at any time.

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    $\begingroup$ As an aside, imo it is important to mention that this is the reason why we cared to talk about $e$ in the first place. When we calculate the derivative of $a^x$, it comes out to be $ka^x$. A natural question to ask is, for what number $a$ is the constant $k=1$. That is the Euler's number $e$, and the constant thus becomes $\ln(e)$. $\endgroup$
    – Physiker
    Commented Jul 19, 2021 at 18:30
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    $\begingroup$ @IndischerPhysiker - of course and that is what I was trying to convey in my first sentence. The value of $\mathrm{e}$ is special as it allows the derivative to be equal to the value. $\endgroup$
    – jalex
    Commented Jul 19, 2021 at 19:28
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Compound interest is exponential decay when you examine it closely.

Look at e in the formula for compound interest: (1 + 1/n)^n

Look at the fraction 1/n. The larger n gets, the smaller the fraction gets. In fact, it gets exponentially smaller. Exponential decay.

In the infinite sum for the compound interest equation above, the larger n gets, the closer it gets to e. If you let n run to infinity, the answer would converge to e. An easy way to see this is to set principal to $1.00, interest rate to 100%, and compounding periods to fractions of one year i.e. for compounded every six months would be (1+ 1/2)^2. Then try higher and higher numbers for n. The return on principal plus principal approaches e. Exponential decay of return.

In fact, when you graph the compound interest formula, it looks like a reflection of e^x ( See Wolfram Alpha on the constant e).

Euler’s number has to do with exponentials. All constants to a time exponential, have a derivative that has a proportionality constant. When the proportionality constant is the number 1, the constant is e.

However, it should be understood that the compound interest formula has a limit of e, whereas, the function e^-x has a limit of zero. And it should be understood that many infinite series can converge to a real number. To understand why e (Euler’s number) is special, see YouTube by 3Blue1Brown called “What’s so special about Euler’s number e?”

“If you invest your money, the rate at which it grows is proportional to the amount of money there at any time. In all of these cases where some variable’s rate of change is proportional to itself, the function describing that variable over time is going to look like some kind of exponential. And even though there are lots of ways to write any exponential function, it is very natural to choose to express these functions as e to the power of some constant times t since that constant carries a very natural meaning. It’s the same as the proportionality constant between the size of the changing variable and the rate of change.”—Grant Sanderson

But if you are asking “Why?” e shows up in biology, math(s), compound interest, radioactive decay, etc., that is like asking why pi shows up in population growth and many other functions when it comes from a circle formula. Most of Science can’t answer Why? They can measure observations, predict from observations with mathematical principles, but cannot yet answer “why?” IMHO, someday they will.

The more interesting question is if radioactive decay has a predictable half-life, how can it be spontaneous i.e. meaning “random decay” without a pattern in the individual atom. Quantum Physics cannot yet answer this. They can measure decay in half-lives, but they also can mathematically explain spontaneous emission. How does something random form a pattern?

BTW, physics cannot really be separated from mathematics, so your question is a physics question.

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    $\begingroup$ (1) $1/n$ does not get exponentially smaller with increasing $n$. (2) Its perfectly reasonable to ask why $\pi$ shows up all over the place, and the answers usually relate to $\pi$ being half of the period of the complex exponential function $e^{ix}$. It's also perfectly reasonable to ask the same question of $e$. Such questions are better phrased as e.g. "how does $e$ arise in this expression?" and are distinct from the "why" questions which science is unable to answer. (3) Random events can be often be described by probability distributions, and can still be studied mathematically. $\endgroup$
    – J. Murray
    Commented Dec 3, 2022 at 20:04
  • $\begingroup$ (4) Physics and mathematics are completely different enterprises which share a common language, so physics and mathematics can absolutely be separated - at least in some cases. Not every mathematical question relates to physics. $\endgroup$
    – J. Murray
    Commented Dec 3, 2022 at 20:04
  • $\begingroup$ You can do math for math’s sake, but you can’t do physics without math and a question about radioactive decay is physics. Euler’s number e is defined by the compound interest formula. It is defined by the infinite fraction that can be created by the formula. It is also defined by the infinite series found by Newton. It is also defined by the exponential proportionality constant. Euler found that e to the x contains the infinite series for sines and cosines multiplied by the imaginary number i. When you have a number that is defined by something, you can’t answer why. $\endgroup$
    – Voyajer
    Commented Dec 4, 2022 at 12:47
  • $\begingroup$ (1) All fractions get smaller by increasing the denominator. Basic arithmetic. Think again. (2) Especially if they have the same exponent as the denominator, they get exponentially smaller. $\endgroup$
    – Voyajer
    Commented Dec 4, 2022 at 12:56
  • $\begingroup$ Exponentially smaller generally means that something goes like $a^{-n}$ for some positive number $a$. $1/n$ does not qualify; you might call it inverse-linear or something to that effect, but exponential has a specific meaning. In any case, any answer to why $e$ is present in the formula for exponential decay is purely one based on convention, since $N(t)=N_0 2^{-t/T}$ works just as well (and actually better when we're talking about decay in terms of half-lives). $\endgroup$
    – J. Murray
    Commented Dec 4, 2022 at 14:54

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