Answers · The mathematics underneath

What is the difference between a half-life and a decay rate?

In shortA half-life is the time for an amount to be multiplied by one half, while a decay rate says how fast that multiplication happens, either continuously or as a per-step factor.

A half-life is a time, not an amount. It is the time required for the quantity to be multiplied by one half, so after one half-life half remains, after two half-lives one quarter remains, and after three one eighth remains. A decay rate is the other side of the same statement: it says how quickly that multiplication by a number below one happens, either as a continuous rate per unit time or as a fixed factor applied each step. The half-life fixes the time scale; the rate fixes the speed, and the two are convertible without changing the curve.

N(t)=N_0\left(\frac12\right)^{t/t_{1/2}}=N_0 e^{-t/\tau},\qquad \tau=\frac{t_{1/2}}{\ln 2},\qquad t_{1/2}=\tau\ln 2.

Take a half-life of 3 minutes. At $t=0$ the amount is 1. At $t=3$ it is $1/2$ . At $t=6$ it is $1/2$ times $1/2$ , which is $1/4$ . At $t=9$ it is $1/4$ times $1/2$ , which is $1/8$ . At $t=12$ it is $1/16$ . The subtraction that gives the same first step would be $1-1/2=1/2$ , then $1/2-1/2=0$ at six minutes. That is wrong because the second three minutes removes half of what is left, not half of the original.

The natural guess is that a half-life is a fixed amount subtracted each step. That guess is linear decay: the same quantity leaves every step, so the graph is a straight line and the value reaches zero in finite time. Half-life decay is multiplicative: the same fraction leaves every step, so the amount is multiplied by one half every three minutes and the graph bends. After six minutes the multiplicative answer is one quarter, after nine one eighth, and after twenty one minutes seven halvings have happened, leaving one part in 128, which is $0.0078125$ of the start. It is small but not zero, and no number of half-lives makes it exactly zero.

FIG 1

half-lives elapsed	fraction remaining	value starting at 1
1	1/2	0.5
2	1/4	0.25
3	1/8	0.125
7	1/128	0.0078125
10	1/1024	0.0009765625

The logarithm makes the equivalence exact. Taking natural logs of $N(t)=N_0(1/2)^{t/t_{1/2}}$ gives $\ln N(t)=\ln N_0-(t/t_{1/2})\ln 2$ , a straight line falling by $\ln 2$ every half-life. Taking natural logs of $N_0 e^{-t/\tau}$ gives $\ln N(t)=\ln N_0-t/\tau$ . Equating the slopes gives $1/\tau=(\ln 2)/t_{1/2}$ , which is the same conversion. The half-life is the time for the log to fall by $\ln 2$ , and the continuous rate is the fall per unit time of the log. They are two readings of one slope.

The two common rates

Physics and finance usually write the decay as $e^{-t/\tau}$ , where $\tau$ is the time constant, the time to fall to $1/e$ of the start. Chemistry and engineering usually write the half-life, the time to fall to $1/2$ . The conversion is simple but easy to invert: $\tau=t_{1/2}/\ln 2$ , so the time constant is always longer than the half-life. With a half-life of 3 minutes and $\ln 2=0.693147$ , $\tau=3/0.693147=4.328$ minutes. The continuous rate is $k=1/\tau=0.231$ per minute, because $0.231$ times 3 is $0.693$ , and $e^{-0.693}=0.5$ .

b=\left(\frac12\right)^{\Delta/t_{1/2}},\qquad t_{1/2}=\frac{\Delta\ln 2}{-\ln b}.

A per-step factor is the third spelling. If a quantity is multiplied by $0.99$ each step, its half-life in steps is $\ln(0.5)/\ln(0.99)=(-0.693147)/(-0.0100503)=68.97$ steps. The round trip is $0.5^{1/68.97}=0.99$ . If the step is one second, the half-life is 68.97 seconds; if the step is one minute, it is 68.97 minutes. The factor alone does not fix a time until the step length is stated. With a half-life of 3 minutes, the per-minute factor is $0.5^{1/3}=0.7937$ , so after one minute $79.37$ per cent remains, and after three minutes $0.7937^3=0.5$ .

When the conversion goes wrong

Two errors account for most confusion. The first is inverting the conversion: $\tau=t_{1/2}\ln 2$ would give 3 times 0.693, which is 2.079 minutes, shorter than the half-life. That cannot be right, because falling to $1/e$ takes longer than falling to $1/2$ . The second is treating a rate quoted as $1/\tau$ as if it were a per-step factor. A continuous rate of 0.231 per minute is not a factor of 0.231; the factor over one minute is $e^{-0.231}=0.794$ , and over three minutes it is $e^{-0.693}=0.5$ .

The fixed half-life depends on the decay being multiplicative. If a process removes a fixed amount each step, the time to halve is not fixed: starting at 1 with a subtraction of 0.1 per step takes five steps to reach 0.5, but starting at 0.6 takes one step. Such a process reaches zero and has no half-life in this sense. If the decay is multiplicative but the step length changes, the per-step factor changes while the half-life in time and the continuous rate do not. That is why a learning rate schedule written as a factor per epoch must be converted using the epoch length before it can be compared with a rate written per second.

Worth keeping

Half-life is the time for a quantity to be multiplied by one half, so two half-lives leave one quarter, not zero.
The continuous time constant is the half-life divided by ln 2, which makes it about 1.4427 times longer than the half-life.
A per-step factor below one has half-life ln(0.5)/ln(factor) steps; for 0.99 that is about 69 steps.

Where this goes next

This question is one step of Logarithms, Growth and Scale, a course that works through its subject from the beginning. The page that teaches this part of it is What Decays, and How Fast, and it reads the whole way through whether or not you have an account.

What an account changes is the practice: problems standing under every idea, marked with a reason rather than a tick, in a copy of the course that is yours.

Read that page The whole course

What is the difference between a half-life and a decay rate?

The two common rates

When the conversion goes wrong

Worth keeping

Where this goes next

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