Saturday, August 25, 2007

Mean lifetime

Mean lifetime
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Given an assembly of elements, the number of which decreases ultimately to zero, the lifetime (also called the mean lifetime) is a certain number that characterizes the rate of reduction ("decay") of the assembly. Specifically, if the individual lifetime of an element of the assembly is the time elapsed between some reference time and the removal of that element from the assembly, the mean lifetime is the arithmetic mean of the individual lifetimes.

Typically, the notion of mean lifetime is used in connection with exponential decay. The remainder of this article confines itself to this particular decay pattern.

[edit] Mean lifetime in exponential decay

The mean lifetime τ of elements in an exponentially decaying assembly is equal to the reciprocal of the decay constant (cf. Exponential decay). Thus, it is the time needed for the assembly to be reduced by a factor of e. It is related to the half-life t1 / 2 by

\tau \cdot \ln 2 = t_{1/2}.\,

Thus the mean lifetime is 44% longer than the half-life, e.g. Polonium-210 has a half-life of 138 days, and a mean lifetime of 200 days.

[edit] Derivation

In exponential decay, the population is governed by the following formula:

N = N_0 e^{-\lambda t} \,

where t is the time, N is the number of elements in the assembly at that time, N0 is the population at the initial reference t = 0, and λ is a parameter characteristic of the decay called the decay constant. The mean lifetime τ is the expected value of the amount of time before an unstable object undergoes a decay. First, we let c be the normalizing factor to convert to a probability space.

1 = \int_{0}^{\infty}c \cdot N_0 e^{-\lambda t}\, dt = c \cdot \frac{N_0}{\lambda}

c = \frac{\lambda}{N_0}.

We see that exponential decay is a scalar multiple of the exponential distribution, which has a well-known expected value. We can compute it here using integration by parts.

\tau = \langle t \rangle = \int_{0}^{\infty} t \cdot c \cdot N_0 e^{-\lambda t}\, dt = \int_{0}^{\infty} \lambda t e^{-\lambda t}\, dt = \frac{1}{\lambda}.

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