Based on astronomy topic what do you mean by an e-folding time of the main stellar population model?

Answer 1

Great question, Akhil! The concept of an "e-folding" time is actually a very general one that occurs in many different areas of physics and astronomy. The common usage of this term arises from the fact that the exponential function (e^x) occurs so widely.

As an made-up example, suppose I have a population of radioactive atoms that decays according to the following function:

N(t) = N₀e^-t/10

where N is the number of atoms, t is the elapsed time in seconds, and N₀ is some positive integer. Notice that when t=0, we have N = N₀, which means that N₀ represents the number of atoms at the start.

Because of the 10 in the denominator, when the time reaches t=10 seconds, the number of atoms left will be N₀e^-1, i.e. the original number of atoms divided by e. Similarly, if you elapsed time by another 10 seconds and put t=20, you would have N₀e^-2, which is the number of atoms at time t=10 but again divided by e. We see that at every multiple of 10 seconds, the number of atoms gets divided by e. Thus, we say that the e-folding time for the number of atoms is 10 seconds.

This same reasoning applies to any scenario where an exponential factor shows up.

Additionally, note that in principle this can happen for any number, not just e. For example, if my number of atoms over time was

N(t) = N₀3^-t/10

then for every 10 seconds that elapsed, the number of atoms would go down by a factor of three, and I might say that the "three-folding" time is 10 seconds. But in practice, scientists rarely talk this way because any general exponential function can be rewritten so that e is the base of the exponent.