Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Given the following function, how does one rewrite the exponential part of the equation into $e^{-L/L_{0}}$, where $L_{0}$ is the decay constant


I know this problem (demi-vie) from school and it is solved usually by setting $f(x)$ to $\frac{1}{2}$.

So for $f(L)=\frac{1}{2}$ this gives: $$ \begin{align*} \frac{1}{32}\frac{b^{2}}{a(b-a)} &=\frac{1}{32}\frac{b}{a}\left(1-\frac{a}{b}\right)^{-1}\\ &= \exp\left(\frac{-2L_{0}}{c}\bigl(2d(b-a)\bigr)^{1/2}\right)\\ \end{align*} $$

$$ \Rightarrow \frac{c}{-2(2d(b-a))^{1/2}}\log\left(\frac{b^{2}}{32a(b-a)}\right) = L_{0} $$

Wikipedia gives another definition of decay constant: $$t_{1/2}=\frac{\log(2)}{\lambda},$$ where $\lambda$ is the decay constant. So the "correct" $L_{0}$ would be:$$\frac{L_{0}}{\log(2)} = \frac{c}{-2(2d(b-a))^{1/2}\log(2)}\log\left(\frac{b^{2}}{32a(b-a)}\right)= L_{0corrected} $$

Now the exp part in $f(L)$ should be rewritten as $e^{-L/L_{0}}$. I don't see how to achieve that. Does somebody see how this is possible? Merci.

share|cite|improve this question
You write $f(x)$, but I can't quite tell where $x$ is in your exponential expression... – J. M. Dec 5 '11 at 1:00
Sorry for the confusion. – VVV Dec 5 '11 at 1:03

1 Answer 1

up vote 1 down vote accepted

I think the problem is that you're including the constants in front of the exponential in your calculations. Decay constants, half-lives etc. only relate to the exponential decay, not to the amplitude. So in this case you simply have


with the characteristic decay length

$$L_0=\frac c{2\left(2d(b-a)\right)^{1/2}}\;,$$


$$f(L)=16\frac{a}{b}\left(1-\frac{a}{b}\right)\exp\left(-\lambda L\right)$$

with the decay rate


share|cite|improve this answer
Thank you, joriki. – VVV Dec 5 '11 at 10:30

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.