2
$\begingroup$

This is sorta homeworkish, yet it still relates to my understanding of the physics behind it as well.

Does anyone know how you determine t_98 (aka the length to contain 98% of the energy) for a photonshower in a scintillator crystal? I've been told it's "simply" to integrate: $$\frac{dE}{dt}=E_0Ct^\alpha e^{-\beta t} \text{ where }t=X/X_0 \\X_0 \text{ being radiation length and }C=\frac{\beta^{\alpha +1}}{\Gamma(\alpha+1)}$$ which integrated from 0 to $\infty$ gives $E=E_0$, but I feel like I should set the top boundary to $t$ so that I could solve for $t$ afterwards, but that gives me an impossible loop of $\Gamma$-functions, so I presume I need to set the top boundary to something numerical - but I do not for the life of me know what - setting top boundary to $\alpha$ gave me a value close to what I presume is the value I'm looking for, but I don't feel like I have any justification for doing so. My googling for the way to do this has only led me to rip out bits of my hair so far, seeing as in most texts they seem to skip the actual step of the calculation and just give the answer (which is always different) or say that one should use GEANT4 to calculate the length.

If someone could guide me to understanding how one does determine this I would be eternally grateful.

$\endgroup$
  • $\begingroup$ Honestly I feel like the top boundary is the value I am supposed to determine, by setting $E=0.98E_0$, but I sorta lack the mathematical know-how to determine the top boundary of the integral. $\endgroup$ – Fidelis Oct 18 '14 at 18:23
  • $\begingroup$ Wolfram alpha gives answers mostly in Gamma-function forms. But in all seriousness there is no reason you need an exact solution. Use a numeric integration. $\endgroup$ – dmckee Oct 19 '14 at 0:11
  • $\begingroup$ That there is what I consider a Gamma-function-loop, seeing as you get Gamma(a+1, bt). I assume you know the answer to how one calculate the length, so is my reasoning correct when I assume that I'm supposed to get my $constants * int(t^a * e^{-bt}) = 0.98$ ? Or is it a mistake to think like that? $\endgroup$ – Fidelis Oct 19 '14 at 0:25
  • $\begingroup$ "so is my reasoning correct when I assume that I'm supposed to get my $constants * int(ta∗e−bt)=0.98$ ?" That's right. But there is no loop in the answer Wolfram gives and more than an integral that gives you $\sin (3x + 2)$ has gotten you into a "trig-loop". You just need the inverse function. Or numerical solutions are fine for this job. I'm afraid that as you move your colleagues will simply assume that you can find a solution to things like this without aid. That's why the sources don't mention it. $\endgroup$ – dmckee Oct 19 '14 at 0:32
  • $\begingroup$ Hopefully someone will actually teach me how to do things like these during the next 3 years of my education, even so it pretty much feels like you're supposed to know everything from the get-go without ever having seen it before. I feel like the Gamma(a+1, bt) makes it "loop", since then I have to integrate from $\text{t to }\infty$ and I'm trying to find t, will take a look at it and see if I can manage to calculate it, thanks for your help yet again dmckee. :) $\endgroup$ – Fidelis Oct 19 '14 at 0:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.