# dirac function

Tags: dirac, function
 P: 13 Hi All, I have a problem in understanding the concept of dirac delta function. Let say I have a function, q(r,z,t) and its defined as q(r,z,t)= δ(t)Q(r,z), where δ(t) is dirac delta function and Q(r,z) is just the spacial distribution. My question are: 1. How can I find the time derivative of this function, that is, $\frac{\partial q(r,z,t)}{\partial t}$? 2. will hankel transformation of $\frac{\partial q(r,z,t)}{\partial t}$ be equal to zero (even when Q(r,z) $\neq$ 0)? Thank you in advance. FM
 HW Helper Sci Advisor Thanks P: 9,146 http://www.physicsforums.com/showthread.php?t=372548 The delta function is actually a distribution, and is not differentiable in the classical sense. In order to consider such differentiation, we have to revert to generalized derivatives. This is done by assuming a certain level of differentiability on f and some vanishing conditions. -- Kreizhn (post #2)
 P: 13 Hi Simon, Thanks for your response. Unfortunately, I'm still not totally clear. Can you please be more explicity. Once again, thank you. FM
HW Helper
 P: 13 Yes, I did, but I didn't fully grasp it. Anyway, this is what I can come up with, please take a look and let me know if it makes (physical) sense. Definition: q(r,z,t)=δ(t)Q(r,z) $\frac{\partial q(r,z,t)}{\partial t} = Q(r,z) \frac{d}{dt}[δ(t)]$ $\frac{\partial q(r,z,t)}{\partial t} = Q(r,z) δ^{'}(t)$ $\frac{\partial q(r,z,t)}{\partial t} = Q(r,z) \frac{t}{t} δ^{'}(t)$ since: $x δ^{'}(x) = -δ(x)$ Hence, $\frac{\partial q(r,z,t)}{\partial t} = -\frac{Q(r,z)}{t} δ(t)$ Thank you for your help FM