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Dirac function 
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#1
Jan713, 03:59 AM

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, [itex]\frac{\partial q(r,z,t)}{\partial t}[/itex]? 2. will hankel transformation of [itex]\frac{\partial q(r,z,t)}{\partial t}[/itex] be equal to zero (even when Q(r,z) [itex] \neq [/itex] 0)? Thank you in advance. FM 


#2
Jan713, 05:48 AM

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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) 


#3
Jan713, 03:50 PM

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 


#4
Jan713, 05:22 PM

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Dirac function
Did you read the link?



#5
Jan813, 05:05 AM

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) [itex]\frac{\partial q(r,z,t)}{\partial t} = Q(r,z) \frac{d}{dt}[δ(t)][/itex] [itex]\frac{\partial q(r,z,t)}{\partial t} = Q(r,z) δ^{'}(t)[/itex] [itex]\frac{\partial q(r,z,t)}{\partial t} = Q(r,z) \frac{t}{t} δ^{'}(t)[/itex] since: [itex]x δ^{'}(x) = δ(x) [/itex] Hence, [itex]\frac{\partial q(r,z,t)}{\partial t} = \frac{Q(r,z)}{t} δ(t)[/itex] Thank you for your help FM 


#6
Jan813, 07:26 AM

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PF Gold
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Did you understand it well enough to grasp what a "distribution" or "generalized function" is?



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