Understanding Dirac Delta Function: Time Derivative & Hankel Transformation

[femiadeyemi]

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 spatial 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

 

[Simon Bridge]

https://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)

 

[femiadeyemi]

Hi Simon,
Thanks for your response. Unfortunately, I'm still not totally clear. Can you please be more explicity.
Once again, thank you.
FM

 

[Simon Bridge]

Did you read the link?

 

[femiadeyemi]

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

 

[HallsofIvy]

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