What is the Definition of the Delta Function?

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SUMMARY

The discussion centers on the Dirac delta function, particularly its role in modeling instantaneous impulses in physics. It is derived from the Heaviside step function, where the delta function acts as the derivative of the Heaviside function, represented as H'(x) = δ(x). The delta function is defined as a linear continuous functional on the space of C∞(ℝ)-functions with compact support, with the property δ(ϕ) = ϕ(0). This concept is crucial for understanding momentum generation in physics, particularly in the context of Gauss' law and Coulomb's law.

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Leo Liu
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TL;DR
See the title please.
I came across it in the derivation of Gauss' law of electric flux from Coulomb's law. I did some research on it, but the wikipedia page about it was slightly confusing. All I know about it is that it models an instantaneous surge by a distribution. However I am still perplexed by this concept. Can someone please explain it to me like I am a 5-year-old with some calculus or direct me to some useful resources other than wiki?
 
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There are multiple approaches to delta function. Here I would start from
Heaviside Step Funiiton
https://en.wikipedia.org/wiki/Heaviside_step_function
We can not differentiate Heaviside Step Function at 0 in usual healthy sense. But Dirac boldly invented a strange delta "function" as the result of its derivative including x=0.
H'(x)=\delta(x)

Say we give 1 Ns impulse to a 1kg body to make it move with 1m/s.
There are many ways to do it, e.g.
Applying Force 1 N during 1 second,
Force 10N during 1/10 second,
Force 100N during 1/100second,
-----
Force 1/x N during x second,
------

We can make time duration x as small as we like with increasing Force 1/x .

There is no limit of x to zero in this sequence because 1/x diverges to infinity. But Dirac say there exists force limit ##\delta(x)##, that is to say, in a instant of time, the infinite force is applied to generate 1 Ns momentum on the body. Delta function is a momentum maker in a instant of time.
\int_{-\infty}^x \delta(\xi) d\xi = H(x)
Heaviside step function shows thus made momentum by Delta function at x=0.
 
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Leo Liu said:
Summary:: See the title please.

Can someone please explain it to me like I am a 5-year-old ...
The Dirac delta function is rarely taught at infant school. Especially these days.
 
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PeroK said:
The Dirac delta function is rarely taught at infant school. Especially these days.
What's the world coming to these days ?! :wink: A B C Dd !
 
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anuttarasammyak said:
There are multiple approaches to delta function. Here I would start from
Heaviside Step Funiiton
https://en.wikipedia.org/wiki/Heaviside_step_function
Thank you. This concept isn't hard to grasp at all. I will look into its connection with Dirac delta.
 
##\delta##-function is a linear continuous functional on the space of ##C^\infty(\mathbb{R})##-functions with compact support. By definition ##\delta(\varphi)=\varphi(0)##.
The formulas like ##H'=\delta## are understood in the generalized sense
$$-\int_{\mathbb{R}}H(x)\varphi'(x)dx=\varphi(0)$$
For details see any textbook on functional analysis. For example https://www.springer.com/gp/book/9783540586548
 
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