# Dirac Delta

1. Aug 9, 2010

### vasel

I'm curious about the use of the Dirac Delta function. I am familiar with the function itself, but have never really seen in used in actual problems. The only problems I've worked with the function are those specifically about the function (ie. Evaluate the Dirac Delta function at x=3).

My question is, how do I recognize the situation in which I can use the Dirac Delta function and how do I go about using it?

Is it possible to use it to determine the density of a "point mass"? How would this be done?

Thanks!

2. Aug 9, 2010

### Dr Lots-o'watts

In some contexts, the Dirac Delta function and the Dirac Comb can be used to approximate light pulses. Perhaps most often where Fourrier analysis is applicable.

3. Aug 9, 2010

### gabbagabbahey

The density of a point mass is an excellent example of the applicability of the Dirac Delta distribution. Let's take a look and see why:

(1) Let's assume there is a point mass $m$ located at $\textbf{r}=\textbf{a}$, what is the total mass enclosed by any volume containing the point $\textbf{r}=\textbf{a}$? What is the total mss enclosed by any volume not containing the point $\textbf{r}=\textbf{a}$?

(2) Now, take a look at the definition of localized volume mass density; $$\rho(\textbf{r})=\frac{dm(\textbf{r})}{d^3 r}[/itex].... in words, localized volume mass density at any given point in space is the mass enclosed by an infinitesimally small volume centered at that point, divided by the volume. This definition should be fairly familiar to you. (3) Let's see how this definition applies to a point mass... clearly, the mass density will be zero everywhere except at $\textbf{r}=\textbf{a}$. There will be a finite amount of mass at that point ($dm$ is finite), but the volume of a point is zero and division by zero leads an indefinite result. So, the density of a point mass is zero everywhere except at its location, where it is infinite (indefinite). Yet, if you integrate the density (add up all the little bits of mass) over any region enclosing $\textbf{r}=\textbf{a}$, you must get the mass of the particle, $m$....a finite result! If you integrate the density over any region not enclosing $\textbf{r}=\textbf{a}$, you will get zero (since there is no mass in that region). Compare these properties to the properties that define the Dirac Delta function, and you should deduce that the density of a point mass is proportional to the 3D- Dirac Delta function. Moreover, the constant of proportionality must be the mass of the point particle: $\rho(\textbf{r})=m\delta^3 (\textbf{r}-\textbf{a})$. Last edited: Aug 9, 2010 4. Aug 9, 2010 ### vasel Gab, that was a very clear and enlightening example. This concept makes a lot more sense to me now that I can see how it can actually be worked into a problem. Thanks so much for your help! - vasel 5. Aug 10, 2010 ### The_Duck Suppose a mass M is at rest at the origin. At time t = 0, I hit it sharply with a hammer, delivering an impulse P. What is the position of the of the mass as a function of time? One way to solve this is to write the force acting on the mass as [tex]F(t) = P \delta(t) = Ma$$

Integrating,

$$P \int \delta(t) dt = Mv$$

$$P \theta(t) = Mv$$ where $$\theta$$ is the http://en.wikipedia.org/wiki/Heaviside_step_function" [Broken].

$$x = \frac{P}{M}t$$ for $$t > 0$$, and $$x = 0$$ for $$t <= 0$$.

You don't need delta functions to do this, but this shows how delta functions are able to measure "instantaneous" or "infinitesimal" things like sharp impulses and point masses.

Last edited by a moderator: May 4, 2017