Jump to content
Sign in to follow this  
mooeypoo

The meaning of the Dirac Delta

Recommended Posts

Hey again guys,

 

I was in a study session today with classmates, and we had a hard time answering a question that involved the dirac delta. But the difficulty wasn't so much the method (we have that in the book and as equations online), but rather with the concept.

 

What is the meaning of the dirac delta, specifically in 3D??

If you need to "imagine" it in 3D - how would you describe it? How does it look? What's the point of it?

 

If anyone can help with this, it'll be greatly appreciated.. we can plug in numbers into equations, but if we don't understand what the concept is, we have no clue what we're doing.

 

Thanks in advance :)

 

~moo

Share this post


Link to post
Share on other sites

I tend to think of it as a "selector". In that it looks through the range that the integral is over, and selects the integrand at a particular value.

 

There are several decent physical examples of where dirac deltas are used. One is in the study of a population of particles. Let x be a particle size (can be volume, or diameter, or mass, anything really). And, consider a solution that is supersaturated. Crystals will condense out of the supersaturation, and whatever tool used to measure the particle size will have a lower limit. In terms of that tool, a particle will spontaneously appear at a size, call it [math]x_l[/math] for limit. One of the terms in the equation that described the distribution of particle sizes will include a term proportional to [math]\delta(x-x_l)[/math] because of the appearance of new particles at size [math]x_l[/math]. In this case, the dirac delta "selects" the size of the newly nucleated particles.

 

Another case would be the limit of instantaneous impulses. Consider a wall that is not perfectly elastic. That is, when something hits the wall it loses a little bit of momentum. The momentum impulse to that something that often be approximated by a term proportional to a dirac delta. Because the momentum changes over an instantaneous moment. Sure, it really isn't instantaneous, but as often happens it really depends on the level of detail you need to know about. In this case, the dirac delta "selects" how much momentum is lost and the time the collision occurs.

 

That second example is also good at another similar interpretation of the dirac delta. It is a very good model for instantaneous changes. It is often considered the "derivative" of the Heaviside function. (The Heaviside function is the one that jumps from 0 to 1 at a specified point).

 

Finally, the dirac delta is also good for being a first (or maybe even zeroth) order approximation to a distribution. Sure, a normal distribution has width, but if you want to get a decent approximation, treating the normal distribution as a dirac delta at the mean isn't terrible. The approximation also makes the integrals much easier to compute!

 

I hope that this is a start. I don't actually know the physics as to what Dirac and colleagues were doing to motivate the creation of the delta function. But, I wrote out how I think of the delta function.

Share this post


Link to post
Share on other sites

I think of it in terms akin to collapsing a wave function. You have a distribution that exists over all space, but you've made a measurement, so you get the answer for one particular location.

Share this post


Link to post
Share on other sites

Wow.

 

Okay, first off -- I really appreciate your responses and your effort. Thanks a lot, really, I am extremely confused over this subject and the professor acts as if it's the most obvious thing in the world.

 

But I still don't quite get it, and I think it might be because the examples are a bit over my head - I don't think I got to that level yet, so, I can't make sense of it like that. :(

 

Is it possible to "imagine" the dirac delta as a spacial function? My friend suggested that if we look at the function on the xyz axis, it might look like those depictions of the black-holes in 3D, with its 'center' located on the point where the function exists.

 

Is that right?

 

I understand that the dirac delta talks about distributions, but I can't quite understand what's the benefit of using that over others. My current homework, for example, have this question (under the subject of the dirac delta):

 

Write the charge density

[math]\rho(\hat{r})[/math]

for a point charge q located on the y-axis at distance d from the origin.

 

So in this case, our distance vector is

 

[math]\hat{r}=<0,d,0>[/math]

 

And since this is under the dirac delta chapter (and looking at similar questions) we understood that the charge density would be represented as:

 

[math]\rho(\hat{r})=q\int_{-\infty}^{\infty}\delta^3(r-\hat{r})d^3r[/math]

 

And since vector d is in the integration domain (of ALL SPACE -- infinity to infinity) then the delta expression equals 1, and the entire expression is, therefore, equal to q.

 

But the question (if, indeed, we did it correctly, which I think we did, but do correct us if we haven't) is -- why use dirac delta in this case at all?? what does it mean to use the dirac delta in this case? I remember there are other ways to calculate surface and volume charge density (I learned it in the previous physics course) that doesn't invovle a dirac delta. Then what use is it? Is it more accurate?

 

What worries me mainly, is that we are studying the methods and equations, but we will ahve to use them in the future and choose them ourselves. So, when I will encounter a question in Electrodynamics -- how do I know when I use the delta function, and when not to? What does the delta function give me as opposed to other methods, so I know how to pick it to solve certain problems..

 

Thanks again guys, and sorry -- this really is confusing to me...

 

~moo

Share this post


Link to post
Share on other sites

Well, firstly, the charge density shouldn't be an integral. A density is the probable amount of something (charge) located in a volume in the space.

 

What you've written there is a charge cumulative distribution function. The integral adds up all the charge that is out there.

 

The charge density function in this case would be [math]\rho(\mathbf{r}) = \delta(\mathbf{r}-\mathbf{\hat{r}})[/math]. (I don't particularly care for the [math]\delta^3[/math] notation, myself. I think that if you are talking about a vector in the argument, then it is obvious what is meant. Maybe it is just me. I also don't like the [math]d^3[/math] notion at the end of integrals, either. I think that [math]d\mathbf{r}[/math] is also more obvious.)

 

That way, when you integrated over the entire space, the delta selects out the charge that located only at that point.

 

Here's a related example: When you launch a probe to land on Mars, we need to know how all the different bodies in the solar system affect the probe via gravity, right? Well, to do that, you integrate over the entire solar system to find all the gravitational forces -- you get one resultant vector that is the net effect of gravity on the probe at that one point in time. With this information, you can plan what amount and in what direction you would want that probe to fire its thrusters.

 

Now, there are ways of making that integral over all the mass in the solar system easier. The delta function is one of the primary ways. The mass of Pluto out there, while small, is going to have an effect on the probe. But what doesn't really matter is if the poles are pointed toward the probe, or away from the probe. The changes in the the local mass density of Pluto are so small and so far away from the probe that they are insignificantly small. In effect, all the probe "sees" is a point source. So, in computing the gravitational effect integral, you just set Pluto equal to a delta function. Something like [math]\rho_{Pluto} = m_{Pluto}\delta(\mathbf{r}-\mathbf{r}_{Pluto})[/math]. The density function of the entire solar system would be [math]\rho_{Solar System} = \rho_{Pluto} + \rho_{Uranus} + \rho_{Neptune} +\rho_{Saturn} + etc.[/math] Where, depending on the need, you may have to treat each density function of each planet differently. And at different times. When the probe is just leaving Earth, you will want to use a detailed density function of Earth but treating Mars as a point source may be okay. Vice versa, getting closer to Mars you will probably want to use a detailed density map of Mars but treat the Earth as a point source. It all changes based on what the needed accuracy is. And how much time you are willing to spend computing to that certain accuracy.

 

So, the delta function is good for treating things as perfect infinitesimal point sources. In the real world, there is no such thing. But, often, a point source approximation is really quite accurate enough, and using more complicated (though more accurate) representations can be a colossal waste of time and computing power -- the added accuracy is unnecessary.

 

And, ultimately, it makes the math easier. Instead of integrating over a function, you just select out the integrand at a certain point. Much faster and easier.

Share this post


Link to post
Share on other sites

Hello,

Actually my lecturere gave me an assignment to prepare a paper on Dirac Delta function but the problem is that I am really having difficulty understanding what it is exactly?

I mean the mathematical definition is given everywhere but what about the physical meaning?

Could u please explain me about it with some very easy examples so that I can get what it really means?

Share this post


Link to post
Share on other sites

roshan, were the examples and explanations above unclear? And if they were, can you ask more specific questions?

Share this post


Link to post
Share on other sites
roshan, were the examples and explanations above unclear? And if they were, can you ask more specific questions?

Bignose, I think I am going to join the request for more examples.

 

I'm over this course already, and I understand enough to solve equations and I *think* get the point, but I still don't QUITE get where this is useful in practice.

 

That is -- why would we need to use this type of distribution (it's very unique looking, at least to me..) for physical cases? I'm not sure I understand why the need for the Dirac Delta exists. It sounds statistical to me, while I always thought physics is quite definite -- if you calculate the mass of an object, why insert a mass distribution that looks like that? How do we know the distribution isn't unique for the object? Where is this type of distribution coming from that it's being used so broadly (as in, we use it for a lot of physical cases)..?

 

That's at least what I would love to know.. from the examples you wrote I understood the basic function of the dirac delta, but I am not sure I know where it's coming from and why it's useful.

 

I hope that also fits roshanisilwal question.. if not, he's welcome to post his own, of course :)

 

 

~moo

Share this post


Link to post
Share on other sites

You may find it helpful to think of it as a limit of another fundtion, that is,

 

[math] \delta(x-x_0) = \lim_{a \to 0} \frac{1}{a \sqrt{\pi}} e^{-(x-x_0)^2/a^2}[/math].

 

It is probably most useful when used to facilitate Green's Functions. If you can solve a differential equation with a Dirac delta source, you can solve it for any source you like simply by integrating this solution (the Green's Function) over space weigthed by the source.

Share this post


Link to post
Share on other sites

The first thing to realize regarding the delta function is that it isn't a function.

 

Looking at the delta function (OK, I'm a physicist at heart) as a limit of some series of functions is a very good approach to visualize what is going on. Bignose (post #2) and Severian (post #9) talked about it terms of a normal distribution. That's a very nice way of looking at it, but of course not the way. Some of the reasons this is particularly useful: It's already normalized for you, the normal distribution is analytic everywhere, and it generalizes to multiple dimensions.

 

That said,

Here's a related example: When you launch a probe to land on Mars, we need to know how all the different bodies in the solar system affect the probe via gravity, right? Well, to do that, you integrate over the entire solar system to find all the gravitational forces -- you get one resultant vector that is the net effect of gravity on the probe at that one point in time. With this information, you can plan what amount and in what direction you would want that probe to fire its thrusters.

 

Now, there are ways of making that integral over all the mass in the solar system easier. The delta function is one of the primary ways.

Nobody does this. For one thing, that is in essence a time-varying mass distribution function the size of the solar system!

 

For another thing, why bother? The errors in timing; vehicle position, velocity, and attitude; thruster alignment and thrust variance; sensor measurements; ... completely overwhelm the contribution of Pluto, or Uranus, or Jupiter for that matter. (Unless of course the probe is close to Pluto or Uranus or Jupiter.) Trajectory planning uses fairly simple (often extremely simple) gravity models. They do rely on precision models of the relevant planetary bodies to have been pre-computed (which they are; we have ephemeris models that predict the locations and orientations of the solar system bodies well into the future).

 

For yet another thing, point masses models are absolutely the wrong thing to use for a vehicle in low Earth, Moon, or Mars orbit. The Earth has a large equatorial bulge, and the Moon and Mars are *lumpy*.

Share this post


Link to post
Share on other sites

 

For yet another thing, point masses models are absolutely the wrong thing to use for a vehicle in low Earth, Moon, or Mars orbit. The Earth has a large equatorial bulge, and the Moon and Mars are *lumpy*.

 

DH, I said that already. I said you can only treat things as point masses when you are far away.

When the probe is just leaving Earth, you will want to use a detailed density function of Earth but treating Mars as a point source may be okay.

Share this post


Link to post
Share on other sites

Distant objects are treated as point masses, but that does not necessitate a Dirac delta formulation. There is no need to do so, and aerospace engineers simply don't. If this thread is still alive next week, I'll ask my coworkers who were trained as aerospace engineers if they even know what the Dirac delta function is, let alone use it. I know the answer to the latter: They don't. I suspect the answer to the former question is the similar.

Share this post


Link to post
Share on other sites

It doesn't really matter if they know what the Dirac Delta function is or not: do they use any point sources in their simulations at all? Because that would be represented by a delta function.

 

I know I do:

 

There is a finite time when a particle hits a solid object or when two particles collide, but I still approximate the momentum impulse and energy change by a delta function because it makes the math easier. Particles that form during a nucleation process in a supersaturated solution don't actually appear at a finite size instantaneous, nevertheless, I still use a delta function to represent the nucleation process because the error introduced is negligible. When there is a reaction occurring on a particle or on a wall or anything else smaller than the discrete volume, the heat source and mass sources and sinks are treated as point sources and sinks -- and hence are delta functions.

 

Anything anywhere that can be treated as a point source is mathematically represented by a delta function.

Share this post


Link to post
Share on other sites

No. A point is a point. Attaching an attribute to a point does not necessarily mean you are using, implicitly or explicitly, delta functions. Delta functions arise when you try, implicitly or explicitly, to take a gradient across the point.

 

There is no need to use delta distributions in modeling point masses because Newton's law of gravitation strictly speaking only pertains to point masses.

Share this post


Link to post
Share on other sites

I'll write out some more specific examples to try to make it clearer:

 

Consider a distribution of particles, described by their volume, v. Let the distribution of particle volumes be denoted [math]f(v)[/math]

 

How that volume distribution changes over time is described by the population balance equation in this form:

 

[math]\frac{\partial f}{\partial t} + \frac{\partial (\dot{V}f)}{\partial v} = h[/math]

 

where [math]\dot{V}[/math] is the growth rate of the particles and h on the right hand side is all birth and death functions (agglomeration, breakage, nucleation).

 

Let's keep it simple and consider only a birth process of nucleation of particles that nucleate at a size [math]v_n[/math]. In this case,

 

[math]h=\dot{n}(S)\delta(v-v_n)[/math]

 

[math]\frac{\partial f}{\partial t} + \frac{\partial (\dot{V}f)}{\partial v} = \dot{n}\delta(v-v_n)[/math]

 

The terms meaning that a change in the number distribution of particles of size v is due to the combined effects of number of particles that grow into that size, the number of particles that grow out of that size, and the number of particles at that size that nucleate.

 

Now, let's use a finite volume method (in the pop balance literature typically called a sectional method) to solve this equation. That is, we are going to set up a number of bins 1,2,3,...,i, integrate the population balance over the volumes covered by each bin, and solve for the number of particles in each bin as a function of time.

 

Let the volumes of particles in bin i be [math](x_i,x_{i-1})[/math]. And let [math]N_i[/math] be the number of particles in each bin.

 

I.e. [math]N_i = \int^{x_i}_{x_{i-1}}f(v)dv[/math]

 

So, let's integrate over the population balance equation

 

[math]\int^{x_i}_{x_{i-1}}\left(

\frac{\partial f}{\partial t} \right)dv=\int^{x_i}_{x_{i-1}} \left( -\frac{\partial (\dot{V}f)}{\partial v} +

\dot{n}\delta(v-v_n) \right) dv [/math]

 

After doing the integrations, you will end up with a set of equations that looks like:

 

[math]\frac{dN_i}{dt} = G_1(N_{i-1}\dot(V)) - G_2(N_i\dot(V))[/math]

 

where the G functions are the growth (I didn't write out all the details, because it gets messy, and usually fixed sized bins yields inaccurate results). They are meant to show that the change in the number of particles in bin i are due to particles from bin i-1 growing into bin i, and the number of particles in bin i growing into particles covered by bin i+1.

 

Except the smallest bin where nucleation results, that will have an equation that looks like:

 

[math]\frac{dN_i}{dt} = - G_2(N_i\dot(V)) + \dot(n)[/math]

 

There is no smaller bin for particles to grow up from, but particles can nucleate into that size.

 

In this equation, the nucleation, which is described by a delta function becomes a source in the equation for number of particles in the smallest bin.

 

Is this totally realistic? No, because particles don't just nucleate are one size only. And, the whole description above is assuming something like a perfectly mixed batch crystalizer -- no inhomogeneity in supersaturation anywhere, no inhomogeneity in the particle size distribution anywhere. But, using a delta function to describe the nucleation events is pretty accurate -- accurate enough in terms of all the other errors in the simulation, and that's why it works.

 

-----------------------------------

 

Let me show another one:

 

I want to start with convection equation in a fluid fluid

 

[math]\rho \frac{\partial \phi}{\partial t} + \rho \mathbf{v} \cdot \nabla \phi = \nabla \cdot D \nabla \phi + S [/math]

 

[math]\phi[/math] is the conserved substance -- this could be temperature, concentration of a solute, etc.

v is the fluid velocity convecting the conserved qty

D is the diffusion rate of the conserved qty

S is the source or sink of the conserved qty

 

Let [math]\phi[/math] be the fluid temperature. Further, consider the flow over a hot-wire anemometer.

 

A hot-wire anemometer is a device inserted into the fluid flow to meter the flow. It works by measuring how much electric current is drawn through it based on the cooling effect of the fluid flow around it. But, it is a very thin piece of wire, and in some situations (high turbulence) you can ignore its disturbances to the fluid flow. But it is a source for heat. If the above equation were written for 2-D x&y, and the anemometer were placed in the fluid along the z direction, there would be a heat source in the fluid that is a point source at the given x & y location of anemometer. That is, the source in the above equation would be [math]S=W\delta{\mathbf{x}-\mathbf{x}_a}[/math] where [math]\mathbf{x}_a[/math] is the location of the wire, and W is the rate of heat coming from the wire.

 

Let's disrectize the above equation using finite volumes. I'm not going to do through the details (you can find them in any good Computational Fluid Dynamics (CFD) text) but the the space gets broken into little squares. The notation is pretty simple in that each discretized equation is written for the finite volume located at P. The cells along the x axis are labeled E (east) and W (west) with typically W being the cell next to P with the smaller x value, and E being the cell next to P with the higher x value. Along the y axis, there is N (north) and S (south), with N being higher y and S being smaller y.

 

You will integrate the above equation over each cell to create a set of discrete equations:

 

[math]\int^{x_E}_{x_W} \int^{y_N}_{y_S}(eqn)dy dx [/math]

 

The discretized equations look like:

 

[math]a_P\phi_P = a_W\phi_W + a_E\phi_E + a_N\phi_N + a_S\phi_S [/math] where the a's for E,W,N,&S take care of all the diffusion and convection. a_P takes care of the time derivative.

 

In words, this is the change of the conserved quantity [math]\phi[/math] in cell P is due to the convection and diffusion through the east face of the cell, the convection and diffusion through the west face, and similarly the north and south faces. The temperature will get convected and diffused around everywhere, except the finite volume where the location of the anemometer is. The discrete equation for that cell looks like:

 

[math]a_P\phi_P = a_W\phi_W + a_E\phi_E + a_N\phi_N + a_S\phi_S + W[/math].

 

Because the integrals above were over space and the heat from the anemometer is treated as a point source that only shows up in one cell. Of a discrete equations for the temperature across the 2-D solution space, only one of them has a source. That's where the delta function comes in -- to treat the heat source as a point source.

 

-----------------

 

I showed two different examples where the point sources have to be described using the delta function so that when you discretize the solution space, the sources show up in the discretized equations correctly. This is where they are useful. Because, they simplify the simulation method by only having source in one discretized cell. Realistically, this may not be correct -- in real life there is no such thing as a perfect point source -- but it may be good enough for the simulation accuracy to be all right.

 

I hope that this explains what I mean better. I have many more examples where they are used all the time, but I don't want to spend the time writing them all out (the above took about an hour as it is) because I think I've explained it in a lot of detail already.

Share this post


Link to post
Share on other sites
The first thing to realize regarding the delta function is that it isn't a function.

 

Looking at the delta function (OK, I'm a physicist at heart) as a limit of some series of functions is a very good approach to visualize what is going on. Bignose (post #2) and Severian (post #9) talked about it terms of a normal distribution. That's a very nice way of looking at it, but of course not the way. Some of the reasons this is particularly useful: It's already normalized for you, the normal distribution is analytic everywhere, and it generalizes to multiple dimensions.

 

This is essentially what I wanted to say. Clearly it's not possible to define an integrable function which has a finite non-zero integral but which is only non-zero at a set of points of measure zero. If you want to know more about the technical aspects of what the delta "function" is, you should actually look up distributions. Assuming you take the correct space of test functions (look up Schwarz space of rapidly reducing functions) then you're guaranteed to do cool stuff like Fourier transforms, and many of the fundamental properties hold as well.

Share this post


Link to post
Share on other sites

Could you please explain dirac delta function as a impulse? Also as stated above that this is rather a measure than a function, could you also explain how we can define dirac delta function as ameasure or distribution?

Share this post


Link to post
Share on other sites

We've already given one example of how to define the Dirac delta distribution in terms of a function: think of it as a limit of a series of functions. The normal probability distribution forms one such series. A couple more are a series of square or triangular pulses. Consider a family of square pulses given by some parameter a: f(x;a)=2/a if -a<x<a and zero elsewhere. Now take the limit as a→0. Tada! The delta distribution.

 

A series of triangular pulses is similar. Imagine a family of unit area isosceles triangles characterized by a single parameter a, with base length 2a and height of 1/a. Place the base on the triangles on the x axis and the apices on the y axis. Now form a family of functions based on these triangles. Here, f(x;a)=(1-|x/a|)/a if -a<x<a and f(x;a)=0 elsewhere.

 

Re impulses: Look at the integrals of these families of functions:

 

[math]F(x;a) = \int_{-\infty}^x f(\xi;a)d\xi[/math]

 

For the square pulses, F(x;a) is zero for x≤-a and one for x≥a. For -a<x<a, F(x;a) forms a linear ramp from 0 to 1. In the limit a→0, F(x) becomes a step function: An impulse response.

Share this post


Link to post
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
Sign in to follow this  

×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.