Dirac Delta Function question(s)

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OK, so my basic understanding of Dirac Delta Function is that it shows the probability of finding a point at [tex](p,q)[/tex] at time [tex]t[/tex]. Dirac Delta is 0 everywhere except for [tex](p_{0},q_{0})[/tex].

So my question comes
Is it possible that a point enters the [tex](p_{0},q_{0})[/tex] and stays there (for some period of time)? Then dirac's delta would have a chunk instead of one point missing, which does not match it's definition.

Correct me if I am wrong please
 
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Andy Resnick
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My understanding for the delta function is that it represents a point source. A point source of *what* depends on the application. But, when solving (for example) an inhomogeneous ODE, the use of Green's functions and the Delta function are a simple way of building up a complete solution.
 
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The probability of finding a particle at some point will be described by some sort of probability distribution function. The dirac delta function (d(x)) is a function that is zero everywhere, except at 0 (or a shifted origin) where its infinite. It ussually describes a point source, as aforementioned. For instance, in the classical point-particle case the density could be described by a delta function (i.e. zero everywhere except infinite at the point where the particle is).
You can also create a delta function that is zero everywhere, and finite at a certain point... this occassionally pops-up in a prob. dist. function... can't think of a particle case though.
 
rbj
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You can also create a delta function that is zero everywhere, and finite at a certain point... this occassionally pops-up in a prob. dist. function... can't think of a particle case though.
problem is that one integrates to nothing. from a physical POV, it's virtually nothing.
 
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rbj, thats really not true at all. the integral of a deltafunction is always finite, which you can see in practice - but fundamentally its true by definition.
At the same time its far from "virtually nothing" because a single source (described by a delta function) can account for all of the phenomena in a given region.
 
nicksauce
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Actually it is true. If f(x) = {0; x!=x0, C; x = x0} for some real number C, then the integral from -infinity to infinity of f(x) is 0.
 
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Well, from the book I read, the dirac delta function is described differently. let me type an approximation here (I have to translate it while I type):


"
Gibbs' ensemble can be represented as a 2-Dimensional cloud of points, which can be described by the function [tex]\rho(q,p,t)[/tex]. It has a simple physical interpretation: it's nothing but the probability of finding a point at time [tex]\displaystyle{t}[/tex] with the coordinates [tex]\displaystyle{q,p}[/tex]. Thus, the function [tex]\displaystyle{\rho}[/tex] imples being zero at all values of [tex]\displaystyle{q,p}[/tex], except for one case [tex]q_{0},p_{0}[/tex]. Functions having the property that they are zero everywhere except for one point are called Dirac Delta Functions and are labeled [tex]\displaystyle{\delta(x)}[/tex]. Function that we need in this case, [tex]\delta(x-x_{0})[/tex] equals zero everywhere except for [tex]x=x_{0}[/tex]. Thus, for time [tex]\displaystyle{t}[/tex]=0, the probabilty [tex]\displaystyle{\rho}[/tex] has the format [tex]\rho=\delta(q-q_{0})\delta(p-p_{0})[/tex].
"

And by the way

[tex]\displaystyle\int^{\infty}_{-\infty} \delta(x)\,dx = 1[/tex]
 
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That is definitely the dirac delta function. at the same time, rho isn't the probability, because at the point qo,po, its infinite instead of one (one is probability version of "hell yes, definitely" of course).
rho, in this case, is the probability density function; i.e. if you integrate it over some amount of space -> you find the probability of finding it in that space.
If you integrate this specific rho over any space that includes q0 p0, you get 1. i.e. if you are looking at any space that includes the point q0,p0, you are certain to find the particle.
Does that clear up things?
 
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That is definitely the dirac delta function. at the same time, rho isn't the probability, because at the point qo,po, its infinite instead of one (one is probability version of "hell yes, definitely" of course).
rho, in this case, is the probability density function; i.e. if you integrate it over some amount of space -> you find the probability of finding it in that space.
If you integrate this specific rho over any space that includes q0 p0, you get 1. i.e. if you are looking at any space that includes the point q0,p0, you are certain to find the particle.
Does that clear up things?
oooh
darn
I've been stupid.
Thanks, it all fits together into my world now! D=
 
Andy Resnick
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The probability of finding a particle at some point will be described by some sort of probability distribution function. The dirac delta function (d(x)) is a function that is zero everywhere, except at 0 (or a shifted origin) where its infinite. It ussually describes a point source, as aforementioned. For instance, in the classical point-particle case the density could be described by a delta function (i.e. zero everywhere except infinite at the point where the particle is).
You can also create a delta function that is zero everywhere, and finite at a certain point... this occassionally pops-up in a prob. dist. function... can't think of a particle case though.
Delta functions of finite height (?) are used a lot in optics- the magnitude of the delta functions are correlated with amplitudes of spatial frequencies. Also, the height can be correlate with intensity.
 
Dirac delta is not a function and thinking about it as if it's a function is completely misleading.

It is a distribution, it is defined in terms of what it does when integrated against a function. The visual representation of it being a big spike is cute but I think ultimately misleading.
 
rbj
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no, Iz, the function that i commented on (one that is zero everywhere, but for one point where it takes on a finite value) has integral of zero. i don't even need to look up your references (i had some role in writing one of them).

Dirac delta is not a function and thinking about it as if it's a function is completely misleading.

It is a distribution, it is defined in terms of what it does when integrated against a function. The visual representation of it being a big spike is cute but I think ultimately misleading.
from a strict mathematical POV, i guess you're right, David, but from an engineering POV, i know for certain that you are not. in fact, ton's of textbooks are printed defining the dirac delta as such a limiting function of some "nascent delta" with unity area that is getting thinner and thinner and, in the limit, becomes virtually zero width. when used to sample decent and continuous functions, this "misleading" manner of thinking about is just fine and better, IMO (and i used to teach electrical engineering in years past) than trying to confuse a bunch of sophmores or juniors with the rigorous mathematical definition because they're gonna think of it as a big spike anyway.

i also think that where the dirac delta is used in Newtonian mechanics, the infinitely thin spike with finite and non-zero area is just fine also.
 
nicksauce
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reilly
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A delta function is a so-called generalized function or distribution. It is not a function by usual standards, but is the limit of normal function, when it is part of an integrand. For example, a delta function is, among many alternatives, the limit of a Gaussian probability distribution as the standard deviation approaches zero.

In quantum mechanics, physicists play fast and loose with delta functions, particularly with products of delta functions. Thus thinking in terms of Gaussian helps greatly.

There's an excellent book by Lighthill, Generalized Functions and Fourier Analysis, somewhat older. But worth the trouble to find and read -- it is at an advanced undergraduate level. He defines, as do most authors on the subject, generalized functions as the limits of ordinary functions. rbj is right except for the "misleading" part.

If you understand things at the level of Lighthill, then looking at a delta as a narrow Gaussian is quite proper, as long as it is used as part of an integrand.

In E&M, a delta function is used as a charge distribution of a point particle -- useful fiction. Further, it is almost always used as the source for Green's Functions -- the Green's Functions obtained this way are identical to those obtained in the old fashioned way, with use of jump conditions -- see any older text on E&M.

When I taught Advanced QM and E&M, I defined a delta function as the derivative of a so-called theta function --
THETA(X) = 0 if X<0; 1 if X>0;
some authors define THETA(0) =1/2.
This is an approach quite common in physics.

Regards,
Reilly Atkinson
 
Andy Resnick
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Dirac delta is not a function and thinking about it as if it's a function is completely misleading.

It is a distribution, it is defined in terms of what it does when integrated against a function. The visual representation of it being a big spike is cute but I think ultimately misleading.
Just becasue it may not have orthodox mathematical properties does not make it less useful for physical representation of certain physical phenomena.
 
Just becasue it may not have orthodox mathematical properties does not make it less useful for physical representation of certain physical phenomena.
Well I have a problem with both parts of what you said. I did not say that it was useless for representing physical phenomena. Nor did I say that it was unorthodox, I said that it was not a function, as in real valued function, which is not the same thing as unorthodox.

I know every one is saying that they usually learned the Dirac Delta to be defined as a limit of a sequence of functions, and that's the first time I saw it too... but it's a pretty stupid way to define it.

Simply defining it by what it does to functions makes much more sense because it's a practical definition, it's telling you exactly what it does. The Dirac Delta is a measure and saying it works by [tex]\int \delta(x)f(x)=f(0)[/tex] is much more simple and direct than considering a sequence of functions with an ill-defined limit.
 
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If you're going to make claims that much of Analysis is wrong, ie that the integral of f(x) is non-zero, where f(x) = {0 x!=0, C x=0}, then you are going to actually have to provide some references that support your claims, rather than references that give us definitions we already know.
if the references say, explicitly that the integral of the delta function is non-zero, i think that supports my claim that the integral of the delta function is non-zero.
Additionally, its true even for a finite delta function; i.e. f(x) = C*(del(x)) [constant C times the delta function of x], then the integral is again non-zero, in this case C.
Again, this is exactly what the references show. For a more academic reference see Boas' Math Methods, or any other math methods book.
 
Hurkyl
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Functions have a value at zero.

Ordinary limits of functions do not satisfy the equation
[tex]
\int_a^b \left( \lim_{t \rightarrow c} f_t(x) \right) g(x) \, dx
=
\lim_{t \rightarrow c} \left( \int_a^b f_t(x) g(x) \, dx\right)
[/tex]


What is to be gained by insisting upon having students think that a dirac delta distribution as something that has a value at zero, or is equal to an ordinary limit of functions?
 
lurflurf
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One can use the Riemann-Stieltjes integral, limits, or Dirac sequences, or distribution theory to justify the Dirac Delta function. One can then deal with bothersome technical points as they arise. The Dirac Delta function is often well used in an informal way.

Dirac's delta function is not a function! (despite the name)
What is wanted in applications is a way to unify local and global sources
Best not to ask questions that ask for trouble like
δ(δ(x))
or
1/δ(x)
 
rbj
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Functions have a value at zero.
every function? log(x) has a value at zero?

Ordinary limits of functions do not satisfy the equation
[tex]
\int_a^b \left( \lim_{t \rightarrow c} f_t(x) \right) g(x) \, dx
=
\lim_{t \rightarrow c} \left( \int_a^b f_t(x) g(x) \, dx\right)
[/tex]


What is to be gained by insisting upon having students think that a dirac delta distribution as something that has a value at zero,
i wouldn't suggest that to students (other than possibly an infinite value in the loose sense).

or is equal to an ordinary limit of functions?
so that we can get some work done in Linear Systems class.

A delta function is a so-called generalized function or distribution. It is not a function by usual standards, but is the limit of normal function, when it is part of an integrand. For example, a delta function is, among many alternatives, the limit of a Gaussian probability distribution as the standard deviation approaches zero.
but calling it a limiting function would not be legit. it is legit to define:

[tex] \delta(t) = \lim_{a \rightarrow 0} \delta_a(t) [/tex]

where

[tex] \int_{-\infty}^{+\infty} \delta_a(t) dt = 1 \quad \forall a>0 [/tex]

and

[tex] \lim_{a \rightarrow 0} \delta_a(t) = 0 \quad \forall t \ne 0 [/tex]

can we say that in an electrical engineering class and still be respected by the mathematicians?

In quantum mechanics, physicists play fast and loose with delta functions,
i guess electrical engineers play even faster and looser with them.

particularly with products of delta functions.
yeah, we don't do that. i dunno how convolving a delta with another would do, but we don't multiply deltas with deltas. if this multiplication was done in the frequency domain, it would corrospond to convolving DC against DC in the time domain, with an undefinied (infinite) result for every lag in the convolution. it's kinda non-sensical.

Thus thinking in terms of Gaussian helps greatly.
i don't know why, as a nascent delta that the Gaussian is better than a rectangular or triangular function of unit area. the weird one would be a sinc() function since, as [itex]a \rightarrow 0[/itex], the 1/x envelope of the sinc() would remain at the same amplitude.

There's an excellent book by Lighthill, Generalized Functions and Fourier Analysis, somewhat older. But worth the trouble to find and read -- it is at an advanced undergraduate level. He defines, as do most authors on the subject, generalized functions as the limits of ordinary functions. rbj is right except for the "misleading" part.
i dunno. i think in an undergraduate EE's first Linear System Theory class (now the name in vogue is "Signals and Systems" after that Oppenhiem and Wilsky text), to fuss with this "delta function is not a real function, but a distribution" stuff only adds cruft and confusion.

If you understand things at the level of Lighthill, then looking at a delta as a narrow Gaussian is quite proper, as long as it is used as part of an integrand.

In E&M, a delta function is used as a charge distribution of a point particle -- useful fiction. Further, it is almost always used as the source for Green's Functions -- the Green's Functions obtained this way are identical to those obtained in the old fashioned way, with use of jump conditions -- see any older text on E&M.

When I taught Advanced QM and E&M, I defined a delta function as the derivative of a so-called theta function --
THETA(X) = 0 if X<0; 1 if X>0;
some authors define THETA(0) =1/2.
This is an approach quite common in physics.
that's what we EEs call a "unit step function" and other folks call the "Heaviside" step function, and we do view it as the integral of the dirac delta. because of the undefined slope of the unit step at 0, we haven't often called the "dirac delta the derivative of the unit step", although it effectively plays that role.
 
Hurkyl
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every function? log(x) has a value at zero?
Well, we had previously been talking about generalized functions whose domain was the real numbers, so I didn't think it necessary to specify that I was talking about functions whose domain was the real numbers.

i wouldn't suggest that to students (other than possibly an infinite value in the loose sense).
But yet, that is what you are suggesting, if you tell them to think of it as a function! I'm not asking one to give a full, rigorous presentation of the theory of distributions; but I am suggesting that, at a minimum, the students should be made aware that they are working with something that isn't a function -- merely similar to a function. And ideally, they should be presented with a list of valid arithmetic manipulations.


but calling it a limiting function would not be legit. it is legit to define:

[tex] \delta(t) = \lim_{a \rightarrow 0} \delta_a(t) [/tex]
...
can we say that in an electrical engineering class and still be respected by the mathematicians?
No. However, if you were to further add that this is not the limit operation they learned in their calculus class, that would be better. And if you specified that the meaning of this new limit operator is that, whenever it appears inside an integral, you're first supposed to pull it outside of the integral (at which point it turns into an ordinary limit), then, IMHO, you should have the respect of mathematicians. (And, more importantly, you avoid alienating any of your students who actually understood their calculus courses, and dare to think about what they're doing)

If this were an ordinary limit, then we would have the (incorrect) equation:

[tex]\int_a^b \delta(x) f(x) \, dx = \int_a^b \left( \lim_{c \rightarrow 0} \delta_c(x) \right) f(x) \, dx [/tex]

whereas actual meaning of the notation is that we should have this (correct) equation:

[tex]\int_a^b \delta(x) f(x) \, dx = \left( \lim_{c \rightarrow 0} \int_a^b \delta_c(x) f(x) \, dx \right) [/tex]



In fact, if we treated the limit as an ordinary one, and treated the integral according to the usual conventions, then we would get
[tex]\int_{-\infty}^{+\infty} \left( \lim_{c \rightarrow 0} \delta_c(x) \right) f(x) \, dx = 0[/tex]
for every function f.
 
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Andy Resnick
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Well I have a problem with both parts of what you said. I did not say that it was useless for representing physical phenomena. Nor did I say that it was unorthodox, I said that it was not a function, as in real valued function, which is not the same thing as unorthodox.

I know every one is saying that they usually learned the Dirac Delta to be defined as a limit of a sequence of functions, and that's the first time I saw it too... but it's a pretty stupid way to define it.

Simply defining it by what it does to functions makes much more sense because it's a practical definition, it's telling you exactly what it does. The Dirac Delta is a measure and saying it works by [tex]\int \delta(x)f(x)=f(0)[/tex] is much more simple and direct than considering a sequence of functions with an ill-defined limit.
Actually, I learned it by the defintion you give, [tex]\int \delta(x)f(x)=f(0)[/tex], rather than as a limit. I too found the limit method of definition to be confusing and long-winded.
 
rbj
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But yet, that is what you are suggesting, if you tell them to think of it as a function! I'm not asking one to give a full, rigorous presentation of the theory of distributions; but I am suggesting that, at a minimum, the students should be made aware that they are working with something that isn't a function -- merely similar to a function. And ideally, they should be presented with a list of valid arithmetic manipulations.


it is legit to define:

[tex] \delta(t) = \lim_{a \rightarrow 0} \delta_a(t) [/tex]

where

[tex] \int_{-\infty}^{+\infty} \delta_a(t) dt = 1 \quad \forall a>0 [/tex]

and

[tex] \lim_{a \rightarrow 0} \delta_a(t) = 0 \quad \forall t \ne 0 [/tex]

can we say that in an electrical engineering class and still be respected by the mathematicians?
No. However, if you were to further add that this is not the limit operation they learned in their calculus class, that would be better. And if you specified that the meaning of this new limit operator is that, whenever it appears inside an integral, you're first supposed to pull it outside of the integral (at which point it turns into an ordinary limit), then, IMHO, you should have the respect of mathematicians. (And, more importantly, you avoid alienating any of your students who actually understood their calculus courses, and dare to think about what they're doing).
Hurk, this is stuff that does not happen in normal calculus courses. it wasn't until i took a course in Real Analysis and first learned of Lebesgue integration that i had any idea that, as functions are strictly defined by mathematicians, that for a real function f(x) that agrees with g(x) almost everywhere, that their integrals were the same. i immediately asked the the math prof about the dirac delta functions (what we EEs called "unit impulse function") and without much else elucidation, he just said that it wasn't a real function. that didn't help explain how it was that we were using this nonfunction like a function in all of our EE communications theory and control systems classes. it had a Fourier and Laplace Transform. it could be integrated (and becomes a unit step function when it is). we knew the sampling property of it and that eventually the impulse function had to end up in an integral for us to extract an actual number using it. we added them together, and even, regarding the Nyquist Sampling Theorem, we have the audacity to say this:

[tex] \sum_{n=-\infty}^{+\infty} \delta(t-n) = \sum_{k=-\infty}^{+\infty} e^{i 2 \pi k t} [/tex]

we say that and use it even though the deltas are not in any integral (but to get the Fourier coefficients, the delta goes into an integral). the DSP works. we understand what happens in the frequency domain when we sample. nothing crashes and we're just fine. even though this is in many signal processing and discrete linear systems texts, i've been told by more than one mathematician that the above equation is "meaningless". i think that it is not.

when physicists use the dirac delta as an impulse function (like an idealized force vs. time function for a very hard and elastic collision) must they be any more rigorous than the EEs are with it. we can't get away with playing fast and loose?
 

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