Dirac Delta Function: Explanation & Usage

[Fredrik]

I can't quite understand what delta-not and delta-one are, Fredrik (apologies, I can't use LaTex currently). Can you explain what they stand for?

I defined them in post #14. Not sure what more I can say, unless you explain what issues you're having with the definition.

I have never encountered such a vague and confusing topic in maths so far

Nothing is butchered quite as badly by mediocre physics text as the Dirac delta and tensors. Not sure which is worse. I remember being a lot more frustrated about the tensors actually, so maybe that's worse.

 

[dreamLord]

That usage of the Dirac function is quite beautiful wolfman, I like it very much. Thanks!

 

[Fredrik]

I have never seen Fredrik's notation, and I can't really make any sense of it. The Dirac delta is never equal to anything besides 0 or infinity.

The second sentence here suggests that you're thinking of $\delta$ as a function that takes numbers as input. It's not defined that way in any rigorous treatments. It's defined either as a distribution (a function that takes "nice enough" functions to real numbers) or as a measure (a function that take subsets of ℝ to non-negative extended real numbers that we can think of as the "sizes" of those sets).

 

[WannabeNewton]

I remember being a lot more frustrated about the tensors actually, so maybe that's worse.

Lol you should see how tensors are defined in Melvin Schwartz' book on Electromagnetism. You will cry. It was some definition in terms of rotations that I had never even seen before, and I first skimmed this book after having done Wald for some time so I was like what the hell is this?!

 

[Jolb]

The second sentence here suggests that you're thinking of $\delta$ as a function that takes numbers as input. It's not defined that way in any rigorous treatments. It's defined either as a distribution (a function that takes "nice enough" functions to real numbers) or as a measure (a function that take subsets of ℝ to non-negative extended real numbers that we can think of as the "sizes" of those sets).

You're right, but this technicality couldn't be any more irrelevant. I gave a definition for the dirac as the limit of functions, and in almost any expression with a Dirac delta distribution, you could approximate the expression to arbitrary accuracy using a function by picking a big enough \alpha in the definition I gave. The subtleties of what you're talking about won't matter at all when the OP is using Griffiths; in fact Griffiths himself says you can treat it like a function in his book.

On the other hand, your definitions are very confusing because you use the notation δ(f) to mean something completely different from δ(f(x)), which is highly nonstandard (at least for people at the level of Griffiths. Maybe it makes sense if you've taken distribution theory, but I'm sure the OP hasn't.) That's what caused the OP to get confused:

Sorry, I thought δ(f(x)) = ∫f(x)δ(x)dx - which I see makes no sense.

 

[Fredrik]

You're right, but this technicality couldn't be any more irrelevant.

It's relevant to people who want to understand the posts on page 1 of this thread better.

On the other hand, your definitions are very confusing because you use the notation δ(f) to mean something completely different from δ(f(x)), which is highly nonstandard (at least for people at the level of Griffiths. Maybe it makes sense if you've taken distribution theory, but I'm sure the OP hasn't.) That's what caused the OP to get confused:

I don't see how it can be that confusing when the definition is that simple and stated that clearly. I guess it has something to do with the fact that people at that level are for some reason used to thinking of f(x) as a function. Of course, f is the function, and f(x) is its value at x. My notation should make sense to people who understand that distinction. It's certainly not "highly non-standard".

Since $\delta$, as I defined it, takes functions to numbers, it doesn't make sense to try to use the number f(x) as input.

 

[Jolb]

It's relevant to people who want to understand the posts on page 1 of this thread better.

And the posts on page 1 you're talking about obviously did more to confuse the OP than to help him.

I guess it has something to do with the fact that people at that level are for some reason used to thinking of f(x) as a function. Of course, f is the function, and f(x) is its value at x. My notation should make sense to people who understand that distinction. It's certainly not "highly non-standard".

Well your notation δ(f) doesn't appear in Griffiths EM or any other undergraduate physics text I've ever used. However, you commonly see δ(f(x)) [I'm not sure if it's in Griffiths EM but it's certainly in Griffiths QM] and you commonly see f(x) abbreviated as f. And, honestly, stressing the distinction between f and f(x) reeks of a mathematician. Most physicists would say f(x) is "a function," and even though they know it's really the value of the function f at the point x, they do identify f(x) with f since confusing the two things is almost impossible for a physicist. Confusion would only arise in more advanced contexts. Similarly, physicists often call δ(x) the "Dirac delta function." This less rigorous but much less unnecessarily fussy approach is the one Griffiths takes, along with most other undergrad physics book authors.

I don't see how it can be that confusing when the definition is that simple and stated that clearly.

Let me try to retrace your definitions to demonstrate where they lead to confusion.

For each real number x, define $\delta_x$ by $\delta_x(f)=f(x)$ for all test functions f. Define the notation $\int f(x)\delta(x-y)dx$ to mean $\delta_y(f)$. This ensures that $\int f(x)\delta(x-y)dx=f(y)$.

And later you say.

Only one of them ($\delta_0$) is equal to $\delta$.

It looks to me like you're defining the symbol δ, but nowhere does it say it's the Dirac delta... But as best I can figure out you're saying that δ₀ is the Dirac delta. (?)

If that were the case then the definition you give for the Dirac delta is:
$\int f(x)\delta(x)dx =: \delta(f)$
So you have a δ(x) in your definition of δ(f)... what is the definition of δ(x)? You give a definition in terms of sampling I guess, but the point is that δ(x) is actually the Dirac delta (at least in the way Dirac and most physicists use it), not δ(f).

-----
Side note: Also, if I'm guessing right that you say δ₀ is the Dirac delta, this seems to imply that only δ(x) ever has anything to do with the Dirac delta, never δ(x-x₀)

 

[Fredrik]

Well your notation δ(f) doesn't appear in Griffiths EM or any other undergraduate physics text I've ever used.

That's because they don't cover the definition of $\delta$, so this is not an argument for using a different notation in the definition.

It looks to me like you're defining the symbol δ, but nowhere does it say it's the Dirac delta... But as best I can figure out you're saying that δ₀ is the Dirac delta. (?)

I defined $\delta$ at the start of that post:

One thing that I think should be mentioned is that when $\delta$ is defined as a function that takes test functions to numbers, the definition can be written as $\delta(f)=f(0)$ for all test functions f.

I think it's 100% clear from the context that the delta I'm referring to is the one that's the topic of this thread. This distribution is what we need to make sense of the notation $\int f(x)\delta(x)dx$.

You quoted something else at the end of the post. I defined a distribution $\delta_x$ for each real number x. By the definition of $\delta$, we have $\delta(f)=f(0)$ for all test functions f. By the definition of $\delta_0$, we have $\delta_0(f)=f(0)$ for all test functions f. So for all test functions f, we have $\delta(f)=f(0)=\delta_0(f)$. This is why I said that $\delta=\delta_0$. They have the same domain and the same value at each point in the domain.

These $\delta_x$ distributions (the set $\{\delta_x|x\in\mathbb R\})$ are what we need to make sense of the notation $\int f(x)\delta(x-y)dx$ where y≠0.

If that were the case then the definition you give for the Dirac delta is:
$\int f(x)\delta(x)dx =: \delta(f)$

Nooooo...That equality defines the notation $\int f(x)\delta(x)dx$ by saying that it's equal to the previously defined $\delta(f)$. The expression $\delta(x)$ is left undefined, just like the $\partial y$ in the definition of $\partial f/\partial y$. The entire expression $\int f(x)\delta(x)dx$ is defined to be equal to $\delta(f)$.

Recall that the notation $\int f(x)\delta(x)dx$ was already in use by physicists before anyone knew how to make sense of it. The point of the definition I'm talking about is that it can be used to assign a meaning to the notation $\int f(x)\delta(x)dx$.

So you have a δ(x) in your definition of δ(f)... what is the definition of δ(x)? You give a definition in terms of sampling I guess, but the point is that δ(x) is actually the Dirac delta (at least in the way Dirac and most physicists use it), not δ(f).

Some of this is based on a misunderstanding of what I was defining, but you have a point about the symbol $\delta$ being used in two places. This should however not be a problem once you understand that we're defining the entire expression $\int f(x)\delta(x)dx$, not the component parts of it. I suppose I could have used a different notation, like $\Delta$ or $D_\delta$, for the delta distribution, but I doubt that it would have made much of a difference.

Side note: Also, if I'm guessing right that you say δ₀ is the Dirac delta, this seems to imply that only δ(x) ever has anything to do with the Dirac delta, never δ(x-x₀)

I'm not sure what the standard terminology is, but it makes sense to me to call each $\delta_x$ with $x\in\mathbb R$ a delta distribution, and to call $\delta_0$ (also denoted by $\delta$) the delta distribution.

 

[Jolb]

That's because they don't cover the definition of $\delta$, so this is not an argument for using a different notation in the definition.

The definition I gave is in many common undergraduate textbooks. Take for example Shankar's Principles of Quantum Mechanics, 2nd Edition, page 61, equation 1.10.19. I'll just restate it here: \delta(x):=\lim_{\alpha\rightarrow\infty}\sqrt{\frac{\alpha}{\pi}}e^{-\alpha x^2}

I think it's 100% clear from the context that the delta I'm referring to is the one that's the topic of this thread. This distribution is what we need to make sense of the notation $\int f(x)\delta(x)dx$.

Absolutely not. I think it's 100% clear in this thread and in Griffiths that δ(x) is the topic, not your δ(f), which you use to define a big expression containing δ(x), and you leave δ(x) undefined.

These $\delta_x$ distributions (the set $\{\delta_x|x\in\mathbb R\})$ are what we need to make sense of the notation $\int f(x)\delta(x-y)dx$ where y≠0.

As I said before, if you use the definition I gave/Shankar gives, you can get arbitrarily close approximations to any expression (as far as Griffiths EM is concerned) with a δ-distribution by replacing it with the function in my/Shankar's definition, provided you pick a large enough \alpha. I think this makes perfect sense as a way to interpret δ(x) and the expression $\int f(x)\delta(x-y)dx$. We don't need fussy and confusing distribution theory.

The rest of your issues are just correcting my confusion of what you had written. Seriously, it is confusing. This is what happens when you go overboard with math, even when you don't accidentally reuse a symbol.

Recall that the notation $\int f(x)\delta(x)dx$ was already in use by physicists before anyone knew how to make sense of it.

Well those physicists (Dirac, Fermi, Feynman, etc.) were able to use it correctly without the full glory of distribution theory. So I think they knew adequately how to make sense of it. That's what's important here, not the formalism.

I'm not sure what the standard terminology is, but it makes sense to me to call each $\delta_x$ with $x\in\mathbb R$ a delta distribution, and to call $\delta_0$ (also denoted by $\delta$) the delta distribution.

To reiterate: δ(f) is not the Dirac delta. δ(x) is a Dirac delta. Two different things. Everything of the form δ(x-x₀) is also a Dirac delta.

This is the way physicists including Griffiths talk about the Dirac delta.

 

[Fredrik]

The definition I gave is in many common undergraduate textbooks. Take for example Shankar's Principles of Quantum Mechanics, 2nd Edition, page 61, equation 1.10.19.

I don't consider that a definition. It's a description of how to think about a concept that's left undefined.

Absolutely not. I think it's 100% clear in this post and in Griffiths that δ(x) is the topic, not your δ(f), which is defined in terms of δ(x), which you leave undefined.

You are 100% wrong about this. My sentence said "...when δ is defined as a function that takes test functions to numbers..." It's impossible to interpret that as anything but the Dirac delta, and it's impossible to think that I'm talking about some other version of it than the one that takes test functions to numbers, because I stated explicitly that that's the version I'm talking about.

We don't need fussy and confusing distribution theory.

If you mean that you don't need it for this type of problem, you're probably right. I haven't really thought about it, but it's not needed for most practical issues. If you mean that it's never needed, your view is pretty naive. If you mean that it wasn't needed in this thread when I posted it, you are wrong, because I only posted to clarify a few points abouts distributions that had already been made in this thread.

Seriously, it is confusing. This is what happens when you go overboard with math, even when you don't accidentally reuse a symbol.

Seriously, a definition is not going overboard.

To reiterate: δ(f) is not the Dirac delta. δ(x) is a Dirac delta. Two different things. Everything of the form δ(x-x₀) is also a Dirac delta.

I can see how one might prefer a non-rigorous presentation of the idea, but it doesn't make sense to argue that this is the right way to do it, while the rigorous definition is the wrong way.

 

[Jolb]

Your method is not wrong, it is just more confusing and goes overboard with fussy mathematics in spite of there being a simpler definition which is perfectly adequate for any calculation in Griffiths EM. What you wrote would be a lot less confusing if you make the correction you suggested: keep a separate symbol for δ(f) so it doesn't get confused with δ(f(x))--if you had done that I probably wouldn't have complained. But even if you had done that, it would have been unnecessarily rigorous for a Griffiths EM student.

Indeed, if we go beyond Griffiths EM into, say, peer-reviewed Math journals, then in some cases it will turn out that my method is wrong and yours is right. But you have to go to quite a high level to find where my method breaks down--such a high level is not something a Griffiths EM student needs to be worrying about.

For the record, there should be an implicit "in the context of Griffiths EM" along with everything I said up to here: certainly there are places where distribution theory is needed, and it is an important aspect of mathematics. But my definition works fine for Griffiths EM and pretty much anything else you do as a physics undergrad (and most physics grad classes too).

Edit: Maybe you don't need to go to "quite a high level" just to find pathological examples. But these pathological examples certainly won't show up in the level of Griffiths EM, and they probably won't pose any real problems until you reach a high level.

 

[pwsnafu]

The definition I gave is in many common undergraduate textbooks. Take for example Shankar's Principles of Quantum Mechanics, 2nd Edition, page 61, equation 1.10.19. I'll just restate it here: \delta(x):=\lim_{\alpha\rightarrow\infty}\sqrt{\frac{\alpha}{\pi}}e^{-\alpha x^2}

I need to point out that that definition is only valid as a weak limit (i.e. in the weak topology of functionals). If you don't explain what a weak limit is, it's a very bad definition.

 

[lurflurf]

The Riemann–Stieltjes integral is a good way to handle these simple matters without the theory of distributions. The Riemann–Stieltjes integral is not in most first year calculus books, so you might need a second year book. We consider sums like

$$\int \! \mathop{f}(x) \, \mathop{dg}=\lim \sum \mathop{f}(x^*_k)(\mathop{g}(x_{k-a}-\mathop{g}(x_k))$$
where for the Riemann integral g(x)=x

Then

$$\int \! \delta (x) \, \mathop{f}(x) \, \mathop{d x}=\int \! \mathop{f}(x) \, \mathop{d u}=\mathop{f}(0)$$
where
u=0 x<0;1x>0
u'=δ

This all arises when we want to model something as concentrated. For example something that happens instantly or substance that is only at one point or surface instead of a region in space.

 

[pwsnafu]

The Riemann–Stieltjes integral is a good way to handle these simple matters without the theory of distributions. The Riemann–Stieltjes integral is not in most first year calculus books, so you might need a second year book. <snip>

The R-S integral is also good because there is a geometric interpretation. See
Gregory L. Bullock, A geometric interpretation of the Riemann-Stieltjes integral,
The American Mathematical Monthly 95 (1988), no. 5, 448–455.

 

[Jolb]

I need to point out that that definition is only valid as a weak limit (i.e. in the weak topology of functionals). If you don't explain what a weak limit is, it's a very bad definition.

Well I didn't go into the formalism of the topology of functionals, because I don't think that would help for a Griffiths EM student. What I said a few times is that "you can get arbitrarily close approximations to any expression with a δ (in Griffiths) by substituting it with my definition, picking a large enough \alpha." This is actually the same thing Shankar says in the section I referenced. I think that in the context of Griffiths, that's perfectly unambiguous and well-defined and is equivalent to whatever fancy mathematical terminology you prefer.