A question about Dirac delta function...


by James MC
Tags: delta, dirac, function
James MC
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#1
Feb3-13, 12:16 AM
P: 168
Hello,

Is this correct:
[tex] \int [f_j(x)\delta (x-x_i) f_k(x)\delta (x-x_i)]dx = f_j(x_i)f_k(x_i)[/tex]
If it is not, what must the left hand side look like in order to obtain the right handside, where the right hand side multiplies two constants?

Thanks!
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marcusl
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#2
Feb3-13, 11:37 AM
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First the range of the integral must include the point x=x_i or all bets are off. Second, it's cleaner to use only one delta function.
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Feb3-13, 12:41 PM
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Quote Quote by marcusl View Post
First the range of the integral must include the point x=x_i or all bets are off. Second, it's cleaner to use only one delta function.
It's not only cleaner, using two delta functions is incorrect!

Things like [itex]\delta(x-x_i)\delta(x-x_i)[/itex] or [itex]\delta^2(x)[/itex] don't exist (as far as I know).

James MC
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#4
Feb3-13, 08:14 PM
P: 168

A question about Dirac delta function...


Okay, thanks. I would be interested to know why exactly [itex]\delta^2(x)[/itex] doesn't exist. Assuming it doesn't, then how do I derive the right hand side of the equals sign? Can the two test functions simply be placed beside each other (as if the spike on the graph were transformed by the test function j and then the second test function k)?

That is, is this correct:
[tex] \int^{\infty}_{-\infty}[f_j(x) f_k(x)\delta (x-x_i)]dx = f_j(x_i)f_k(x_i)[/tex]
(Marcus: integral range is now explicit.)
pwsnafu
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#5
Feb3-13, 08:32 PM
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Quote Quote by James MC View Post
Okay, thanks. I would be interested to know why exactly [itex]\delta^2(x)[/itex] doesn't exist.
We just don't know what it should be. Unlike standard multiplication of real functions, there is no "canonical" multiplication of generalized functions.

Assuming it doesn't, then how do I derive the right hand side of the equals sign? Can the two test functions simply be placed beside each other (as if the spike on the graph were transformed by the test function j and then the second test function k)?

That is, is this correct:
[tex] \int^{\infty}_{-\infty}[f_j(x) f_k(x)\delta (x-x_i)]dx = f_j(x_i)f_k(x_i)[/tex]
(Marcus: integral range is now explicit.)
Yes. If you let ##g(x) = f_j(x) \, f_k(x)## then this is trivial.
James MC
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#6
Feb3-13, 09:12 PM
P: 168
Thank you pwsnafu. I just have a quick follow up question: can one of the test functions be a function from x to a vector quantity? I understand that if ##f_j(x_i)## is a scalar function, say, 2, then the height of the spike on the graph doubles, from ##\infty## to ##2\infty##. But I'm not sure what happens to the spike if we multiply by a vector, is this well defined?

I'm not sure of a good example, but imagine for some reason you knew the mass (scalar) and acceleration (vector) of a Newtonian particle, and you wanted to determine its force (vector). Let's say you also wanted to show off your grasp of delta functions in the process; does this make sense:
[tex] \int^{\infty}_{-\infty}[m(x) \vec{a}(x)\delta (x-x_i)]dx = m_i*\vec{a}_i = \vec{F}_i[/tex]
...where ##\vec{F}_i## is the force on the particle located at ##x_i##, and ##m_i*\vec{a}_i## is that particle's mass times its instantaneous acceleration?
pwsnafu
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Feb3-13, 09:56 PM
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Quote Quote by James MC View Post
Thank you pwsnafu. I just have a quick follow up question: can one of the test functions be a function from x to a vector quantity?
Never seen it in the literature.

I understand that if ##f_j(x_i)## is a scalar function, say, 2, then the height of the spike on the graph doubles, from ##\infty## to ##2\infty##.
You don't understand at all. To begin with ##2\infty = \infty## on the extended reals.
We aren't doing calculus in the ordinals.

But I'm not sure what happens to the spike if we multiply by a vector, is this well defined?
The "spike" is a heuristic. Mathematically, you are not "integrating a function with an infinitely tall spike", because that equals 0 or is divergent (depending on your definition of integral).

I'm not sure of a good example, but imagine for some reason you knew the mass (scalar) and acceleration (vector) of a Newtonian particle, and you wanted to determine its force (vector). Let's say you also wanted to show off your grasp of delta functions in the process; does this make sense:
[tex] \int^{\infty}_{-\infty}[m(x) \vec{a}(x)\delta (x-x_i)]dx = m_i*\vec{a}_i = \vec{F}_i[/tex]
...where ##\vec{F}_i## is the force on the particle located at ##x_i##, and ##m_i*\vec{a}_i## is that particle's mass times its instantaneous acceleration?
Assuming ##\mathbf{F}:\mathbf{R}^n\to\mathbf{R}^m##, you would need define
##\int_{\mathbb{R}^n} \mathbf{F}(\mathbb{x}) \, \delta(\mathbf{x}-\mathbf{x}_i) \, d\mathbf{x} := \mathbf{F}(\mathbf{x}_i)##
and develop the theory from scratch. ##\mathbf{R}^m## is not field for ##m>1##, so that is no longer a functional.

Note that in the literature you'll sometimes see ##\delta^3(x-x_i)##, they mean "the Dirac with variable in 3D space" and not "Dirac cubed".
James MC
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#8
Feb3-13, 11:10 PM
P: 168
Quote Quote by pwsnafu View Post
Never seen it in the literature.
Interesting. After all, I take it vector integration is well defined for ordinary distributions, and commonly used in vector calculus. I wonder what the difference is here? This has come up elsewhere, for consider post #9, where elfmotat formulates Newton's gravitation feild law in DDF terms, where rather than an acceleration vector, there is a position vector inside the integrand. Is there a difference here or is his equation wrong? (See: http://www.physicsforums.com/showpos...10&postcount=9)
Quote Quote by pwsnafu View Post
You don't understand at all. To begin with ##2\infty = \infty## on the extended reals.
We aren't doing calculus in the ordinals.
Technically yes, though I thought this was a pretty standard piece of picture thinking used in the teaching of Dirac delta functions. In fact I just got that from 12.30 minutes in, on this teaching clip: http://www.khanacademy.org/math/diff...delta-function
Quote Quote by pwsnafu View Post
Assuming ##\mathbf{F}:\mathbf{R}^n\to\mathbf{R}^m##, you would need define
##\int_{\mathbb{R}^n} \mathbf{F}(\mathbb{x}) \, \delta(\mathbf{x}-\mathbf{x}_i) \, d\mathbf{x} := \mathbf{F}(\mathbf{x}_i)##
and develop the theory from scratch. ##\mathbf{R}^m## is not field for ##m>1##, so that is no longer a functional.
What is gained by defining a delta function for the force F (or for the field g(x) in Elfmotat's gravitation equation)? Assuming that one function can invoke vector quantities, then we know (via the sifting property) this:
[tex] \int^{\infty}_{-\infty}[m(x) \vec{a}(x)\delta (x-x_i)]dx = m_i*\vec{a}_i[/tex]
And we also know that the right hand side is equivalent to the right result (that is ##\vec{F}_i##). I don't understand how this result is blocked if we haven't yet defined a DDF for the force.
pwsnafu
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#9
Feb4-13, 12:44 AM
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Quote Quote by James MC View Post
Interesting. After all, I take it vector integration is well defined for ordinary distributions, and commonly used in vector calculus. I wonder what the difference is here? This has come up elsewhere, for consider post #9, where elfmotat formulates Newton's gravitation feild law in DDF terms, where rather than an acceleration vector, there is a position vector inside the integrand. Is there a difference here or is his equation wrong? (See: http://www.physicsforums.com/showpos...10&postcount=9)
I'm not a physicist. I can't parse what you wrote in the link.
A generalised function is a functional: by definition that means the co-domain is the base field, i.e. ℝ1. If you want other spaces you'll need to do it yourself.

What is gained by defining a delta function for the force F (or for the field g(x) in Elfmotat's gravitation equation)? Assuming that one function can invoke vector quantities, then we know (via the sifting property) this:
[tex] \int^{\infty}_{-\infty}[m(x) \vec{a}(x)\delta (x-x_i)]dx = m_i*\vec{a}_i[/tex]
And we also know that the right hand side is equivalent to the right result (that is ##\vec{F}_i##). I don't understand how this result is blocked if we haven't yet defined a DDF for the force.
You misread my post. I never said F is the force. It's an arbitrary function ##\mathbb{R}^n\to\mathbb{R}^m##. I just chose F.
elfmotat
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#10
Feb4-13, 01:18 AM
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Hello again James!

The equation I wrote for g(x) in the thread you mention is actually one-dimensional, though it is easily generalized to three dimensions. It tells you the gravitational field at x due to a mass-density function [itex]\rho=\sum_i m_i~ \delta(\zeta -x_i)[/itex]. I.e. it tells you the gravitational field at point x due to a finite number of point particles of mass [itex]m_i[/itex] located at the points [itex]x_i[/itex].

The analogous vector equation you're looking for is:

[tex]\mathbf{g}(\mathbf{x})=G\int \sum_i \frac{\mathbf{x}-\mathbf{r}}{|\mathbf{x}-\mathbf{r}|^3}~m_i~\delta(\mathbf{r}-\mathbf{x}_i)~\mathrm{d}^3\mathbf{r}[/tex]

This tells you the gravitational field at the position vector [itex]\mathbf{x}[/itex] due to a number of point particles of mass [itex]m_i[/itex] located at the position vectors [itex]\mathbf{x}_i[/itex].

Note that this equation is well known and can be found here: http://en.wikipedia.org/wiki/Gauss's...Newton.27s_law
pwsnafu
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Feb4-13, 01:35 AM
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James: you need to be very careful because a "vector function" could mean ℝ3 → ℝ, or ℝ3 → ℝ3 depending on the authors.

elfmotat: you've bolded g, so I'm assuming it's co-domain is ℝ3. If that's true, I'm guessing you are applying Dirac component-wise, is that correct?
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Feb4-13, 01:45 AM
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Quote Quote by pwsnafu View Post
elfmotat: you've bolded g, so I'm assuming it's co-domain is ℝ3. If that's true, I'm guessing you are applying Dirac component-wise, is that correct?
Yes, that's right. More explicitly, I'm using:

[tex]\int \delta (\mathbf{r}-\mathbf{x}_i)\mathrm{d}^3 \mathbf{r}=\iiint \delta (x-x_i) \delta (y-y_i) \delta (z-z_i)~\mathrm{d}x\mathrm{d}y\mathrm{d}z[/tex]

Where [itex]\mathbf{r}=(x,y,z)[/itex] and [itex]\mathbf{x}_i=(x_i,y_i,z_i)[/itex]. (Though this assumes we're working in Cartesian coordinates.)
James MC
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#13
Feb4-13, 06:41 AM
P: 168
Thanks elfmotat! I was wondering if one just needed more spatial dimensions. I'm curious about your formulation, is it important that you have the sum exactly where you put it? For example, are these three equivalent:
[tex] \mathbf{g}(\mathbf{x})=G\int \sum_i \frac{\mathbf{x}-\mathbf{r}}{|\mathbf{x}-\mathbf{r}|^3}~m_i~\delta(\mathbf{r}-\mathbf{x}_i)~\mathrm{d}^3\mathbf{r}[/tex]
[tex] \mathbf{g}(\mathbf{x})=G \sum_i\int \frac{\mathbf{x}-\mathbf{r}}{|\mathbf{x}-\mathbf{r}|^3}~m_i~\delta(\mathbf{r}-\mathbf{x}_i)~\mathrm{d}^3\mathbf{r}[/tex]
[tex] \mathbf{g}(\mathbf{x})=G\int \frac{\mathbf{x}-\mathbf{r}}{|\mathbf{x}-\mathbf{r}|^3}\sum_i ~m_i~\delta(\mathbf{r}-\mathbf{x}_i)~\mathrm{d}^3\mathbf{r}[/tex]
(Reason I ask: I've always used the second equation to represent the many particle law, and I use the third equation to show that by swapping in the sum, one can recover the form of the single particle law in a manner that is relevant to proving mass addditivity. So I'm wondering if they're both equivalent to your formulation (i.e. the first equation).)

And now relating this to the inertial law, am I right in thinking that to account for acceleration in DDF form, I just need to make clear that I'm working in three dimensions:
[tex] \int m(\mathbf{x}) \mathbf{a}(\mathbf{x})\delta (\mathbf{x}-\mathbf{x}_i)]d^3\mathbf{x} = m_i*\mathbf{a}_i = \mathbf{F}_i[/tex]
And for the total force ##F_T## on multiple particles:
[tex] \sum_i \int m_i(\mathbf{x}) \mathbf{a}_i(\mathbf{x})\delta (\mathbf{x}-\mathbf{x}_i)]d^3\mathbf{x} = \mathbf{F}_T[/tex]
Thanks for the terminological hint pwsnafu, and sorry for the misunderstanding about F - I see what you mean now.
elfmotat
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Feb4-13, 12:16 PM
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Quote Quote by James MC View Post
Thanks elfmotat! I was wondering if one just needed more spatial dimensions. I'm curious about your formulation, is it important that you have the sum exactly where you put it? For example, are these three equivalent:
[tex] \mathbf{g}(\mathbf{x})=G\int \sum_i \frac{\mathbf{x}-\mathbf{r}}{|\mathbf{x}-\mathbf{r}|^3}~m_i~\delta(\mathbf{r}-\mathbf{x}_i)~\mathrm{d}^3\mathbf{r}[/tex]
[tex] \mathbf{g}(\mathbf{x})=G \sum_i\int \frac{\mathbf{x}-\mathbf{r}}{|\mathbf{x}-\mathbf{r}|^3}~m_i~\delta(\mathbf{r}-\mathbf{x}_i)~\mathrm{d}^3\mathbf{r}[/tex]
[tex] \mathbf{g}(\mathbf{x})=G\int \frac{\mathbf{x}-\mathbf{r}}{|\mathbf{x}-\mathbf{r}|^3}\sum_i ~m_i~\delta(\mathbf{r}-\mathbf{x}_i)~\mathrm{d}^3\mathbf{r}[/tex]
(Reason I ask: I've always used the second equation to represent the many particle law, and I use the third equation to show that by swapping in the sum, one can recover the form of the single particle law in a manner that is relevant to proving mass addditivity. So I'm wondering if they're both equivalent to your formulation (i.e. the first equation).)
These are all equivalent.

Quote Quote by James MC View Post
And now relating this to the inertial law, am I right in thinking that to account for acceleration in DDF form, I just need to make clear that I'm working in three dimensions:
[tex] \int m(\mathbf{x}) \mathbf{a}(\mathbf{x})\delta (\mathbf{x}-\mathbf{x}_i)]d^3\mathbf{x} = m_i*\mathbf{a}_i = \mathbf{F}_i[/tex]
And for the total force ##F_T## on multiple particles:
[tex] \sum_i \int m_i(\mathbf{x}) \mathbf{a}_i(\mathbf{x})\delta (\mathbf{x}-\mathbf{x}_i)]d^3\mathbf{x} = \mathbf{F}_T[/tex]
Thanks for the terminological hint pwsnafu, and sorry for the misunderstanding about F - I see what you mean now.
I'm a bit confused as to why you're trying to write F=ma with Dirac deltas. Perhaps you could explain your thought process a bit?
James MC
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#15
Feb4-13, 09:23 PM
P: 168
I do appreciate you trying to help me, Elfmotat. The background to all this is somewhat involved. What follows is a trade off between brevity and coherence. I aim to meet you somewhere in the middle. Okay here goes...

The project concerns the principle of mass additivity, which states that the mass of a composite object is the sum of the masses of its elementary parts. In most textbook discussions of Newtonian mechanics, this principle is simply presupposed, despite playing important roles (e.g. in the proof that centres of mass obey Newton's laws). However, some textbooks are more rigorous and set out to prove mass additivity, to show that "mass additivity is a consequence of Newtonian laws". I have attached an example (taken from Kibble & Berkshire's 'Classical Mechanics' (p12)).

The proof is only ever given in the context of inertial mass, and involves something like the following move. Where ##F_T## is the total force on a number of point masses indexed by i (due to an external source, which I do not specify, so as to simplify the equations), we begin with:
[tex] \sum_i m_i\mathbf{a}_i = \mathbf{F}_T[/tex]
It is then assumed that the particles indexed by i compose a composite body and that the force on the composite is equivalent to the total force on the parts (##F_T=F_c##). A physical situation is then assumed in which the accelerations of the particles indexed by i are identical, which allows the following transformation:
[tex] \mathbf{a}_i\sum_i m_i = \mathbf{F}_c[/tex]
It is then assumed that the acceleration of the composite ##\mathbf{a}_c## just is the acceleration of its parts so that:
[tex] \mathbf{a}_c\sum_i m_i = \mathbf{F}_c[/tex]
Which is taken to show that given the force and acceleration of the composite, the composite's mass must be additive, that is:
[tex] \sum_i m_i = m_c[/tex]
Hence mass is additive.

There is an objection to this proof, which focuses on the simplification that the acceleration of the parts are all identical. One way to show that this simplification is problematic is to show that the analogous simplification required to prove gravitational mass additivity is unphysical. The analogous simplification requires, not identical accelerations, but identical positions. The proof starts with the fundamental gravitation law:
[tex] \mathbf{g}(\mathbf{x})=G\sum_i \frac{\mathbf{x}-\mathbf{x}_i}{|\mathbf{x}-\mathbf{x}_i|^3}~m_i~[/tex]
Now the analogue of the acceleration simplification: the positions of the particles indexed by i are identical, so that the following transformation is possible:
[tex] \mathbf{g}(\mathbf{x})=G \frac{\mathbf{x}-\mathbf{x}_1}{|\mathbf{x}-\mathbf{x}_1|^3}\sum_i~m_i~[/tex]

Assuming that the composite determines the same field as the parts (##\mathbf{g}(\mathbf{x})=\mathbf{g}_c(\mathbf{x})##) and that the composite is positioned where its parts are positioned, we can derive gravitational mass additivity in the same way. But, so the objection goes, the proof fails because it assumes the unphysical assumption of particle position overlap. This enables one to object to the inertial proof with the following argument:

(1) If inertial mass additivity is a consequence of Newton's inertial laws then gravitational mass additivity is a consequence of Newton's gravitation law.
(2) It is not the case that gravitational mass additivity is a consequence of Newton's gravitation law. (From above.)
(3) Therefore, it is not the case that inertial mass additivity is a consequence of Newton's inertial laws.

Okay, so we've seen (i) a standard textbook proof for mass additivity and (ii) an objection to that proof. There are surely many ways of responding to the objection. I am looking at one type of response. This response states that the simplifications are only a product of the simple formalism chosen in the proofs. The response sets out to show that if we can formulate the equations so that they can cope with composites whose parts have varying positions and accelerations then a more general proof is possible. Enter Dirac delta functions.

The gravitational proof can be formulated as follows:

Fundamental many-particle law:
[tex] \mathbf{g}(\mathbf{x})=G \sum_i\int \frac{\mathbf{x}-\mathbf{r}}{|\mathbf{x}-\mathbf{r}|^3}~m_i~\delta(\mathbf{r}-\mathbf{x}_i)~\mathrm{d}^3\mathbf{r}[/tex]
Recover the form of the single particle law:
[tex] \mathbf{g}(\mathbf{x})=G\int \frac{\mathbf{x}-\mathbf{r}}{|\mathbf{x}-\mathbf{r}|^3}\sum_i ~m_i~\delta(\mathbf{r}-\mathbf{x}_i)~\mathrm{d}^3\mathbf{r}[/tex]
Assume that the field determined by the composite is the field determined by its parts:
[tex] \mathbf{g}_c(\mathbf{x})=G\int \frac{\mathbf{x}-\mathbf{r}}{|\mathbf{x}-\mathbf{r}|^3}\sum_i ~m_i~\delta(\mathbf{r}-\mathbf{x}_i)~\mathrm{d}^3\mathbf{r}[/tex]
Infer gravitational mass additivity.

Then in the inertial case:

Fundamental many-particle law:
[tex] \sum_i \int m_i(\mathbf{x}) \mathbf{a}_i(\mathbf{x})\delta (\mathbf{x}-\mathbf{x}_i)d^3\mathbf{x} = \mathbf{F}_T[/tex]
Recover the form of the single particle law (this involves deriving ##\mathbf{a}_c(\mathbf{x})##, the acceleration distribution of the composite, from the acceleration distributions of the parts, which I'll just assume here):
[tex] \int \mathbf{a}_c(\mathbf{x})\sum_i m_i(\mathbf{x}) \delta (\mathbf{x}-\mathbf{x}_i)d^3\mathbf{x} = \mathbf{F}_T[/tex]
Assume that the force on the composite is the force on the parts:
[tex] \int \mathbf{a}_c(\mathbf{x})\sum_i m_i(\mathbf{x}) \delta (\mathbf{x}-\mathbf{x}_i)d^3\mathbf{x} = \mathbf{F}_c[/tex]
Infer inertial mass additivity.

Sorry about the length of the post. I suspect people are having trouble with my inertial laws because I'm not defining the external force that determines the acceleration of all the parts? I thought leaving the external force undefined simplifies the equations, but that may be the thing that's causing problems.

Anyway, I'll stop now. I am very interested in your thoughts on this!
Attached Files
File Type: pdf Kibble and Berkshire.pdf (248.7 KB, 2 views)
elfmotat
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#16
Feb4-13, 10:14 PM
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When talking about a "composite" body, it is assumed that there are internal forces which keep the individual particles at the same positions relative to each other. This is known as a "rigid body." So when an external force is applied to the (center of the) composite the force is "transferred" to all of the individual particles, resulting in each particle obtaining the same acceleration vector.

In the gravitational case each individual particle in a composite body certainly experiences a different gravitational force. This results in what is known as "tidal force." In most cases, it is assumed that the composite is either (1) too small to be significantly affected by tidal forces, or (2) the internal forces in the composite offset tidal forces so that the body isn't deformed.

I suspect your troubles are coming from these simplifying assumptions.
James MC
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#17
Feb4-13, 10:44 PM
P: 168
I'm not assuming that composite bodies must be rigid bodies, so I am under no constraint, as far as I can tell, to keep them from deforming. Are you saying that this is precisely what the problem is? That is, are you saying that because I do not stipulate a mechanism that keeps the composite rigid, the total force ##F_T## on all the particles is not well defined (whether or not we use deltas)? This would locate the problem at the very first equation! (I take it there is no corresponding problem in the gravity case, given that we are only defining a potential at a point due to many particles; and not a distributed force.)
elfmotat
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Feb4-13, 10:57 PM
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Quote Quote by James MC View Post
I'm not assuming that composite bodies must be rigid bodies, so I am under no constraint, as far as I can tell, to keep them from deforming. Are you saying that this is precisely what the problem is? That is, are you saying that because I do not stipulate a mechanism that keeps the composite rigid, the total force ##F_T## on all the particles is not well defined (whether or not we use deltas)? This would locate the problem at the very first equation! (I take it there is no corresponding problem in the gravity case, given that we are only defining a potential at a point due to many particles; and not a distributed force.)
The total force can still be defined by equation (1). What's impossible is the assumption that all particles will have identical accelerations. For that to be possible you need internal forces to offset the tidal forces.


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