What do physikers think about the Dirac Delta function?
robert bristow-johnson
(this will read easier with a mono-spaced font.)
I realize that this is a mathematical question, but I would like to hear a
POV from physicists. I think I know the electrical engineer's perspective
but I'm expecting that folks here might tend to be more mathematically
rigorous in their attitude (maybe that's a false assumption?).
The Dirac impulse function can be defined in an myriad of ways, but the most
common is as a limit of
{ 1 |x| < 0.5
{
unit pulse: p(x) = { 0.5 |x| = 0.5
{
{ 0 |x| > 0.5
of course, the integral of p(t/tau) is tau.
dirac impulse: d(t) = lim 1/tau * p(t/tau)
tau->0
More rigorous engineers skip the above and define the dirac impulse only in
terms of its salient properties:
d(t) = 0 for all |t| > 0
and
+inf
integral{ d(t) dt} = 1
-inf
from which it is easy to determine that
+inf
integral{ f(t) * d(t) dt} = f(0)
-inf
and
+inf
integral{ f(t) * d(t-T) dt} = f(T)
-inf
which is probably the most useful single property to remember about d(t).
So the rigorous engineers would insist that it doesn't even make any sense
to write down d(t) without surrounding it with an integral. Okay, I can
live with that.
But the problem came up when I took a course in Real Analysis and we learned
about Lebesgue measure and Lebesque integration (as opposed to Riemann
integration) and, more specifically, Fatou's Lemma which states that if two
functions, f(t) and g(t) are equal "almost everywhere" (as I recall, that
means "everywhere except perhaps at a countable number of discrete points),
then the integrals (over the same limits) of f(t) and of g(t) are the same.
But engineers (and I would think physicists, but that's why I'm asking) must
decidedly take issue with that since, if f(t) = d(t) and g(t) = 0,
f(t) = g(t) everywhere except at one discrete point and
+a
integral{ d(t) dt} = 1
-a
a > 0
+a
integral{ 0 dt} = 0
-a
and 0 is usually not the same value as 1 so Fatou's Lemma is violated.
I've heard the term "distribution" with regard to d(t) brought up, but it
sure wasn't a satisfying explanation of this contradiction. Is there a
favorite means of thinking of or dealing with this that physicists tend to
gravitate toward?
thanks in advance.
r b-j
Danny Ross Lunsford
> (this will read easier with a mono-spaced font.)
>
> I realize that this is a mathematical question, but I would like to hear a
> POV from physicists. I think I know the electrical engineer's perspective
> but I'm expecting that folks here might tend to be more mathematically
> rigorous in their attitude (maybe that's a false assumption?).
The physical attitude is that the appearence of a distribution is a
shorthand for integration.
-drl
Hendrik van Hees
robert bristow-johnson wrote:
>=20
> (this will read easier with a mono-spaced font.)
>=20
> I realize that this is a mathematical question, but I would like to
> hear a
> POV from physicists. I think I know the electrical engineer's
> perspective but I'm expecting that folks here might tend to be more
> mathematically rigorous in their attitude (maybe that's a false
> assumption?).
>=20
> The Dirac impulse function can be defined in an myriad of ways, but
> the most common is as a limit of
>=20
> { 1 |x| < 0.5
> {
> unit pulse: p(x) =3D { 0.5 |x| =3D 0.5
> {
> { 0 |x| > 0.5
>=20
>=20
> of course, the integral of p(t/tau) is tau.
>=20
>=20
> dirac impulse: d(t) =3D lim 1/tau * p(t/tau)
> tau->0
>=20
>=20
That's one of many possible definitions of the delta distribution.
Contrary to many other physicists I always insest on calling this
construct the delta-distribution, and I always tell everybody (and I
don't care, whether he wants to know it or not ;-)) that it is no
function. It is also important to emphasize, that the limit tau->0 is a
limit "in the weak sense", i.e., it is defined by taking the integral
over a test function, which has to be choson within an appropriate
function space of "sufficiently well behaved functions", for instance
Schwarz's space of functions of "rapidly falling modulus" or C_{\infty}
functions with compact support, or whatever is appropriate to your
application in physics.=20
> More rigorous engineers skip the above and define the dirac impulse
> only in terms of its salient properties:
>=20
> d(t) =3D 0 for all |t| > 0
>=20
> and
> +inf
> integral{ d(t) dt} =3D 1
> -inf
>=20
> from which it is easy to determine that
>=20
> +inf
> integral{ f(t) * d(t) dt} =3D f(0)
> -inf
>=20
> and
> +inf
> integral{ f(t) * d(t-T) dt} =3D f(T)
> -inf
>=20
> which is probably the most useful single property to remember about
> d(t).
That's not more rigorous, but more sloppy, and I'd never tell this any
student. As Einstein said, one should explain things as easy as
possible but not more easy!
If you are genius like Dirac himself hand have the feeling, how to
handle such sloppily defined objects correctly, you may be very
sucessful in doing so, but we "normal mortals" should think about a
careful definition.
> So the rigorous engineers would insist that it doesn't even make any
> sense
> to write down d(t) without surrounding it with an integral. Okay, I
> can live with that.
There they are completely right. The delta distribution is a very clever
trick to write an unbounded linear functional as an integral.
>=20
> But the problem came up when I took a course in Real Analysis and we
> learned about Lebesgue measure and Lebesque integration (as opposed to
> Riemann integration) and, more specifically, Fatou's Lemma which
> states that if two functions, f(t) and g(t) are equal "almost
> everywhere" (as I recall, that means "everywhere except perhaps at a
> countable number of discrete points), then the integrals (over the
> same limits) of f(t) and of g(t) are the same.
That's the reason, why the delta-distribution is not a function and you
never ever should call it a function. I was very confused with that,
when I learned calculus and physics in the first semesters at
university. The physicists always handled this "delta function" with
gread virtuosity, while our calculus professor carefully proved that
such a thing can never exist, for exactly the reason you gave above.
When I asked him about the "delta function", he told me, that it is a
generalized function, and one should better call it by its right name,
namly a distribution, and so I did from then on, and all confusion was
gone (by far not all confusions between physicists and mathematicians
are so easily solved as this ;-)).
> I've heard the term "distribution" with regard to d(t) brought up, but
> it
> sure wasn't a satisfying explanation of this contradiction. Is there
> a favorite means of thinking of or dealing with this that physicists
> tend to gravitate toward?
So, I guess my answer is of not much help for you :-((.=20
But you can always think about the delta-distribution to be given by a
weak limit in the sense, you have written down above. You can also take
smooth functions as a "delta series", for instance a normalized Gau=DF
distribution, with the mean value at zero and the standard deviation
(width) as a parameter. The delta-distribution then is the weak limit
of the delta-distribution with the width taken to 0.
"Weak limit" has to be understood as follows: You take a test function,
for instance a smooth function with compact suppurt f, multiply it by
the Gau=DF distribution and letting then the width of the distribution
tend to 0. For all smooth functions with compact support this procedure
will give f(0), which defines the delta distribution as a functional in
the function space C_0^{\infty}(R) by f|->f(0), i.e., each function in
this function space is mapped to its value at the argument 0.
--=20
Hendrik van Hees Cyclotron Institute=20
Phone: +1 979/845-1411 Texas A&M University=20
Fax: +1 979/845-1899 Cyclotron Institute, MS-3366
http://theory.gsi.de/~vanhees/ College Station, TX 77843-3366
r...@maths.tcd.ie
robert bristow-johnson <r...@surfglobal.net> writes:
>More rigorous engineers skip the above and define the dirac impulse only in
>terms of its salient properties:
> d(t) = 0 for all |t| > 0
>and
> +inf
> integral{ d(t) dt} = 1
> -inf
>from which it is easy to determine that
> +inf
> integral{ f(t) * d(t) dt} = f(0)
> -inf
>and
> +inf
> integral{ f(t) * d(t-T) dt} = f(T)
> -inf
>which is probably the most useful single property to remember about d(t).
>So the rigorous engineers would insist that it doesn't even make any sense
>to write down d(t) without surrounding it with an integral. Okay, I can
>live with that.
If we want to understand what's happening with the delta function,
it helps to introduce the notion of a functional. This is something
which takes a function and outputs a number. For example, taking
the integral of a function gives you a number, so the operation
which maps a function f to the integral of f is a functional. Similarly,
the operation which maps f to f(0), that is, which merely evaluates the
function at 0, is a functional. We might also, for example, map
f to f'(0), that is, to its derivative at zero.
These are examples of linear functionals, which are our favourite
kinds of functionals. If S[f] is a linear functional of f, then
S[af+g] = a*S[f] + S[g], where a is a number and f and g are
functions. Linear functionals are mathematical objects which
can be studied on their own terms, but they can often be
written in a more familiar way.
For example, we can take the functional which maps f to its
value at zero and write it as:
+infty
integral delta(x)f(x) dx
-infty
Similarly, the functional which maps f to its derivative at zero:
+infty
- integral delta'(x)f(x) dx
-infty
where delta' is the derivative of the delta function. In that sense,
the rigorous engineers are right; the integral sign needs to be stuck
to the delta function before we can identify a well-defined mathematical
object that we're talking about.
>But the problem came up when I took a course in Real Analysis and we learned
>about Lebesgue measure and Lebesque integration (as opposed to Riemann
>integration) and, more specifically, Fatou's Lemma which states that if two
>functions, f(t) and g(t) are equal "almost everywhere" (as I recall, that
>means "everywhere except perhaps at a countable number of discrete points),
>then the integrals (over the same limits) of f(t) and of g(t) are the same.
>But engineers (and I would think physicists, but that's why I'm asking) must
>decidedly take issue with that since, if f(t) = d(t) and g(t) = 0,
>f(t) = g(t) everywhere except at one discrete point and
> +a
> integral{ d(t) dt} = 1
> -a
> a > 0
> +a
> integral{ 0 dt} = 0
> -a
>and 0 is usually not the same value as 1 so Fatou's Lemma is violated.
>I've heard the term "distribution" with regard to d(t) brought up, but it
>sure wasn't a satisfying explanation of this contradiction. Is there a
>favorite means of thinking of or dealing with this that physicists tend to
>gravitate toward?
If you don't want to think about the delta functions in terms of
linear functionals, there's another option, which is to think of
everything you're doing as being "in some kind of a limit". That is,
suppose that you replace delta(x) with a function that looks like
G_A(x)=N(A)*exp(-Ax^2), where N(A) is a normalization constant to ensure
that the integral is one. Then, for large values of A, the function
looks like a Gaussian curve with a very narrow very tall peak. For
functions f which don't change too fast near x=0, the relationship:
integral f(x)*G_A(x) dx = f(0)
will approximately hold.
Since, for all real positive values of A, G_A is a real function,
Fatou's Lemma is never violated, but you are restricting your analysis
to those functions f which don't change very much over the width
of your Gaussian curve. As you make A larger, the set of functions
for which this analysis is good increases. You have to suppose
that the ideal delta function is something that you are never
rigorously dealing with, but that you are effectively asking what
kind of behaviour you would expect if you let A go to infinity.
I would guess that most physicists have something like this in
the back of their mind when dealing with delta functions, because,
when you're dealing with quantum field theory, it's often necessary
to keep track (at least in the back of your head) of what the "order
of limits" is; that is, each delta function you're dealing with
might have a different A parameter, and the equations you get
at the end will depend on which ones are approaching delta
functions more quickly. The parameters are often physically
relevant things, for example the reciprocal of the spacing
between atoms in a lattice, which can give you some intuition about
what the relevant order of limits should be.
I should mention that this way of dealing with delta functions
reflects the type of functions with which physicists are likely
to deal. For example, electric fields will hopefully be nice and
smooth, so we can treat our delta functions as described above.
On the other hand, if we come across the function f(x)=1 if x
is rational and 0 otherwise, then the "limit" interpretation
is going to fall apart, while the linear functional approach
(and Fatou's Lemma) will remain valid.
R.
Danny Ross Lunsford
> I realize that this is a mathematical question, but I would like to hear a
> POV from physicists. I think I know the electrical engineer's perspective
> but I'm expecting that folks here might tend to be more mathematically
> rigorous in their attitude (maybe that's a false assumption?).
More on the physical side...
Something like Maxwell's equations doesn't really deal with charges. It
deals with a smooth "current density" - to get a number - charge - out
you have to integrate. Since integration implies a domain, the physical
validity is in principle only valid over regions and not point-wise.
A delta-function is an abstraction of the idea of a density to a
point-function.
-drl
robert egri
robert bristow-johnson <r...@surfglobal.net> wrote in message news:<BC80EFBD.9A0A%r...@surfglobal.net>...
> [...]
> +inf
> integral{ f(t) * d(t-T) dt} = f(T)
> -inf
>
> which is probably the most useful single property to remember about d(t).
>
> So the rigorous engineers would insist that it doesn't even make any sense
> to write down d(t) without surrounding it with an integral. Okay, I can
> live with that.
>
[...]
I would disagree with that being the most useful; it is true that it
describes the ideal effect of "sampling" that makes it memorable in
signal processing but I would argue the Lighthill's definition is more
relevant, namely that delta(t) is to be understood not as a "function"
but as equivalent classes of function sequences {g_n(x)}, such that
lim integral g_n(x) f(x) dx = f(0). In this sense, it is not very
important what your sampling function is specifically, as long as it
is zero most everywhere, and it still has unit area. This is exactly
what you want from a good sampler, i.e, not to be sensitive to the
actual shape and still provide nearly the same value of the signal
being sampled.
robert egri
robert bristow-johnson <r...@surfglobal.net> wrote in message news:<BC80EFBD.9A0A%r...@surfglobal.net>...
[...]
> But the problem came up when I took a course in Real Analysis and we learned
> about Lebesgue measure and Lebesque integration (as opposed to Riemann
> integration) and, more specifically, Fatou's Lemma which states that if two
> functions, f(t) and g(t) are equal "almost everywhere" (as I recall, that
> means "everywhere except perhaps at a countable number of discrete points),
> then the integrals (over the same limits) of f(t) and of g(t) are the same.
[...]
Picking sum nits this is not "Fatou's lemma"; according to Riesz &
Szokefalvi-Nagy, page 39, Fatou's lemma says that:
If the functions f_n(x), non-negative and summable in (a,b), tend
almost everywhere to a function f(x), and if furthermore the sequence
of values
integral{_a^b} f_n(x)dx
is bounded, then the function f(x) is summable and
integral{_a^b} f(x)dx <= lim inf{_a^b} integral f_n(x)dx.
See also: P. Fatou: Series trigonometrique et series de Taylor, Acta
Math. 1906, pp335-400
("Summable" is Lebesgue's term for integrable.)
rge
Cl.Massé
"robert bristow-johnson" <r...@surfglobal.net> a écrit dans le message de
news:BC80EFBD.9A0A%r...@surfglobal.net...
> But the problem came up when I took a course in Real Analysis and we
> learned about Lebesgue measure and Lebesque integration (as opposed to
> Riemann integration) and, more specifically, Fatou's Lemma which
> states that if two functions, f(t) and g(t) are equal "almost
> everywhere" (as I recall, that means "everywhere except perhaps at a
> countable number of discrete points), then the integrals (over the
> same limits) of f(t) and of g(t) are the same.
>
> But engineers (and I would think physicists, but that's why I'm
> asking) must decidedly take issue with that since, if f(t) = d(t) and
> g(t) = 0, f(t) = g(t) everywhere except at one discrete point and
>
> +a
> integral{ d(t) dt} = 1
> -a
> a > 0
> +a
> integral{ 0 dt} = 0
> -a
>
> and 0 is usually not the same value as 1 so Fatou's Lemma is violated.
d(t) isn't a real function, so the lemma doesn't apply to it. d(t) is
the limit of a series of real functions, but this limit doesn't belong
to the set of the real functions, in other words, this set is open.
That is analogous to the openness of R, which can be closed by adding an
infinity or two. d(t) is just an infinity, that could be defined as d(0)
= infinity, d(t) = 0 elsewhere. You know that the limit of 0 x infinity
isn't defined.
> I've heard the term "distribution" with regard to d(t) brought up, but
> it sure wasn't a satisfying explanation of this contradiction. Is
> there a favorite means of thinking of or dealing with this that
> physicists tend to gravitate toward?
The set of distributions is an extension of the set of real functions.
As this set is a vector space, it suffices to introduce the d function
and all its derivatives, and all the elements are deduced by linear
combinations.
--
~~~~ clmasse at free dot fr
Liberty, Equality, Profitability.
Aaron Bergman
robert bristow-johnson <r...@surfglobal.net> wrote:
> (this will read easier with a mono-spaced font.)
>
> I realize that this is a mathematical question, but I would like to hear a
> POV from physicists. I think I know the electrical engineer's perspective
> but I'm expecting that folks here might tend to be more mathematically
> rigorous in their attitude (maybe that's a false assumption?).
There's a perfectly rigorous definition of a delta function in terms of
distributions. See the book _Functional Analysis_ by Reed and Simon for
an exposition.
>
> The Dirac impulse function can be defined in an myriad of ways, but the most
> common is as a limit of
>
> { 1 |x| < 0.5
> {
> unit pulse: p(x) = { 0.5 |x| = 0.5
> {
> { 0 |x| > 0.5
>
>
> of course, the integral of p(t/tau) is tau.
>
>
> dirac impulse: d(t) = lim 1/tau * p(t/tau)
> tau->0
There's a general definition of a Dirac sequence that encodes this sort
of definition. Lang's book _Complex Analysis_ discusses these.
>
> More rigorous engineers skip the above and define the dirac impulse only in
> terms of its salient properties:
>
[...]
> +inf
> integral{ f(t) * d(t-T) dt} = f(T)
> -inf
>
> which is probably the most useful single property to remember about d(t).
This ends up being the distribution definition of the delta function.
> So the rigorous engineers would insist that it doesn't even make any sense
> to write down d(t) without surrounding it with an integral. Okay, I can
> live with that.
That, again, is essentially the mathematical definition.
> But the problem came up when I took a course in Real Analysis and we learned
> about Lebesgue measure and Lebesque integration (as opposed to Riemann
> integration) and, more specifically, Fatou's Lemma which states that if two
> functions, f(t) and g(t) are equal "almost everywhere" (as I recall, that
> means "everywhere except perhaps at a countable number of discrete points),
> then the integrals (over the same limits) of f(t) and of g(t) are the same.
>
> But engineers (and I would think physicists, but that's why I'm asking) must
> decidedly take issue with that since, if f(t) = d(t) and g(t) = 0,
> f(t) = g(t) everywhere except at one discrete point and
>
> +a
> integral{ d(t) dt} = 1
> -a
> a > 0
> +a
> integral{ 0 dt} = 0
> -a
>
> and 0 is usually not the same value as 1 so Fatou's Lemma is violated.
>
> I've heard the term "distribution" with regard to d(t) brought up, but it
> sure wasn't a satisfying explanation of this contradiction. Is there a
> favorite means of thinking of or dealing with this that physicists tend to
> gravitate toward?
There is no contradiction because the delta function isn't a function.
It is a distribution. I'll leave it to someone else to go through the
definition if they hav the energy. A distribution is basically something
that takes in a function (of rapid decrease at infinity, say) and gives
out a number. And function gives a distribution by doing the integral,
but there are more distributions than functions. The delta 'function' is
a particularly simply distribution that isn't a function.
Aaron
robert bristow-johnson
Lunsford at antima...@yahoo.NOSE-PAM.com wrote on 03/22/2004 19:25:
> robert bristow-johnson wrote:
>>
>> I realize that this is a mathematical question, but I would like to hear a
>> POV from physicists. I think I know the electrical engineer's perspective
>> but I'm expecting that folks here might tend to be more mathematically
>> rigorous in their attitude (maybe that's a false assumption?).
>
> The physical attitude is that the appearence of a distribution is a
> shorthand for integration.
Okay, fine. Then how does that answer the apparent contradiction to Fatou's
Lemma:
+a
integral{ d(t) dt} = 1
-a
a > 0
+a
integral{ 0 dt} = 0
-a
?
where d(t) = 0 a.e.
How do mathematical physicists (J.B. ???) deal with that?
r b-j
Ohne
It is safe to say that the majority of physicists do not know Lebesque
integration, and so would never become confused by technical questions
such as yours. That being said, the "contradiction" you bring up is
vacuous for the simple reason that the Dirac delta function is not, in
fact, a function.
Regards,
Ohne
ueb
nothing to do with nature. It is good for system theory.
Ulrich
Ohne
Creighton Hogg
> "robert bristow-johnson" <r...@surfglobal.net> a écrit dans le message de
> news:BC80EFBD.9A0A%r...@surfglobal.net...
> > it sure wasn't a satisfying explanation of this contradiction. Is
> > there a favorite means of thinking of or dealing with this that
> > physicists tend to gravitate toward?
>
> The set of distributions is an extension of the set of real functions.
> As this set is a vector space, it suffices to introduce the d function
> and all its derivatives, and all the elements are deduced by linear
> combinations.
How do you define derivatives for distributions? The delta distribution
isn't even continuous, so how do you define its derivatives?
robert bristow-johnson
Al maybe?) for compiling my response to different posters of the same thread
into one. I'm doing it again because I don't want to post 6 or 8 times at
once. I just want to make clear, that I am not a troll and am not trying to
"be difficult" in this thread.
r b-j
ueb wrote:
> The Dirac impulse is an idealized standard function, and has
> nothing to do with nature. It is good for system theory.
Which is what I am most familiar with. Once in a while we get in fights on
comp.dsp about the Dirac impulse (and the Shannon/Nyquist/Whitaker Sampling
and Reconstruction Theorem) and I have taken a certain position in that
argument (which depends on the Dirac Impulse never quite getting to the
limit so that "tau" is always a little bigger than 0 although "tau" is
allowed to get arbitrarily close). Anyway, rather than get the same old
"party line" given to me from the mathematicians, I thought maybe physicists
would have the same kind of problem with the Dirac impulse function.
Personally, I don't have that much of a conceptual problem, just a problem
with someone claiming that I cannot practically treat it like one of its
close approximations (say with tau = 10^-43 second) "because d(t) isn't a
real function". I believe that this objection is not relevant, but I'm
trying to get more information and POV from another closely related
discipline.
Basically, what I've been saying is that, since the area of d(t) is the
dimensionless unity, the dimension of the dependent variable of d(t) (the
height of the impulse) must be the reciprocal of the dimension of the
independent variable of d(t). If the independent variable, t, is "time",
then the height of d(t) is in units of 1/time. But invariably, there will
be those who cry foul "because the Dirac Delta function isn't a function".
For a linear system that has the same species of animal coming out as going
in (commonly voltage in and voltage out), then the impulse response h(t) of
that linear system must have dimension of its dependent variable as 1/time.
r...@maths.tcd.ie wrote:
> If we want to understand what's happening with the delta function,
> it helps to introduce the notion of a functional. This is something
> which takes a function and outputs a number.
I know what a functional is. The way they taught me in Metric Spaces and
Functional Analysis was even more general. A functional maps a point in a
metric space to a number and a linear functional maps a point in a normed
linear space to a number in such a way that superposition applies. That
point in a metric space could be a function, a finite or infinite discrete
sequence, a sequence of functions, or something else that I can't even think
of, just as long as there is a legit distance metric. I understand that,
with a fixed T, that
+inf
integral{ f(t) * d(t-T) dt}
-inf
is a linear functional.
Ohne wrote:
It sounds like tautology. What you are saying is that "What you think is
not true because it is defined to be not true." And I am begging the
question, why?
Aaron Bergman wrote:
> There is no contradiction because the delta function isn't a function.
> It is a distribution. I'll leave it to someone else to go through the
> definition if they hav the energy. A distribution is basically something
> that takes in a function (of rapid decrease at infinity, say) and gives
> out a number.
Like a functional. Fine, I know about functionals. But the Dirac Delta is
a functional only inside the integral expression, like:
+inf
integral{ f(t) * d(t) dt} = f(0)
-inf
That is a linear functional that sends the function f(t) to the number f(0).
d(t) is not a "functional" all by itself. Is the nomenclature then that
"d(t)" outside of the integral is a "distribution"?
> And function gives a distribution by doing the integral,
> but there are more distributions than functions. The delta 'function' is
> a particularly simply distribution that isn't a function.
So the set that contains all distributions also contains, as a subset, all
functions. But that subset is definitely smaller. Sorta like saying that
there are more real numbers than rational numbers?
robert egri wrote:
> Picking sum nits this is not "Fatou's lemma"; according to Riesz &
> Szokefalvi-Nagy, page 39, Fatou's lemma says that:
You're right, I guess. My 23 year old class notes have it as a lemma of
Fatou's Lemma but, as I look it up in my H.L. Royden text, it comes
immediately before (p. 80) Fatou's Lemma. So this is correcting a 2 decade
misrecollection of mine. I guess there is no fancy theorem or lemma that
this fact from real analysis is named after.
Cl.Massé wrote:
> d(t) isn't a real function, so the lemma doesn't apply to it. d(t) is
> the limit of a series of real functions, but this limit doesn't belong
> to the set of the real functions, in other words, this set is open.
This is the most salient statement here. I guess then my question is, other
than the fact that "they" have simply declared it to be so, that "d(t) isn't
a real function", why not? It's defined for all t except t = 0 where it is
infinite, but in such a way that you can integrate it even across t=0. The
real function 1/sqrt(|t|) is not defined for t=0 either (it is +infinite)
but I can integrate it with any finite limits of integration, even across
t=0.
All of the functions that approach d(t) in the limit are legit and real
functions that have integral of 1. Why must the limit, d(t), not belong in
that set? Why must the set be considered open?
> That is analogous to the openness of R, which can be closed by adding an
> infinity or two. d(t) is just an infinity, that could be defined as d(0)
> = infinity, d(t) = 0 elsewhere. You know that the limit of 0 x infinity
> isn't defined.
It depends on how it is expressed. Simply saying that "0 x inf is not
defined" is well accepted, but when you add the word "limit", it really
depends on what you mean.
certainly lim 1/n * n is defined and it is 1.
n->inf
saying lim 0 * n is defined and it is 0.
n->inf
So I am not sure what you mean.
finally...
Danny Ross Lunsford wrote:
> More on the physical side...
>
> Something like Maxwell's equations doesn't really deal with charges. It
> deals with a smooth "current density" - to get a number - charge - out
> you have to integrate. Since integration implies a domain, the physical
> validity is in principle only valid over regions and not point-wise.
>
> A delta-function is an abstraction of the idea of a density to a
> point-function.
This sounds like how electrical engineers think of it also. It's an
abstraction that doesn't really exist in nature. I just think of it as one
of the approximating limit functions with a finite width that is thinner
than I'll ever need it to be in my physical context (say, an electrical
circuit).
Again, thanks for all responses.
r b-j
Jerzy Karczmarczuk
Creighton Hogg wrote:
> How do you define derivatives for distributions? The delta distribution
> isn't even continuous, so how do you define its derivatives?
Oh dear...
Since C.H. quotes a posting about distributions, he should have seen that
such distribution are meant as functionals, as *integral operators*. So,
take the def. of the appropriate convolution: INT{f(x)*delta(y-x)}dx as
your starting point, replace delta(...) by delta'(...), and assuming that
by a "formalized <<fiat>>" you can apply the integration by parts, you see
that the derivative delta'(x) "extracts" -f'(0), and it can be iterated to
produce higher derivatives, since the "integrated parts" vanish.
If you are not satisfied, do it using a regular approximation of delta, and
perform the manipulation *before* taking the limit. And believe that you
are not committing any sin...
If you are still not satisfied, perhaps just treat delta ALGEBRAICALLY, not
analytically, within the frame of the operational calculus: Heaviside, then
extended and somewhat ordered by Mikusinski, etc. Here taking the derivative
means just the "multiplication by the differentiating operator". Mikusinski
(quite old, but newer than Heaviside's) formalism is more or less playing
with such distributions in the space of Laplace transforms of functions.
They behave as algebraic entities, and the question "what does it REALLY mean"
need not be posed.
But all this is standard stuff, ALL this can be found in standard books.
Jerzy Karczmarczuk
Aaron Bergman
In article <BC87ED24.9C85%r...@surfglobal.net>,
robert bristow-johnson <r...@surfglobal.net> wrote:
> Aaron Bergman wrote:
>
> > There is no contradiction because the delta function isn't a function.
> > It is a distribution. I'll leave it to someone else to go through the
> > definition if they hav the energy. A distribution is basically something
> > that takes in a function (of rapid decrease at infinity, say) and gives
> > out a number.
>
> Like a functional. Fine, I know about functionals. But the Dirac Delta is
> a functional only inside the integral expression, like:
>
> +inf
> integral{ f(t) * d(t) dt} = f(0)
> -inf
>
> That is a linear functional that sends the function f(t) to the number f(0).
>
> d(t) is not a "functional" all by itself. Is the nomenclature then that
> "d(t)" outside of the integral is a "distribution"?
\delta(t) is just a symbol. Inside an integral, it says that this
integral is actually a distribution. By treating \delta(t) as if it were
a function, we can derive various rules for manipulating the symbol. For
example, we can derive formulae for things like \delta(f(x)). We do this
by writing:
\int dx \delta(f(x)) g(x)
substitute y = f(x)
\int dy |f'(x)|^{-1} \delta(y) g(x)
So, this picks out the zeroes of f(x), call them x_i and gives
\sum g(x_i) / |f'(x_i)|
= \sum \int dx (\delta(x - x_i) / |f'(x)|) g(x)
Thus, we've derived the 'formula'
\delta(f(x)) = \sum \delta(x - x_i) / |f'(x)|
In some sense, this can be thought of as a definition of \delta(f(x)).
We define that symbol so that it acts as if it were a real function
under the integral sign. In another post, I mention how one should
define the symbol \delta'(x).
However, \delta(x) really isn't a function. For example, writing
something like \delta(x)\delta(x) doesn't mean anything, but squaring is
a perfectly legitimate thing to do to a real function.
> > And function gives a distribution by doing the integral,
> > but there are more distributions than functions. The delta 'function' is
> > a particularly simply distribution that isn't a function.
>
> So the set that contains all distributions also contains, as a subset, all
> functions. But that subset is definitely smaller. Sorta like saying that
> there are more real numbers than rational numbers?
Be careful. Those statements aren't the same. The set of functions is
naturally embedded in the set of distributions. I'm not sure about the
cardinality of the sets, however. One always have to be careful when
dealing with infinite sets.
Aaron
Jerzy Karczmarczuk
Aaron Bergman wrote:
> \delta(t) is just a symbol. Inside an integral, it says that this
> integral is actually a distribution.
...
> However, \delta(x) really isn't a function. For example, writing
> something like \delta(x)\delta(x) doesn't mean anything, but squaring is
> a perfectly legitimate thing to do to a real function.
I would *NOT* recommend saying that something is "just a symbol". In a sense,
everything you put down is a symbol. But delta is - conceptually - a decently
defined mathematical entity, without such, and similar definitions, there would
be no modern theoretical physics.
Its properties are defined, are known, are - sometimes - frustrating, but it
might be useful to admit that Mathematics, which is a harsh mistress, sometimes
is unbelievably tolerant...
I believe that I mentioned this once here.
delta(x)*delta(x) is not something 'meaningless', although it must be treated
<<cum grano salis>>, since this is equivalent to delta(x)*delta(0), and the
second factor is, well... you see yourself that this is an abomination.
Nevertheless in the collision theory, where you start with amplitudes based on
normalized quantum states, and during the final stage you compute the cross-
sections normalized to the unit volumes, unit time-slices, etc., you have to
admit that
1. The amplitude A contains delta4(p-p') : the momentum conservation between the
initial and the final state.
2. You have to compute A*A in order to get the probabilities. So you *GET* this
awful delta4(0).
3. But you divide it by (V*T) where V is the total volume and T - the total time
of the interaction. They are infinite. Together they are equivalent to
delta4(0) times some (2*pi)^4 or similar [[I forget always whether one should
multiply or divide without checking it up ...]]. And the Morality is back.
Jerzy Karczmarczuk
Aaron Bergman
<Pine.LNX.4.44.040324...@dill.hep.wisc.edu>,
Creighton Hogg <wch...@hep.wisc.edu> wrote:
The derivative of a delta function is the functional that takes the
function f(x) and gives out f'(0) (give or take a sign). This just comes
from integration by parts.
Aaron
Patrick Van Esch
>
> Basically, what I've been saying is that, since the area of d(t) is the
> dimensionless unity, the dimension of the dependent variable of d(t) (the
> height of the impulse) must be the reciprocal of the dimension of the
> independent variable of d(t).
Yes, but that's satisfied by the dirac delta function:
delta(ax+b) = 1/abs(a) delta(x+b/a)
>
> Ohne wrote:
>
> > It is safe to say that the majority of physicists do not know Lebesque
> > integration, and so would never become confused by technical questions
> > such as yours.
I don't know what famous physicist said: "If the fact that your
airplane stays in the air depends on the difference between a Riemann
integral and a Lebesgue integral, I don't want to fly in it ! "
>
> Like a functional. Fine, I know about functionals. But the Dirac Delta is
> a functional only inside the integral expression, like:
>
> +inf
> integral{ f(t) * d(t) dt} = f(0)
> -inf
>
> That is a linear functional that sends the function f(t) to the number f(0).
>
> d(t) is not a "functional" all by itself. Is the nomenclature then that
> "d(t)" outside of the integral is a "distribution"?
What is meant is that d(t) by itself has no meaning. When you think
about it, whenever you use a delta function, it is an intermediate
expression that sooner or later has to be put in an integral to give a
result (something to be compared to experiment, a number). Moreover,
you always have to have an expression linear in d(t). For example, it
is hard to work with (d(t))^2.
>
> So the set that contains all distributions also contains, as a subset, all
> functions. But that subset is definitely smaller. Sorta like saying that
> there are more real numbers than rational numbers?
Exactly !
>
>
> Cl.Massé wrote:
>
> > d(t) isn't a real function, so the lemma doesn't apply to it. d(t) is
> > the limit of a series of real functions, but this limit doesn't belong
> > to the set of the real functions, in other words, this set is open.
>
> This is the most salient statement here. I guess then my question is, other
> than the fact that "they" have simply declared it to be so, that "d(t) isn't
> a real function", why not? It's defined for all t except t = 0 where it is
> infinite, but in such a way that you can integrate it even across t=0.
One cannot really say that d(t) is DEFINED to be infinite in t=0.
Real functions are mappings from R to R so infinity is not part of it.
However, the LIMIT equal to infinity can have a meaning, as in 1/x^2,
which is not defined in 0, but has +Infinity as a limit. But you
cannot put the infinity there in by hand, it comes out of the
neighbourhood of the undefined point.
If you want to extend the definition of real functions as mappings
from R to R-bar (closure of R, containing + and - Infinity), you screw
up quite a lot of structure of what a real function is. But moreover
you do not solve the problem for the deltafunction:
if delta(0) = +Infinity, then what is doubledelta(t) = 2 delta(t) ?
doubledelta(0) = +Infinity too... so doubledelta(t) = delta(t), so
their integrals should be the same ? but one is 1 and the other is
2...
>
> All of the functions that approach d(t) in the limit are legit and real
> functions that have integral of 1. Why must the limit, d(t), not belong in
> that set? Why must the set be considered open?
The set of mappings from R->R IS open, it is not defined as such. It
is a bit as saying: why must 5 - 8 be considered negative ?
> It depends on how it is expressed. Simply saying that "0 x inf is not
> defined" is well accepted, but when you add the word "limit", it really
> depends on what you mean.
>
> certainly lim 1/n * n is defined and it is 1.
> n->inf
>
>
> saying lim 0 * n is defined and it is 0.
> n->inf
>
> So I am not sure what you mean.
You are tacitly assuming that limit(n->inf) (A(n) B(n)) = limit(A(n))
limit(B(n)), which is a theorem with conditions, and doesn't apply
here because the conditions are not valid.
The problem is that when you write the above, you have a whole
sequence, and you see HOW you you are approaching infinity. Once you
replace this sequence with its limiting VALUE, you have lost the
information of how you approach infinity.
Exactly the same happens with the deltafunction: its integral = 1
comes only from the fact that you know a sequence of functions with
integral 1. Once you have the limiting case only, you don't know
anymore how you got there and this integral property is gone.
cheers,
Patrick.
robert bristow-johnson
In article c23e597b.04032...@posting.google.com, Patrick Van E=
sch
at van...@ill.fr wrote on 03/26/2004 15:48:
> robert bristow-johnson <r...@surfglobal.net> wrote in message
> news:<BC87ED24.9C85%r...@surfglobal.net>...
>>=20
>> Basically, what I've been saying is that, since the area of d(t) is th=
e
>> dimensionless unity, the dimension of the dependent variable of d(t) (=
the
>> height of the impulse) must be the reciprocal of the dimension of the
>> independent variable of d(t).
>=20
> Yes, but that's satisfied by the dirac delta function:
>=20
> delta(ax+b) =3D 1/abs(a) delta(x+b/a)
you're preaching to the choir here. but some have objected to the dimens=
ion
of "a" getting carried to the outside. i think they would just say that =
"a"
can only be a number.
=20
>> Ohne wrote:
>>=20
>>> It is safe to say that the majority of physicists do not know Lebesqu=
e
>>> integration, and so would never become confused by technical question=
s
>>> such as yours.=20
>=20
> I don't know what famous physicist said: "If the fact that your
> airplane stays in the air depends on the difference between a Riemann
> integral and a Lebesgue integral, I don't want to fly in it ! "
i thought it was R.W. Hamming (of "Hamming Window" namesake). i do not
know, without looking it up, what his degrees were in, but i always
considered him to be more of an EE. he could have been a physicist, but =
i
think he taught in electrical engineering.
>> Like a functional. Fine, I know about functionals. But the Dirac Del=
ta is
>> a functional only inside the integral expression, like:
>>=20
>> +inf
>> integral{ f(t) * d(t) dt} =3D f(0)
>> -inf
>>=20
>> That is a linear functional that sends the function f(t) to the number=
f(0).
>>=20
>> d(t) is not a "functional" all by itself. Is the nomenclature then th=
at
>> "d(t)" outside of the integral is a "distribution"?
>=20
> What is meant is that d(t) by itself has no meaning.
well, that's the debate. or, at least, a part of it. engineers toss it
around, without the integral, a lot. e.g. in the sampling theorem (at le=
ast
according to me :-), we say that:
+inf +inf
q(t) =3D T * SUM d(t - k*T) =3D SUM exp(j*2*pi*n*t/T)
k=3D-inf n=3D-inf
now, admittedly, in order to get the the Fourier series coefficients
(they're all the dimensionless 1), the deltas end up in an integral. but
signal processing engineers toss these d(t) functions around a lot withou=
t
integrating them. ultimately, they probably *do* wind up in an integral
(perhaps the convolution integral to get the output of some linear system=
).
> When you think
> about it, whenever you use a delta function, it is an intermediate
> expression that sooner or later has to be put in an integral to give a
> result (something to be compared to experiment, a number). Moreover,
> you always have to have an expression linear in d(t).
it gets ugly if you don't.
>> Cl.Mass=E9 wrote:
>>=20
>>> d(t) isn't a real function, so the lemma doesn't apply to it. d(t) i=
s
>>> the limit of a series of real functions, but this limit doesn't belon=
g
>>> to the set of the real functions, in other words, this set is open.
>>=20
>> This is the most salient statement here. I guess then my question is,=
other
>> than the fact that "they" have simply declared it to be so, that "d(t)=
isn't
>> a real function", why not? It's defined for all t except t =3D 0 wher=
e it is
>> infinite, but in such a way that you can integrate it even across t=3D=
0.
>=20
> One cannot really say that d(t) is DEFINED to be infinite in t=3D0.
right, it just winds up being infinite.
=20
> You are tacitly assuming that limit(n->inf) (A(n) B(n)) =3D limit(A(n))
> limit(B(n)), which is a theorem with conditions, and doesn't apply
> here because the conditions are not valid.
i didn't believe i was assuming that.
> The problem is that when you write the above, you have a whole
> sequence, and you see HOW you you are approaching infinity.
yup.
> Once you replace this sequence with its limiting VALUE, you have lost t=
he
> information of how you approach infinity.
yup.
> Exactly the same happens with the deltafunction: its integral =3D 1
> comes only from the fact that you know a sequence of functions with
> integral 1. Once you have the limiting case only, you don't know
> anymore how you got there and this integral property is gone.
i thought that we are defining that this property (the integral is 1) is
part and parcel to the Dirac delta function. agreed, if you only defined
d(t) to be 0 for |t|>0 and infinite for t=3D0 without any specification o=
f how
you get there, certainly you cannot make any conclusion to what the integ=
ral
is.
anyway, perhaps that is what i am fundamentally missing. whether it's a
unit area rectangular pulse or a unit are gaussian pulse or a unit area
sin(x)/x pulse, i've always considered that part of the *definition* of t=
he
Dirac Delta function (as a "function"), is *how* it gets there, i.e. what=
is
happening in the limit. of course, if that information is thrown away, y=
ou
cannot tell what the integral is and it could just as well be zero which
would satisfy that Lebesgue integration property (that i incorrectly
recalled to be a consequence of Fatou's Lemma).
r b-j
Patrick Van Esch
> > I don't know what famous physicist said: "If the fact that your
> > airplane stays in the air depends on the difference between a Riemann
> > integral and a Lebesgue integral, I don't want to fly in it ! "
>
> i thought it was R.W. Hamming (of "Hamming Window" namesake). i do not
> know, without looking it up, what his degrees were in, but i always
> considered him to be more of an EE. he could have been a physicist, but =
> i
> think he taught in electrical engineering.
Well, an EE is a kind of applied physicist :-)
cheers,
Patrick.
Cl.Massé
> > d(t) isn't a real function, so the lemma doesn't apply to it. d(t)
> > is the limit of a series of real functions, but this limit doesn't
> > belong to the set of the real functions, in other words, this set is
> > open.
"robert bristow-johnson" <r...@surfglobal.net> a écrit dans le message de
news:BC87ED24.9C85%r...@surfglobal.net...
> This is the most salient statement here. I guess then my question is,
> other than the fact that "they" have simply declared it to be so, that
> "d(t) isn't a real function", why not? It's defined for all t except
> t = 0 where it is infinite, but in such a way that you can integrate
> it even across t=0. The real function 1/sqrt(|t|) is not defined for
> t=0 either (it is +infinite) but I can integrate it with any finite
> limits of integration, even across t=0.
f1(t) = 1/sqrt(|t|) isn't a real function over R, although its integral
over ]0, +inf) is defined through a limit. You can define a function
f2(t) over R that is equal to 1/sqrt(|t|) for any finite number, and
f2(0) = 0. It has the same integral, yet the lemna doesn't apply because
f1 and f2 have different definition domains. Defining f3(0) = 1 and the
same value elsewhere, the lemna applies to f2 and f3.
> All of the functions that approach d(t) in the limit are legit and
> real functions that have integral of 1. Why must the limit, d(t), not
> belong in that set? Why must the set be considered open?
An open set is just a set where the limit of all the (convergent)
sequences in it doesn't belong to it. As d(t) isn't a real function
over R (it hasn't a real value at 0), and is a limit of a sequence of
real functions over R, the set is open. Another example is an "open"
interval. There are series in it with a limit equal the one bound. The
set of the limits of all series in a set is the closure of this set.
This terminology is defined for a topology, that is, a structure that
says which points are near, loosely speaking. So the set of all
distributions is the closure of the set of all real functions over R.
> > That is analogous to the openness of R, which can be closed by
> > adding an infinity or two. d(t) is just an infinity, that could be
> > defined as d(0) = infinity, d(t) = 0 elsewhere. You know that the
> > limit of 0 x infinity isn't defined.
> It depends on how it is expressed. Simply saying that "0 x inf is not
> defined" is well accepted, but when you add the word "limit", it
> really depends on what you mean.
>
> certainly lim 1/n * n is defined and it is 1.
> n->inf
>
>
> saying lim 0 * n is defined and it is 0.
> n->inf
>
> So I am not sure what you mean.
It isn't defined because it may have different values, including 0 and
infinity. d(t) and a d(t) both are "infinite" at 0, but their integral
over R are different. It is possible to define a sequence of real
functions with a support tending to {0}, but with an integral tending to
any real, and even to infinity.
All that is more mathematical language, but topology is much used in
physics. And to say the truth, delta functions don't exist in Nature,
it is an idealization handy for calculations. For example, a signal
whose spectrum (Fourier transform) is a delta function has an infinite
extend, or support, in time, which doesn't occur.
Charles Francis
In article <BC80EFBD.9A0A%r...@surfglobal.net>, robert bristow-johnson
<r...@surfglobal.net> writes
>
> +a
> integral{ d(t) dt} = 1
> -a
> a > 0
> +a
> integral{ 0 dt} = 0
> -a
>
>and 0 is usually not the same value as 1 so Fatou's Lemma is violate
d(0) is undefined, so Fatou's lemma does not apply
Regards
--
Charles Francis
p.ki...@imperial.ac.uk
> Hello Robert,
Of course. Except I'd guess that the majority of physicists, if
confronted with an ambiguous situation involving a delta function
would most likely replace the delta with a peaked function which
has an appropriate limit (e.g. a normalised gaussian), and
do the calculation followed by the limit as the gaussian width
tends to zero.
The "Dirac delta function" might not be a function, but what most
physicists implicitly use when implimenting it *is* -- it's
just a very narrow peaked function of some kind, one so narrow
that everything else looks flat on the scale of its width.
If I had to worry about Lebesque integration every time I needed
some kind of narrow peaked thingy to bung into an equation, I'd
never get anything done. ;-)
--
---------------------------------+---------------------------------
Dr. Paul Kinsler
Blackett Laboratory (QOLS) (ph) +44-20-759-47520 (fax) 47714
Imperial College London, Dr.Paul...@physics.org
SW7 2BW, United Kingdom. http://www.qols.ph.ic.ac.uk/~kinsle/
Cl.Massé
> but signal processing engineers toss these d(t) functions around a lot
> without integrating them.
And they get a distribution, which isn't a real function. But for any
practical calculation, the result should be a real that is to be
compared with a physical measurement, which is a real. Therefore, an
integration is mandatory to get this result, since it is the only
operation on a distribution that yields a real.
Jerzy Karczmarczuk
> The "Dirac delta function" might not be a function, but what most
> physicists implicitly use when implimenting it *is* -- it's
> just a very narrow peaked function of some kind, one so narrow
> that everything else looks flat on the scale of its width.
I disagree. *MOST* physicists are confronted with the beast in the
context of conservation laws, and similar 'sharp' situations, and
the "virtual smoothing" would do more harm than good. They use delta
as a distribution, as the identity integral operator. They eliminate
some integration variables thanks to it. They represent it *exactly*
through its Fourier transform [(and then do anything they wish, for
example they evaluate the Fourier integral by some stationary phase
approximation, etc.)].
Certainly, there are people who must visualise everything for themselves,
and it helps to "see" the shape of the derivatives of delta: the first
is a positive/negative wiggle, the second is an infinitly high "volcano"
with an infinitely deep "crater" on top, etc., but for all the effective
calculations it is better to stick to the "formal" intuition rather than
to the "visual" one.
(Unless you are Richard Feynman, but they became recently rather rare...)
Jerzy Karczmarczuk
robert bristow-johnson
at p.ki...@imperial.ac.uk wrote on 03/29/2004 07:31:
> Ohne <oen...@hotmail.com> wrote:
>
>> It is safe to say that the majority of physicists do not know Lebesque
>> integration,
BTW, i would guess that the majority of physicists have been exposed to the
concepts of Lebesgue measure and Lebesque integration. if this EE has in
college, i would think physicists would also.
>> and so would never become confused by technical questions
>> such as yours. That being said, the "contradiction" you bring up is
>> vacuous for the simple reason that the Dirac delta function is not, in
>> fact, a function.
>
> Of course. Except I'd guess that the majority of physicists, if
> confronted with an ambiguous situation involving a delta function
> would most likely replace the delta with a peaked function which
> has an appropriate limit (e.g. a normalised gaussian), and
> do the calculation followed by the limit as the gaussian width
> tends to zero.
this is almost precisely the attitude of engineers toward the Dirac Delta
function. we might be cruder and use a thin rectangular spike instead of a
gaussian.
> The "Dirac delta function" might not be a function, but what most
> physicists implicitly use when implimenting it *is* -- it's
> just a very narrow peaked function of some kind, one so narrow
> that everything else looks flat on the scale of its width.
>
> If I had to worry about Lebesque integration every time I needed
> some kind of narrow peaked thingy to bung into an equation, I'd
> never get anything done. ;-)
likewise. :-)
r b-j
Charles Francis
<kar...@info.unicaen.fr> writes
>p.ki...@imperial.ac.uk wrote:
>
>> The "Dirac delta function" might not be a function, but what most
>> physicists implicitly use when implimenting it *is* -- it's
>> just a very narrow peaked function of some kind, one so narrow
>> that everything else looks flat on the scale of its width.
>
>context of conservation laws, and similar 'sharp' situations, and
>the "virtual smoothing" would do more harm than good.
empirical physicists are not confronted with sharp data, but with
margins of error. The idealisation of a perfect measurement is no less
fictitious than the idea that the delta is a function.
>Certainly, there are people who must visualise everything for themselves,
>and it helps to "see" the shape of the derivatives of delta: the first
>is a positive/negative wiggle, the second is an infinitly high "volcano"
>with an infinitely deep "crater" on top, etc., but for all the effective
>calculations it is better to stick to the "formal" intuition rather than
>to the "visual" one.
>(Unless you are Richard Feynman, but they became recently rather rare...)
And I wonder why they are so rare. They were far more populous 200 years
ago. Perhaps it is due to too much sound advice not to emulate them.
Best ignore it (the advice) imv.
Regards
--
Charles Francis
Andy Kowar
van...@ill.fr (Patrick Van Esch) wrote in message news:<c23e597b.04032...@posting.google.com>...
> robert bristow-johnson <r...@surfglobal.net> wrote in message news:<BC87ED24.9C85%r...@surfglobal.net>...
> >
> > Basically, what I've been saying is that, since the area of d(t) is the
> > dimensionless unity, the dimension of the dependent variable of d(t) (the
> > height of the impulse) must be the reciprocal of the dimension of the
> > independent variable of d(t).
>
> Yes, but that's satisfied by the dirac delta function:
>
> delta(ax+b) = 1/abs(a) delta(x+b/a)
>
>
> >
> > Ohne wrote:
> >
> > > It is safe to say that the majority of physicists do not know Lebesque
> > > integration, and so would never become confused by technical questions
> > > such as yours.
>
> I don't know what famous physicist said: "If the fact that your
> airplane stays in the air depends on the difference between a Riemann
> integral and a Lebesgue integral, I don't want to fly in it ! "
Actually, the Lebesque integral is less useful in physics the Rieman
integral as Lebesque integral does not exist in some important cases
(like in signal theory). The airplane you mention might actually fly
but without a radio.:-)
Some see studying undoubtedly mathematically very pretty theory of
Lebesque integral as useless because in the end one can compute less
of practical stuff. Rieman-Stjelties integral in turn is probably the
most useful one in physics.
Going back to the original question, iI cannot say what or how others
think about Dirac delta, but personally, when trying to visualize the
whole concept, I follow rather the Mikusinski's approach to
distributions as limits of series of functions. It is similar to
extending rational to real numbers. I feel the interpretation of Dirac
delta as a functional (mathematically equivalent to Mikusinki's) less
intuitive in most of cases.
AK
>
>
>[...]
> cheers,
> Patrick.
gr...@math.niu.edu
This is not the reason Fatou doesn't apply. In fact, all that Fatou
requires is a function that is defined almost everywhere. The problem
is that the delta 'function' is not a function, even one defined
only almost everywhere.
Furthermore, what usually goes as Fatou's lemma goes as follows:
If f_n is a sequence of non-negative valued measurable functions, then
int lim f_n <= lim int f_n
assuming the limit of f_n exists pointwise almost everywhere. (It is
actually allowable to use limit infima in this to avoid assuming limits
exist).
For any sequence defining a delta function, the pointwise limit of f_n
will be 0 almost everywhere (usually everywhere except 0). Hence,
int lim f_n =0 while lim int f_n =1. Since 0<=1, there is no violation
of Fatou.
--Dan Grubb
Charles Francis
<ank...@yahoo.com> writes
>Going back to the original question, iI cannot say what or how others
>think about Dirac delta, but personally, when trying to visualize the
>whole concept, I follow rather the Mikusinski's approach to
>distributions as limits of series of functions. It is similar to
>extending rational to real numbers. I feel the interpretation of Dirac
>delta as a functional (mathematically equivalent to Mikusinki's) less
>intuitive in most of cases.
>
>A
This works very well. One way to make it rigorous is treat a function as
an infinite dimensional vector by taking a series of n-dimensional
vector spaces and let n go to infinity. In each finite dimensional space
the delta is well defined and the integral is replaced by a sum. I have
been using this method to study the construction of qft, and it turns
out that diverging loop integrals are automatically regularised. The
divergence doesn't appear in the finite dimensional model, and only
comes in as a result of a wrong order of taking limits.
Regards
--
Charles Francis
robert bristow-johnson
> Charles Francis <cha...@clef.demon.co.uk> wrote in message
> news:<ovT3vtwv...@clef.demon.co.uk>...
>> In article <BC80EFBD.9A0A%r...@surfglobal.net>, robert bristow-johnson
>> <r...@surfglobal.net> writes
>>>
>>> +a
>>> integral{ d(t) dt} = 1
>>> -a
>>> a > 0
>>> +a
>>> integral{ 0 dt} = 0
>>> -a
>>>
>>> and 0 is usually not the same value as 1 so Fatou's Lemma is violate
>>
>>
>> d(0) is undefined, so Fatou's lemma does not apply
>
> This is not the reason Fatou doesn't apply. In fact, all that Fatou
> requires is a function that is defined almost everywhere. The problem
> is that the delta 'function' is not a function, even one defined
> only almost everywhere.
>
> Furthermore, what usually goes as Fatou's lemma goes as follows:
hey guys, it has already been pointed out by Robert Egri and acknowledged by
me that i misascribed this to Fatou's Lemma. it was a note in my old class
notes from a class in Real Analysis that i took in the late 70s. in the
book ("Real Analysis" H.L.Royden), this "proposition" is first stated on p.
80, one section before Fatou. sorry for the misattribution.
just FWIW, i think the explanation that has helped me the most (none has
completely cut it, but i recognize that the problem probably exists more
with my inability to get my mind wrapped around this) is that, perhaps, for
a function candidate to qualify as a "true" function, it cannot be allowed
to "remember" how it may have been defined from a limit of real functions.
it can only be defined in terms of how it maps real numbers to other real
numbers.
i.e. there are a lot (an infinite number) of possibilities for some f(x) = 0
for |x|>0 and infinite (or undefined) for x=0. for f(x) to be a normal real
function (as defined by the powers that be), we cannot add to its definition
any more than the function mapping (as stated above). we cannot add to its
definition that it was a limit of some other functions that all had an area
of 1. only the basic real-to-real mapping is allowed.
it *still* seems to me a semantic. whether we call it a "function" or a
"distribution" or a "farg", it is what it is. we evidently can integrate it
and get a non-zero real number, whatever we call it. the fact that it is
not *physically* realizable was never the issue for me. we have concepts in
mathematics, like infinities and complex numbers, that do not exist in
quantities of natural phenomena (i think that's true, but i'm sure someone
will set me straight if it ain't), but they conceptually exist in
mathematical usage.
r b-j
Patrick Van Esch
robert bristow-johnson <r...@surfglobal.net> wrote in message news:<BC91EE14.A129%r...@surfglobal.net>...
[...]
>
> just FWIW, i think the explanation that has helped me the most (none has
> completely cut it, but i recognize that the problem probably exists more
> with my inability to get my mind wrapped around this) is that, perhaps, for
> a function candidate to qualify as a "true" function, it cannot be allowed
> to "remember" how it may have been defined from a limit of real functions.
> it can only be defined in terms of how it maps real numbers to other real
> numbers.
Yes, that is exactly what is meant by a real function ! It is a
MAPPING from R to R, so you should be able 1) to plug in any x of R
(eventually a subset of R, the domain of f) and get out y = f(x) and
2) the prescription in 1) FULLY defines the function (and that's what
fails with the dirac delta).
>
> i.e. there are a lot (an infinite number) of possibilities for some f(x) = 0
> for |x|>0 and infinite (or undefined) for x=0. for f(x) to be a normal real
> function (as defined by the powers that be), we cannot add to its definition
> any more than the function mapping (as stated above). we cannot add to its
> definition that it was a limit of some other functions that all had an area
> of 1. only the basic real-to-real mapping is allowed.
Exactly.
>
> it *still* seems to me a semantic. whether we call it a "function" or a
> "distribution" or a "farg", it is what it is. we evidently can integrate it
> and get a non-zero real number, whatever we call it.
Yes, of course. But in order to be able to talk to eachother, the
powers that be defined a function as given above, and you should use
another word for something else. You're right that other
constructions which are "function-like" but need more than just a
mapping description x->f(x) are also valid constructions but people
would be very confused if we called it function. That's not just a
matter of words. The properties of a function (a mapping) lead to
several theorems that apply to functions. If you now change that
definition, then all those theorems are not valid anymore. That's why
we should stick to certain terms.
For example: if you have two functions f and g, it is evident what's
meant with f x g. For distributions it is a lot nastier.
cheers,
patrick.
gr...@math.niu.edu
Sorry I didn't see that. I didn't mean to harp.
>
> just FWIW, i think the explanation that has helped me the most (none has
> completely cut it, but i recognize that the problem probably exists more
> with my inability to get my mind wrapped around this) is that, perhaps, for
> a function candidate to qualify as a "true" function, it cannot be allowed
> to "remember" how it may have been defined from a limit of real functions.
> it can only be defined in terms of how it maps real numbers to other real
> numbers.
>
> i.e. there are a lot (an infinite number) of possibilities for some f(x) = 0
> for |x|>0 and infinite (or undefined) for x=0. for f(x) to be a normal real
> function (as defined by the powers that be), we cannot add to its definition
> any more than the function mapping (as stated above). we cannot add to its
> definition that it was a limit of some other functions that all had an area
> of 1. only the basic real-to-real mapping is allowed.
>
No. This is not what is going on. There are not an infinite number of
ways a function can be infinite at a point. A function is a function
is a function. It takes in some things from some set and spits out
things from another (or possibly the same) set. The 'dirac' delta is
not an extended real-valued function on the real line. (In other
words,
it does not take real numbers as arguments and spit out
real-or-infinite
values.) It is a distribution.
Now, what is a distribution? It is a function (some would prefer to
call it an operator) that takes a function and spits out a real
number.
There are some continuity requirements concerning derivatives and all
and some details about which functions can be operated on, i.e. 'the
test
functions', but that is the essence. Oh, distributions are also
supposed
to be linear, i.e. if we operate on a linear combination of functions,
the result is the linear combination of the results of operating on
the
individual functions.
Thus, if L is a distribution, we want L(af+bg)=aL(f) +bL(g) where
f and g are test functions and a,b are real (or complex) numbers.
It's important to know that the test functions are all infinitely
differentiable functions, and any derivative of a test function
is again a test function. Hence, if f is a test function, we can
evaluate f^(n)(x), the n^th derivative of f at any point x. We also
require that test functions be 'rapidly decreasing' at infinity.
Perhaps some examples are in order. First, let k be any integrable
function. We can define a distribution by setting L_k (f) to be
the integral of f(x)k(x) from -\infty to \infty. This is a linear
operator that takes test functions to numbers and has the required
continuity. *By convention*, this distribution and the function k
are identified, even though they are different creatures. By
properties
of the integral, two k's that differ only on a set of measure 0
are identified as distributions.
The Dirac Delta is the distribution that takes a test function f and
spits out the value f(0). That's it. Now, this distribution cannot
be obtained from any k as in the previous paragraph, so the delta
is not a function (by the previous abuse of terminology). Another
distribution that cannot be obtained from a k is the 'derivative
of the delta', where L(f)=-f'(0).
Unfortunately, people like to confuse matters and think of the delta
as a function and think of the integral of some delta(x)f(x) as giving
f(0). Not strictly true, but good enough if you are careful not to
give it too much credence.
It is possible to write the action of many distributions by finding a
sequence k_n of integrable functions and writing L(f) as the limit of
L_(k_n) (f). For the delta, gaussians, and box functions, etc. can be
used for such sequences 'converging' to the delta. This is not
pointwise
convergence of functions!!! To be strict, it is a 'weak' convergence
of linear maps.
It should be pointed out that there are functions that converge to the
derivative of the delta as in the previous paragraph. Unlike for the
delta, though, these functions can be chosen to converge pointwise
*everywhere* to the zero function (even at x=0).
---Dan Grubb