What Happens to Fourier Transforms in Curved Spacetime?
Jay R. Yablon
example, of a vector J^u, according to:
J^u(k) = $ d^4x J^u(x) exp[-i p_s x^s] (1)
Does this change at all in curved spacetime? In particular, is it
necessary to generalize the Fourier Kernel exp[-i p_s x^s], to something
else, for example:
exp[-i p_s x^s] --> exp[-i p_s x^s + F(x)] (2)
where F(x) is some function of the coordinates x^u?
Or, does (1) still suffice?
Thanks,
Jay.
____________________________
Jay R. Yablon
Email: jya...@nycap.rr.com
co-moderator: sci.physics.foundations
Weblog: http://jayryablon.wordpress.com/
Web Site: http://home.roadrunner.com/~jry/FermionMass.htm
Ilja
> In flat spacetime, one may take the forward Fourier transform, for
> example, of a vector J^u, according to:
>
> J^u(k) = $ d^4x J^u(x) exp[-i p_s x^s] � �(1)
>
> Does this change at all in curved spacetime? �In particular, is it
> necessary to generalize the Fourier Kernel exp[-i p_s x^s], to something
> else, for example:
>
> exp[-i p_s x^s] --> exp[-i p_s x^s + F(x)] � (2)
>
> where F(x) is some function of the coordinates x^u?
>
> Or, does (1) still suffice?
Worse, there is no nice generalization of Fourier transforms to
curved background.
Gordon Stangler
Although I had that feeling, I would like some more information/
proof/ your intuition on this statement.
Tom Roberts
Worse still, I believe that no such generalization can be defined, such that it
applies to any general Lorentzian manifold. The problem, of course, is that in
general integration is not well defined over a finite region of a manifold, much
less over the entire manifold.
Tom Roberts
Jay R. Yablon
"Tom Roberts" <tjro...@sbcglobal.net> wrote in message
news:fs-dnaHkfr7...@giganews.com...
It is good to think about this problem not in terms of "Fourier
transforms" per se, but rather, to think about the (forward) Fourier
transform as an integral over all of a manifold of a particular function
times the kernel exp[-i p_s x^s]. That is, you are correct IMHO to
think about this in terms of integrals in curved spacetime. We can then
turn to the separate question of what the Fourier kernel must be in
curved spacetime.
For integrals ($) of some function F(x) in curved spacetime, we
generally have:
$ F(x) sqrt(-g) d^4x (1)
Now, it seems to me that the mathematicians can tell us the conditions
under which the integral (1) can and cannot be evaluated in a
well-defined way. It would seem to me to depend upon such things as 1)
the behavior of F(x), 2) the character of F(x) (is it a scalar, a
vector, a tensor, or what, exactly, is F(x)) 3) the behavior of the
metric tensor g_uv (and its determinant g), and possibly 4) the boundary
conditions on both F(x) and g_uv at t,x,y,x approaching infinity.
I would be very curious about the conditions under which (1) is and is
not well-defined. Then, knowing those conditions, perhaps we can talk
about a particular integral, namely the Fourier transform.
Jay
Jay R. Yablon
Let me add to the mix of this discussion, part of a post by "Omega
Cubed" at sci.physics.relativity (yes, amazingly, there are still some
people there who are not trolls), as well as my reply, all below. I
would ask, please, for your comments on this as well as what I posted a
little earlier. Jay.
"Omega Cubed" <omega...@netzero.net> wrote in message
news:Hz4Em.27537$AB3....@newsfe15.ams2...
> On 2009-10-22, Jay R. Yablon <jya...@nycap.rr.com> wrote:
>> In flat spacetime, one may take the forward Fourier transform, for
>> example, of a vector J^u, according to:
>>
>> J^u(k) = $ d^4x J^u(x) exp[-i p_s x^s] (1)
>>
>> Does this change at all in curved spacetime? In particular, is it
>> necessary to generalize the Fourier Kernel exp[-i p_s x^s], to
>> something
>> else, for example:
>>
>> exp[-i p_s x^s] --> exp[-i p_s x^s + F(x)] (2)
>>
>> where F(x) is some function of the coordinates x^u?
>>
>> Or, does (1) still suffice?
>
> You can't define the Fourier transform in curved space, in general.
> Especially not for tensorial/vectorial quantities.
>
> You can define a notion of the Fourier transform IF you have a global
> coordinate system which covers the entire manifold AND if you use the
> "natural" volume form for the coordinate system (i.e. the pull back of
> the Euclidean volume form) in the integral. But that basically means
> that you are doing the standard flat-space Fourier transform without
> regard to geometry.
>
> But such definitions will, of course, depend heavily on the
> coordinates used, and is not natural. Therefore they are not
> particularly useful in practical situations.
>
> Now, there are two notions that can allow some generalization of the
> Fourier transform to a manifold (that I know of).
>
> I. Harmonic analysis on groups.
>
> The Fourier transform can be defined on topological groups. The idea
> is that R^n is a topological group under translations. If you consider
> all group homomorphisms from R^n into the circle group, you end up
> with another group, called the dual group of R^n. Turns out that the
> dual group of R^n is homomorphic to R^n itself. This is how we end up
> with the phase-space vs. physical space description.
>
> To illustrate this further, if instead of R^n, we start with the
> n-dimensional torus T^n. One will see that the group homomorphisms
> into circle groups can be characterized by the period of each of the
> dimensions on the torus. Since group homomorphisms from the circle to
> the circle must be of the form theta -> k theta where k is an integer,
> we see that the dual group of T^n is Z^n, the n-dimensional integers.
> This shows that over T^n, a Fourier series can be defined, whereas
> over R^n it is no longer a discrete sum.
>
> In general, this can also be done for a large class of locally compact
> abelian groups. (I've heard that it can also be done for certain
> non-abelian groups, but that's beyond my knowledge and interest in so
> far as harmonic analysis is concerned.)
>
> I probably explained very poorly the idea here. You can read a bit
> more about it (and get some decent references) on the Wiki page
> http://en.wikipedia.org/wiki/Pontrjagin_duality
>
> The point I am trying get across here is that this definition is (a)
> very algebraic and (b) depends on having a group structure on the
> fundamental underlying manifold. Generically, of course, a Riemannian
> manifold will not be a topological group. And if you are interested in
> general relativity, the evolving metric throws a big monkey wrench
> into this procedure.
. . .
Thank you for this helpful post. I am going to focus on what you wrote
above, as that seems most pertinent to what I am considering.
Let's go to your link at
http://en.wikipedia.org/wiki/Pontrjagin_duality#Fourier_transform.
They generalize the forward Fourier transform as follows:
f-hat(xi) = $ f(x) xi(x)-bar d mu(x) (1)
mu(x) is a Haar measure, which at
http://en.wikipedia.org/wiki/Haar_measure is said to be "is a way to
assign an 'invariant volume' to subsets of locally compact topological
groups and subsequently define an integral for functions on those
groups."
If the group in question is R^4 and it is locally compact, then I
interpret this to mean that:
d mu(x) = sqrt(-g(x)) d^4x (2)
Is this so?
Next, f(x) is the function for which we are taking the Fourier
transform. No rocket science needed here.
Finally, I interpret xi(x)-bar, especially given that the inverse
transform uses xi(x), to be the generalization of the Fourier Kernel
U-bar=exp[-i p_s x^s], where I use U to denote that this is a unitary.
Thus, I take this all to mean that:
xi(x)-bar xi(x) = 1 (3)
just as
U-bar U =exp[-i p_s x^s] exp[i p_s x^s] = 1 (4)
This ensures that the inverse of the forward transform yields the
original function, which seems to be a condition precedent to defining
any sensible type of Fourier transform.
Thus, with xi = p = energy/momentum, I would write (1) to define the
forward Fourier transform in the form:
f-hat(p) = $ f(x) U-bar sqrt(-g(x)) d^4x (5)
using a unitary matrix which I would define by:
U == exp [i (p_s x^s + alpha(x) )] (6)
where alpha(x) is some unspecified function of the spacetime coordinates
x^u. (6) above is a completely general unitary kernel, which for the
alpha(x) = 0 reduces to the usual Fourier kernel of flat spacetime.
>From what I read here, this would be well-defined on "subsets of locally
compact topological groups and subsequently define an integral for
functions on those groups." Am I on the right track here?
Also, how would one characterize the requirements that must be satisfied
by any curved R^4 spacetime for which (1) and (5) does make sense (i.e.,
can be evaluated)? Must it be a subset? Or, can it be the whole
manifold so long as that manifold (and perhaps it boundaries) satisfy
certain conditions?
Thanks,
Jay
Oh No
You immediately hit a problem. The kernel is a coordinate dependent
quantity. It does not exist in any coordinate free way in curved
spacetime.
Regards
--
Charles Francis
moderator sci.physics.foundations.
charles (dot) e (dot) h (dot) francis (at) googlemail.com (remove spaces and
braces)
Rich L.
This may be an off the wall comment by someone who is definitely not
as theoretically grounded in higher math as some others here, but I
wonder if you are asking the wrong question?
The Fourier transform works because in many situations the set of sine
(and cosine) waves with appropriate boundary conditions forms a
complete, orthogonal and normalizable set of functions. That is, any
arbitrary, and not too pathological, function can be approximated to
arbitrary precision with a sum of these sine waves. It is exactly
like representing a vector in terms of amplitudes of an orthonormal
set of basis vectors that span the space, only the "space" is of
functions and the "basis vectors" are the (infinite set of) sine/
cosine functions. In the case of a non-Euclidian space, sine and
cosine functions may very well not the the correct, or at least most
appropriate, "basis" to use. The eigenfunctions that are solutions to
many differential equations (with appropriate restrictions that I'm
not qualified to list, but would have to look up) tend to form such
complete, orthogonal and normalizable sets of functions. If I recall
correctly these make up what are called a Hilbert Space (based on a
recollection from a math class 30+ years ago). My thought is that if
you are trying to do EM in the curved space, for example, that the
appropriate functions should be the EM modes (solutions to Maxwell's
Equations, or the relativistic equivalents) in that curved space.
These solutions very likely would not be ordinary sine/cosine
functions. If you are doing something other than EM, the differential
equations for that subject, and the boundary conditions, might yield a
more appropriate "basis" set of functions to do an expansion of your
solutions.
Charles' comment about not being able to define a coordinate
independent integration (needed for normalizing and calculating the
amplitudes of each component function) may be a more fundamental issue
for Lorentzian space, I haven't a clue myself.
Rich L.
Ilja
To prove such things is not possible - at most one can proof that
some properties of the Fourier transformation cannot hold anymore.
But one can of course fix some coordinates (irrelevant of their
physical meaning) and do the Fourier analysis in these coordinates.
The math would work but it would not have much physical interest,
because the Fourier transforms would differ in different coordinates,
and if the coordinates are only local, they would not allow to recover
the original function.
Let's try this: There are interesting generalization of the Fourier
transform in the theory of more general abelian groups.
Abelian groups are basically Q, R, S^1, Z, Z_p, rational numbers,
but also some more extravagant things like p-adic numbers.
For all of them them one can define a Fourier transformation,
which transforms functions on a given abelian group into
functions on some dual abelian group. So the structure
of an underlying abelian group is important, decisive for
the game to work.
Standard application in physics is the application to the
abelian group of translations R^3.
On a curved background, you have no group of translations.
Another way to generalize is to non-abelian groups. There
is already no dual group, but harmonic analysis on works
nicely on homogeneous spaces and the groups them-self.
Again, general metrics do not define homogeneous spaces.
Integration is not a problem, there is, last but not least,
the natural measure sqrt{-g} d^4x. The problem is the
failure of symmetry.
Gordon Stangler
> Let's try this: There are interesting generalization of the Fourier
> transform in the theory of more general abelian groups.
>
> Abelian groups are basically Q, R, S^1, Z, Z_p, rational numbers,
> but also some more extravagant things like p-adic numbers.
> For all of them them one can define a Fourier transformation,
> which transforms functions on a given abelian group into
> functions on some dual abelian group. So the structure
> of an underlying abelian group is important, decisive for
> the game to work.
>
> Standard application in physics is the application to the
> abelian group of translations R^3.
What's an abelian group? What is S^1, and p-adic numbers? Sorry, I
am unfamiliar with these terms.
Jay R. Yablon
"Ilja" <ilja.sc...@googlemail.com> wrote in message
news:dacaf5f5-04f7-492f...@v15g2000prn.googlegroups.com...
. . .
>
> Integration is not a problem, there is, last but not least,
> the natural measure sqrt{-g} d^4x. The problem is the
> failure of symmetry.
>
I want for the moment just to ask you about this one point.
I am fully agreed that sqrt{-g} d^4x needs to be the volume element for
the *forward* transform into momentum space. sqrt(-g(x^u)) is, of
course, a function of the spacetime coordinates.
What, however, do you envision as the volume element of the *inverse*
transform from momentum space back to spacetime?
Also, would you agree as a general principle, even in curved spacetime,
that the inverse of the forward transform of F(x) must be the original
F(x)?
As an aside, I did an arxiv paper on Non-Euclidean momentum space three
years ago, which was the last time a took a close, hard look at momentum
space as a geometrical construct. That is viewable at
http://arxiv.org/abs/hep-ph/0610377. I plan to revisit that paper to
see if anything jumps out at me.
Thanks,
Jay
Bob_for_short
>
> You immediately hit a problem. The kernel is a coordinate dependent
> quantity. It does not exist in any coordinate free way in curved
> spacetime.
>
necessity.
See A. Logunov's RTG for help.
======================================= MODERATOR'S COMMENT:
Approved with a caveat: Please provide a link to the Logunov work that you have in mind so that people have some meat to work with. Othetrwise, this looks like a drive-by slam at GTR. JRY
Tom Roberts
> What, however, do you envision as the volume element of the *inverse*
> transform from momentum space back to spacetime?
In the usual formulation in flat spacetime, the momentum representation
of a function on configuration space is in essence its Fourier
transform. The invariant volume element in momentum representation is
d^4p/E [E = p^0; c=1]
This is not really the Fourier transform of a function on spacetime, but
rather the F.T. of a function of particle coordinates. For massive
particles E>0, but for photons one has an infrared divergence in the
volume element. There are surely additional caveats....
I have not a clue how this might "generalize" to curved manifolds....
Tom Roberts
Jay R. Yablon
"Tom Roberts" <tjrobe...@sbcglobal.net> wrote in message
news:ttGdnQYZCrw...@giganews.com...
In flat spacetime, omitting the 2pi factors just to keep it simple, the
Dirac delta is, essentially, the inverse FT of 1:
delta^4 (x) = $ d^4k (1) exp i[k_s x^s] (1)
The volume element in curved spacetime is sqrt(-g(x)) d^4 x, where g is
a function of x. In momentum space, the volume element must therefore
be a corresponding sqrt(-g(p)) d^4 p, where the metric determinant g is
now a function of momentum.
Now -- let us suppose that we can use the Fourier kernel exp i[k_s x^s]
in curved spacetime as well. In fact, I do not believe that this is so
in general, but I am working on a proof that this is so *in the specific
case of gauge theory in curved spacetime*. That is, if you use the FT
in gauge theory, then even in curved spacetime, the FT uses the Fourier
kernel exp -i[k_s x^s]. In particular, suppose that the Fourier kernel
is something other than this, say, to be perfectly general, exp -i[k_s
x^s - alpha (x,p)] where alpha (x,p) is a completely arbitrary function
of spacetime and / or energy-momentum. The only constraint imposed by
this form, is that we retain unitarity, so that the inverse FT of the FT
is still the original function. (Anything else, it seems, would make no
sense.)
The essence of the proof is based on the fact that:
exp -i[k_s x^s - alpha (x,p)] = exp i alpha (x,p) exp -i k_s x^s (2)
is just like a gauge transformation and in a theory that is invariant
under gauge transformations (and, it turns out, general coordinate
transformations), there is no difference between using the kernel exp
i[k_s x^s], and using some other kernel exp -i[k_s x^s - alpha (x,p)].
Again, this is not true as a general proposition, but I believe I will
be able to offer a proof that it is true in gauge theory. In fact, this
would mean that another benefit of gauge theory is that it permits you
to simplify Fourier transforms in curved spacetime by using the same
kernel exp -i k_s x^s without change.
If I can show this -- and I'll ask you to bear with me for now and I
hope to have this posted later in the week -- then let's now look at the
Dirac delta. In flat spacetime, the delta is given by (1) above. In
curved spacetime, one must use the volume element sqrt(-g(p)) d^4 p
rather than just d^4 p. So, if I am correct that the kernel remains
exp i[k_s x^s] for gauge theory (that is, that the gauge theory remains
unaltered in its content using any other other kernel exp -i[k_s x^s -
alpha (x,p)]), then in curved spacetime (1) becomes:
sqrt(-g(x)) = $ d^4k sqrt(-g(p)) exp i[k_s x^s] (3)
That is, this integral is now the inverse FT of sqrt(-g(p)). Now, let's
take some function D(p) in momentum space. To keep something specific
in mind, think of D(p) as a propagator. Then, the inverse Fourier
transform in curved spacetime is:
sqrt(-g(x)) * D(x) = $ d^4k sqrt(-g(p)) D(p) exp i[k_s x^s] (4)
That is, in spacetime, this is now the *convolution* of the metric
tensor determinant with the propagator.
Now, let's consider flat spacetime. The metric tensor sqrt(-g(x)) = 1.
Therefore, its forward Fourier transform:
$ d^4x sqrt(-g(x)) exp -i[k_s x^s] = delta^4 (p) = sqrt(-g(p)) (5)
is a momentum space delta function. That, then, is the sqrt(-g(p))
which shows up in (3) and (4), if we are in flat spacetime.
So, what I believe fundamentally happens in curved spacetime -- in
addition to the fact which I have not discussed here that the
derivatives in spacetime no longer commute which means that one has to
take care with commuting momenta with one another -- is that the metric
tensor determinant ends up getting convolved with physical quantities
that in flat spacetime we can consider without convolution, or,
equivalently, that we can multiply by a perfect unit impulse in momentum
space.
That is, if you go from flat to curved spacetime in the context of a
theory that is invariant under "exp i alpha" gauge transformations, then
sqrt(-g(x)) ends up getting convolved with other physical quantities of
interest which, in flat spacetime, we do not have to convolve (or,
implicitly, quantities which we are implicitly convolving with 1).
More to follow, in detail. This is just an outline of what I am
presently trying to develop.
Jay
Ilja
> "Ilja" <ilja.schmel...@googlemail.com> wrote in message
> > Integration is not a problem, there is, last but not least,
> > the natural measure sqrt{-g} d^4x. The problem is the
> > failure of symmetry.
> I am fully agreed that sqrt{-g} d^4x needs to be the volume element for
> the *forward* transform into momentum space. �sqrt(-g(x^u)) is, of
> course, a function of the spacetime coordinates.
>
> What, however, do you envision as the volume element of the *inverse*
> transform from momentum space back to spacetime?
On the momentum variables p_m we have the inverse metric g^mn
and the corresponding determinant is simply 1/sqrt{-g}.
The measure is not the problem, but there is not even a forward
transformation, even less a backward one.
> Also, would you agree as a general principle, even in curved spacetime,
> that the inverse of the forward transform of F(x) must be the original
> F(x)?
Whatever else would not be correctly named "backward transformation".
Jay R. Yablon
> On 24 Okt., 10:00, "Jay R. Yablon" <jyab...@nycap.rr.com> wrote:
>> "Ilja" <ilja.schmel...@googlemail.com> wrote in message
>> > Integration is not a problem, there is, last but not least,
>> > the natural measure sqrt{-g} d^4x. The problem is the
>> > failure of symmetry.
>
>> I am fully agreed that sqrt{-g} d^4x needs to be the volume element
>> for
>> the *forward* transform into momentum space. sqrt(-g(x^u)) is, of
>> course, a function of the spacetime coordinates.
>>
>> What, however, do you envision as the volume element of the *inverse*
>> transform from momentum space back to spacetime?
>
> On the momentum variables p_m we have the inverse metric g^mn
> and the corresponding determinant is simply 1/sqrt{-g}.
But, of what is the inverse metric g^mn a function? g^mn(x)? Or,
g^mn(p)?
I say this especially in light of the forward and inverse equations sown
at http://en.wikipedia.org/wiki/Pontrjagin_duality#Fourier_transform,
where the Haar measures (corresponding the with volume element
sqrt(-g)d^4x for the forward transform), are given in general as the
forward volume element d mu(x) and the backward volume element d nu(xi).
I originally thought about using 1/sqrt{-g}, same as you, but thinking
about this more an reading the above link, it seems to me that the d^4 p
multiplier needs to be a function of p, not x. Also, to maintain
general coordinate invariance multiplying d^4 p, you need sqrt(-g),
rather that 1/sqrt(-g) in any case, think about the Jacobeans.
While I can be persuaded otherwise by a good argument to the contrary, I
think the correct answer is, that the volume element of the *inverse*
transform needs to be g^mn(p).
See also my recent post in this thread in reply to Tom Roberts, which
fills in some more details of how I am trying to develop Fourier
transformations in curved spacetime (in the context of gauge theory,
which simplifies the kernel and general coordinate transformation of the
kernel because the theory is invariant under various manipulations of
the kernel), where I believe that the real fundamental result is
understanding that in curved spacetime, the sqrt(-g(x)) term get
*convolved* with the physical quantities of interest Q, according to
sqrt(-g)*Q, which quantities Q in flat spacetime need no convolution
(or, are implicitly convolved simply with sqrt(-g(x))=1).
>
> The measure is not the problem, but there is not even a forward
> transformation, even less a backward one.
>
>> Also, would you agree as a general principle, even in curved
>> spacetime,
>> that the inverse of the forward transform of F(x) must be the
>> original
>> F(x)?
>
> Whatever else would not be correctly named "backward transformation".
So, we seem to be agreed on that, i.e., IF one were to develop a curved
spacetime generalization of the FT, one condition would be the necessity
that a forward followed by a backward transformation needs to be the
identity (which then operates as one constraint on whatever we might
develop). Indeed, a "transformation" cannot be even named as such,
unless there is an inverse which leads back to the original. The whole
practical mathematical purpose of a transform, is that a calculation in
one system is "hard" and in another system is "easier," so you go from
hard to easy, do the calculation, then transform back. The very
simplest, elementary example of this, is multiplying by adding
logarithms.
Jay.
Jay R. Yablon
"Jay R. Yablon" <jya...@nycap.rr.com> wrote in message
news:7kjqf8F...@mid.individual.net...
On third thought, you may be right that 1/sqrt{-g}is the appropriate
volume element for the inverse transform. I did some calculations of
the FT followed by the inverse, and kept getting a -g overall factor,
and so needed to "normalize" by 1/-g to make F^-1 F = 1 Then, I thought
further about how momentum and positions spaces are conjugates, and so
should be inversed. The metric goes contravariant to g^mn as you point
out, and the volume element d^4p=d_0 d_1 d_2 d_3 goes covariant and
everything inverts.
Jay
X-Phy
> What's an abelian group? �What is S^1, and p-adic numbers? �Sorry, I
> am unfamiliar with these terms.
An abelian group is a commutative group, where ab = ba for any
elements a and b of the group.
S^1 is the one dimentional sphere, here it is the unit complex circle
e^ix where x is a real.
X-Phy
> The Fourier transform works because in many situations the set of sine
> (and cosine) waves with appropriate boundary conditions forms a
> complete, orthogonal and normalizable set of functions. �ｿｽThat is, any
> arbitrary, and not too pathological, function can be approximated to
> arbitrary precision with a sum of these sine waves. �ｿｽIt is exactly
> like representing a vector in terms of amplitudes of an orthonormal
> set of basis vectors that span the space, only the "space" is of
> functions and the "basis vectors" are the (infinite set of) sine/
> cosine functions. �ｿｽIn the case of a non-Euclidian space, sine and
> cosine functions may very well not the the correct, or at least most
> appropriate, "basis" to use. �ｿｽThe eigenfunctions that are solutions to
> many differential equations (with appropriate restrictions that I'm
> not qualified to list, but would have to look up) tend to form such
> complete, orthogonal and normalizable sets of functions. �ｿｽIf I recall
> correctly these make up what are called a Hilbert Space (based on a
> recollection from a math class 30+ years ago). �ｿｽMy thought is that if
> you are trying to do EM in the curved space, for example, that the
> appropriate functions should be the EM modes (solutions to Maxwell's
> Equations, or the relativistic equivalents) in that curved space.
> These solutions very likely would not be ordinary sine/cosine
> functions. �ｿｽIf you are doing something other than EM, the differential
> equations for that subject, and the boundary conditions, might yield a
> more appropriate "basis" set of functions to do an expansion of your
> solutions.
I'll expand on that idea. The set of solutions may be chosen so that
they are the customary sine/cosine or exponential functions in flat
space-time far from the curved region. For that to work, a set of
time slices must be chosen, which are flat asymptotically, and the
solutions will depend on time in the general case where the curvature
also depends on time.
X-Phy
> You immediately hit a problem. The kernel is a coordinate dependent
> quantity. It does not exist in any coordinate free way in curved
> spacetime.
Coordinate dependency may be seen in the expansion in spherical
harmonics, which is FT in the case of curved coordinates but flat
space. As spherical harmonics can in turn be expanded in plane wave
functions, there is a kernel transformation associated to the
coordinate change.
Ilja
> "Ilja" <ilja.schmel...@googlemail.com> wrote in message
> g^mn(p)?
It is a function of x. And for the FT to work one would indeed like to
have something depending on p.
FT simply doesn't work on curved space. At least I see no way
to obtain something workable.
Gordon Stangler
Thank you. Is a commutative group the same thing as an abelian
group? If so, why the special name?
X-Phy
> Thank you. �Is a commutative group the same thing as an abelian
> group? �If so, why the special name?
Yes, as a tribute to the mathematician Niels Henrik Abel.
Gordon Stangler
Ah. Thank you for the information. It has been most helpful.