Register to reply

Metric tensor transformed into momentum space?

by Jay R. Yablon
Tags: metric, momentum, space, tensor, transformed
Share this thread:
Jay R. Yablon
#1
Nov4-06, 03:39 PM
P: n/a
The metric tensor g_uv is typically understood to be a function of
spacetime coordinates, that is, g_uv(x^u).

Has anyone ever considered a metric tensor Fourier transformed into
momentum space,

g_uv(p^u) = int {d^4x exp (-2 pi i p x) g_uv(x^u)}?

And, is there any reason why it might not make sense to consider this?
Above, int is an integral sign and exp is exponentiation.

Note that the inverse transformation, of course, is:

g_uv(x^u) = int {d^4p exp (+2 pi i p x) g_uv(p^u)}

which includes the factor int d^4p which after reduction becomes int
(dp^2/p^2) = ln p^2 which, evaluated between a mass m and infinity oo,
becomes the logarithmically-divergent term ln (oo/m^2) to which a cutoff
oo-->M^2 is assigned, such that ln (oo/m^2) --> ln (M^2/m^2), which is
then "swept under the rug" into the definition of the running charge
during renormalization.

Thanks,

Jay.
_____________________________
Jay R. Yablon
Email: jyablon@nycap.rr.com

Phys.Org News Partner Physics news on Phys.org
Engineers develop new sensor to detect tiny individual nanoparticles
Tiny particles have big potential in debate over nuclear proliferation
Ray tracing and beyond
carlip-nospam@physics.ucdavis.edu
#2
Nov4-06, 03:39 PM
P: n/a
Jay R. Yablon <jyablon@nycap.rr.com> wrote:
> The metric tensor g_uv is typically understood to be a function of
> spacetime coordinates, that is, g_uv(x^u).


> Has anyone ever considered a metric tensor Fourier transformed into
> momentum space,


> g_uv(p^u) = int {d^4x exp (-2 pi i p x) g_uv(x^u)}?


In what coordinate system? The quantity in your exponent is not
well-behaved under coordinate transformations. Neither is your
integral -- only the integral of a scalar density (or equivalently
a four-form, in four dimensions) is independent of the choice of
coordinates.

Steve Carlip

thomas_larsson_01@hotmail.com
#3
Nov4-06, 03:39 PM
P: n/a
Jay R. Yablon skrev:
>
> Has anyone ever considered a metric tensor Fourier transformed into
> momentum space,
>


If you want, you can expand the functions over spacetime in a
plane wave basis, or any other complete basis for that matter.
Locally at least, there might be global problems. In the
vincinity of Minkowski space, with g = eta + h, it is even
useful. The small term h then describes the graviton, which
has the dispersion relation p^2 = 0.

Far away from Minkowski space, the Fourier components will
transform in a weird way under coordinate transformations, and
a momentum space picture appears useless to me.


Jay R. Yablon
#4
Nov4-06, 03:39 PM
P: n/a
Metric tensor transformed into momentum space?

<carlip-nospam@physics.ucdavis.edu> wrote in message
news:eg6pj2$4bl$1@skeeter.ucdavis.edu...
> Jay R. Yablon <jyablon@nycap.rr.com> wrote:
>> The metric tensor g_uv is typically understood to be a function
>> of
>> spacetime coordinates, that is, g_uv(x^u).

>
>> Has anyone ever considered a metric tensor Fourier transformed into
>> momentum space,

>
>> g_uv(p^u) = int {d^4x exp (-2 pi i p x) g_uv(x^u)}?

>
> In what coordinate system? The quantity in your exponent is not
> well-behaved under coordinate transformations. Neither is your
> integral -- only the integral of a scalar density (or equivalently
> a four-form, in four dimensions) is independent of the choice of
> coordinates.
>
> Steve Carlip


Yes, agreed. Please permit me to modify this and be more specific.

Think of, say, scattering experiments where one has fermion with an
initial momentum p, final p' and a photon momentum q. Suppose we have a
metric tensor which is a function of all three, g_uv(p',q,p)
(similarly, say, to the Gamma_u(p',q,p) which are used to represent
perturbations at the fermion/photon vertex -- I am thinking specifically
of an anticommutator defined by g_uv = .5{Gamma_u Gamma_v} and trying
to use this as a metric tensor because this does reduce to the Minkowski
metric n_uv for q-->0, that is, g_uv(p,0,p)=n_uv). Suppose we specify a
"momentum differential element" dp^u, (u=0,1,2,3) which may be
represented in some coordinate system of our choosing. Is there any
reason we cannot define an "invariant mass interval" dm in momentum
space, such that:

dm(p',q,p)^2 = g_uv(p',q,p) dp^u dp^v (1)

where in momentum space, the usual spacetime elements are replaced by
ds(x^u)-->dm(p',q,p); dx^u-->dp^u; g_uv(x^u) --> g_uv(p',q,p), and the
Fourier transform, in keeping with what you said above about using a
well-behaved four-form operating on a scalar density, is:

dm(p',q,p)^ = int {d^4x exp (-2 pi i p x) ds(x^u)^2}? (2)

Thus, in the simple case, if we were to choose Cartesian coordinates,
the q-->0; g_uv(p,0,p)-->n_uv expression for (1) would be:

dm(p,0,p)^2 = dE^2 - dp_x^2 - dp_y^2 - dp_z^2 (3)

Thanks.

Jay R. Yablon
_____________________________
Jay R. Yablon
Email: jyablon@nycap.rr.com

Oh No
#5
Nov4-06, 03:39 PM
P: n/a
Thus spake carlip-nospam@physics.ucdavis.edu
>Jay R. Yablon <jyablon@nycap.rr.com> wrote:
>> The metric tensor g_uv is typically understood to be a function of
>> spacetime coordinates, that is, g_uv(x^u).

>
>> Has anyone ever considered a metric tensor Fourier transformed into
>> momentum space,

>
>> g_uv(p^u) = int {d^4x exp (-2 pi i p x) g_uv(x^u)}?


I consider something rather like that as the momentum space wave
function for free electrons, positrons and photons, at time x^0=0

F_m(p) = (1/2pi)^3 sum_r yta(r) int {d^3x exp (ipx) f_n(0,x) u_n(p,r)}

Here u_n is either a Dirac spinor with r = 1,2 and yta(r) = 1, or, for
the photon, it is a timelike, longitudinal or transverse unit vector
with yta(0)=-1, and yta(r) = 1 for r=1,2,3.

>In what coordinate system? The quantity in your exponent is not
>well-behaved under coordinate transformations. Neither is your
>integral -- only the integral of a scalar density (or equivalently
>a four-form, in four dimensions) is independent of the choice of
>coordinates.


You once asked me that question, and the answer was that my model was
rather incompletely developed. But I was grateful for the question as it
pushed me to work things out a bit better. The answer, briefly, goes
like this.

The wave function is calculated using quantum coordinates, which make it
uniquely possible to write down wave functions and retain radially plane
waves. I hold that this is an essential part of quantum theory. The
feature of quantum coordinates which makes this possible is that they
satisfy a metric

dS^2 = dt^2 - dr^2 - g^2(r)dO^2

This is not the physical metric, but in keeping with the orthodox
interpretation of quantum theory probability amplitudes are not physical
and are merely used in the calculation of probabilities.

The metric for quantum coordinates is

ds^2 = a^2(t)(4(dt^2-dr^2) - sin^2(r)(dO^2)/4)

The factors of 2^2 have to do with the way the teleconnection yields a
square law for cosmological redshift in terms of the expansion
parameter, and I think they are fundamentally related to spin half for
the electron. I suppose that means I should really reformulate qed
completely in this metric, but in practice I revert to a flat space
approximation for qed and get rid of them.

Regards

--
Charles Francis
substitute charles for NotI to email



Register to reply