Metric tensor transformed into momentum space?by Jay R. Yablon Tags: metric, momentum, space, tensor, transformed 
#1
Nov406, 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 logarithmicallydivergent 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 


#2
Nov406, 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 wellbehaved under coordinate transformations. Neither is your integral  only the integral of a scalar density (or equivalently a fourform, in four dimensions) is independent of the choice of coordinates. Steve Carlip 


#3
Nov406, 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. 


#4
Nov406, 03:39 PM

P: n/a

Metric tensor transformed into momentum space?
<carlipnospam@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 > wellbehaved under coordinate transformations. Neither is your > integral  only the integral of a scalar density (or equivalently > a fourform, 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 wellbehaved fourform 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 


#5
Nov406, 03:39 PM

P: n/a

Thus spake carlipnospam@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 >wellbehaved under coordinate transformations. Neither is your >integral  only the integral of a scalar density (or equivalently >a fourform, 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^2dr^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 


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