View Full Version : Loop quantum gravity and the cosmological constant problem
melroysoares@hotmail.com
Dec24-04, 06:49 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Can someone tell me (or point me to a reference) as to how Loop Quantum\nGravity solves the cosmological\nconstant problem?\nThanks\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Can someone tell me (or point me to a reference) as to how Loop Quantum
Gravity solves the cosmological
constant problem?
Thanks
melroysoares@hotmail.com
Dec25-04, 12:16 PM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>2 cups flour\nOnion, garlic\nSalt\npepper\ngarlic powder\ncayenne pepper\nhot sauce, etc.\nOil for frying\n\nMix milk, eggs, hot sauce in a bowl, add chopped onion and garlic.\nSeason the meat liberally, and marinate for several hours.\nPlace seasoned flour in a paper or plastic shopping bag,\ndrop pieces in a few a time, shake to coat thoroughly,\nthen deep fry in hot oil (350°) for about 15 minutes.\nDrain and place on paper towels.\n\n\n\nMiscarriage with Mustard Greens\n\nWhy waste it? Otherwise, and in general, use ham or salt pork to season greens.\nThe technique of smothering greens can be used with many vegetables;\ngreen beans work especially well. Meat is not necessary every day, don?t\nbe afraid to alter any dish to vegetarian tastes.\n\n1 premature baby, born dead\nLarge bunch of mustard greens\n2 white onions, 1 cup chopped celery\nVegetable oil (or hog fat)\nSalt, pepper, garlic, etc.\n\nLightly brown onions, celery, garlic and meat in large heavy pot.\nAdd a little water and the greens (which should be thoroughly cleaned and washed).\nSmother slowly for at least 2 hours, adding small amounts of water\nwhen it starts to stick.\nStir frequently.\nWhen ready - serve with rice, grilled smoked sausage, green salad, and iced tea.\nCoffee and apple pie then brandy.\n\n\n\nMaternity Ward Pot Luck Dinner\n\nIf you can?t get anything fresh from the hospital, nursery, or morgue;\nyou can at least get rid of all the leftovers in your refrigerator.\n\n1 - 2 lbs. cubed meat (human flesh, chicken, turkey, beef...)\n1 -2 lbs. coarsely chopped vegetables\n(carrots, potatoes, turnips, cauliflower, cabbage...)\nBell pepper\nonions\ngarlic\nginger\nsalt pepper, etc.\nOlive oil\nbutter\n\nBrown the meat and s\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>2 cups flour
Onion, garlic
Salt
pepper
garlic powder
cayenne pepper
hot sauce, etc.
Oil for frying
Mix milk, eggs, hot sauce in a bowl, add chopped onion and garlic.
Season the meat liberally, and marinate for several hours.
Place seasoned flour in a paper or plastic shopping bag,
drop pieces in a few a time, shake to coat thoroughly,
then deep fry in hot oil (350°) for about 15 minutes.
Drain and place on paper towels.
Miscarriage with Mustard Greens
Why waste it? Otherwise, and in general, use ham or salt pork to season greens.
The technique of smothering greens can be used with many vegetables;
green beans work especially well. Meat is not necessary every day, don?t
be afraid to alter any dish to vegetarian tastes.
1 premature baby, born dead
Large bunch of mustard greens
2 white onions, 1 cup chopped celery
Vegetable oil (or hog fat)
Salt, pepper, garlic, etc.
Lightly brown onions, celery, garlic and meat in large heavy pot.
Add a little water and the greens (which should be thoroughly cleaned and washed).
Smother slowly for at least 2 hours, adding small amounts of water
when it starts to stick.
Stir frequently.
When ready - serve with rice, grilled smoked sausage, green salad, and iced tea.
Coffee and apple pie then brandy.
Maternity Ward Pot Luck Dinner
If you can?t get anything fresh from the hospital, nursery, or morgue;
you can at least get rid of all the leftovers in your refrigerator.
1 - 2 lbs. cubed meat (human flesh, chicken, turkey, beef...)
1 -2 lbs. coarsely chopped vegetables
(carrots, potatoes, turnips, cauliflower, cabbage...)
Bell pepper
onions
garlic
ginger
salt pepper, etc.
Olive oil
butter
Brown the meat and s
melroysoares@hotmail.com
Dec25-04, 02:42 PM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>salad, macaroni and cheese (homemade) and iced tea...\n\n\n\nSpaghetti with Real Italian Meatballs\n\nIf you don?t have an expendable bambino on hand,\nyou can use a pound of ground pork instead.\nThe secret to great meatballs, is to use very lean meat.\n\n1 lb. ground flesh; human or pork\n3 lb. ground beef\n1 cup finely chopped onions\n7 - 12 cloves garlic\n1 cup seasoned bread crumbs\n˝ cup milk, 2 eggs\nOregano\nbasil\nsalt\npepper\nItalian seasoning, etc.\nTomato gravy (see index)\nFresh or at least freshly cooked spaghetti or other pasta\n\nMix the ground meats together in a large bowl,\nthen mix each of the other ingredients.\nMake balls about the size of a baby?s fist\n(there should be one lying around for reference).\nBake at 400°for about 25 minutes -\nor you could fry them in olive oil.\nPlace the meatballs in the tomato gravy, and simmer for several hours.\nServe on spaghetti.\nAccompany with green salad, garlic bread and red wine.\n\n\n\nNewborn Parmesan\n\nThis classic Sicilian cuisine can easily be turned into Eggplant Parmesan\nIf you are planning a vegetarian meal. Or you could just as well use veal -\nafter all, you have to be careful - Sicilians are touchy about their young\nfamily members...\n\n6 newborn or veal cutlets\nTomato gravy (see index)\n4 cups mozzarella, 1cup parm\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>salad, macaroni and cheese (homemade) and iced tea...
Spaghetti with Real Italian Meatballs
If you don?t have an expendable bambino on hand,
you can use a pound of ground pork instead.
The secret to great meatballs, is to use very lean meat.
1 lb. ground flesh; human or pork
3 lb. ground beef
1 cup finely chopped onions
7 - 12 cloves garlic
1 cup seasoned bread crumbs
˝ cup milk, 2 eggs
Oregano
basil
salt
pepper
Italian seasoning, etc.
Tomato gravy (see index)
Fresh or at least freshly cooked spaghetti or other pasta
Mix the ground meats together in a large bowl,
then mix each of the other ingredients.
Make balls about the size of a baby?s fist
(there should be one lying around for reference).
Bake at 400°for about 25 minutes -
or you could fry them in olive oil.
Place the meatballs in the tomato gravy, and simmer for several hours.
Serve on spaghetti.
Accompany with green salad, garlic bread and red wine.
Newborn Parmesan
This classic Sicilian cuisine can easily be turned into Eggplant Parmesan
If you are planning a vegetarian meal. Or you could just as well use veal -
after all, you have to be careful - Sicilians are touchy about their young
family members...
6 newborn or veal cutlets
Tomato gravy (see index)
4 cups mozzarella, 1cup parm
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>melroysoares@hotmail.com Wrote:\n> Can someone tell me (or point me to a reference) as to how Loop\n> Quantum\n> Gravity solves the cosmological constant problem?\n> Thanks\n\nFrom the spin foam approach, as in:\n\nQuantum symmetry, the cosmological constant and Planck scale\nphenomenology\nGiovanni Amelino-Camelia, Lee Smolin, Artem Starodubtsev\nhttp://arxiv.org/abs/hep-th/0306134\n\nrelying on a deformation parameter for the symmetry group,\nwhich gives the cosmological constant a positive value.\n\nHappy holidays\nKea :blushing:\n\n------------------------------------------------------------------------\nThis post submitted through the LaTeX-enabled physicsforums.com\nTo view this post with LaTeX images:\nhttp://www.physicsforums.com/showthread.php?t=57704#post411525\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>melroysoares@hotmail.com Wrote:
> Can someone tell me (or point me to a reference) as to how Loop
> Quantum
> Gravity solves the cosmological constant problem?
> Thanks
From the spin foam approach, as in:
Quantum symmetry, the cosmological constant and Planck scale
phenomenology
Giovanni Amelino-Camelia, Lee Smolin, Artem Starodubtsev
http://arxiv.org/abs/http://www.arxiv.org/abs/hep-th/0306134
relying on a deformation parameter for the symmetry group,
which gives the cosmological constant a positive value.
Happy holidays
Kea :blushing:
------------------------------------------------------------------------
This post submitted through the LaTeX-enabled physicsforums.com
To view this post with LaTeX images:
http://www.physicsforums.com/showthread.php?t=57704#post411525
whopkins@csd.uwm.edu
Jan19-05, 11:46 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Classical Yang-Mills theory already has a non-zero cosmological\nconstant, when formulated in terms of fibre bundles. The curvature\nscalar, for a non-abelian Yang Mills theory, reduces to the form:\nR - 1/4 F^2 + 1/4 f^2\nwhere the F\'s are the field strength, and the f\'s the structure\nconstants.\n\nSo, anything that incorporates the above representation of Yang-Mills\nwill, likewise, have a cosmological constant in it. The so-called\n"loop quantum gravity" is actually a misnomer. It should read "loop\nYang-Mills theory", since it\'s actually a theory of the quantization of\nYang-Mills fields, not of gravity specifically (except in as far as (as\nan afterthought) gravity can be put into Yang-Mills form). If you put\nthis theory up on a fibre bundle, a non-zero cosmological constant\ncomes out of the last term.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Classical Yang-Mills theory already has a non-zero cosmological
constant, when formulated in terms of fibre bundles. The curvature
scalar, for a non-abelian Yang Mills theory, reduces to the form:
R - 1/4 F^2 + 1/4 f^2
where the F's are the field strength, and the f's the structure
constants.
So, anything that incorporates the above representation of Yang-Mills
will, likewise, have a cosmological constant in it. The so-called
"loop quantum gravity" is actually a misnomer. It should read "loop
Yang-Mills theory", since it's actually a theory of the quantization of
Yang-Mills fields, not of gravity specifically (except in as far as (as
an afterthought) gravity can be put into Yang-Mills form). If you put
this theory up on a fibre bundle, a non-zero cosmological constant
comes out of the last term.
Ken S. Tucker
Jan23-05, 08:57 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\nMr. whopk\n(nice to talk again)\ncould you be a bit more specific about the variables\n"F" and "f", maybe mathematical and physical, but\nat my technician level please.\nKen\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Mr. whopk
(nice to talk again)
could you be a bit more specific about the variables
"F" and "f", maybe mathematical and physical, but
at my technician level please.
Ken
whopkins@csd.uwm.edu
Jan27-05, 09:44 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\nThe Yang-Mills field generalizes the electromagnetic field, whose\npotentials are given by the four vector A_m, (A_0 = phi, (A_1,A_2,A_3)\n= A), up to multiples of c. The Yang-Mills field is A^a_m, the index a\nranging over the dimensions of the symmetry group\'s Lie algebra. For\nthe electromagnetic field there\'s only one index a, so only one set of\npotentials. For U(1) x SU(2) x SU(3), a ranges over 12 values (8 for\nSU(3), 3 for SU(2), 1 for U(1)).\n\nThe field strength F^a_{mn} generalizes the electromagnetic field\ntensor, but is given by a non-linear expression in general:\nF^a_{mn} = d_m A^a_n - d_n A^a_m + sum f^a_{bc} A^b_m\nA^c_n\n(d_m here means partial with respect to the coordinate\nx^m)\nwhere the f\'s are the structure constants of the Lie algebra. In the\ncase of electromagnetism, U(1) is Abelian and for all Abelian groups,\nthe corresponding Lie algebra has structure constants f = 0. For\nSU(2), the Lie algebra is given by a basis\n[T1,T2] = T3; [T2,T3] = T1; [T3,T1] = T2\nand the only non-zero structure constants are:\nf^1_{23} = f^2_{31} = f^3_{12} = 1\nf^1_{32} = f^2_{13} = f^3_{21} = -1.\nSo, then there\'s a part to the field strength that\'s quadratic in the\npotentials. Arranging the 3 components of the SU(2) potential as a\nvector (A^1_m, A^2_m, A^3_m) = A_m, then the equation for F is written\nas:\nF_{mn} = d_m A_n - d_m A_n + A_m x A_n.\nFor SU(3) the structure constants are more elaborate and numerous.\n\nA metric on the Lie algebra will be adjoint invariant when it lowers\nthe index on f resulting in a completely anti-symmetric f_{abc}. For\nSU(2), the only such metrics are all of the form:\nk_{ab} = k delta_{ab}\nin which case, one finds\nf_{123} = f_{231} = f_{312} = k = -f_{321} =\n-f_{213} = -f_{132}.\nLikewise, for SU(3) only one metric, up to a multiple, exists which has\nthe corresponding property. For U(1), the structure constants are 0,\nand the metric isn\'t determined by this requirement.\n\nThe square F^2 is then\nsum k_{ab} F^a_{mn} F^b_{pq} g^{mp} g^{nq}\nwhere g is the spacetime metric. The square f^2 is\nf^2 = sum (f^a_{bc} f^b_{ad} k^{cd})\nwhere k^{cd} is the inverse of the metric k_{ab} (in the SU(2) example\nabove, k^{ab} = (1/k) delta^{ab}).\n\nThe "Kaluza-Klein" manifold is constructed on top of the ordinary 4-D\nmanifold by adding the extra dimensions of the Lie algebra and\nextending the metric to incorporate it. So, if h_{AB} denotes the\nmetric for the total space, its components with respect to the metrics\n(g,k) and the potentials A will be:\nh_{mn} = g_{mn} + sum (k_{ab} A^a_m A^b_n)\nh_{mb} = sum (k_{ab} A^a_m)\nh_{an} = sum (k_{ab} A^b_n)\nh_{ab} = k_{ab}.\nCalculating the curvature scalar for h, denoting it R_h, yields the\nresult\nR_h = R - 1/4 F^2 + 1/4 f^2\nwhich corresponds to the curvature scalar for the original 4-D\nmanifold, plus the contribution F^2 which mimicks the effect of a\nYang-Mills field, and a contribution f^2 which gives you a cosmological\nconstant.\n\nA particle moving along a geodesic will have velocity components (v_m;\nv_a), the v_a\'s giving you the components of its charge. The charge of\nthe particle is not a single number but a Lie vector (with 12\ncomponents for U(1) x SU(2) x SU(3)). So, the geodesic equations for\nthe particle will break down into a part corresponding to the geodesic\nof the original manifold, plus an extra part, analogous to the Lorentz\nforce, corresponding to the Yang-Mills field. The force will be\nproportional to the sum (v_a F^a_{mn}), so is actually best considered\nas the scalar product (v).(F_{mn}) of the charge vector by the\n12-component vector F_{mn}. There will be an extra equation for the\ntime variation of v, as well, when there are non-zero structure\nconstants. The resulting equations are called Wong\'s equations. The\nchange in the charge vector (v) corresponds to flavor changing\ninteractions, e.g. an electron transforming into a neutrino.\n\nHowever, Wong\'s equations only results from the geodesic equations if\nan adjoint invariant metric was used for the Lie algebra. In that\ncase, the total magnitude of the charge:\n|v|^2 = sum (k^{ab} v_a v_b)\nwill be a constant of motion. So, the charge can then be thought of as\nbeing composed of a charge magnitude plus a "complexion", which is its\norientation. Two sources with the same orientation will repel, and\nwill attract if they have opposite complexions. In general their\nmutual force will be proportional to the product of their charge\nmagnitudes and the cosine of the angle between their complexions. The\nconservation of charge magnitude means that the Yang-Mills force is not\ncreating or destroying the charge, per se, but merely rotating its\ncomplexion. (The field, itself, carries off the difference between the\ninitial and final charge).\n\nSo, this naturally raises the question of whether a metric exists for\nU(1) x SU(2) x SU(3) that makes all the particles have an equal charge\nmagnitude. The answer is no -- unless:\n(1) you add in an extra U(1) for a force corresponding to the\nquantum number\n(Baryon - Lepton)\n(2) you include right-handed neutrinos and left-handed\nanti-neutrinos,\nwhich were excluded by the Standard .\nIn this case, there are actually two sets of quadratic combinations\nthat yield constant values, as indicated previously; one involving\nSU(3) and (Baryon - Lepton) and the other involving SU(2) and\n(Hypercharge - (Baryon - Lepton)/2), this latter quantity being the\nright-handed analogue of the isospin charge in SU(2). The occurrence\nof the last quantity strongly suggests there is a second SU(2) lurking\naround, which it is a part of, and the quadratic invariant is just IL^2\n+ IR^2, where IL, IR are the totals of the charges corresponding to the\noriginal SU(2) and the other, still unseen, one. Likewise, the (Baryon\n- Lepton) number fits in nicely with the rest of SU(3) to give you\nU(3), and the corresponding invariant is just the sum of the squares of\nthe 9 components of the resulting U(3) charge.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>The Yang-Mills field generalizes the electromagnetic field, whose
potentials are given by the four vector A_m, (A_0 = \phi, (A_1,A_2,A_3)= A), up to multiples of c. The Yang-Mills field is A^{a_m}, the index a
ranging over the dimensions of the symmetry group's Lie algebra. For
the electromagnetic field there's only one index a, so only one set of
potentials. For U(1) x SU(2) x SU(3), a ranges over 12 values (8 for
SU(3), 3 for SU(2), 1 for U(1)).
The field strength F^{a_}{mn} generalizes the electromagnetic field
tensor, but is given by a non-linear expression in general:
F^{a_}{mn} = d_m A^{a_n} - d_n A^{a_m} + sum f^{a_}{bc} A^{b_m}A^{c_n}(d_m here means partial with respect to the coordinate
x^m)
where the f's are the structure constants of the Lie algebra. In the
case of electromagnetism, U(1) is Abelian and for all Abelian groups,
the corresponding Lie algebra has structure constants f = . For
SU(2), the Lie algebra is given by a basis
[T1,T2] = T3; [T2,T3] = T1; [T3,T1] = T2
and the only non-zero structure constants are:
f^{1_}{23} = f^{2_}{31} = f^{3_}{12} = 1f^{1_}{32} = f^{2_}{13} = f^{3_}{21} = -1.
So, then there's a part to the field strength that's quadratic in the
potentials. Arranging the 3 components of the SU(2) potential as a
vector (A^{1_m}, A^{2_m}, A^{3_m}) = A_m, then the equation for F is written
as:
F_{mn} = d_m A_n - d_m A_n + A_m x A_n.
For SU(3) the structure constants are more elaborate and numerous.
A metric on the Lie algebra will be adjoint invariant when it lowers
the index on f resulting in a completely anti-symmetric f_{abc}. For
SU(2), the only such metrics are all of the form:
k_{ab} = k \delta_{ab}
in which case, one finds
f_{123} = f_{231} = f_{312} = k = -f_{321} =-f_{213} = -f_{132}.
Likewise, for SU(3) only one metric, up to a multiple, exists which has
the corresponding property. For U(1), the structure constants are 0,
and the metric isn't determined by this requirement.
The square F^2 is then
sum k_{ab} F^{a_}{mn} F^{b_}{pq} g^{mp} g^{nq}
where g is the spacetime metric. The square f^2 is
f^2 = sum (f^{a_}{bc} f^{b_}{ad} k^{cd})
where k^{cd} is the inverse of the metric k_{ab} (in the SU(2) example
above, k^{ab} = (1/k) \delta^{ab}).
The "Kaluza-Klein" manifold is constructed on top of the ordinary 4-D
manifold by adding the extra dimensions of the Lie algebra and
extending the metric to incorporate it. So, if h_{AB} denotes the
metric for the total space, its components with respect to the metrics
(g,k) and the potentials A will be:
h_{mn} = g_{mn} + sum (k_{ab} A^{a_m} A^{b_n})h_{mb} = sum (k_{ab} A^{a_m})h_{an} = sum (k_{ab} A^{b_n})h_{ab} = k_{ab}.
Calculating the curvature scalar for h, denoting it R_h, yields the
result
R_h = R - 1/4 F^2 + 1/4 f^2
which corresponds to the curvature scalar for the original 4-D
manifold, plus the contribution F^2 which mimicks the effect of a
Yang-Mills field, and a contribution f^2 which gives you a cosmological
constant.
A particle moving along a geodesic will have velocity components (v_m;v_a), the v_a's giving you the components of its charge. The charge of
the particle is not a single number but a Lie vector (with 12
components for U(1) x SU(2) x SU(3)). So, the geodesic equations for
the particle will break down into a part corresponding to the geodesic
of the original manifold, plus an extra part, analogous to the Lorentz
force, corresponding to the Yang-Mills field. The force will be
proportional to the sum (v_a F^{a_}{mn}), so is actually best considered
as the scalar product (v).(F_{mn}) of the charge vector by the
12-component vector F_{mn}. There will be an extra equation for the
time variation of v, as well, when there are non-zero structure
constants. The resulting equations are called Wong's equations. The
change in the charge vector (v) corresponds to flavor changing
interactions, e.g. an electron transforming into a neutrino.
However, Wong's equations only results from the geodesic equations if
an adjoint invariant metric was used for the Lie algebra. In that
case, the total magnitude of the charge:
|v|^2 = sum (k^{ab} v_a v_b)
will be a constant of motion. So, the charge can then be thought of as
being composed of a charge magnitude plus a "complexion", which is its
orientation. Two sources with the same orientation will repel, and
will attract if they have opposite complexions. In general their
mutual force will be proportional to the product of their charge
magnitudes and the cosine of the angle between their complexions. The
conservation of charge magnitude means that the Yang-Mills force is not
creating or destroying the charge, per se, but merely rotating its
complexion. (The field, itself, carries off the difference between the
initial and final charge).
So, this naturally raises the question of whether a metric exists for
U(1) x SU(2) x SU(3) that makes all the particles have an equal charge
magnitude. The answer is no -- unless:
(1) you add in an extra U(1) for a force corresponding to the
quantum number
(Baryon - Lepton)
(2) you include right-handed neutrinos and left-handed
anti-neutrinos,
which were excluded by the Standard .
In this case, there are actually two sets of quadratic combinations
that yield constant values, as indicated previously; one involving
SU(3) and (Baryon - Lepton) and the other involving SU(2) and
(Hypercharge - (Baryon - Lepton)/2), this latter quantity being the
right-handed analogue of the isospin charge in SU(2). The occurrence
of the last quantity strongly suggests there is a second SU(2) lurking
around, which it is a part of, and the quadratic invariant is just IL^2+ IR^2, where IL, IR are the totals of the charges corresponding to the
original SU(2) and the other, still unseen, one. Likewise, the (Baryon
- Lepton) number fits in nicely with the rest of SU(3) to give you
U(3), and the corresponding invariant is just the sum of the squares of
the 9 components of the resulting U(3) charge.
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