View Full Version : classical em field (was Re: Weinberg on GR
Buzurg Shagird
Apr7-04, 08:44 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resize=yes,status=no,wi dth=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>Charles Francis <charles@clef.demon.co.uk> wrote in message news:<c4kf67\n$1ma$1@lfa222122.richmond.edu>...\n\ n> I don\'t know quite how you work this out. There are four photon\n> polarisation states, not just two. The time like and longitudinal states\n> must always have the same amplitude, and may be considered as a\n> lightlike states, but although cross sections for this state are zero\n> and some treatments discard it, it is responsible for the Coulomb force,\n> and if it is discarded it becomes impossible to model the classical em\n> field.\n\nThis is perhaps a really stupid question -- how does one model the\nclassical em field starting from the quantum theory? If you start from\na path integral and look at the hbar -> 0 limit, you get the classical\naction, but you mean something else above.\n\n-S.\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>Charles Francis <charles@clef.demon.co.uk> wrote in message news:<c4kf67
$1ma$1@lfa222122.richmond.edu>...
> I don't know quite how you work this out. There are four photon
> polarisation states, not just two. The time like and longitudinal states
> must always have the same amplitude, and may be considered as a
> lightlike states, but although cross sections for this state are zero
> and some treatments discard it, it is responsible for the Coulomb force,
> and if it is discarded it becomes impossible to model the classical em
> field.
This is perhaps a really stupid question -- how does one model the
classical em field starting from the quantum theory? If you start from
a path integral and look at the \hbar -> limit, you get the classical
action, but you mean something else above.
-S.
Eric A. Forgy
Apr7-04, 08:45 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resize=yes,status=no,wi dth=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>Hi Doug,\n\nDoug Sweetser <sweetser@alum.mit.edu> wrote:\n> Hello:\n>\n> This is the closest that string theory has ever gotten to things I\n> think about :-) Let me point out a few key differences.\n\nThis reminds me of a Valentine\'s day box of chocolates my mom bought\nme when I was a little kid. It had a cartoon picture of a cowboy\nsinging to his dog, (imagine a southern twang) "Oh lord, it\'s hard to\nbe humble *gulp* When you\'re perfect in every way." I loved that :)\n\nI, for one, would feel much more motivated to discuss this stuff with\nyou if you could back down a bit and admit you don\'t know what you\'re\ntalking about. I freely admit that I don\'t know what I\'m talking about\nall the time :) It\'s no fun discussing something when the other person\nis convinced they are correct and won\'t consider the possibility there\nis room for improvement in their ideas.\n\n> >> > g = Sym(DA)\n> >>\n> >> This is a good symmetric tensor. What needs to be checked is whether\n> >> it always has the right signature; this seems nontrivial.>\n> >\n> >Yeah. There seems to be a lot of magic that has to happen to make this\n> >idea viable. I am losing my optimism :)\n>\n> In my work, Sym(DA) is definitely not the metric tensor, so the issue\n> of signature is moot.\n>\n> ***\n> A small epiphany: it sounds like both Arnold and Eric (prehaps Urs)\n> think that Sym(DA) should play the role of the metric tensor: it is\n> second rank, symmetric and claimed to be related to gravity. I can see\n> why that would be utterly confusing. Note, _I_ never said that was its\n> role, the idea didn\'t occur to me :-)\n\nI know you never said this, but maybe you should start thinking about\nit. There are (apparently) good reasons for taking seriously a rank-2\ntensor P = g + F, where g IS the metric tensor that defines dot\nproduct, etc.\n\nIf the metric obtained from\n\ng\n= Sym(DA)\n= DA - dA\n\ncould be made viable as a physical theory somehow, I think that would\nbe pretty interesting and would bolster the credibility of some of the\nthings you are talking about.\n\n> Sym(DA) is about the fields that\n> together compose gravity, it is not the tool to make a scalar given two\n> vectors. The metric comes out of solving the force equation for a\n> potential that solves the field equations. There is a mathematical\n> link between force and curvature, although it is not well known\n> (Tristan Needham, "Visual Complex Analysis", p. 241-247).\n> ***\n\nOK. Let\'s consider this for a moment. Your proposed action can be\nwritten in an index free manner via\n\nS = 1/2 int_M (P,P) vol - int_M (P,J) vol\n\nwhere P = DA = g + F. Note that both ( , ) and vol depend on the\nmetric tensor. If your g is NOT the metric tensor, then both ( , ) and\nvol will be independent of A and finding your equations of motion is\nalmost exactly like that of Maxwell\'s equations.\n\nVarying S with respect to A, we get\n\nD^dag P = J\n\nwhere D^dag is the adjoint of D, i.e.\n\n(Du,w) = (u,D^dag w).\n\nI don\'t know about the other equations. In EM, we\'d have\n\ndF = 0\nd^dag F = J,\n\nso perhaps we want\n\nDF = 0\n\nas well. Then, your equations of motion may turn out to be\n\nDF = 0\nD^dag F = J.\n\nI\'m not motivated enough to check this, but I\'m guessing you\'ll end up\nwith something like this. On second thought, this could be trouble\nbecause D^2 involves curvature.\n\n> Instead Sym(DA) is a field strength tensor very similar to EM. As you\n> know, the antisymmetric field F is a nice way to write out what the\n> classical fields we experience, E and B. In potential notation, it\n> would be something like:\n>\n> E^i = -d^0 A^i - d^i A^0\n> B^k = d^i A^j - d^j A^i\n>\n> A pretty darn similar game happens for g, but there are a few sign\n> flips, a new field appears along the diagonal, and one needs to include\n> the Christoffel symbols:\n>\n> e^i = +d^0 A^i - d^i A^0 - 2 L_w^0i A^w\n> b^k = -d^i A^j - d^j A^i - 2 L_w^ij A^w\n> g^u = d^u A^u - L_w^uu A^w\n>\n> All told, 16 parts, although g^0 is the one we almost always deal with\n> in our classical lives. I just love to look at these three fields e,\n> b, and g, because they look like EM with a connection. Elegant, no?\n\nNot at first sight, no :)\n\n> Something funny happens at this point. I think people trained in\n> general relativity want me to say what metric I am using here. The\n> thing is, no one asks a similar question about the potential A^u.\n\nThen you have been talking to the wrong people. The potential A alone\nis definitely not sufficient for writing down Maxwell\'s equations. The\nlast time I checked, the scattering from a coffee cup was different\nthan the scattering from a donut. The potential A is a 1-form that is\nindependent of the metric. If A were the only thing you needed, then\nMaxwell\'s equations would be a topological field theory and we\nwouldn\'t be able to tell a coffee cup from a donut by looking at it.\nIn the EM action, we\'ve got\n\nS = 1/2 int_M (F,F) vol - int_M (A,J) vol\n\nwhere F = dA. The variation of S with respect to A brings in the\nmetric via ( , ) and vol and the resulting equation of motion is\n\nd^dag A = J.\n\nThe metric is used to define d^dag. So if no one mentions what metric\nthey are using, it is because they are assuming flat background\nMinkowskii space. It is still crucial to the theory whether they state\nit or not.\n\n> I believe I should be able to proceded without specifying either the\n> metric or the potential. I will only have one 4-potential and one\n> metric, both of which have derivatives in the tensor P and second\n> derivatives in the field equations. I want to keep things as general\n> as possible for as long as I can.\n\nThe catch is, as you mentioned Professor Baez must have pointed out\nages ago, if your g is NOT the metric tensor, then you have a\nbackground dependent theory because ( , ) and vol are fixed once you\nwrite down your action. Perhaps this doesn\'t bother you.\n\n> >Yes, I agree 100% and I think that Doug also agrees whether he\n> >realizes it or not :) I saw him state recently in another post that he\n> >varies A_u AND A_u;v, but in order to vary A_u;v separately from A_u\n> >means that he is really varying D. So maybe he does have enough\n> >degrees of freedom after all (which I think he does, he just needs to\n> >get his terminology straight).\n>\n> I am trying to crib from EM. For the classical Lagrange density, one\n> varies the potential A^u, and that induces a variation in A^u,v. When\n> one does a similar thing for P, the induced variation is in A^u;v which\n> contains the connection. I still feel insecure about how to talk about\n> that induced variation in the connection.\n\nThe term A^u;v involves both A and the connection D. It sounds like,\nup to now, you have been wanting to keep D (along with the metric\ntensor) fixed and only vary A. If so, then you are dead on arrival.\nYou simply won\'t have enough degrees of freedom regardless of how you\nwant to wriggle out of it by saying that you don\'t have a gauge theory\nso everything is alright.\n\n> For the record, I am not trying to recreate the Einstein-Maxwell field\n> equations. I want to keep Maxwell and punt the other that has resisted\n> all efforts at quantization.\n>\n>\n> Here are my equations of motion:\n>\n> Jq^u - Jm^u = A^u;v_;v\n>\n> It is easy to find a solution which is an inverse distance _squared_.\n> This is the solution bright people find quickly and know they must move\n> on. Why? A force involves the derivative of the potential. Take the\n> derivative of an inverse distance squared and the resulting force will\n> not be an inverse distance squared, therefore it ain\'t physical. I\n> think this is the reason no one works with this approach: it gets the\n> distance dependance of the force law flat out obviously wrong. Darn.\n>\n> Here is the mistake that is made. The physical gravity law is\n> classical. Classical laws break spacetime symmetry.\n\nDo you mean "non-relativistic" instead of "classical"?\n\n> I will need to stay away from the NG action because I have no sense of\n> what an ambient spacetime is. That is the deep end of the pool.\n\nThis probably isn\'t a bad idea. I\'m slowly convincing myself that\nstring theory can get away from a dependence on a background target\nspace. In LQG, they began with a manifold and then embedded the\ncombinatorial spin network into the manifold. After turning the crank,\nit turns out that the original manifold wasn\'t even needed and you\ncould have just started out with the combinatorial object. In a\nsimilar way, string theory starts with a worldsheet and an embedding\nof the worldsheet into some target space (note: a 2d worldsheet is\npretty much a combinatorial object too as far as its topology is\nconcerned). I think that, in the end, the embedding can probably be\ngotten rid of and we can deal with the worldsheet intrinsically. At\nthis point, this is mere speculation (motivated by some good reasons)\nand I DEFINITELY don\'t know what I\'m talking about :)\n\nBest regards,\nEric\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>Hi Doug,
Doug Sweetser <sweetser@alum.mit.edu> wrote:
> Hello:
>
> This is the closest that string theory has ever gotten to things I
> think about :-) Let me point out a few key differences.
This reminds me of a Valentine's day box of chocolates my mom bought
me when I was a little kid. It had a cartoon picture of a cowboy
singing to his dog, (imagine a southern twang) "Oh lord, it's hard to
be humble *gulp* When you're perfect in every way." I loved that :)
I, for one, would feel much more motivated to discuss this stuff with
you if you could back down a bit and admit you don't know what you're
talking about. I freely admit that I don't know what I'm talking about
all the time :) It's no fun discussing something when the other person
is convinced they are correct and won't consider the possibility there
is room for improvement in their ideas.
> >> > g = Sym(DA)
> >>
> >> This is a good symmetric tensor. What needs to be checked is whether
> >> it always has the right signature; this seems nontrivial.>
> >
> >Yeah. There seems to be a lot of magic that has to happen to make this
> >idea viable. I am losing my optimism :)
>
> In my work, Sym(DA) is definitely not the metric tensor, so the issue
> of signature is moot.
>
> ***
> A small epiphany: it sounds like both Arnold and Eric (prehaps Urs)
> think that Sym(DA) should play the role of the metric tensor: it is
> second rank, symmetric and claimed to be related to gravity. I can see
> why that would be utterly confusing. Note, _I_ never said that was its
> role, the idea didn't occur to me :-)
I know you never said this, but maybe you should start thinking about
it. There are (apparently) good reasons for taking seriously a rank-2
tensor P = g + F, where g IS the metric tensor that defines dot
product, etc.
If the metric obtained from
g
= Sym(DA)
= DA - dA
could be made viable as a physical theory somehow, I think that would
be pretty interesting and would bolster the credibility of some of the
things you are talking about.
> Sym(DA) is about the fields that
> together compose gravity, it is not the tool to make a scalar given two
> vectors. The metric comes out of solving the force equation for a
> potential that solves the field equations. There is a mathematical
> link between force and curvature, although it is not well known
> (Tristan Needham, "Visual Complex Analysis", p. 241-247).
> ***
OK. Let's consider this for a moment. Your proposed action can be
written in an index free manner via
S = 1/2 \int_M (P,P)[/itex] vol - \int_M (P,J) vol
where P = DA = g + F. Note that both ( , ) and vol depend on the
metric tensor. If your g is NOT the metric tensor, then both ( , ) and
vol will be independent of A and finding your equations of motion is
almost exactly like that of Maxwell's equations.
Varying S with respect to A, we get
D^{dag} P = J
where D^{dag} is the adjoint of D, i.e.
(Du,w) = (u,D^{dag} w).
I don't know about the other equations. In EM, we'd have
dF =
d^{dag} F = J,
so perhaps we want
DF =
as well. Then, your equations of motion may turn out to be
DF =
D^{dag} F = J.
I'm not motivated enough to check this, but I'm guessing you'll end up
with something like this. On second thought, this could be trouble
because D^2 involves curvature.
> Instead Sym(DA) is a field strength tensor very similar to EM. As you
> know, the antisymmetric field F is a nice way to write out what the
> classical fields we experience, E and B. In potential notation, it
> would be something like:
>
> E^i = -d^0 A^i - d^i A^0
> B^k = d^i A^j - d^j A^i
>
> A pretty darn similar game happens for g, but there are a few sign
> flips, a new field appears along the diagonal, and one needs to include
> the Christoffel symbols:
>
> e^i = +d^0 A^i - d^i A^0 - 2 L_w^0i A^w
> b^k = -d^i A^j - d^j A^i - 2 L_w^ij A^w
> g^u = d^u A^u - L_w^uu A^w
>
> All told, 16 parts, although g^0 is the one we almost always deal with
> in our classical lives. I just love to look at these three fields e,
> b, and g, because they look like EM with a connection. Elegant, no?
Not at first sight, no :)
> Something funny happens at this point. I think people trained in
> general relativity want me to say what metric I am using here. The
> thing is, no one asks a similar question about the potential A^u.
Then you have been talking to the wrong people. The potential A alone
is definitely not sufficient for writing down Maxwell's equations. The
last time I checked, the scattering from a coffee cup was different
than the scattering from a donut. The potential A is a 1-form that is
independent of the metric. If A were the only thing you needed, then
Maxwell's equations would be a topological field theory and we
wouldn't be able to tell a coffee cup from a donut by looking at it.
In the EM action, we've got
S = 1/2 \int_M (F,F) vol - \int_M (A,J) vol
where F = dA. The variation of S with respect to A brings in the
metric via ( , ) and vol and the resulting equation of motion is
d^{dag} A = J.
The metric is used to define d^{dag}. So if no one mentions what metric
they are using, it is because they are assuming flat background
Minkowskii space. It is still crucial to the theory whether they state
it or not.
> I believe I should be able to proceded without specifying either the
> metric or the potential. I will only have one 4-potential and one
> metric, both of which have derivatives in the tensor P and second
> derivatives in the field equations. I want to keep things as general
> as possible for as long as I can.
The catch is, as you mentioned Professor Baez must have pointed out
ages ago, if your g is NOT the metric tensor, then you have a
background dependent theory because ( , ) and vol are fixed once you
write down your action. Perhaps this doesn't bother you.
> >Yes, I agree 100% and I think that Doug also agrees whether he
> >realizes it or not :) I saw him state recently in another post that he
> >varies [itex]A_u AND A_u;v, but in order to vary A_u;v separately from A_u
> >means that he is really varying D. So maybe he does have enough
> >degrees of freedom after all (which I think he does, he just needs to
> >get his terminology straight).
>
> I am trying to crib from EM. For the classical Lagrange density, one
> varies the potential A^u, and that induces a variation in A^u,v. When
> one does a similar thing for P, the induced variation is in A^u;v which
> contains the connection. I still feel insecure about how to talk about
> that induced variation in the connection.
The term A^u;v involves both A and the connection D. It sounds like,
up to now, you have been wanting to keep D (along with the metric
tensor) fixed and only vary A. If so, then you are dead on arrival.
You simply won't have enough degrees of freedom regardless of how you
want to wriggle out of it by saying that you don't have a gauge theory
so everything is alright.
> For the record, I am not trying to recreate the Einstein-Maxwell field
> equations. I want to keep Maxwell and punt the other that has resisted
> all efforts at quantization.
>
>
> Here are my equations of motion:
>
> Jq^u - Jm^u = A^u;v_;v
>
> It is easy to find a solution which is an inverse distance _squared_.
> This is the solution bright people find quickly and know they must move
> on. Why? A force involves the derivative of the potential. Take the
> derivative of an inverse distance squared and the resulting force will
> not be an inverse distance squared, therefore it ain't physical. I
> think this is the reason no one works with this approach: it gets the
> distance dependance of the force law flat out obviously wrong. Darn.
>
> Here is the mistake that is made. The physical gravity law is
> classical. Classical laws break spacetime symmetry.
Do you mean "non-relativistic" instead of "classical"?
> I will need to stay away from the NG action because I have no sense of
> what an ambient spacetime is. That is the deep end of the pool.
This probably isn't a bad idea. I'm slowly convincing myself that
string theory can get away from a dependence on a background target
space. In LQG, they began with a manifold and then embedded the
combinatorial spin network into the manifold. After turning the crank,
it turns out that the original manifold wasn't even needed and you
could have just started out with the combinatorial object. In a
similar way, string theory starts with a worldsheet and an embedding
of the worldsheet into some target space (note: a 2d worldsheet is
pretty much a combinatorial object too as far as its topology is
concerned). I think that, in the end, the embedding can probably be
gotten rid of and we can deal with the worldsheet intrinsically. At
this point, this is mere speculation (motivated by some good reasons)
and I DEFINITELY don't know what I'm talking about :)
Best regards,
Eric
Doug Sweetser
Apr7-04, 09:02 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resize=yes,status=no,wi dth=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\n\nHello Eric:\n\nSomethings I don\'t know, like string theory, despite a few years of\nreading posts in SPR. I have been wrong and learned a lot on SPR: the\nimportance of commas versus semicolons (although the notation is out of\nfashion), the proper way to write an action in flat or curved\nspacetime, and the trace of a mixed tensor, just to name three specific\nexamples.\n\nSomethings I and Mathematica do understand: how to go from the Lagrange\ndensity discussed in this thread to the Watt/Misner exponential metric.\nThe notebook can be downloaded from here:\n\nhttp://theworld.com/~sweetser/quaternions/notebooks/lubos.nb\n\nI do show considerably less flexibility when I have a calculation\nworked out and confirmed by a symbolic math package. I will readily\nadmit that I do not know how to discuss this path with grace. "The\nrain in Spain falls mainly in France." Oops.\n\n[Why the odd notebook name? I promised Lubos if at any time in the\nnext ten years any proposal to explain how gravity works gains\nuniversal acceptance using more than four dimensions, I would give him\n$100. This is a conditional grant, not a wager. For me, it is the\ntangible calculations in that notebook that give me the confidence to\nwrite out check #1850 to Mr. Motl, and I will sign and date it once\ndiscussions in SPR reflect consensus that someone on his team has won\nin whatever dimensions more than four.]\n\n\n\\> There are (apparently) good reasons for taking seriously a rank-2\n\\> tensor P = g + F, where g IS the metric tensor that defines dot\n\\> product, etc.\n\nEnjoy that research direction. The details of my own work are\nincompatible with this approach. It is the signature of Sym(DA) issue.\nOn the path between the Lagrange density and the exponential metric, I\nhave a Sym(DA) that applies only to weak gravity fields. It looks like\nthis:\n\nSym(DA) = k/sigma^2 I(4)\n\nwhere: k is a small spring constant\nsigma^2 is a Lorentz invariant distance, dR^2 - (c dt)^2\nI(4) is a 4D identity matrix\n\nThis is the way a particular potential changes. It has the correct\ndistance dependence needed for a force law. It is not a metric,\nthe signature would be flat out wrong. Your work, should you go that\ndirection, will need to guarantee the correct signature of Sym(DA) as\nArnold pointed out, and you expressed concern over that critical\ndetail. My efforts cannot have Sym(DA) be a metric tensor. The metric\ntensor is derived from a solution to the force equation.\n\n\n\\> OK. Let\'s consider this for a moment. Your proposed action can be\n\\> written in an index free manner via\n\\>\n\\> S = 1/2 int_M (P,P) vol - int_M (P,J) vol\n\nThis doesn\'t look correct. There are two problems with the coupling\nterm. First, P is a second rank tensor, J is rank 1. If this was just\na typo (P in the place of A), there still would be an error. I am used\nto seeing the same sign for the contraction of the second rank tensor\nand the coupling term. No matter, I\'ll use your conventions. For EM,\nlike charges repel. For the action you wrote, like charges repel. For\nthis reason alone, the above action can never explain how gravity works\nsince like charges attract. Here is how I would write it in an index\nfree manner using your conventions:\n\nS = 1/2 int_M (P,P) vol - int_M (A,Jq - Jm) vol\n\nwhere: Jq is the electric current density\nJm is the mass current density (mass density * G^(1/2))\n\nThis will be similar to Maxwell, but different.\n\nDP != 0\nD^dag P = Jq - Jm\n\nEssentially Jm decreases Jq a wee bit (like inertia always does, slows\nthings down. The "wee" indicates that Jm will often be thirteen orders\nof magnitude smaller than Jq, and we only know q to ten significant\nfigures).\n\n\\> On second thought, this could be trouble because D^2 involves\n\\> curvature.\n\nSounds like a chance to have an equation that involves the derivative\nof a metric. That would be good. What I get for the first component\nis:\n\nrhoq - rhom = phi^;u_;u\n\nOne could look for a metric that solves this equation under certain\nconditions. In a vacuum, phi^;u_;u would equal zero. This expression\ninvolves a third order derivative of the metric tensor. If the field\nis weak, a second order derivative may be enough to describe spacetime\ncurvature, or phi^;u_,u = 0. Under static conditions, the Watt/Misner\nmetric solves that equation (proof added to the lubos.nb notebook).\n\n\n\\> The catch is, as you mentioned Professor Baez must have pointed out\n\\> ages ago, if your g is NOT the metric tensor, then you have a\n\\> background dependent theory because ( , ) and vol are fixed once you\n\\> write down your action. Perhaps this doesn\'t bother you.\n\nThis issue is my top concern. I can understand this reflex reaction:\ngravity is a metric theory, so we must vary the metric field. Reflexes\nare good things: they prevent burns after all :-)\n\nSometimes it is worth challenging a reflex, but it is not easy work.\nLet\'s start from something I know you are familiar with, because every\nbook on general relativity introduces this topic the same way. There\nis a 4-potential, call it A^u. Now all you want to do is take its\nderivative, A^u,v. At this point they say HA! Wrongo! A^u,v does not\ntransform as a tensor. One needs the Christoffel symbol of the second\nkind:\n\nA^u;v = A^u,v - L_w^uv A^w\n\nThis is a contravariant derivative that transforms like a tensor.\n\nI want to think about such derivatives from a completely different\nperspective: that of symmetry. Imagine someone did a measurement of a\nchange in a potential field, and they determined a particular component\n- say A^1;3 - was 41. How much of that 41 came from A^u,v? How much\ncame from L_w^uv A^w? If A^u,v was zero, then it would all come from\nthe connection. If the connection happened to be zero, it would all\ncome from the change in the potential. There is a symmetry principle\noperating in the definition of covariant derivatives.\n\nDoes this symmetry principle already have a name?\n\nThis symmetry principle is absolutely central to my work. Any thoughts\nabout this symmetry?\n\n\ndoug\nquaternions.com\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>Hello Eric:
Somethings I don't know, like string theory, despite a few years of
reading posts in SPR. I have been wrong and learned a lot on SPR: the
importance of commas versus semicolons (although the notation is out of
fashion), the proper way to write an action in flat or curved
spacetime, and the trace of a mixed tensor, just to name three specific
examples.
Somethings I and Mathematica do understand: how to go from the Lagrange
density discussed in this thread to the Watt/Misner exponential metric.
The notebook can be downloaded from here:
http://theworld.com/~sweetser/quaternions/notebooks/lubos.nb
I do show considerably less flexibility when I have a calculation
worked out and confirmed by a symbolic math package. I will readily
admit that I do not know how to discuss this path with grace. "The
rain in Spain falls mainly in France." Oops.
[Why the odd notebook name? I promised Lubos if at any time in the
next ten years any proposal to explain how gravity works gains
universal acceptance using more than four dimensions, I would give him
$100. This is a conditional grant, not a wager. For me, it is the
tangible calculations in that notebook that give me the confidence to
write out check #1850 to Mr. Motl, and I will sign and date it once
discussions in SPR reflect consensus that someone on his team has won
in whatever dimensions more than four.]
> There are (apparently) good reasons for taking seriously a rank-2
> tensor P = g + F, where g IS the metric tensor that defines dot
> product, etc.
Enjoy that research direction. The details of my own work are
incompatible with this approach. It is the signature of Sym(DA) issue.
On the path between the Lagrange density and the exponential metric, I
have a Sym(DA) that applies only to weak gravity fields. It looks like
this:
Sym(DA) = k/\sigma^2 I(4)
where: k is a small spring constant
\sigma^2 is a Lorentz invariant distance, dR^2 - (c dt)^2
I(4) is a 4D identity matrix
This is the way a particular potential changes. It has the correct
distance dependence needed for a force law. It is not a metric,
the signature would be flat out wrong. Your work, should you go that
direction, will need to guarantee the correct signature of Sym(DA) as
Arnold pointed out, and you expressed concern over that critical
detail. My efforts cannot have Sym(DA) be a metric tensor. The metric
tensor is derived from a solution to the force equation.
> OK. Let's consider this for a moment. Your proposed action can be
> written in an index free manner via
>
> S = 1/2 \int_M (P,P) vol - \int_M (P,J) vol
This doesn't look correct. There are two problems with the coupling
term. First, P is a second rank tensor, J is rank 1. If this was just
a typo (P in the place of A), there still would be an error. I am used
to seeing the same sign for the contraction of the second rank tensor
and the coupling term. No matter, I'll use your conventions. For EM,
like charges repel. For the action you wrote, like charges repel. For
this reason alone, the above action can never explain how gravity works
since like charges attract. Here is how I would write it in an index
free manner using your conventions:
S = 1/2 \int_M (P,P)[/itex] vol - \int_M (A,Jq - Jm) vol
where: Jq is the electric current density
Jm is the mass current density (mass density * G^(1/2))
This will be similar to Maxwell, but different.
DP !=
D^{dag} P = Jq - Jm
Essentially Jm decreases Jq a wee bit (like inertia always does, slows
things down. The "wee" indicates that Jm will often be thirteen orders
of magnitude smaller than Jq, and we only know q to ten significant
figures).
> On second thought, this could be trouble because [itex]D^2 involves
> curvature.
Sounds like a chance to have an equation that involves the derivative
of a metric. That would be good. What I get for the first component
is:
rhoq - rhom = \phi^;u_;u
One could look for a metric that solves this equation under certain
conditions. In a vacuum, \phi^;u_;u would equal zero. This expression
involves a third order derivative of the metric tensor. If the field
is weak, a second order derivative may be enough to describe spacetime
curvature, or \phi^;u_,u = . Under static conditions, the Watt/Misner
metric solves that equation (proof added to the lubos.nb notebook).
> The catch is, as you mentioned Professor Baez must have pointed out
> ages ago, if your g is NOT the metric tensor, then you have a
> background dependent theory because ( , ) and vol are fixed once you
> write down your action. Perhaps this doesn't bother you.
This issue is my top concern. I can understand this reflex reaction:
gravity is a metric theory, so we must vary the metric field. Reflexes
are good things: they prevent burns after all :-)
Sometimes it is worth challenging a reflex, but it is not easy work.
Let's start from something I know you are familiar with, because every
book on general relativity introduces this topic the same way. There
is a 4-potential, call it A^u. Now all you want to do is take its
derivative, A^u,v. At this point they say HA! Wrongo! A^u,v does not
transform as a tensor. One needs the Christoffel symbol of the second
kind:
A^u;v = A^u,v - L_w^uv A^w
This is a contravariant derivative that transforms like a tensor.
I want to think about such derivatives from a completely different
perspective: that of symmetry. Imagine someone did a measurement of a
change in a potential field, and they determined a particular component
- say A^1;3 - was 41. How much of that 41 came from A^u,v? How much
came from L_w^uv A^w? If A^u,v was zero, then it would all come from
the connection. If the connection happened to be zero, it would all
come from the change in the potential. There is a symmetry principle
operating in the definition of covariant derivatives.
Does this symmetry principle already have a name?
This symmetry principle is absolutely central to my work. Any thoughts
about this symmetry?
doug
quaternions.com
Arnold Neumaier
Apr7-04, 09:02 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resize=yes,status=no,wi dth=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\n\nEric A. Forgy wrote:\n\\>\n\\> Arnold Neumaier \\<Arnold.Neumaier@univie.ac.at\\> wrote:\n\\>\n\\>\\>And one has to check that in the well understood special cases\n\\>\\>(weak field) one recovers exactly the traditional formulas.\n\\>\n\\>\n\\> Definitely and I am too lazy to do that in any detail :) I got far\n\\> enough to convince Urs that the action was at least independent of\n\\> coordinates (which wasn\'t obvious the way I presented it the first\n\\> time), but then he asked, "What are the equations of motion you get\n\\> after varying A and D?" I quickly gave up after a few pages of\n\\> computation :) I did get far enough to see that the possibility of\n\\> having a nice weak field emerging was not completely out of the\n\\> question.\n\nBut if you are too lazy to complete, others will not even begin...\n\nRemember the saying, success is 1% inspiration and 99% perspiration!\nOne has to go many many ways leading into dead ends before one\nencounters one thet really leads to something new...\n\nOthers will start being interedsted in your ideas only when you showed\nthat it actually worked in some interesting case.\n\n\nArnld Neumaier\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>Eric A. Forgy wrote:
>
> Arnold Neumaier <Arnold.Neumaier@univie.ac.at> wrote:
>
>>And one has to check that in the well understood special cases
>>(weak field) one recovers exactly the traditional formulas.
>
>
> Definitely and I am too lazy to do that in any detail :) I got far
> enough to convince Urs that the action was at least independent of
> coordinates (which wasn't obvious the way I presented it the first
> time), but then he asked, "What are the equations of motion you get
> after varying A and D?" I quickly gave up after a few pages of
> computation :) I did get far enough to see that the possibility of
> having a nice weak field emerging was not completely out of the
> question.
But if you are too lazy to complete, others will not even begin...
Remember the saying, success is 1% inspiration and 99% perspiration!
One has to go many many ways leading into dead ends before one
encounters one thet really leads to something new...
Others will start being interedsted in your ideas only when you showed
that it actually worked in some interesting case.
Arnld Neumaier
Charles Francis
Apr7-04, 09:15 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resize=yes,status=no,wi dth=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>In article <c4re2j\\$dd7\\$1@pcls4.std.com>, Doug Sweetser\n<sweetser@alum.mit.edu> writes\n>\n>\n>\n>Hello Charles:\n>\n>> Yes, this is true. The amplitude of lightlike states is indeterminate.\n>> This is a gauge invariance. I like to split gauge invariance into\n>> different causes. This one is a true physical invariance, the\n>> lightlike states have to be there because you can transform the\n>> transverse states into them but you cannot give an amplitude for them.\n>> Other gauge invariances are simply mathematical and have no such\n>> metaphysical interpretation.\n>\n>If one had a model that was not invariant under a gauge transformation,\n>then these mathematical invariances would no longer be there, dodging\n>metaphysical interpretation. Both Maxwell and general relativity are\n>gauge theories, so your point applies to them. Given their collective\n>success, there is an enormous bias in the physics community in favor of\n>models that have gauge symmetry. Yet your line of reasoning indicates\n>gauge invariant proposals will always have this as an issue. It may be\n>time to work with simple models that are not gauge invariant so that\n>all states are real parts of Nature.\n>\nThis is why I distinguish the indeterminacy of the amplitude of\nlightlike states from other gauge invariance. Frequently the gauge\ninvariance is just a trick in mathematics, like "think of a number, ...\ntake away the number you just thought of. I find no particular\nattraction in that form of gauge invariance. But the lightlike photon\nstates have a physical effect, the Coulomb force, so we must conclude\nthat they are real even though we cannot put an amplitude on them.\n\n\nRegards\n\n--\nCharles Francis\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>In article <c4re2j$dd7$1@pcls4.std.com>, Doug Sweetser
<sweetser@alum.mit.edu> writes
>
>
>
>Hello Charles:
>
>> Yes, this is true. The amplitude of lightlike states is indeterminate.
>> This is a gauge invariance. I like to split gauge invariance into
>> different causes. This one is a true physical invariance, the
>> lightlike states have to be there because you can transform the
>> transverse states into them but you cannot give an amplitude for them.
>> Other gauge invariances are simply mathematical and have no such
>> metaphysical interpretation.
>
>If one had a model that was not invariant under a gauge transformation,
>then these mathematical invariances would no longer be there, dodging
>metaphysical interpretation. Both Maxwell and general relativity are
>gauge theories, so your point applies to them. Given their collective
>success, there is an enormous bias in the physics community in favor of
>models that have gauge symmetry. Yet your line of reasoning indicates
>gauge invariant proposals will always have this as an issue. It may be
>time to work with simple models that are not gauge invariant so that
>all states are real parts of Nature.
>
This is why I distinguish the indeterminacy of the amplitude of
lightlike states from other gauge invariance. Frequently the gauge
invariance is just a trick in mathematics, like "think of a number, ...
take away the number you just thought of. I find no particular
attraction in that form of gauge invariance. But the lightlike photon
states have a physical effect, the Coulomb force, so we must conclude
that they are real even though we cannot put an amplitude on them.
Regards
--
Charles Francis
Arnold Neumaier
Apr7-04, 09:18 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>Charles Francis wrote:\n> In article <4067FE57.8010804@univie.ac.at>, Arnold Neumaier\n> <Arnold.Neumaier@univie.ac.at> writes\n>\n>>\n>> The e/m field is more than photons, and the gravitational field is more\n>> than gravitons. Photons have just two degrees of freedom, but the e/m\n>> field has three, as you can easily convince yourself. A static Coloumb\n>> field is given by the coset defined by the 4-vector (V,0,0,0) under\n>> the gauge group. This is never transverse, and has nothing to do with\n>> photons. But is is doubtless electromagetic...\n>>\n> I don\'t know quite how you work this out. There are four photon\n> polarisation states, not just two. The time like and longitudinal states\n> must always have the same amplitude, and may be considered as a\n> lightlike states, but although cross sections for this state are zero\n> and some treatments discard it, it is responsible for the Coulomb force,\n> and if it is discarded it becomes impossible to model the classical em\n> field.\n\nIn a momentum rerpresentation, the e/m potential is given by a function\nA(p), but two potentials are equivalent if their difference is parallel\nto p. (This is gauge equivalence). Thus actually one works with cosets\nmod p; taking the cosets mod p eats one of the 4 degrees of freedom.\n\nIf one looks at electromagnetic waves one imposes additionally the\nfree Maxwell equation, which amounts to requiring that A(p) is orthogonal\nto p in the Minkowski metrik. This eats a second dof, leaving two\ntransverse dofs. The latter can be spanned by left and right circular\nwaves, which are the classical version of photons of helicity -1 and 1.\nNow quantize...\n\n\nArnold Neumaier\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>Charles Francis wrote:
> In article <4067FE57.8010804@univie.ac.at>, Arnold Neumaier
> <Arnold.Neumaier@univie.ac.at> writes
>
>>
>> The e/m field is more than photons, and the gravitational field is more
>> than gravitons. Photons have just two degrees of freedom, but the e/m
>> field has three, as you can easily convince yourself. A static Coloumb
>> field is given by the coset defined by the 4-vector (V,0,0,0) under
>> the gauge group. This is never transverse, and has nothing to do with
>> photons. But is is doubtless electromagetic...
>>
> I don't know quite how you work this out. There are four photon
> polarisation states, not just two. The time like and longitudinal states
> must always have the same amplitude, and may be considered as a
> lightlike states, but although cross sections for this state are zero
> and some treatments discard it, it is responsible for the Coulomb force,
> and if it is discarded it becomes impossible to model the classical em
> field.
In a momentum rerpresentation, the e/m potential is given by a function
A(p), but two potentials are equivalent if their difference is parallel
to p. (This is gauge equivalence). Thus actually one works with cosets
mod p; taking the cosets mod p eats one of the 4 degrees of freedom.
If one looks at electromagnetic waves one imposes additionally the
free Maxwell equation, which amounts to requiring that A(p) is orthogonal
to p in the Minkowski metrik. This eats a second dof, leaving two
transverse dofs. The latter can be spanned by left and right circular
waves, which are the classical version of photons of helicity -1 and 1.
Now quantize...
Arnold Neumaier
Arnold Neumaier
Apr7-04, 09:18 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>carlip@no-physics-spam.ucdavis.edu wrote:\n> Arnold Neumaier <Arnold.Neumaier@univie.ac.at> wrote:\n>\n>\n>>The e/m field is more than photons, and the gravitational field is more\n>>than gravitons. Photons have just two degrees of freedom, but the e/m\n>>field has three, as you can easily convince yourself. A static Coloumb\n>>field is given by the coset defined by the 4-vector (V,0,0,0) under\n>>the gauge group. This is never transverse, and has nothing to do with\n>>photons. But is is doubtless electromagetic...\n>\n>\n> But is it a ``degree of freedom?\'\' This depends on definitions. The\n> Coulomb part of the field is completely determined by the constraints;\n> it has no independent equations of motion, and is certainly not a\n> *dynamical* degree of freedom.\n\nIt is a degree of freedom in the sense that it is needed to specify the\nelectromagnetic field. If you write them as\np^2 A(p) = j(p)\n(or rather its Fourier transform), one sses that it is the solution\nof a dynamical equation. One gets, of course, _retarded_ Coulomb\nterms, which have a dynamics.\n\nAlso, in terms of perturbative QFT, the Coulomb field is made up of\nvirtual photons, which have 3 dof...\n\n\nArnold Neumaier\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>carlip@no-physics-spam.ucdavis.edu wrote:
> Arnold Neumaier <Arnold.Neumaier@univie.ac.at> wrote:
>
>
>>The e/m field is more than photons, and the gravitational field is more
>>than gravitons. Photons have just two degrees of freedom, but the e/m
>>field has three, as you can easily convince yourself. A static Coloumb
>>field is given by the coset defined by the 4-vector (V,0,0,0) under
>>the gauge group. This is never transverse, and has nothing to do with
>>photons. But is is doubtless electromagetic...
>
>
> But is it a ``degree of freedom?'' This depends on definitions. The
> Coulomb part of the field is completely determined by the constraints;
> it has no independent equations of motion, and is certainly not a
> *dynamical* degree of freedom.
It is a degree of freedom in the sense that it is needed to specify the
electromagnetic field. If you write them as
p^2 A(p) = j(p)
(or rather its Fourier transform), one sses that it is the solution
of a dynamical equation. One gets, of course, _retarded_ Coulomb
terms, which have a dynamics.
Also, in terms of perturbative QFT, the Coulomb field is made up of
virtual photons, which have 3 dof...
Arnold Neumaier
Charles Francis
Apr7-04, 09:25 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\n\nIn article <4072819D.4010604@univie.ac.at>, Arnold Neumaier\n<Arnold.Neumaier@univie.ac.at> writes\n>Charles Francis wrote:\n>> In article <4067FE57.8010804@univie.ac.at>, Arnold Neumaier\n>> <Arnold.Neumaier@univie.ac.at> writes\n>>\n>>>\n>>> The e/m field is more than photons, and the gravitational field is more\n>>> than gravitons. Photons have just two degrees of freedom, but the e/m\n>>> field has three, as you can easily convince yourself. A static Coloumb\n>>> field is given by the coset defined by the 4-vector (V,0,0,0) under\n>>> the gauge group. This is never transverse, and has nothing to do with\n>>> photons. But is is doubtless electromagetic...\n>>>\n>> I don\'t know quite how you work this out. There are four photon\n>> polarisation states, not just two. The time like and longitudinal states\n>> must always have the same amplitude, and may be considered as a\n>> lightlike states, but although cross sections for this state are zero\n>> and some treatments discard it, it is responsible for the Coulomb force,\n>> and if it is discarded it becomes impossible to model the classical em\n>> field.\n>\n>In a momentum rerpresentation, the e/m potential is given by a function\n>A(p), but two potentials are equivalent if their difference is parallel\n>to p. (This is gauge equivalence). Thus actually one works with cosets\n>mod p; taking the cosets mod p eats one of the 4 degrees of freedom.\n>\n>If one looks at electromagnetic waves one imposes additionally the\n>free Maxwell equation, which amounts to requiring that A(p) is orthogonal\n>to p in the Minkowski metrik. This eats a second dof, leaving two\n>transverse dofs. The latter can be spanned by left and right circular\n>waves, which are the classical version of photons of helicity -1 and 1.\n>Now quantize...\n>\n\nYou can\'t do that. You have lost sight of the time-like states. The\nlogical order is to start from the fundamental and derive the classical\nproperties. To do that you have to include four creation and\nannihilation operators and then derive the classical correspondence.\n\n\n\nRegards\n\n--\nCharles Francis\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>In article <4072819D.4010604@univie.ac.at>, Arnold Neumaier
<Arnold.Neumaier@univie.ac.at> writes
>Charles Francis wrote:
>> In article <4067FE57.8010804@univie.ac.at>, Arnold Neumaier
>> <Arnold.Neumaier@univie.ac.at> writes
>>
>>>
>>> The e/m field is more than photons, and the gravitational field is more
>>> than gravitons. Photons have just two degrees of freedom, but the e/m
>>> field has three, as you can easily convince yourself. A static Coloumb
>>> field is given by the coset defined by the 4-vector (V,0,0,0) under
>>> the gauge group. This is never transverse, and has nothing to do with
>>> photons. But is is doubtless electromagetic...
>>>
>> I don't know quite how you work this out. There are four photon
>> polarisation states, not just two. The time like and longitudinal states
>> must always have the same amplitude, and may be considered as a
>> lightlike states, but although cross sections for this state are zero
>> and some treatments discard it, it is responsible for the Coulomb force,
>> and if it is discarded it becomes impossible to model the classical em
>> field.
>
>In a momentum rerpresentation, the e/m potential is given by a function
>A(p), but two potentials are equivalent if their difference is parallel
>to p. (This is gauge equivalence). Thus actually one works with cosets
>mod p; taking the cosets mod p eats one of the 4 degrees of freedom.
>
>If one looks at electromagnetic waves one imposes additionally the
>free Maxwell equation, which amounts to requiring that A(p) is orthogonal
>to p in the Minkowski metrik. This eats a second dof, leaving two
>transverse dofs. The latter can be spanned by left and right circular
>waves, which are the classical version of photons of helicity -1 and 1.
>Now quantize...
>
You can't do that. You have lost sight of the time-like states. The
logical order is to start from the fundamental and derive the classical
properties. To do that you have to include four creation and
annihilation operators and then derive the classical correspondence.
Regards
--
Charles Francis
Doug Sweetser
Apr7-04, 03:14 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>Hello Steve:\n\n> That\'s not correct. Inertial mass is a complicated quantity, and\n> is not, in general, a Lorentz scalar.\n\nYou are right. The inertial mass is complicated, made of many\ncontributions, so will not have a simple transformation properties.\n\n\n>> What would prove or disprove my model was a direct test of gravity\n>> energy gravitates. Since gravity fields are very weak, I don\'t think\n>> this has been done.\n>\n> First of all, there is very clear experimental evidence that\n> gravitational energy contributes E/c^2 to *passive* gravitational\n> mass, that is, to the response of an object to gravity. This\n> comes from searches for the ``Nordtvedt effect\'\' in the Moon\'s\n> orbit; there is a thorough discussion in section 8.1 of Will\'s\n> book, _Theory and experiment in gravitational physics_.\n\nThanks for this reference. This is a test of the weak equivalence\nprinciple. The task is to figure out for a particular metric proposal\nhow it effects the WEP-violating parameter eta:\n\nm_passive/m_inertial = 1 + sum eta E/m_inertial c^2\n\nwhere all the possible contributions to E are summed.\n\nFor the Nordtvedt effect, eta can be expressed in terms of the ten PPN\nparameters;\n\neta = 4 beta - gamma - 3 - 10/3 xi - alpha_1 Will\n+ 2/3 alpha_2 - 2/3 zeta_1 - 1/3 zeta_2 eq. 8.9\n\nFor the metric I work with, beta = gamma = 1, the other eight are zero.\nThus eta = 0. That is consistent with the data.\n\n\n> Second, as I learned recently, there is also strong experimental\n> evidence for the equality of the Moon\'s passive gravitational\n> mass and its *active* gravitational mass. This also comes from\n> Lunar laser ranging; see Bartlett and Van Buren, Phys. Rev. Lett.\n> 57 (1986) 21. Bartlett and Van Buren find that the Moon\'s active\n> and passive gravitational mass are equal to a precision of about\n> 4x10^-12. Since the gravitational binding energy of the Moon\n> contributes 2x10^-11 of its mass, this is good enough to show\n> that gravitational energy gravitates, contributing E/c^2 to\n> active gravitational mass at an accuracy of about 20%.\n\nThere is nothing in my work that would force for a given body\nm_active/m_passive to be anything but 1. Will makes it more precise:\n\nm_active/m_passive = 1 + 1/2 zeta_3 (E_e/m_p) eq. 9.27\n\nwhere E_e is nuclear electrostatic energy.\n\nSince zeta_3 is zero for my metric, this ratio is equal to one. That\nagain is consistent with the data.\n\n\nBy using the PPN parameters, I can show that my metric agrees with the\ntwo experiments you cited. That is a very good thing. I like\nequations and data, they are the trump cards of physics.\n\nI am weak with the words, as my error in the transformation properties\nof the inertial mass shows. What I don\'t understand is how either of\nthe cited experiments could _even in theory_ say anything about\ngravitational energy gravitating. The tests determines a ratio.\nThey cannot determine all the separate parts that go into the numerator\nand denominators. All that can say is that whatever goes into the\nnumerator goes into the denominator in an equal way. The three forms\nof mass that are play a role in gravity, m_inertial, m_active and\nm_passive are complicated. These experiment tells of their\nrelationship to each other, nothing more or less.\n\nIn defense of the authors, I did not find the phrase "gravitational\nenergy gravitates" in either paper. I also did not see it in the\napplicable sections of Will\'s book. I certainly may have overlooked\nit, but I usually underline such things.\n\n\ndoug\nquaternions.com\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>Hello Steve:
> That's not correct. Inertial mass is a complicated quantity, and
> is not, in general, a Lorentz scalar.
You are right. The inertial mass is complicated, made of many
contributions, so will not have a simple transformation properties.
>> What would prove or disprove my model was a direct test of gravity
>> energy gravitates. Since gravity fields are very weak, I don't think
>> this has been done.
>
> First of all, there is very clear experimental evidence that
> gravitational energy contributes E/c^2 to *passive* gravitational
> mass, that is, to the response of an object to gravity. This
> comes from searches for the ``Nordtvedt effect'' in the Moon's
> orbit; there is a thorough discussion in section 8.1 of Will's
> book, _Theory and experiment in gravitational physics_.
Thanks for this reference. This is a test of the weak equivalence
principle. The task is to figure out for a particular metric proposal
how it effects the WEP-violating parameter \eta:
m_{passive}/m_{inertial} = 1 +[/itex] sum \eta E/m_{inertial} c^2
where all the possible contributions to E are summed.
For the Nordtvedt effect, \eta can be expressed in terms of the ten PPN
parameters;
\eta = 4 \beta - \gamma - 3 - 10/3 \xi - \alpha_1 Will
+ 2/3 \alpha_2 - 2/3 \zeta_1 - 1/3 \zeta_2 eq. 8.9
For the metric I work with, \beta = \gamma = 1, the other eight are zero.
Thus \eta = . That is consistent with the data.
> Second, as I learned recently, there is also strong experimental
> evidence for the equality of the Moon's passive gravitational
> mass and its *active* gravitational mass. This also comes from
> Lunar laser ranging; see Bartlett and Van Buren, Phys. Rev. Lett.
> 57 (1986) 21. Bartlett and Van Buren find that the Moon's active
> and passive gravitational mass are equal to a precision of about
> [itex]4x10^-12. Since the gravitational binding energy of the Moon
> contributes 2x10^-11 of its mass, this is good enough to show
> that gravitational energy gravitates, contributing E/c^2 to
> active gravitational mass at an accuracy of about 20%.
There is nothing in my work that would force for a given body
m_{active}/m_{passive} to be anything but 1. Will makes it more precise:
m_{active}/m_{passive} = 1 + 1/2 \zeta_3 (E_e/m_p) eq. 9.27
where E_e is nuclear electrostatic energy.
Since \zeta_3 is zero for my metric, this ratio is equal to one. That
again is consistent with the data.
By using the PPN parameters, I can show that my metric agrees with the
two experiments you cited. That is a very good thing. I like
equations and data, they are the trump cards of physics.
I am weak with the words, as my error in the transformation properties
of the inertial mass shows. What I don't understand is how either of
the cited experiments could _even in theory_ say anything about
gravitational energy gravitating. The tests determines a ratio.
They cannot determine all the separate parts that go into the numerator
and denominators. All that can say is that whatever goes into the
numerator goes into the denominator in an equal way. The three forms
of mass that are play a role in gravity, m_{inertial}, m_{active} and
m_{passive} are complicated. These experiment tells of their
relationship to each other, nothing more or less.
In defense of the authors, I did not find the phrase "gravitational
energy gravitates" in either paper. I also did not see it in the
applicable sections of Will's book. I certainly may have overlooked
it, but I usually underline such things.
doug
quaternions.com
Arnold Neumaier
Apr8-04, 05:50 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\n\nBuzurg Shagird wrote:\n\n> This is perhaps a really stupid question -- how does one model the\n> classical em field starting from the quantum theory?\n\nBy coherent states.\nSee, e.g., the quantum optics book by Mandel and Wolf.\n\n\nArnold Neumaier\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>Buzurg Shagird wrote:
> This is perhaps a really stupid question -- how does one model the
> classical em field starting from the quantum theory?
By coherent states.
See, e.g., the quantum optics book by Mandel and Wolf.
Arnold Neumaier
Buzurg Shagird
Apr8-04, 02:27 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>Arnold Neumaier <Arnold.Neumaier@univie.ac.at> wrote in message news:<40745D29.8030905@univie.ac.at>...\n> Buzurg Shagird wrote:\n>\n> > This is perhaps a really stupid question -- how does one model the\n> > classical em field starting from the quantum theory?\n>\n> By coherent states.\n> See, e.g., the quantum optics book by Mandel and Wolf.\n\nThanks. This is the answer I was looking for. I had seen that book\nbut not carefully enough, so I didn\'t realise you could get the\nclassical field that way ...\n\nDo you know if it is possible to do similar things for non-Abelian\ngauge fields? I know people try to (perhaps successfully) construct\ncoherent states in non-Abelian gauge theory -- does that lead to\nclassical Yang-Mills theory? Sorry if this is a really trivial\nquestion.\n\nAnd no offence meant, but I didn\'t understand Charles Francis\'\nderivation.\n\n-S.\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>Arnold Neumaier <Arnold.Neumaier@univie.ac.at> wrote in message news:<40745D29.8030905@univie.ac.at>...
> Buzurg Shagird wrote:
>
> > This is perhaps a really stupid question -- how does one model the
> > classical em field starting from the quantum theory?
>
> By coherent states.
> See, e.g., the quantum optics book by Mandel and Wolf.
Thanks. This is the answer I was looking for. I had seen that book
but not carefully enough, so I didn't realise you could get the
classical field that way ...
Do you know if it is possible to do similar things for non-Abelian
gauge fields? I know people try to (perhaps successfully) construct
coherent states in non-Abelian gauge theory -- does that lead to
classical Yang-Mills theory? Sorry if this is a really trivial
question.
And no offence meant, but I didn't understand Charles Francis'
derivation.
-S.
<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>Charles Francis <charles@clef.demon.co.uk> wrote in message news:<c4v8rq\\$dju\\$1@lfa222122.richmond.edu>...\ n> Although many text books say that the classical limit of a quantum\n> theory is given by the hbar -> 0 limit this is a blunder of the first\n> magnitude. It is clearly meaningless to let hbar -> 0, since hbar is a\n> constant of nature. The poor student who expects to believe his text\n> books may be forgiven, but the author who is supposed to have learned\n> his subject may not. I generally treat the authors of such books as\n> cranks with no understanding of quantum theory whatsoever.\n>\n\nThey are just saying that the reason we don\'t notice quantum effects\nin every day life is because hbar is so small. Therefore quantum\nmechanics should reduce to classical mechanics as you take hbar -> 0.\nIt\'s similar to the fact that we don\'t notice relativistic effects\nbecause c is so large. Therefore, special relativity should reduce to\nclassical mechanics as you take c -> infinity. Obviously, no one is\nsuggesting that you can actually change the values of hbar or c. Dirac\ncame up with a formula where quantum mechanics brackets are related to\nclassical Poisson brackets by dividing by ihbar.\n\nDavid\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>Charles Francis <charles@clef.demon.co.uk> wrote in message news:<c4v8rq$dju$1@lfa222122.richmond.edu>...
> Although many text books say that the classical limit of a quantum
> theory is given by the \hbar -> limit this is a blunder of the first
> magnitude. It is clearly meaningless to let \hbar -> 0, since \hbar is a
> constant of nature. The poor student who expects to believe his text
> books may be forgiven, but the author who is supposed to have learned
> his subject may not. I generally treat the authors of such books as
> cranks with no understanding of quantum theory whatsoever.
>
They are just saying that the reason we don't notice quantum effects
in every day life is because \hbar is so small. Therefore quantum
mechanics should reduce to classical mechanics as you take \hbar -> .
It's similar to the fact that we don't notice relativistic effects
because c is so large. Therefore, special relativity should reduce to
classical mechanics as you take c -> infinity. Obviously, no one is
suggesting that you can actually change the values of \hbar or c. Dirac
came up with a formula where quantum mechanics brackets are related to
classical Poisson brackets by dividing by ihbar.
David
Doug Sweetser
Apr11-04, 11:44 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>\n\n\nHello:\n\nI have a cool thing to add about this symmetry stuff. When one uses\nthe word symmetry, people expect to see a Lagrange density, then a\nfield which when added in changes nothing. Lets do that. Here is my\nunified field Lagrange density for a vacuum:\n\nL = int d^4 (sqrt (-det g)) A^u;v A_u;v\n\nNow consider an arbitrary field B^u which has this property:\n\nB^u,v = L_w^uv B^w\n\nor B^u,v - L_w^uv B^w = 0\n\nIs is clear why A^u;v = (A^u + B^u)^;v? Then\n\nL = int d^4 (sqrt (-det g)) (A^u + B^u)^;v (A_u + B_u)_;v\n\nThat looks like a well defined symmetry.\n\n\ndoug\nquaternions.com\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>Hello:
I have a cool thing to add about this symmetry stuff. When one uses
the word symmetry, people expect to see a Lagrange density, then a
field which when added in changes nothing. Lets do that. Here is my
unified field Lagrange density for a vacuum:
L = \int d^4 (\sqrt (-det g)) A^u;v A_u;v
Now consider an arbitrary field B^u which has this property:
B^u,v = L_w^uv B^w
or B^u,v - L_w^uv B^w =
Is is clear why A^u;v = (A^u + B^u)^;v? Then
L = \int d^4 (\sqrt (-det g)) (A^u + B^u)^;v (A_u + B_u)_;v
That looks like a well defined symmetry.
doug
quaternions.com
Arnold Neumaier
Apr14-04, 03:17 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>Buzurg Shagird wrote:\n> Arnold Neumaier <Arnold.Neumaier@univie.ac.at> wrote in message news:<40745D29.8030905@univie.ac.at>...\n>\n>>Buzu rg Shagird wrote:\n>>\n>>\n>>>This is perhaps a really stupid question -- how does one model the\n>>>classical em field starting from the quantum theory?\n>>\n>>By coherent states.\n>>See, e.g., the quantum optics book by Mandel and Wolf.\n>\n>\n> Thanks. This is the answer I was looking for. I had seen that book\n> but not carefully enough, so I didn\'t realise you could get the\n> classical field that way ...\n\nActually it is somewhat worse than my simple answer suggests.\nCoherent states give you the free classical field only.\nThis suffices for quantum optics.\nThe interacting case is poorly understood.\n\n> Do you know if it is possible to do similar things for non-Abelian\n> gauge fields? I know people try to (perhaps successfully) construct\n> coherent states in non-Abelian gauge theory -- does that lead to\n> classical Yang-Mills theory?\n\nIn principle, yes, but I haven\'t seen it anywhere.\nThe problem is that no one really understands the relations between\nclassical field theory and quantum field theory.\nSo one only finds occasional remarks, usually without reference\nto a more careful discussion.\n\n\nArnold Neumaier\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>Buzurg Shagird wrote:
> Arnold Neumaier <Arnold.Neumaier@univie.ac.at> wrote in message news:<40745D29.8030905@univie.ac.at>...
>
>>Buzurg Shagird wrote:
>>
>>
>>>This is perhaps a really stupid question -- how does one model the
>>>classical em field starting from the quantum theory?
>>
>>By coherent states.
>>See, e.g., the quantum optics book by Mandel and Wolf.
>
>
> Thanks. This is the answer I was looking for. I had seen that book
> but not carefully enough, so I didn't realise you could get the
> classical field that way ...
Actually it is somewhat worse than my simple answer suggests.
Coherent states give you the free classical field only.
This suffices for quantum optics.
The interacting case is poorly understood.
> Do you know if it is possible to do similar things for non-Abelian
> gauge fields? I know people try to (perhaps successfully) construct
> coherent states in non-Abelian gauge theory -- does that lead to
> classical Yang-Mills theory?
In principle, yes, but I haven't seen it anywhere.
The problem is that no one really understands the relations between
classical field theory and quantum field theory.
So one only finds occasional remarks, usually without reference
to a more careful discussion.
Arnold Neumaier
Eric A. Forgy
Apr14-04, 05:50 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>Hi Doug,\n\nYour refusal to tone down the rhetoric has greatly reduced my interest\nin discussing things with you, but I thought I would at least point\nout how obvious this is when looked at in the index-free version of\nyour proposal. You\'ve essentially got a twice covariant tensor P\nobtained from a 1-form potential A via\n\nP = DA.\n\nYour source-free action looks like\n\nS = int_M (P,P) vol.\n\nYou\'re basically saying (I know you know this) that if you have a\n1-form B such that\n\nDB = 0,\n\nthen adding B to A is like a gauge transform, i.e. if\n\nA\' = A + B\n\nthen\n\nDA\' = DA\n\nand A,A\' give the same physics. Think about the meaning of the\nequation DB = 0 though. Is there any solution besides B = 0?\n\nBest regards,\nEric\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>Hi Doug,
Your refusal to tone down the rhetoric has greatly reduced my interest
in discussing things with you, but I thought I would at least point
out how obvious this is when looked at in the index-free version of
your proposal. You've essentially got a twice covariant tensor P
obtained from a 1-form potential A via
P = DA[/itex].
Your source-free action looks like
S = \int_M (P,P) vol.
You're basically saying (I know you know this) that if you have a
1-form B such that
[itex]DB = 0,
then adding B to A is like a gauge transform, i.e. if
A' = A + B
then
DA' = DA
and A,A' give the same physics. Think about the meaning of the
equation DB = though. Is there any solution besides B = ?
Best regards,
Eric
Charles Francis
Apr15-04, 02:57 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>In article <53ca460a.0404081058.13447ad9@posting.google.com >, Ulmo\n<ulmo@cheerful.com> writes\n>Charles Francis <charles@clef.demon.co.uk> wrote in message news:<c4v8rq\\$dju\\$1@lfa222122.richmond.edu>...\ n>> Although many text books say that the classical limit of a quantum\n>> theory is given by the hbar -> 0 limit this is a blunder of the first\n>> magnitude. It is clearly meaningless to let hbar -> 0, since hbar is a\n>> constant of nature. The poor student who expects to believe his text\n>> books may be forgiven, but the author who is supposed to have learned\n>> his subject may not. I generally treat the authors of such books as\n>> cranks with no understanding of quantum theory whatsoever.\n>>\n>\n>They are just saying that the reason we don\'t notice quantum effects\n>in every day life is because hbar is so small. Therefore quantum\n>mechanics should reduce to classical mechanics as you take hbar -> 0.\n>It\'s similar to the fact that we don\'t notice relativistic effects\n>because c is so large. Therefore, special relativity should reduce to\n>classical mechanics as you take c -> infinity. Obviously, no one is\n>suggesting that you can actually change the values of hbar or c.\n\nWould that it were so. But as far as I have been able to judge from what\nis said, many people simply apply a formula, without thinking that it\ndoes not make sense, just as they apply quantisation as a recipe without\nthinking that it does not follow mathematical logic.\n\n\n\nRegards\n\n--\nCharles Francis\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>In article <53ca460a.0404081058.13447ad9@posting.google.com>, Ulmo
<ulmo@cheerful.com> writes
>Charles Francis <charles@clef.demon.co.uk> wrote in message news:<c4v8rq$dju$1@lfa222122.richmond.edu>...
>> Although many text books say that the classical limit of a quantum
>> theory is given by the \hbar -> limit this is a blunder of the first
>> magnitude. It is clearly meaningless to let \hbar -> 0, since \hbar is a
>> constant of nature. The poor student who expects to believe his text
>> books may be forgiven, but the author who is supposed to have learned
>> his subject may not. I generally treat the authors of such books as
>> cranks with no understanding of quantum theory whatsoever.
>>
>
>They are just saying that the reason we don't notice quantum effects
>in every day life is because \hbar is so small. Therefore quantum
>mechanics should reduce to classical mechanics as you take \hbar -> .
>It's similar to the fact that we don't notice relativistic effects
>because c is so large. Therefore, special relativity should reduce to
>classical mechanics as you take c -> infinity. Obviously, no one is
>suggesting that you can actually change the values of \hbar or c.
Would that it were so. But as far as I have been able to judge from what
is said, many people simply apply a formula, without thinking that it
does not make sense, just as they apply quantisation as a recipe without
thinking that it does not follow mathematical logic.
Regards
--
Charles Francis
carlip@no-physics-spam.ucdavis.edu
Apr15-04, 04:03 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\n\nDoug Sweetser <sweetser@alum.mit.edu> wrote:\n\nIn an earlier post:\n\n> >> What would prove or disprove my model was a direct test of gravity\n> >> energy gravitates. Since gravity fields are very weak, I don\'t think\n> >> this has been done.\n\nI replied\n\n> > First of all, there is very clear experimental evidence that\n> > gravitational energy contributes E/c^2 to *passive* gravitational\n> > mass [...]\n\n> > Second, as I learned recently, there is also strong experimental\n> > evidence for the equality of the Moon\'s passive gravitational\n> > mass and its *active* gravitational mass. [...]\n\nIn case it wasn\'t clear: experimentally, gravitational energy\ncontributes E/c^2 to passive gravitational mass. Experimentally,\npassive gravitational mass is equal to active gravitational mass.\nTherefore, experimentally, gravitational energy contributes E/c^2\nto active gravitational mass. That *means* that gravitational\nenergy gravitates, the condition that you said would disprove\nyour model.\n\n> There is nothing in my work that would force for a given body\n> m_active/m_passive to be anything but 1. Will makes it more\n> precise:\n\n> m_active/m_passive = 1 + 1/2 zeta_3 (E_e/m_p) eq. 9.27\n\n> where E_e is nuclear electrostatic energy.\n\nYou are taking this out of context. This gives *one* contribution\nto m_active/m_passive, namely the contribution that would arise if\nthe electrostatic contribution to m_active differed from the\nelectrostatic contribution to m_passive. There is another term\nof the same form that would come in if the gravitational contribution\nto m_active differed from the gravitational contribution to m_passive.\n\n> By using the PPN parameters, I can show that my metric agrees with\n> the two experiments you cited.\n\nIf that were true, then your theory would imply that gravitational\nenergy gravitates. Does it, or doesn\'t it?\n\nSteve Carlip\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>Doug Sweetser <sweetser@alum.mit.edu> wrote:
In an earlier post:
> >> What would prove or disprove my model was a direct test of gravity
> >> energy gravitates. Since gravity fields are very weak, I don't think
> >> this has been done.
I replied
> > First of all, there is very clear experimental evidence that
> > gravitational energy contributes E/c^2 to *passive* gravitational
> > mass [...]
> > Second, as I learned recently, there is also strong experimental
> > evidence for the equality of the Moon's passive gravitational
> > mass and its *active* gravitational mass. [...]
In case it wasn't clear: experimentally, gravitational energy
contributes E/c^2 to passive gravitational mass. Experimentally,
passive gravitational mass is equal to active gravitational mass.
Therefore, experimentally, gravitational energy contributes E/c^2
to active gravitational mass. That *means* that gravitational
energy gravitates, the condition that you said would disprove
your model.
> There is nothing in my work that would force for a given body
> m_{active}/m_{passive} to be anything but 1. Will makes it more
> precise:
> m_{active}/m_{passive} = 1 + 1/2 \zeta_3 (E_e/m_p) eq. 9.27
> where E_e is nuclear electrostatic energy.
You are taking this out of context. This gives *one* contribution
to m_{active}/m_{passive}, namely the contribution that would arise if
the electrostatic contribution to m_{active} differed from the
electrostatic contribution to m_{passive}. There is another term
of the same form that would come in if the gravitational contribution
to m_{active} differed from the gravitational contribution to m_{passive}.
> By using the PPN parameters, I can show that my metric agrees with
> the two experiments you cited.
If that were true, then your theory would imply that gravitational
energy gravitates. Does it, or doesn't it?
Steve Carlip
Arnold Neumaier
Apr15-04, 11:38 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>Charles Francis wrote:\n> In article <53ca460a.0404081058.13447ad9@posting.google.com >, Ulmo\n> <ulmo@cheerful.com> writes\n>\n>>Charles Francis <charles@clef.demon.co.uk> wrote in message news:<c4v8rq\\$dju\\$1@lfa222122.richmond.edu>...\ n>>\n>>>Although many text books say that the classical limit of a quantum\n>>>theory is given by the hbar -> 0 limit this is a blunder of the first\n>>>magnitude. It is clearly meaningless to let hbar -> 0, since hbar is a\n>>>constant of nature. The poor student who expects to believe his text\n>>>books may be forgiven, but the author who is supposed to have learned\n>>>his subject may not. I generally treat the authors of such books as\n>>>cranks with no understanding of quantum theory whatsoever.\n>>>\n>>\n>>They are just saying that the reason we don\'t notice quantum effects\n>>in every day life is because hbar is so small. Therefore quantum\n>>mechanics should reduce to classical mechanics as you take hbar -> 0.\n>>It\'s similar to the fact that we don\'t notice relativistic effects\n>>because c is so large. Therefore, special relativity should reduce to\n>>classical mechanics as you take c -> infinity. Obviously, no one is\n>>suggesting that you can actually change the values of hbar or c.\n>\n> Would that it were so. But as far as I have been able to judge from what\n> is said, many people simply apply a formula, without thinking that it\n> does not make sense, just as they apply quantisation as a recipe without\n> thinking that it does not follow mathematical logic.\n\n\nWhat really happens is that hbar is a free parameter in a family\nof quantum theories. One particular choice of them corresponds to\nphysical reality, and the limit hbar to zero corresponds to a classical\nworld. Of course, since the real world is known not to be classical,\none cannot take the limit in reality, where hbar is fixed.\n\nConstants of nature are always free parameters in the corresponding\nmathematical model, where they may usually take any reasonable value,\nand hence may have limits.\n\nTalking of a limit is always a mathematical procedure, and hence\nmakes only sense in a mathematical model. But whenever this model\nhas a parameter for which a limiting situation is well defined,\nthis makes perfect sense and is as far from being a\n\'blunder of the first magnitude\' as one can imagine.\n\nDon\'t think the many excellent people were dudes!\nWhenever there is a discrepancy between tradition and one\'s own\nunderstanding, the reason lies in the vast majority of cases\nin the latter.\n\n\nArnold Neumaier\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>Charles Francis wrote:
> In article <53ca460a.0404081058.13447ad9@posting.google.com>, Ulmo
> <ulmo@cheerful.com> writes
>
>>Charles Francis <charles@clef.demon.co.uk> wrote in message news:<c4v8rq$dju$1@lfa222122.richmond.edu>...
>>
>>>Although many text books say that the classical limit of a quantum
>>>theory is given by the \hbar -> limit this is a blunder of the first
>>>magnitude. It is clearly meaningless to let \hbar -> 0, since \hbar is a
>>>constant of nature. The poor student who expects to believe his text
>>>books may be forgiven, but the author who is supposed to have learned
>>>his subject may not. I generally treat the authors of such books as
>>>cranks with no understanding of quantum theory whatsoever.
>>>
>>
>>They are just saying that the reason we don't notice quantum effects
>>in every day life is because \hbar is so small. Therefore quantum
>>mechanics should reduce to classical mechanics as you take \hbar -> .
>>It's similar to the fact that we don't notice relativistic effects
>>because c is so large. Therefore, special relativity should reduce to
>>classical mechanics as you take c -> infinity. Obviously, no one is
>>suggesting that you can actually change the values of \hbar or c.
>
> Would that it were so. But as far as I have been able to judge from what
> is said, many people simply apply a formula, without thinking that it
> does not make sense, just as they apply quantisation as a recipe without
> thinking that it does not follow mathematical logic.
What really happens is that \hbar is a free parameter in a family
of quantum theories. One particular choice of them corresponds to
physical reality, and the limit \hbar to zero corresponds to a classical
world. Of course, since the real world is known not to be classical,
one cannot take the limit in reality, where \hbar is fixed.
Constants of nature are always free parameters in the corresponding
mathematical model, where they may usually take any reasonable value,
and hence may have limits.
Talking of a limit is always a mathematical procedure, and hence
makes only sense in a mathematical model. But whenever this model
has a parameter for which a limiting situation is well defined,
this makes perfect sense and is as far from being a
'blunder of the first magnitude' as one can imagine.
Don't think the many excellent people were dudes!
Whenever there is a discrepancy between tradition and one's own
understanding, the reason lies in the vast majority of cases
in the latter.
Arnold Neumaier
Doug Sweetser
Apr16-04, 02:28 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>Hello Steve:\n\nThis _sounds_ reasonable (ie I understand your line of reasoning):\n\n> In case it wasn\'t clear: experimentally, gravitational energy\n> contributes E/c^2 to passive gravitational mass.\n\nThis point was not claimed in the section on the Nordtvedt effect or in\nBartlett and Van Buren paper. If that is due to my miss-reading, I\nappologize. Given the technical importance, I would think these\nauthors would cite it directly, instead of being oblique.\n\n\n>> There is nothing in my work that would force for a given body\n>> m_active/m_passive to be anything but 1. Will makes it more\n>> precise:\n>\n>> m_active/m_passive = 1 + 1/2 zeta_3 (E_e/m_p) eq. 9.27\n>\n>> where E_e is nuclear electrostatic energy.\n>\n> You are taking this out of context.\n\nTo be fair, it should be noted this was what Will was talking about.\nYour point is correct: we should think of every darn contribution\npossible. One obvious one is gravitational energy. What we know from\nexperiments, and what we agree has been shown is:\n\nm_active/m_passive = 1\n\nFrom an algebraic viewpoint, there are three possibilities why this can\nbe: either zeta_3 is zero or the sum over all possible forms of energy\nmakes zero contributions to this ratio, or both. I am not advocating\nanything right now, just pointing out an ambiguity. This may be why\nthe authors did not write "gravitational energy gravitates," eq. 9.27\ndoes reveal those cards.\n\n\n>> By using the PPN parameters, I can show that my metric agrees with\n>> the two experiments you cited.\n>\n> If that were true,\n\n??? What I mean specifically is this:\n\neta = 4 beta - gamma - 3 - 10/3 xi - alpha_1 Will\n+ 2/3 alpha_2 - 2/3 zeta_1 - 1/3 zeta_2 eq. 8.9\n\nfor my proposal eta = 1\n\n\nm_active/m_passive = 1 + 1/2 zeta_3 (E_e/m_p) eq. 9.27\n\nfor my proposal m_active/m_passive = 1\n\nI dislike going beyond equations 8.9 and 9.27. Even if we generalize\n9.27 to include other forms of energy, my theory will always predict\nthat the ratio is equal to one.\n\n> your theory would imply that gravitational energy gravitates. Does\n> it, or doesn\'t it?\n\nA nice simple, direct question, yet I feel like I am on a witness stand\nbeing asked a trick question. What is true, confirmed by experiment,\nis 8.9 and 9.27 are equal to one. The metric I work with is consistent\nwith that.\n\nLike the electromagnetic field not being a source of electrical charge,\nthe gravitational field cannot be a source of mass charge.\n\n\ndoug\nquaternions.com\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>Hello Steve:
This _sounds_ reasonable (ie I understand your line of reasoning):
> In case it wasn't clear: experimentally, gravitational energy
> contributes E/c^2 to passive gravitational mass.
This point was not claimed in the section on the Nordtvedt effect or in
Bartlett and Van Buren paper. If that is due to my miss-reading, I
appologize. Given the technical importance, I would think these
authors would cite it directly, instead of being oblique.
>> There is nothing in my work that would force for a given body
>> m_{active}/m_{passive} to be anything but 1. Will makes it more
>> precise:
>
>> m_{active}/m_{passive} = 1 + 1/2 \zeta_3 (E_e/m_p) eq. 9.27
>
>> where E_e is nuclear electrostatic energy.
>
> You are taking this out of context.
To be fair, it should be noted this was what Will was talking about.
Your point is correct: we should think of every darn contribution
possible. One obvious one is gravitational energy. What we know from
experiments, and what we agree has been shown is:
m_{active}/m_{passive} = 1
From an algebraic viewpoint, there are three possibilities why this can
be: either \zeta_3 is zero or the sum over all possible forms of energy
makes zero contributions to this ratio, or both. I am not advocating
anything right now, just pointing out an ambiguity. This may be why
the authors did not write "gravitational energy gravitates," eq. 9.27
does reveal those cards.
>> By using the PPN parameters, I can show that my metric agrees with
>> the two experiments you cited.
>
> If that were true,
??? What I mean specifically is this:
\eta = 4 \beta - \gamma - 3 - 10/3 \xi - \alpha_1[/itex] Will
+ 2/3 \alpha_2 - 2/3 \zeta_1 - 1/3 \zeta_2 eq. 8.9
for my proposal \eta = 1m_{active}/m_{passive} = 1 + 1/2 \zeta_3 (E_e/m_p) eq. 9.27
for my proposal [itex]m_{active}/m_{passive} = 1
I dislike going beyond equations 8.9 and 9.27. Even if we generalize
9.27 to include other forms of energy, my theory will always predict
that the ratio is equal to one.
> your theory would imply that gravitational energy gravitates. Does
> it, or doesn't it?
A nice simple, direct question, yet I feel like I am on a witness stand
being asked a trick question. What is true, confirmed by experiment,
is 8.9 and 9.27 are equal to one. The metric I work with is consistent
with that.
Like the electromagnetic field not being a source of electrical charge,
the gravitational field cannot be a source of mass charge.
doug
quaternions.com
Charles Francis
Apr16-04, 05:51 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\n\nIn article <c5ma9m\\$1bh\\$1@lfa222122.richmond.edu>, Arnold Neumaier\n<Arnold.Neumaier@univie.ac.at> writes\n>Charles Francis wrote:\n>> In article <53ca460a.0404081058.13447ad9@posting.google.com >, Ulmo\n>><ulmo@cheerful.com> writes\n>>\n>>>Charles Francis <charles@clef.demon.co.uk> wrote in message\n>>>news:<c4v8rq\\$dju\\$1@lfa222122.richm ond.edu>...\n>>>\n>>>>Although many text books say that the classical limit of a quantum\n>>>>theory is given by the hbar -> 0 limit this is a blunder of the first\n>>>>magnitude. It is clearly meaningless to let hbar -> 0, since hbar is a\n>>>>constant of nature. The poor student who expects to believe his text\n>>>>books may be forgiven, but the author who is supposed to have learned\n>>>>his subject may not. I generally treat the authors of such books as\n>>>>cranks with no understanding of quantum theory whatsoever.\n>>>>\n>>>\n>>>They are just saying that the reason we don\'t notice quantum effects\n>>>in every day life is because hbar is so small. Therefore quantum\n>>>mechanics should reduce to classical mechanics as you take hbar -> 0.\n>>>It\'s similar to the fact that we don\'t notice relativistic effects\n>>>because c is so large. Therefore, special relativity should reduce to\n>>>classical mechanics as you take c -> infinity. Obviously, no one is\n>>>suggesting that you can actually change the values of hbar or c.\n\n>> Would that it were so. But as far as I have been able to judge from\n>>what is said, many people simply apply a formula, without thinking\n>>that it does not make sense, just as they apply quantisation as a\n>>recipe without thinking that it does not follow mathematical logic.\n>\n>\n>What really happens is that hbar is a free parameter in a family\n>of quantum theories. One particular choice of them corresponds to\n>physical reality, and the limit hbar to zero corresponds to a classical\n>world. Of course, since the real world is known not to be classical,\n>one cannot take the limit in reality, where hbar is fixed.\n\nAs if to illustrate what I was saying! What really happens is that the\nclassical correspondence appears statistically from the expectation of\nobservables when there is a a large number, N, of quantum particles.\nhbar/N ->0 is a perfectly sensible, both physically and mathematically.\n\n\n\nRegards\n\n--\nCharles Francis\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>In article <c5ma9m$1bh$1@lfa222122.richmond.edu>, Arnold Neumaier
<Arnold.Neumaier@univie.ac.at> writes
>Charles Francis wrote:
>> In article <53ca460a.0404081058.13447ad9@posting.google.com>, Ulmo
>><ulmo@cheerful.com> writes
>>
>>>Charles Francis <charles@clef.demon.co.uk> wrote in message
>>>news:<c4v8rq$dju$1@lfa222122.richmond.edu>...
>>>
>>>>Although many text books say that the classical limit of a quantum
>>>>theory is given by the \hbar -> limit this is a blunder of the first
>>>>magnitude. It is clearly meaningless to let \hbar -> 0, since \hbar is a
>>>>constant of nature. The poor student who expects to believe his text
>>>>books may be forgiven, but the author who is supposed to have learned
>>>>his subject may not. I generally treat the authors of such books as
>>>>cranks with no understanding of quantum theory whatsoever.
>>>>
>>>
>>>They are just saying that the reason we don't notice quantum effects
>>>in every day life is because \hbar is so small. Therefore quantum
>>>mechanics should reduce to classical mechanics as you take \hbar -> .
>>>It's similar to the fact that we don't notice relativistic effects
>>>because c is so large. Therefore, special relativity should reduce to
>>>classical mechanics as you take c -> infinity. Obviously, no one is
>>>suggesting that you can actually change the values of \hbar or c.
>> Would that it were so. But as far as I have been able to judge from
>>what is said, many people simply apply a formula, without thinking
>>that it does not make sense, just as they apply quantisation as a
>>recipe without thinking that it does not follow mathematical logic.
>
>
>What really happens is that \hbar is a free parameter in a family
>of quantum theories. One particular choice of them corresponds to
>physical reality, and the limit \hbar to zero corresponds to a classical
>world. Of course, since the real world is known not to be classical,
>one cannot take the limit in reality, where \hbar is fixed.
As if to illustrate what I was saying! What really happens is that the
classical correspondence appears statistically from the expectation of
observables when there is a a large number, N, of quantum particles.
\hbar/N ->0 is a perfectly sensible, both physically and mathematically.
Regards
--
Charles Francis
Arnold Neumaier
Apr19-04, 01:54 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>Charles Francis wrote:\n\n>>What really happens is that hbar is a free parameter in a family\n>>of quantum theories. One particular choice of them corresponds to\n>>physical reality, and the limit hbar to zero corresponds to a classical\n>>world. Of course, since the real world is known not to be classical,\n>>one cannot take the limit in reality, where hbar is fixed.\n>\n>\n> As if to illustrate what I was saying! What really happens is that the\n> classical correspondence appears statistically from the expectation of\n> observables when there is a a large number, N, of quantum particles.\n> hbar/N ->0 is a perfectly sensible, both physically and mathematically.\n\nThis is a _different_ (thermodynamic) limit, not the classical\nlimit. It does not give you classical multiparticle physics\na la Boltzmann. And certain quantum phenomena such as\nsuperconductivity survive the thermodynamic limit, while they\nseem impossible in the limit hbar -> 0. Peolple talking about\nthe latter usually mean what they say.\n\n\nArnold Neumaier\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>Charles Francis wrote:
>>What really happens is that \hbar is a free parameter in a family
>>of quantum theories. One particular choice of them corresponds to
>>physical reality, and the limit \hbar to zero corresponds to a classical
>>world. Of course, since the real world is known not to be classical,
>>one cannot take the limit in reality, where \hbar is fixed.
>
>
> As if to illustrate what I was saying! What really happens is that the
> classical correspondence appears statistically from the expectation of
> observables when there is a a large number, N, of quantum particles.
> \hbar/N ->0 is a perfectly sensible, both physically and mathematically.
This is a _different_ (thermodynamic) limit, not the classical
limit. It does not give you classical multiparticle physics
a la Boltzmann. And certain quantum phenomena such as
superconductivity survive the thermodynamic limit, while they
seem impossible in the limit \hbar -> . Peolple talking about
the latter usually mean what they say.
Arnold Neumaier
Doug Sweetser
Apr19-04, 02:30 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>Hello Eric:\n\n> if\n>\n> A\' = A + B\n>\n> then\n>\n> DA\' = DA\n>\n> and A,A\' give the same physics. Think about the meaning of the\n> equation DB = 0 though. Is there any solution besides B = 0?\n\nThis certainly is a key question. If the only solution is B = 0, then\nit would be a trivial result. It is a solution, no doubt, but not the\none and only.\n\nYour question is only partially formed. One has to state both the\nmetric and the potential to evaluate DB.\n\n1. Let the metric be the flat Minkowski metric in Euclidean\ncoordinates. The connection is zero. The only family of solutions are\npotentials that are constants, with no dependence on t, x, y, or z.\nThe Minkowski metric is too dull to find an interesting potential\nsolution.\n\n2. Let the metric be the Watt/Misner exponential metric. One can\ncalculate the connection (L_w^uv), and then search for a potential such\nthat each A^u;v = L_w^uv A^w.\n\nI have yet to do this calculation, too focused on problems with my\ncomputer. It looks like a possibly non-trivial answer to your\nquestion, but I don\'t have the specifics, and specifics matter. So I\nwill work on this in the upcoming weeks, and post here when I have some\nresults, either for or against (case 1 establishes the trivial\npotential, but there appears to be more possibilities for case 2).\n\nLet me explain the source of my "rhetoric". A half dozen times I have\n"bet the theory." What that means is I have formed a specific\ntechnical question, then said if I cannot find an answer, I can forget\nabout this body of work and say so publicly. Two examples: there were\nfield equations as an inspired guess, so I bet I had to find the\nLagrange density that generated those equations, and there was an\nequation, d_u L_w^u0 A^w = 0, that my metric/potential combination had\nto solve. At both times, I had not worked at this level of technical\ndetail. I learned by stumbling. Yet in those two specific cases, it\ndid work out. That is my experience. Your question is significant\nenough for me to bet the theory again.\n\nJohn Baez only said one thing about this work: it presumes a background\nmetric. I think that statement is correct until I can demonstrate how\nthis symmetry works. It would be even better if someone had the magic\nwords to say someone else already has found this. A covariant\nderivative of a 4-potential is a combination of the change in the\n4-potential plus the change in the metric related to that 4-potential.\nOne cannot tell how much of the change in the covariant derivative\ncomes from those two sources until more information is supplied about\nthe 4-potential or the metric.\n\n\ndoug\nquaternions.com\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>Hello Eric:
> if
>
> A' = A + B
>
> then
>
> DA' = DA
>
> and A,A' give the same physics. Think about the meaning of the
> equation DB = though. Is there any solution besides B = ?
This certainly is a key question. If the only solution is B = 0, then
it would be a trivial result. It is a solution, no doubt, but not the
one and only.
Your question is only partially formed. One has to state both the
metric and the potential to evaluate DB.
1. Let the metric be the flat Minkowski metric in Euclidean
coordinates. The connection is zero. The only family of solutions are
potentials that are constants, with no dependence on t, x, y, or z.
The Minkowski metric is too dull to find an interesting potential
solution.
2. Let the metric be the Watt/Misner exponential metric. One can
calculate the connection (L_w^uv), and then search for a potential such
that each A^u;v = L_w^uv A^w.
I have yet to do this calculation, too focused on problems with my
computer. It looks like a possibly non-trivial answer to your
question, but I don't have the specifics, and specifics matter. So I
will work on this in the upcoming weeks, and post here when I have some
results, either for or against (case 1 establishes the trivial
potential, but there appears to be more possibilities for case 2).
Let me explain the source of my "rhetoric". A half dozen times I have
"bet the theory." What that means is I have formed a specific
technical question, then said if I cannot find an answer, I can forget
about this body of work and say so publicly. Two examples: there were
field equations as an inspired guess, so I bet I had to find the
Lagrange density that generated those equations, and there was an
equation, d_u L_w^u0 A^w = 0, that my metric/potential combination had
to solve. At both times, I had not worked at this level of technical
detail. I learned by stumbling. Yet in those two specific cases, it
did work out. That is my experience. Your question is significant
enough for me to bet the theory again.
John Baez only said one thing about this work: it presumes a background
metric. I think that statement is correct until I can demonstrate how
this symmetry works. It would be even better if someone had the magic
words to say someone else already has found this. A covariant
derivative of a 4-potential is a combination of the change in the
4-potential plus the change in the metric related to that 4-potential.
One cannot tell how much of the change in the covariant derivative
comes from those two sources until more information is supplied about
the 4-potential or the metric.
doug
quaternions.com
Charles Francis
Apr20-04, 02:35 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>In article <c613ov\\$d16\\$1@lfa222122.richmond.edu>, Arnold Neumaier\n<Arnold.Neumaier@univie.ac.at> writes\n>Charles Francis wrote:\n>\n>>>What really happens is that hbar is a free parameter in a family\n>>>of quantum theories. One particular choice of them corresponds to\n>>>physical reality, and the limit hbar to zero corresponds to a classical\n>>>world. Of course, since the real world is known not to be classical,\n>>>one cannot take the limit in reality, where hbar is fixed.\n>> As if to illustrate what I was saying! What really happens is that\n>>the classical correspondence appears statistically from the\n>>expectation of observables when there is a a large number, N, of\n>>quantum particles. hbar/N ->0 is a perfectly sensible, both\n>>physically and mathematically.\n>\n>This is a _different_ (thermodynamic) limit, not the classical\n>limit.\n\nI should perhaps have been a little more accurate. The motion of a\nmassive classical particle is the motion of a load of quantum particles.\nThis is statistical and hence determinate (to required accuracy). Since\nit is determinate the state at each instant is, from the point of view\nof qm, (almost) continuously measured and is modelled as a sequence of\nmeasured states indefinitely close together. You do indeed get classical\nmultiparticle physics from taking this limit correctly.\n\n> And certain quantum phenomena such as\n>superconductivity survive the thermodynamic limit, while they\n>seem impossible in the limit hbar -> 0.\n\nI do not mean to say you cannot get other forms of physics too.\n\n> Peolple talking about\n>the latter usually mean what they say.\n>\nI doubt they mean anything sensible at all.\n\n\n\nRegards\n\n--\nCharles Francis\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>In article <c613ov$d16$1@lfa222122.richmond.edu>, Arnold Neumaier
<Arnold.Neumaier@univie.ac.at> writes
>Charles Francis wrote:
>
>>>What really happens is that \hbar is a free parameter in a family
>>>of quantum theories. One particular choice of them corresponds to
>>>physical reality, and the limit \hbar to zero corresponds to a classical
>>>world. Of course, since the real world is known not to be classical,
>>>one cannot take the limit in reality, where \hbar is fixed.
>> As if to illustrate what I was saying! What really happens is that
>>the classical correspondence appears statistically from the
>>expectation of observables when there is a a large number, N, of
>>quantum particles. \hbar/N ->0 is a perfectly sensible, both
>>physically and mathematically.
>
>This is a _different_ (thermodynamic) limit, not the classical
>limit.
I should perhaps have been a little more accurate. The motion of a
massive classical particle is the motion of a load of quantum particles.
This is statistical and hence determinate (to required accuracy). Since
it is determinate the state at each instant is, from the point of view
of qm, (almost) continuously measured and is modelled as a sequence of
measured states indefinitely close together. You do indeed get classical
multiparticle physics from taking this limit correctly.
> And certain quantum phenomena such as
>superconductivity survive the thermodynamic limit, while they
>seem impossible in the limit \hbar -> .
I do not mean to say you cannot get other forms of physics too.
> Peolple talking about
>the latter usually mean what they say.
>
I doubt they mean anything sensible at all.
Regards
--
Charles Francis
Hendrik van Hees
Apr24-04, 12:15 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>Arnold Neumaier wrote:\n\n> This is a _different_ (thermodynamic) limit, not the classical\n> limit. It does not give you classical multiparticle physics\n> a la Boltzmann. And certain quantum phenomena such as\n> superconductivity survive the thermodynamic limit, while they\n> seem impossible in the limit hbar -> 0. Peolple talking about\n> the latter usually mean what they say.\n\nFrom a naive many-body physicist\'s point of view, the Boltzmann equation\ncan be obtained from (real time) quantum field theory by course\ngraining the Wigner transform of the off-diagonal elements of the\nSchwinger-Keldysh two-point Green\'s function over macroscopically\nsmall, microscopically big space-time volumes.\n\nIt\'s only necessary, that for the considered physical process this\ntime-scale argument makes any sense.\n\n--\nHendrik van Hees Cyclotron Institute\nPhone: +1 979/845-1411 Texas A&M University\nFax: +1 979/845-1899 Cyclotron Institute, MS-3366\nhttp://theory.gsi.de/~vanhees/ College Station, TX 77843-3366\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>Arnold Neumaier wrote:
> This is a _different_ (thermodynamic) limit, not the classical
> limit. It does not give you classical multiparticle physics
> a la Boltzmann. And certain quantum phenomena such as
> superconductivity survive the thermodynamic limit, while they
> seem impossible in the limit \hbar -> . Peolple talking about
> the latter usually mean what they say.
From a naive many-body physicist's point of view, the Boltzmann equation
can be obtained from (real time) quantum field theory by course
graining the Wigner transform of the off-diagonal elements of the
Schwinger-Keldysh two-point Green's function over macroscopically
small, microscopically big space-time volumes.
It's only necessary, that for the considered physical process this
time-scale argument makes any sense.
--
Hendrik van Hees Cyclotron Institute
Phone: +1 979/845-1411 Texas A&M University
Fax: +1 979/845-1899 Cyclotron Institute, MS-3366
http://theory.gsi.de/~vanhees/ College Station, TX 77843-3366
Arnold Neumaier
Apr28-04, 02:46 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>Hendrik van Hees wrote:\n> Arnold Neumaier wrote:\n>\n>>This is a _different_ (thermodynamic) limit, not the classical\n>>limit. It does not give you classical multiparticle physics\n>>a la Boltzmann. And certain quantum phenomena such as\n>>superconductivity survive the thermodynamic limit, while they\n>>seem impossible in the limit hbar -> 0. Peolple talking about\n>>the latter usually mean what they say.\n>\n>\n> From a naive many-body physicist\'s point of view, the Boltzmann equation\n> can be obtained from (real time) quantum field theory by course\n> graining the Wigner transform of the off-diagonal elements of the\n> Schwinger-Keldysh two-point Green\'s function over macroscopically\n> small, microscopically big space-time volumes.\n>\n> It\'s only necessary, that for the considered physical process this\n> time-scale argument makes any sense.\n>\n\nYes, but it gives the _quantum_ Boltzmann equation for\nWigner functions, which need not be positive. To get the classical\nBoltzmann equation for a nonnegative phase space density,\nyou still need to do the limit hbar to zero.\n\n\nArnold Neumaier\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>Hendrik van Hees wrote:
> Arnold Neumaier wrote:
>
>>This is a _different_ (thermodynamic) limit, not the classical
>>limit. It does not give you classical multiparticle physics
>>a la Boltzmann. And certain quantum phenomena such as
>>superconductivity survive the thermodynamic limit, while they
>>seem impossible in the limit \hbar -> . Peolple talking about
>>the latter usually mean what they say.
>
>
> From a naive many-body physicist's point of view, the Boltzmann equation
> can be obtained from (real time) quantum field theory by course
> graining the Wigner transform of the off-diagonal elements of the
> Schwinger-Keldysh two-point Green's function over macroscopically
> small, microscopically big space-time volumes.
>
> It's only necessary, that for the considered physical process this
> time-scale argument makes any sense.
>
Yes, but it gives the _quantum_ Boltzmann equation for
Wigner functions, which need not be positive. To get the classical
Boltzmann equation for a nonnegative phase space density,
you still need to do the limit \hbar to zero.
Arnold Neumaier
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