<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\nAre Higgs particles all the same mass?\nDoes a proton have more Higgs particles associated with its rest mass\nthan an electron has associated with its rest mass?\nAnd does the mass of all Higgs particles equal the total rest mass of\nthe universe? Do Higgs particles have short lifetimes like other\nparticles in the vacuum?\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>Are Higgs particles all the same mass?
Does a proton have more Higgs particles associated with its rest mass
than an electron has associated with its rest mass?
And does the mass of all Higgs particles equal the total rest mass of
the universe? Do Higgs particles have short lifetimes like other
particles in the vacuum?
Alfred Einstead
Jul22-04, 05:16 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>\nalistair@goforit64.fsnet.co.uk (alistair) wrote:\n> Are Higgs particles all the same mass?\n\nAll the particles in the standard model are massless. The\nappearance of mass is arrived at only as a side effect of the\nway the particle interacts with the Higgs.\n\nSo, there\'s two pictures. The one "before symmetry breaking"\nhas massless particles interacting with a Higgs that has 4\nmodes. By the time you factor in the mass-production that\narises as a by-product of the interacting with the Higgs field,\nto get the picture "after symmetry breaking" there\'s only 1\nHiggs mode left. And it\'s massive.\n\nSo the question doesn\'t apply directly. The translation\nbetween the two pictures will equate the other 3 Higgs\nmodes with one mode each of the 3 massive field carriers,\none for the Z, one for the W and one for the anti-W.\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>alistair@goforit64.fsnet.co.uk (alistair) wrote:
> Are Higgs particles all the same mass?
All the particles in the standard model are massless. The
appearance of mass is arrived at only as a side effect of the
way the particle interacts with the Higgs.
So, there's two pictures. The one "before symmetry breaking"
has massless particles interacting with a Higgs that has 4
modes. By the time you factor in the mass-production that
arises as a by-product of the interacting with the Higgs field,
to get the picture "after symmetry breaking" there's only 1
Higgs mode left. And it's massive.
So the question doesn't apply directly. The translation
between the two pictures will equate the other 3 Higgs
modes with one mode each of the 3 massive field carriers,
one for the Z, one for the W and one for the anti-W.
kurious
Jul22-04, 02:28 PM
Alistair said:
>Are Higgs particles all the same mass?
>Does a proton have more Higgs particles associated with its rest mass
>than an electron has associated with its rest mass?
>And does the mass of all Higgs particles equal the total rest mass of
>the universe? Do Higgs particles have short lifetimes like other
>particles in the vacuum?
Here are two great replies from the quantum physics section of "physics forums":
VANESCH said:
You ( Alistair ) seem to think that Higgs particles are a kind of marbles that you put in a proton bag to make up its mass. It is absolutely not the case.
What happens, in fact, is that the Higgs particle is a very heavy one, but - like all particles - corresponds in fact to a quantum field.
The problem people had when they were building a quantum theory of elementary particles is that the usual way of giving a mass to a fermion, namely by introducing a term m^2 psi-dagger psi in the lagrangian, didn't work because these terms do not respect the required symmetries of the theory (this is a bit a complicated issue). However, it turned out that the introduction of a scalar particle (field), that was subject to a funny potential such that in the ground state (lowest energy) the field values would NOT be 0, could solve several problems. Indeed, the non-zero value of that field, coupled with an interaction term between that field and, say, the electron field, gave a term in the lagrangian which DID respect the symmetries required, but mimicked, at low energies, as a term that was essentially the same as a mass term.
It also solved another problem: there were 2 theorems in quantum field theory that made life hard. The first one was by 't Hooft, and said that if you want to have a renormalizable (calculable) theory, your interactions need to be described by fields such that the lagrangian obeys, what is called, a gauge symmetry. The problem with a gauge symmetry was that gauge particles have to be massless. People tried to find tricks around it, but Goldstone proved an annoying theorem (Goldstone's theorem), that said that you always have to have massless particles associated with the degrees of freedom of a gauge symmetry, the socalled "goldstone bosons". As these were not observed, that was annoying.
And then the Higgs mechanism was invented. It gave terms that mimicked mass to the gauge particles, it eliminated the goldstone bosons and gave, by its interactions, a mimicked mass term to the fermions.
cheers,
Patrick.
ZEFRAM C said:
Patrick gave a wonderful overview of the Higgs mechanism, but the answers to the questions posted don't immediately follow. Here they are:
1) All Higgs particles DO have the same mass, as they are the quanta of the same Higgs field. Just like all electrons have the same mass.
2) Neither electrons nor protons have Higgs particles "associated" with them. What happens is that the theory treats all fermions as massless, but the "vacuum" we know has in fact the peculiar Higgs background, which makes it appear that they have mass. All particles, except the photon and gluons, are assumed to interact with the Higgs field, and they develop masses proportional to the strength of this coupling, which for the fermions is arbitrary and independent of other constants (it is not so for the W and Z). The electron's coupling to the Higgs is smaller than the proton's, resulting in a smaller apparent mass for the electron.
3) There are very few actual Higgs particles in the universe, as they are unstable. There is a difference between the Higgs background (which is non-zero even in the ABSENCE of Higgs particles), and the quanta of the Higgs field, which are excitations of the field above the background. The former does not contribute to mass of the universe (actually this is a tricky topic which deals with physics currently beyond me). It is the latter that are traditional "particles" and have mass, etc, but they are unstable.
4) The Higgs particles have indeed a very short lifetime. They can decay directly into any pair of fermions, or a pair of W's or Z's. Since the decay constant is proportional to the coupling, the odds of it decaying into heavy particles are greater than for light particles. The lifetime of the Higgs should be pretty small like the t quark. The ATLAS experiment at Cern will try to produce real Higgs particles and observe them via their decays into pairs of Z (with subsequent decays of both Z)or t quarks.
Oz
Jul23-04, 06:35 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>\nkurious <alistair@goforit64.fsnet.co.uk> writes\n>However,\n>it turned out that the introduction of a scalar particle (field), that\n>was subject to a funny potential such that in the ground state (lowest\n>energy) the field values would NOT be 0, could solve several problems.\n>Indeed, the non-zero value of that field, coupled with an interaction\n>term between that field and, say, the electron field, gave a term in\n>the lagrangian which DID respect the symmetries required, but mimicked,\n>at low energies, as a term that was essentially the same as a mass\n>term.\n\nFine. Well, I\'m sure it is because everyone says so.\n\nHow are we to see a box full of photons?\n\nWe can increase the mass of the box by shoving in more photons.\nBut from what you have said, the implication is that photons do not\ninteract with the higgs field, hence they are massless. How do we get\nmass being created by adding massless photons to the box?\n\n--\nOz\nThis post is worth absolutely nothing and is probably fallacious.\n\nBTOPENWORLD address about to cease. DEMON address no longer in use.\n>>Use oz@farmeroz.port995.com<<\nozacoohdb@despammed.com still functions.\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>kurious <alistair@goforit64.fsnet.co.uk> writes
>However,
>it turned out that the introduction of a scalar particle (field), that
>was subject to a funny potential such that in the ground state (lowest
>energy) the field values would NOT be 0, could solve several problems.
>Indeed, the non-zero value of that field, coupled with an interaction
>term between that field and, say, the electron field, gave a term in
>the lagrangian which DID respect the symmetries required, but mimicked,
>at low energies, as a term that was essentially the same as a mass
>term.
Fine. Well, I'm sure it is because everyone says so.
How are we to see a box full of photons?
We can increase the mass of the box by shoving in more photons.
But from what you have said, the implication is that photons do not
interact with the higgs field, hence they are massless. How do we get
mass being created by adding massless photons to the box?
--
Oz
This post is worth absolutely nothing and is probably fallacious.
BTOPENWORLD address about to cease. DEMON address no longer in use.
>>Use oz@farmeroz.port995.com<<
ozacoohdb@despammed.com still functions.
Creighton Hogg
Jul23-04, 11:32 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\nOn 23 Jul 2004, Oz wrote:\n\n>\n> kurious <alistair@goforit64.fsnet.co.uk> writes\n> >However,\n> >it turned out that the introduction of a scalar particle (field), that\n> >was subject to a funny potential such that in the ground state (lowest\n> >energy) the field values would NOT be 0, could solve several problems.\n> >Indeed, the non-zero value of that field, coupled with an interaction\n> >term between that field and, say, the electron field, gave a term in\n> >the lagrangian which DID respect the symmetries required, but mimicked,\n> >at low energies, as a term that was essentially the same as a mass\n> >term.\n>\n> Fine. Well, I\'m sure it is because everyone says so.\n>\n> How are we to see a box full of photons?\n>\n> We can increase the mass of the box by shoving in more photons.\n> But from what you have said, the implication is that photons do not\n> interact with the higgs field, hence they are massless. How do we get\n> mass being created by adding massless photons to the box?\n\nAh! You get a mass increase in the box when adding photons because the\ninvariant mass of the system is increasing.\nA single photon is massless. Two photons decaying from an electron at\nrest, for example, will have an invariant mass equal to the rest mass of\nthe electron. The invariant mass is the magnitude of the sum of the four\nmomenta of all the objects in the system.\nHope that helps.\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>On 23 Jul 2004, Oz wrote:
>
> kurious <alistair@goforit64.fsnet.co.uk> writes
> >However,
> >it turned out that the introduction of a scalar particle (field), that
> >was subject to a funny potential such that in the ground state (lowest
> >energy) the field values would NOT be 0, could solve several problems.
> >Indeed, the non-zero value of that field, coupled with an interaction
> >term between that field and, say, the electron field, gave a term in
> >the lagrangian which DID respect the symmetries required, but mimicked,
> >at low energies, as a term that was essentially the same as a mass
> >term.
>
> Fine. Well, I'm sure it is because everyone says so.
>
> How are we to see a box full of photons?
>
> We can increase the mass of the box by shoving in more photons.
> But from what you have said, the implication is that photons do not
> interact with the higgs field, hence they are massless. How do we get
> mass being created by adding massless photons to the box?
Ah! You get a mass increase in the box when adding photons because the
invariant mass of the system is increasing.
A single photon is massless. Two photons decaying from an electron at
rest, for example, will have an invariant mass equal to the rest mass of
the electron. The invariant mass is the magnitude of the sum of the four
momenta of all the objects in the system.
Hope that helps.
Oz
Jul23-04, 03:18 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>\nCreighton Hogg <wchogg@hep.wisc.edu> writes\n\n>>Oz\n>> We can increase the mass of the box by shoving in more photons.\n>> But from what you have said, the implication is that photons do not\n>> interact with the higgs field, hence they are massless. How do we get\n>> mass being created by adding massless photons to the box?\n>\n>Ah! You get a mass increase in the box when adding photons because the\n>invariant mass of the system is increasing.\n\nThat\'s what I thought.\n\n>A single photon is massless.\n\nYup.\n\n>Two photons decaying from an electron at\n>rest, for example, will have an invariant mass equal to the rest mass of\n>the electron.\n\nThis doesn\'t make sense.\n\n>The invariant mass is the magnitude of the sum of the four\n>momenta of all the objects in the system.\n\nTrouble is the 4-momentum of a photon is zero.\n\n--\nOz\nThis post is worth absolutely nothing and is probably fallacious.\n\nBTOPENWORLD address about to cease. DEMON address no longer in use.\n>>Use oz@farmeroz.port995.com<<\nozacoohdb@despammed.com still functions.\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>Creighton Hogg <wchogg@hep.wisc.edu> writes
>>Oz
>> We can increase the mass of the box by shoving in more photons.
>> But from what you have said, the implication is that photons do not
>> interact with the higgs field, hence they are massless. How do we get
>> mass being created by adding massless photons to the box?
>
>Ah! You get a mass increase in the box when adding photons because the
>invariant mass of the system is increasing.
That's what I thought.
>A single photon is massless.
Yup.
>Two photons decaying from an electron at
>rest, for example, will have an invariant mass equal to the rest mass of
>the electron.
This doesn't make sense.
>The invariant mass is the magnitude of the sum of the four
>momenta of all the objects in the system.
Trouble is the 4-momentum of a photon is zero.
--
Oz
This post is worth absolutely nothing and is probably fallacious.
BTOPENWORLD address about to cease. DEMON address no longer in use.
>>Use oz@farmeroz.port995.com<<
ozacoohdb@despammed.com still functions.
Greg Egan
Jul25-04, 09:16 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>\nIn article <OWFs9OFdfUABFwet@farmeroz.port995.com>, Oz\n<oz@farmeroz.port995.com> wrote:\n\n> Creighton Hogg <wchogg@hep.wisc.edu> writes\n>\n[snip]\n>\n> >Two photons decaying from an electron at\n> >rest, for example, will have an invariant mass equal to the rest mass of\n> >the electron.\n>\n> This doesn\'t make sense.\n>\n> >The invariant mass is the magnitude of the sum of the four\n> >momenta of all the objects in the system.\n>\n> Trouble is the 4-momentum of a photon is zero.\n\nNo, the 4-momentum of a photon has a *magnitude* of zero, but with a\nLorentzian metric (unlike a Euclidean one) having a magnitude of zero\ndoesn\'t imply that the vector itself is zero.\n\nIn orthogonal coordinates in spacetime, a vector v with components\n(v_t,v_x,v_y,v_z) has a squared magnitude of:\n\n|v|^2 = - v_t^2 + v_x^2 + v_y^2 + v_z^2\n\nSome people reverse the signs in the RHS; I\'m using the convention such\nthat timelike vectors, e.g. (1,0,0,0) have -ve squared magnitudes, and\nspacelike vectors, e.g. (0,1,0,0), have +ve squared magnitudes.\n\nThe 4-momentum of a photon is always a non-zero vector with a squared\nmagnitude of zero. Such vectors are known as "null" vectors or\n"lightlike" vectors. It\'s easy to construct such vectors: e.g.\n(1,1,0,0) is a null vector for a photon that is moving spatially purely\nin the x-direction in our chosen coordinate system.\n\nNow, an electron at rest in our coordinate system will have a 4-momentum\nof\n\np_e = (m_e, 0, 0, 0)\n\nwhere m_e is the mass of the electron, and we\'re using units such that\nthe speed of light is 1. The squared magnitude of this is:\n\n|p_e|^2 = -m_e^2\n\nA positron that\'s also at rest in our coordinate system would have:\n\np_p = (m_e, 0, 0, 0)\n\nand the total 4-momentum of the system would be:\n\np_total = (2m_e, 0, 0, 0)\n\nwith\n\n|p_total|^2 = -4m_e^2\n\nIf the electron and positron interact to produce two photons heading in\nthe +ve and -ve x directions, those two photons will have four-momenta of:\n\np_{photon 1} = (m_e, m_e, 0, 0)\np_{photon 2} = (m_e, -m_e, 0, 0)\n\nEach photon\'s 4-momentum individually has a magnitude of zero, but:\n\np_e + p_p = p_total = p_{photon 1} + p_{photon 2}\n\ni.e. conservation of total 4-momentum is required.\n\nAnd in general, the total 4-momentum of any large number of photons won\'t\nbe a null vector. (One obvious exception are the photons in a flash of\nlight that are all moving in the same direction.)\n\n\nGreg Egan\n\nEmail address (remove name of animal and add standard punctuation):\ngregegan netspace zebra net au\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 <OWFs9OFdfUABFwet@farmeroz.port995.com>, Oz
<oz@farmeroz.port995.com> wrote:
> Creighton Hogg <wchogg@hep.wisc.edu> writes
>
[snip]
>
> >Two photons decaying from an electron at
> >rest, for example, will have an invariant mass equal to the rest mass of
> >the electron.
>
> This doesn't make sense.
>
> >The invariant mass is the magnitude of the sum of the four
> >momenta of all the objects in the system.
>
> Trouble is the 4-momentum of a photon is zero.
No, the 4-momentum of a photon has a *magnitude* of zero, but with a
Lorentzian metric (unlike a Euclidean one) having a magnitude of zero
doesn't imply that the vector itself is zero.
In orthogonal coordinates in spacetime, a vector v with components
(v_t,v_x,v_y,v_z) has a squared magnitude of:
|v|^2 = - v_t^2 + v_x^2 + v_y^2 + v_z^2
Some people reverse the signs in the RHS; I'm using the convention such
that timelike vectors, e.g. (1,0,0,0) have -ve squared magnitudes, and
spacelike vectors, e.g. (0,1,0,0), have +ve squared magnitudes.
The 4-momentum of a photon is always a non-zero vector with a squared
magnitude of zero. Such vectors are known as "null" vectors or
"lightlike" vectors. It's easy to construct such vectors: e.g.
(1,1,0,0) is a null vector for a photon that is moving spatially purely
in the x-direction in our chosen coordinate system.
Now, an electron at rest in our coordinate system will have a 4-momentum
of
p_e = (m_e, 0, 0, 0)
where m_e is the mass of the electron, and we're using units such that
the speed of light is 1. The squared magnitude of this is:
|p_e|^2 = -m_e^2
A positron that's also at rest in our coordinate system would have:
p_p = (m_e, 0, 0, 0)
and the total 4-momentum of the system would be:
p_{total} = (2m_e, 0, 0, 0)
with
|p_{total}|^2 = -4m_e^2
If the electron and positron interact to produce two photons heading in
the +ve and -ve x directions, those two photons will have four-momenta of:
i.e. conservation of total 4-momentum is required.
And in general, the total 4-momentum of any large number of photons won't
be a null vector. (One obvious exception are the photons in a flash of
light that are all moving in the same direction.)
Greg Egan
Email address (remove name of animal and add standard punctuation):
gregegan netspace zebra net au
Franz Heymann
Jul25-04, 09:16 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\n"Creighton Hogg" <wchogg@hep.wisc.edu> wrote in message\nnews:Pine.LNX.4.44.0407231027310.3660-100000@erodium.hep.wisc.edu...\n>\n\n[snip]\n\n> Ah! You get a mass increase in the box when adding photons because\nthe\n> invariant mass of the system is increasing.\n> A single photon is massless. Two photons decaying from an electron\nat\n> rest, for example, will have an invariant mass equal to the rest\nmass of\n> the electron.\n\nElectrons are totally stable as far as we know, and even if they were\nnot, the decay products would not be two photons. You need a little\nrewrite here. {:-((\nPerhaps you meant to say the decay of positronium at rest?\n\n[snip]\n\nFranz\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>"Creighton Hogg" <wchogg@hep.wisc.edu> wrote in message
news:Pine.LNX.4.44.0407231027310.3660-100000@erodium.hep.wisc.edu...
>
[snip]
> Ah! You get a mass increase in the box when adding photons because
the
> invariant mass of the system is increasing.
> A single photon is massless. Two photons decaying from an electron
at
> rest, for example, will have an invariant mass equal to the rest
mass of
> the electron.
Electrons are totally stable as far as we know, and even if they were
not, the decay products would not be two photons. You need a little
rewrite here. {:-((
Perhaps you meant to say the decay of positronium at rest?
[snip]
Franz
Creighton Hogg
Jul25-04, 09:16 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\n\n\n\n\nOn 23 Jul 2004, Oz wrote:\n\n>\n> Creighton Hogg <wchogg@hep.wisc.edu> writes\n>\n> >>Oz\n> >> We can increase the mass of the box by shoving in more photons.\n> >> But from what you have said, the implication is that photons do not\n> >> interact with the higgs field, hence they are massless. How do we get\n> >> mass being created by adding massless photons to the box?\n> >\n> >Ah! You get a mass increase in the box when adding photons because the\n> >invariant mass of the system is increasing.\n<snip>\n> >The invariant mass is the magnitude of the sum of the four\n> >momenta of all the objects in the system.\n>\n> Trouble is the 4-momentum of a photon is zero.\n\nNo, the *magnitude* of the 4-momentum of a single photon is zero. The\nvector is not zero unless the photon doesn\'t exist.\n\nTake a single electron at rest decaying into two photons, for simplicity\nsake let\'s assume that the photons are emitted along the x-axis.\nThen the 4-momentum of the electron is (m_e,0,0,0)\nAnd the 4-momenta of the two photons is then\n(m_e/2,m_e/2,0,0) and (m_e/2,-m_e/2)\nThe magnitude for an individual photon is ((m_e)^2 - (m_e)^2)/4 = 0\nbut for the *system* it is (m_e)^2 - 0^2 = m_e^2, the same as the\nelectron.\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>On 23 Jul 2004, Oz wrote:
>
> Creighton Hogg <wchogg@hep.wisc.edu> writes
>
> >>Oz
> >> We can increase the mass of the box by shoving in more photons.
> >> But from what you have said, the implication is that photons do not
> >> interact with the higgs field, hence they are massless. How do we get
> >> mass being created by adding massless photons to the box?
> >
> >Ah! You get a mass increase in the box when adding photons because the
> >invariant mass of the system is increasing.
<snip>
> >The invariant mass is the magnitude of the sum of the four
> >momenta of all the objects in the system.
>
> Trouble is the 4-momentum of a photon is zero.
No, the *magnitude* of the 4-momentum of a single photon is zero. The
vector is not zero unless the photon doesn't exist.
Take a single electron at rest decaying into two photons, for simplicity
sake let's assume that the photons are emitted along the x-axis.
Then the 4-momentum of the electron is (m_e,0,0,0)
And the 4-momenta of the two photons is then
(m_e/2,m_e/2,0,0) and (m_e/2,-m_e/2)
The magnitude for an individual photon is ((m_e)^2 - (m_e)^2)/4 =
but for the *system* it is (m_e)^2 - 0^2 = m_e^2, the same as the
electron.
Franz Heymann
Jul25-04, 11:03 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"Oz" <oz@farmeroz.port995.com> wrote in message\nnews:OWFs9OFdfUABFwet@farmeroz.port995.com...\n>\n> Creighton Hogg <wchogg@hep.wisc.edu> writes\n>\n> >>Oz\n> >> We can increase the mass of the box by shoving in more photons.\n> >> But from what you have said, the implication is that photons do\nnot\n> >> interact with the higgs field, hence they are massless. How do we\nget\n> >> mass being created by adding massless photons to the box?\n> >\n> >Ah! You get a mass increase in the box when adding photons because\nthe\n> >invariant mass of the system is increasing.\n>\n> That\'s what I thought.\n>\n> >A single photon is massless.\n>\n> Yup.\n>\n> >Two photons decaying from an electron at\n> >rest, for example, will have an invariant mass equal to the rest\nmass of\n> >the electron.\n>\n> This doesn\'t make sense.\n>\n> >The invariant mass is the magnitude of the sum of the four\n> >momenta of all the objects in the system.\n>\n> Trouble is the 4-momentum of a photon is zero.\n\nNot at all. A photon will have non-zero values for any or all of its\nspace-like components of 4-momentum. Its time-like component will\nhave the value required to make the *magnitude* of the 4-momentum\nequal to zero.\n\nFranz\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>"Oz" <oz@farmeroz.port995.com> wrote in message
news:OWFs9OFdfUABFwet@farmeroz.port995.com...
>
> Creighton Hogg <wchogg@hep.wisc.edu> writes
>
> >>Oz
> >> We can increase the mass of the box by shoving in more photons.
> >> But from what you have said, the implication is that photons do
not
> >> interact with the higgs field, hence they are massless. How do we
get
> >> mass being created by adding massless photons to the box?
> >
> >Ah! You get a mass increase in the box when adding photons because
the
> >invariant mass of the system is increasing.
>
> That's what I thought.
>
> >A single photon is massless.
>
> Yup.
>
> >Two photons decaying from an electron at
> >rest, for example, will have an invariant mass equal to the rest
mass of
> >the electron.
>
> This doesn't make sense.
>
> >The invariant mass is the magnitude of the sum of the four
> >momenta of all the objects in the system.
>
> Trouble is the 4-momentum of a photon is zero.
Not at all. A photon will have non-zero values for any or all of its
space-like components of 4-momentum. Its time-like component will
have the value required to make the *magnitude* of the 4-momentum
equal to zero.
Franz
Creighton Hogg
Jul25-04, 11:09 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\nOn 25 Jul 2004, Franz Heymann wrote:\n\n>\n>\n>\n> "Creighton Hogg" <wchogg@hep.wisc.edu> wrote in message\n> news:Pine.LNX.4.44.0407231027310.3660-100000@erodium.hep.wisc.edu...\n> >\n>\n> [snip]\n>\n> > Ah! You get a mass increase in the box when adding photons because\n> the\n> > invariant mass of the system is increasing.\n> > A single photon is massless. Two photons decaying from an electron\n> at\n> > rest, for example, will have an invariant mass equal to the rest\n> mass of\n> > the electron.\n>\n> Electrons are totally stable as far as we know, and even if they were\n> not, the decay products would not be two photons. You need a little\n> rewrite here. {:-((\n> Perhaps you meant to say the decay of positronium at rest?\n\nD\'oh! In my defense, I was rather out of it last week. What I really\nwas trying to say was Pi_0, which has a 2-photon decay mode. Why I\nsaid electron, I have no idea.\nThanks for noticing that.\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>On 25 Jul 2004, Franz Heymann wrote:
>
>
>
> "Creighton Hogg" <wchogg@hep.wisc.edu> wrote in message
> news:Pine.LNX.4.44.0407231027310.3660-100000@erodium.hep.wisc.edu...
> >
>
> [snip]
>
> > Ah! You get a mass increase in the box when adding photons because
> the
> > invariant mass of the system is increasing.
> > A single photon is massless. Two photons decaying from an electron
> at
> > rest, for example, will have an invariant mass equal to the rest
> mass of
> > the electron.
>
> Electrons are totally stable as far as we know, and even if they were
> not, the decay products would not be two photons. You need a little
> rewrite here. {:-((
> Perhaps you meant to say the decay of positronium at rest?
D'oh! In my defense, I was rather out of it last week. What I really
was trying to say was \Pi_0, which has a 2-photon decay mode. Why I
said electron, I have no idea.
Thanks for noticing that.
Oz
Jul26-04, 04:41 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\nCreighton Hogg <wchogg@hep.wisc.edu> writes\n>Take a single electron at rest decaying into two photons, for simplicity\n>sake let\'s assume that the photons are emitted along the x-axis.\n>Then the 4-momentum of the electron is (m_e,0,0,0)\n>And the 4-momenta of the two photons is then\n>(m_e/2,m_e/2,0,0) and (m_e/2,-m_e/2)\n>The magnitude for an individual photon is ((m_e)^2 - (m_e)^2)/4 = 0\n>but for the *system* it is (m_e)^2 - 0^2 = m_e^2, the same as the\n>electron.\n\nOf course. I should have spotted that and not asked.\nThe fact that an electron cannot do this is not relevant to the\nexplanation.\n\nSo am I to take it that the effective mass of this box of photons is\nindeed just the sum of the energies of the photons. Surely that must be\nso.\n\nNow in GR we need to take the pressure into account as well to find the\ngravitational bending. Clearly there is a pressure inside the box due to\nthe gas of photons.\n\nNow, my GR is pretty hazy, and this is an odd example anyway. Baez\ndescribes G=T as giving the evolution of a ball of dust. This doesn\'t\nseem to take pressure into account (OK he simplified for dumbos like\nme), but if these is a little more depth to his description than appears\nat first sight a pressure would suggest a propensity to expand, and so\nshould be a negative term. Trouble is I have forgotten.\n\nHmm.... But whilst there may be pressure inside the box, there is none\noutside. Does this mean that the spacetime bending outside the box will\nonly take into account the internal *energy*, and the momenta simply\ncancel out?\n\nSurely that must be so, because externally there should be no difference\nbetween a box full of electrons and a boxfull of equivalent photons as\nper your example above. How would we know otherwise?\n\nOK, that\'s been clarifying (assuming you reply in the affirmative).\n\n===================\n\nOk, now back to my original question....\n\nIf higgs doesn\'t couple to photons, where does the mass of the boxfull\nof photons derive from?\n\nOne is tempted to surmise that it actually couples to the energy and the\nmomentum, but because these are precisely equal and opposite the net\neffect for a free photon is zero. Unfortunately there is no obvious\nreason why a boxfull of them should have mass, taken individually.\n\nOh, of course, there is an instant obvious answer.\n\nIts non-trivial to fill a box with photons. You can *only* do it if the\nwavelength is an integral number of the box dimensions (give half a\nlambda). That is they must be standing waves in the box. Their net\nmomentum will this be zero. That\'s not quite what I mean. The momentum\nof the EM standing wave will be zero (since I don\'t believe in photons).\nThis wave will by definition have energy, but zero momentum.\n\nOoops, that does answer the question, but doesn\'t say that photons don\'t\ncouple with the higgs field. On the contrary it says it does.\n\nWhich means it must also couple with momentum (negatively, if you see\nwhat I mean).\n\nHmmm.\n\nI suppose that must be so. If higgs didn\'t couple with momentum then we\nwould have some oddities under boosts. By coupling to both, boosts\ncancel out and the net result depends only on rest mass (really rest\nenergy), and that\'s invariant. I don\'t quite know why, but this is\nslightly uncomfortable making. Obviously it must all fall out in the\nwash or better people than I would have spotted a flaw.\n\nThere\'s something else odd about the higgs field. I have a \'permanent\nquestion\' as to why two masses with a wavelength of lambda should, when\ntaken together (that is from a long distance) have a wavelength of\nlambda/2. Its hard in a linear system, to get two wavelengths to make\none at twice the frequency, much easier in a non-linear system. I rather\nhoped solitons would fix this, but it comes in here as well.\n\nI need to ponder this before asking more questions.\n\n--\nOz\nThis post is worth absolutely nothing and is probably fallacious.\n\nBTOPENWORLD address about to cease. DEMON address no longer in use.\n>>Use oz@farmeroz.port995.com<<\nozacoohdb@despammed.com still functions.\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>Creighton Hogg <wchogg@hep.wisc.edu> writes
>Take a single electron at rest decaying into two photons, for simplicity
>sake let's assume that the photons are emitted along the x-axis.
>Then the 4-momentum of the electron is (m_e,0,0,0)
>And the 4-momenta of the two photons is then
>(m_e/2,m_e/2,0,0) and (m_e/2,-m_e/2)
>The magnitude for an individual photon is ((m_e)^2 - (m_e)^2)/4 =
>but for the *system* it is (m_e)^2 - 0^2 = m_e^2, the same as the
>electron.
Of course. I should have spotted that and not asked.
The fact that an electron cannot do this is not relevant to the
explanation.
So am I to take it that the effective mass of this box of photons is
indeed just the sum of the energies of the photons. Surely that must be
so.
Now in GR we need to take the pressure into account as well to find the
gravitational bending. Clearly there is a pressure inside the box due to
the gas of photons.
Now, my GR is pretty hazy, and this is an odd example anyway. Baez
describes G=T as giving the evolution of a ball of dust. This doesn't
seem to take pressure into account (OK he simplified for dumbos like
me), but if these is a little more depth to his description than appears
at first sight a pressure would suggest a propensity to expand, and so
should be a negative term. Trouble is I have forgotten.
Hmm.... But whilst there may be pressure inside the box, there is none
outside. Does this mean that the spacetime bending outside the box will
only take into account the internal *energy*, and the momenta simply
cancel out?
Surely that must be so, because externally there should be no difference
between a box full of electrons and a boxfull of equivalent photons as
per your example above. How would we know otherwise?
OK, that's been clarifying (assuming you reply in the affirmative).
===================
Ok, now back to my original question....
If higgs doesn't couple to photons, where does the mass of the boxfull
of photons derive from?
One is tempted to surmise that it actually couples to the energy and the
momentum, but because these are precisely equal and opposite the net
effect for a free photon is zero. Unfortunately there is no obvious
reason why a boxfull of them should have mass, taken individually.
Oh, of course, there is an instant obvious answer.
Its non-trivial to fill a box with photons. You can *only* do it if the
wavelength is an integral number of the box dimensions (give half a
\lambda). That is they must be standing waves in the box. Their net
momentum will this be zero. That's not quite what I mean. The momentum
of the EM standing wave will be zero (since I don't believe in photons).
This wave will by definition have energy, but zero momentum.
Ooops, that does answer the question, but doesn't say that photons don't
couple with the higgs field. On the contrary it says it does.
Which means it must also couple with momentum (negatively, if you see
what I mean).
Hmmm.
I suppose that must be so. If higgs didn't couple with momentum then we
would have some oddities under boosts. By coupling to both, boosts
cancel out and the net result depends only on rest mass (really rest
energy), and that's invariant. I don't quite know why, but this is
slightly uncomfortable making. Obviously it must all fall out in the
wash or better people than I would have spotted a flaw.
There's something else odd about the higgs field. I have a 'permanent
question' as to why two masses with a wavelength of \lambda should, when
taken together (that is from a long distance) have a wavelength of
\lambda/2. Its hard in a linear system, to get two wavelengths to make
one at twice the frequency, much easier in a non-linear system. I rather
hoped solitons would fix this, but it comes in here as well.
I need to ponder this before asking more questions.
--
Oz
This post is worth absolutely nothing and is probably fallacious.
BTOPENWORLD address about to cease. DEMON address no longer in use.
>>Use oz@farmeroz.port995.com<<
ozacoohdb@despammed.com still functions.
Creighton Hogg
Jul26-04, 07:12 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>\nOn 26 Jul 2004, Oz wrote:\n\n>\n>\n>\n> Creighton Hogg <wchogg@hep.wisc.edu> writes\n> >Take a single electron at rest decaying into two photons, for simplicity\n> >sake let\'s assume that the photons are emitted along the x-axis.\n> >Then the 4-momentum of the electron is (m_e,0,0,0)\n> >And the 4-momenta of the two photons is then\n> >(m_e/2,m_e/2,0,0) and (m_e/2,-m_e/2)\n> >The magnitude for an individual photon is ((m_e)^2 - (m_e)^2)/4 = 0\n> >but for the *system* it is (m_e)^2 - 0^2 = m_e^2, the same as the\n> >electron.\n>\n> Of course. I should have spotted that and not asked.\n> The fact that an electron cannot do this is not relevant to the\n> explanation.\n\nIndeed. As I told Franz, I was rather tired when I wrote this on friday.\nI meant Pi_0, but once I said electron once I didn\'t think about it till\nFranz called me on it.\n\n> ===================\n>\n> Ok, now back to my original question....\n>\n> If higgs doesn\'t couple to photons, where does the mass of the boxfull\n> of photons derive from?\n\nI can\'t think of a good way to say it other than, "because that\'s how\ninvariant mass works". Hopefully someone else could come up with\nsomething more enlightening.\n\n> One is tempted to surmise that it actually couples to the energy and the\n> momentum, but because these are precisely equal and opposite the net\n> effect for a free photon is zero. Unfortunately there is no obvious\n> reason why a boxfull of them should have mass, taken individually.\n>\n> Oh, of course, there is an instant obvious answer.\n>\n> Its non-trivial to fill a box with photons. You can *only* do it if the\n> wavelength is an integral number of the box dimensions (give half a\n> lambda). That is they must be standing waves in the box. Their net\n> momentum will this be zero. That\'s not quite what I mean. The momentum\n> of the EM standing wave will be zero (since I don\'t believe in photons).\n> This wave will by definition have energy, but zero momentum.\n>\n> Ooops, that does answer the question, but doesn\'t say that photons don\'t\n> couple with the higgs field. On the contrary it says it does.\n\nNuh uh. This analysis has nothing to do with the coupling of the Higgs\nwith the photon. Do the symmetry breaking yourself and you\'ll find that\nthere\'s three bosons that couple to the Higgs and one that doesn\'t. The\none that doesn\'t is the photon. The only interaction between Higgs and\nphotons is thru higher order terms such as triangle diagrams. These are\nof course not tree level diagrams and have a suppressed cross-section.\nHiggs->photon photon decays are interesting to phenomenology and are a\npossible way to detect the Higgs, since the signal is so clean and\ndistinctive.\n\n> Which means it must also couple with momentum (negatively, if you see\n> what I mean).\n>\n> Hmmm.\n>\n> I suppose that must be so. If higgs didn\'t couple with momentum then we\n> would have some oddities under boosts. By coupling to both, boosts\n> cancel out and the net result depends only on rest mass (really rest\n> energy), and that\'s invariant. I don\'t quite know why, but this is\n> slightly uncomfortable making. Obviously it must all fall out in the\n> wash or better people than I would have spotted a flaw.\n\nIt doesn\'t really work this way though. The only thing that,\nhypothetically, would couple to momentum and energy is the graviton.\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>On 26 Jul 2004, Oz wrote:
>
>
>
> Creighton Hogg <wchogg@hep.wisc.edu> writes
> >Take a single electron at rest decaying into two photons, for simplicity
> >sake let's assume that the photons are emitted along the x-axis.
> >Then the 4-momentum of the electron is (m_e,0,0,0)
> >And the 4-momenta of the two photons is then
> >(m_e/2,m_e/2,0,0) and (m_e/2,-m_e/2)
> >The magnitude for an individual photon is ((m_e)^2 - (m_e)^2)/4 =
> >but for the *system* it is (m_e)^2 - 0^2 = m_e^2, the same as the
> >electron.
>
> Of course. I should have spotted that and not asked.
> The fact that an electron cannot do this is not relevant to the
> explanation.
Indeed. As I told Franz, I was rather tired when I wrote this on friday.
I meant \Pi_0, but once I said electron once I didn't think about it till
Franz called me on it.
> ===================
>
> Ok, now back to my original question....
>
> If higgs doesn't couple to photons, where does the mass of the boxfull
> of photons derive from?
I can't think of a good way to say it other than, "because that's how
invariant mass works". Hopefully someone else could come up with
something more enlightening.
> One is tempted to surmise that it actually couples to the energy and the
> momentum, but because these are precisely equal and opposite the net
> effect for a free photon is zero. Unfortunately there is no obvious
> reason why a boxfull of them should have mass, taken individually.
>
> Oh, of course, there is an instant obvious answer.
>
> Its non-trivial to fill a box with photons. You can *only* do it if the
> wavelength is an integral number of the box dimensions (give half a
> \lambda). That is they must be standing waves in the box. Their net
> momentum will this be zero. That's not quite what I mean. The momentum
> of the EM standing wave will be zero (since I don't believe in photons).
> This wave will by definition have energy, but zero momentum.
>
> Ooops, that does answer the question, but doesn't say that photons don't
> couple with the higgs field. On the contrary it says it does.
Nuh uh. This analysis has nothing to do with the coupling of the Higgs
with the photon. Do the symmetry breaking yourself and you'll find that
there's three bosons that couple to the Higgs and one that doesn't. The
one that doesn't is the photon. The only interaction between Higgs and
photons is thru higher order terms such as triangle diagrams. These are
of course not tree level diagrams and have a suppressed cross-section.
Higgs->photon photon decays are interesting to phenomenology and are a
possible way to detect the Higgs, since the signal is so clean and
distinctive.
> Which means it must also couple with momentum (negatively, if you see
> what I mean).
>
> Hmmm.
>
> I suppose that must be so. If higgs didn't couple with momentum then we
> would have some oddities under boosts. By coupling to both, boosts
> cancel out and the net result depends only on rest mass (really rest
> energy), and that's invariant. I don't quite know why, but this is
> slightly uncomfortable making. Obviously it must all fall out in the
> wash or better people than I would have spotted a flaw.
It doesn't really work this way though. The only thing that,
hypothetically, would couple to momentum and energy is the graviton.
Thomas Dent
Jul26-04, 11:32 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\n\nOz <oz@farmeroz.port995.com> wrote\n\n> How are we to see a box full of photons?\n>\n> We can increase the mass of the box by shoving in more photons.\n> But from what you have said, the implication is that photons do not\n> interact with the higgs field, hence they are massless. How do we get\n> mass being created by adding massless photons to the box?\n\nThis problem is due to two (or many) different uses of the word\n"mass". The mass that the Higgs is supposed to produce is the mass\nappearing in the Lagrangian and in the wave equation, which is a\ntotally invariant numerical quantity which we can call m. For a\nphysical (detectable) particle we have m^2 = E^2 - p.p (forgetting\nabout factors of c). For a single particle, this turns out to be the\nsame as what relativists call the "invariant mass".\n\nNow, there is the "mass" that is talked about by relativists as the\n0-component of the energy-momentum 4-vector. That quantity p^0 will\nchange as a massive (m/=0) particle is accelerated and is non-zero\neven for particles with zero mass (m=0).\n\nAnd there is the so-called "invariant mass" for systems of many\nparticles, which is just equal to the sum of p^0 in the centre-of-mass\nframe (I think).\n\nThe only thing the Higgs does for you is give you the mass m of\nelementary particles. The other "masses" are a combination of m and\nother kinematic variables.\n\nComposite particles such as the proton can have a large mass even\nthough they may be made of massless (m=0) constituents (gluons and\nquarks). (The quarks in question do have small masses m_q, but much\nsmaller than the proton mass scale.) Then the mass of the proton is\nmade up of the energy of the constituent fields which can be thought\nof as constituents moving rapidly inside the proton. Roughly, the\nproton mass can be thought of as the "invariant mass" for all the\ngluons and quarks inside it. (It is more complicated than that because\nthere is also a quark condensate in the vacuum which gives the proton\nmore energy even though it cannot be seen as the kinetic energy of\nanything...)\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>Oz <oz@farmeroz.port995.com> wrote
> How are we to see a box full of photons?
>
> We can increase the mass of the box by shoving in more photons.
> But from what you have said, the implication is that photons do not
> interact with the higgs field, hence they are massless. How do we get
> mass being created by adding massless photons to the box?
This problem is due to two (or many) different uses of the word
"mass". The mass that the Higgs is supposed to produce is the mass
appearing in the Lagrangian and in the wave equation, which is a
totally invariant numerical quantity which we can call m. For a
physical (detectable) particle we have m^2 = E^2 - p.p (forgetting
about factors of c). For a single particle, this turns out to be the
same as what relativists call the "invariant mass".
Now, there is the "mass" that is talked about by relativists as the
0-component of the energy-momentum 4-vector. That quantity p^0 will
change as a massive (m/=0) particle is accelerated and is non-zero
even for particles with zero mass (m=0).
And there is the so-called "invariant mass" for systems of many
particles, which is just equal to the sum of p^0 in the centre-of-mass
frame (I think).
The only thing the Higgs does for you is give you the mass m of
elementary particles. The other "masses" are a combination of m and
other kinematic variables.
Composite particles such as the proton can have a large mass even
though they may be made of massless (m=0) constituents (gluons and
quarks). (The quarks in question do have small masses m_q, but much
smaller than the proton mass scale.) Then the mass of the proton is
made up of the energy of the constituent fields which can be thought
of as constituents moving rapidly inside the proton. Roughly, the
proton mass can be thought of as the "invariant mass" for all the
gluons and quarks inside it. (It is more complicated than that because
there is also a quark condensate in the vacuum which gives the proton
more energy even though it cannot be seen as the kinetic energy of
anything...)
Oz
Jul26-04, 11:32 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\nCreighton Hogg <wchogg@hep.wisc.edu> writes\n>\n>On 26 Jul 2004, Oz wrote:\n>\n>> Ok, now back to my original question....\n>>\n>> If higgs doesn\'t couple to photons, where does the mass of the boxfull\n>> of photons derive from?\n>\n>I can\'t think of a good way to say it other than, "because that\'s how\n>invariant mass works". Hopefully someone else could come up with\n>something more enlightening.\n\nI\'m perfectly happy about a boxfull of photons having a (rest) mass.\nI\'m unclear about how the higgs field generates this mass though.\n\n>> Its non-trivial to fill a box with photons. You can *only* do it if the\n>> wavelength is an integral number of the box dimensions (give half a\n>> lambda). That is they must be standing waves in the box. Their net\n>> momentum will this be zero. That\'s not quite what I mean. The momentum\n>> of the EM standing wave will be zero (since I don\'t believe in photons).\n>> This wave will by definition have energy, but zero momentum.\n>>\n>> Ooops, that does answer the question, but doesn\'t say that photons don\'t\n>> couple with the higgs field. On the contrary it says it does.\n>\n>Nuh uh. This analysis has nothing to do with the coupling of the Higgs\n>with the photon.\n\nHow else to generate mass though?\n\n>Do the symmetry breaking yourself and you\'ll find that\n>there\'s three bosons that couple to the Higgs and one that doesn\'t.\n\nI\'m sure that\'s true, everyone says so.\n\n>The\n>one that doesn\'t is the photon. The only interaction between Higgs and\n>photons is thru higher order terms such as triangle diagrams. These are\n>of course not tree level diagrams and have a suppressed cross-section.\n\nEr, um ....\n\nYou talking about the equivalent of virtual electron pairs?\nWould this give a high enough amplitude to generate that much mass?\nSurely not as we would see photons as having more restmass?\n\n>Higgs->photon photon decays are interesting to phenomenology and are a\n>possible way to detect the Higgs, since the signal is so clean and\n>distinctive.\n\nSome photons!\n\n>> I suppose that must be so. If higgs didn\'t couple with momentum then we\n>> would have some oddities under boosts. By coupling to both, boosts\n>> cancel out and the net result depends only on rest mass (really rest\n>> energy), and that\'s invariant. I don\'t quite know why, but this is\n>> slightly uncomfortable making. Obviously it must all fall out in the\n>> wash or better people than I would have spotted a flaw.\n>\n>It doesn\'t really work this way though. The only thing that,\n>hypothetically, would couple to momentum and energy is the graviton.\n\nThe higgs is a scalar field as I have been told.\nHmm, what does that mean?\n\nI guess it means that its independent of any boosts.\nBut I find it hard to see why it doesn\'t in some sense define a frame.\nIn particular the rest frame for any given particle.\nIts quite hard to visualise it, I will need to think a bit.\n\nAt first glance it would seem to embody some concept that groups\nparticles so a group looks like a single particle from a distance. No, I\nhaven\'t got it clearly.\n\n--\nOz\nThis post is worth absolutely nothing and is probably fallacious.\n\nBTOPENWORLD address about to cease. DEMON address no longer in use.\n>>Use oz@farmeroz.port995.com<<\nozacoohdb@despammed.com still functions.\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>Creighton Hogg <wchogg@hep.wisc.edu> writes
>
>On 26 Jul 2004, Oz wrote:
>
>> Ok, now back to my original question....
>>
>> If higgs doesn't couple to photons, where does the mass of the boxfull
>> of photons derive from?
>
>I can't think of a good way to say it other than, "because that's how
>invariant mass works". Hopefully someone else could come up with
>something more enlightening.
I'm perfectly happy about a boxfull of photons having a (rest) mass.
I'm unclear about how the higgs field generates this mass though.
>> Its non-trivial to fill a box with photons. You can *only* do it if the
>> wavelength is an integral number of the box dimensions (give half a
>> \lambda). That is they must be standing waves in the box. Their net
>> momentum will this be zero. That's not quite what I mean. The momentum
>> of the EM standing wave will be zero (since I don't believe in photons).
>> This wave will by definition have energy, but zero momentum.
>>
>> Ooops, that does answer the question, but doesn't say that photons don't
>> couple with the higgs field. On the contrary it says it does.
>
>Nuh uh. This analysis has nothing to do with the coupling of the Higgs
>with the photon.
How else to generate mass though?
>Do the symmetry breaking yourself and you'll find that
>there's three bosons that couple to the Higgs and one that doesn't.
I'm sure that's true, everyone says so.
>The
>one that doesn't is the photon. The only interaction between Higgs and
>photons is thru higher order terms such as triangle diagrams. These are
>of course not tree level diagrams and have a suppressed cross-section.
Er, um ....
You talking about the equivalent of virtual electron pairs?
Would this give a high enough amplitude to generate that much mass?
Surely not as we would see photons as having more restmass?
>Higgs->photon photon decays are interesting to phenomenology and are a
>possible way to detect the Higgs, since the signal is so clean and
>distinctive.
Some photons!
>> I suppose that must be so. If higgs didn't couple with momentum then we
>> would have some oddities under boosts. By coupling to both, boosts
>> cancel out and the net result depends only on rest mass (really rest
>> energy), and that's invariant. I don't quite know why, but this is
>> slightly uncomfortable making. Obviously it must all fall out in the
>> wash or better people than I would have spotted a flaw.
>
>It doesn't really work this way though. The only thing that,
>hypothetically, would couple to momentum and energy is the graviton.
The higgs is a scalar field as I have been told.
Hmm, what does that mean?
I guess it means that its independent of any boosts.
But I find it hard to see why it doesn't in some sense define a frame.
In particular the rest frame for any given particle.
Its quite hard to visualise it, I will need to think a bit.
At first glance it would seem to embody some concept that groups
particles so a group looks like a single particle from a distance. No, I
haven't got it clearly.
--
Oz
This post is worth absolutely nothing and is probably fallacious.
BTOPENWORLD address about to cease. DEMON address no longer in use.
>>Use oz@farmeroz.port995.com<<
ozacoohdb@despammed.com still functions.
Creighton Hogg
Jul26-04, 01:07 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>\n\n\nOn 26 Jul 2004, Oz wrote:\n\n>\n> Creighton Hogg <wchogg@hep.wisc.edu> writes\n> >\n> >On 26 Jul 2004, Oz wrote:\n> >\n> >> Ok, now back to my original question....\n> >>\n> >> If higgs doesn\'t couple to photons, where does the mass of the boxfull\n> >> of photons derive from?\n> >\n> >I can\'t think of a good way to say it other than, "because that\'s how\n> >invariant mass works". Hopefully someone else could come up with\n> >something more enlightening.\n>\n> I\'m perfectly happy about a boxfull of photons having a (rest) mass.\n> I\'m unclear about how the higgs field generates this mass though.\n\nIt doesn\'t generate the mass.\n\n> >> Its non-trivial to fill a box with photons. You can *only* do it if the\n> >> wavelength is an integral number of the box dimensions (give half a\n> >> lambda). That is they must be standing waves in the box. Their net\n> >> momentum will this be zero. That\'s not quite what I mean. The momentum\n> >> of the EM standing wave will be zero (since I don\'t believe in photons).\n> >> This wave will by definition have energy, but zero momentum.\n> >>\n> >> Ooops, that does answer the question, but doesn\'t say that photons don\'t\n> >> couple with the higgs field. On the contrary it says it does.\n> >\n> >Nuh uh. This analysis has nothing to do with the coupling of the Higgs\n> >with the photon.\n>\n> How else to generate mass though?\n\nThere\'s more to mass beyond coupling to the Higgs vacuum expectation\nvalue. That\'s not the only way.\n\n> >Do the symmetry breaking yourself and you\'ll find that\n> >there\'s three bosons that couple to the Higgs and one that doesn\'t.\n>\n> I\'m sure that\'s true, everyone says so.\n>\n> >The\n> >one that doesn\'t is the photon. The only interaction between Higgs and\n> >photons is thru higher order terms such as triangle diagrams. These are\n> >of course not tree level diagrams and have a suppressed cross-section.\n>\n> Er, um ....\n>\n> You talking about the equivalent of virtual electron pairs?\n> Would this give a high enough amplitude to generate that much mass?\n> Surely not as we would see photons as having more restmass?\n\nNope. It\'s a different type of coupling. The massive electroweak bosons\ngain an effective mass because of their coupling to the Higgs, which then\nacquires a VEV. This supplies a mass term without making the theory\nnon-renormalizable.\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>On 26 Jul 2004, Oz wrote:
>
> Creighton Hogg <wchogg@hep.wisc.edu> writes
> >
> >On 26 Jul 2004, Oz wrote:
> >
> >> Ok, now back to my original question....
> >>
> >> If higgs doesn't couple to photons, where does the mass of the boxfull
> >> of photons derive from?
> >
> >I can't think of a good way to say it other than, "because that's how
> >invariant mass works". Hopefully someone else could come up with
> >something more enlightening.
>
> I'm perfectly happy about a boxfull of photons having a (rest) mass.
> I'm unclear about how the higgs field generates this mass though.
It doesn't generate the mass.
> >> Its non-trivial to fill a box with photons. You can *only* do it if the
> >> wavelength is an integral number of the box dimensions (give half a
> >> \lambda). That is they must be standing waves in the box. Their net
> >> momentum will this be zero. That's not quite what I mean. The momentum
> >> of the EM standing wave will be zero (since I don't believe in photons).
> >> This wave will by definition have energy, but zero momentum.
> >>
> >> Ooops, that does answer the question, but doesn't say that photons don't
> >> couple with the higgs field. On the contrary it says it does.
> >
> >Nuh uh. This analysis has nothing to do with the coupling of the Higgs
> >with the photon.
>
> How else to generate mass though?
There's more to mass beyond coupling to the Higgs vacuum expectation
value. That's not the only way.
> >Do the symmetry breaking yourself and you'll find that
> >there's three bosons that couple to the Higgs and one that doesn't.
>
> I'm sure that's true, everyone says so.
>
> >The
> >one that doesn't is the photon. The only interaction between Higgs and
> >photons is thru higher order terms such as triangle diagrams. These are
> >of course not tree level diagrams and have a suppressed cross-section.
>
> Er, um ....
>
> You talking about the equivalent of virtual electron pairs?
> Would this give a high enough amplitude to generate that much mass?
> Surely not as we would see photons as having more restmass?
Nope. It's a different type of coupling. The massive electroweak bosons
gain an effective mass because of their coupling to the Higgs, which then
acquires a VEV. This supplies a mass term without making the theory
non-renormalizable.
Oz
Jul27-04, 08:51 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\nCreighton Hogg <wchogg@hep.wisc.edu> writes\n\n>There\'s more to mass beyond coupling to the Higgs vacuum expectation\n>value. That\'s not the only way.\n\nOh.\nThat blows my chain of thought right out of the water then!\n\n>Nope. It\'s a different type of coupling. The massive electroweak bosons\n>gain an effective mass because of their coupling to the Higgs, which then\n>acquires a VEV. This supplies a mass term without making the theory\n>non-renormalizable.\n\nOK, so what you are saying is that these three massless particles gain\nmass by coupling to the higgs field. Nothing else.\n\nSo it doesn\'t explain (say) masses of electrons or quarks.\n\n--\nOz\nThis post is worth absolutely nothing and is probably fallacious.\n\nBTOPENWORLD address about to cease. DEMON address no longer in use.\n>>Use oz@farmeroz.port995.com<<\nozacoohdb@despammed.com still functions.\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>Creighton Hogg <wchogg@hep.wisc.edu> writes
>There's more to mass beyond coupling to the Higgs vacuum expectation
>value. That's not the only way.
Oh.
That blows my chain of thought right out of the water then!
>Nope. It's a different type of coupling. The massive electroweak bosons
>gain an effective mass because of their coupling to the Higgs, which then
>acquires a VEV. This supplies a mass term without making the theory
>non-renormalizable.
OK, so what you are saying is that these three massless particles gain
mass by coupling to the higgs field. Nothing else.
So it doesn't explain (say) masses of electrons or quarks.
--
Oz
This post is worth absolutely nothing and is probably fallacious.
BTOPENWORLD address about to cease. DEMON address no longer in use.
>>Use oz@farmeroz.port995.com<<
ozacoohdb@despammed.com still functions.
zefram_c
Jul27-04, 08:52 AM
Trouble is the 4-momentum of a photon is zero.
Ah! It is the magnitude of the 4-momentum |p|^2 = (p^0)^2 - \sum (p^i)^2 (i=1..3) that is zero for a photon. If the 4-p is zero, there is no particle :smile:
To exemplify: two photons may have 4-momenta of (1,1,0,0) and (1,-1,0,0). When you add their 4-momenta together, you get (2,0,0,0). This could represent a particle at rest with invariant mass 2. (eg. a neutral pion). You could also write that sum as (2,0,0,0) = (1,0.5,0,0) + (1,-0.5,0,0). This represents two particles of invariant mass m^2 = 1^2 - 0.5^2 = 0.75 - two electrons going back to back.
Franz Heymann
Jul27-04, 01:50 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>\n\n\n"Thomas Dent" <tdent@auth.gr> wrote in message\nnews:cb504c2c.0407260634.31c5406f@posting.google.com...\n>\n>\n>\n>\n> Oz <oz@farmeroz.port995.com> wrote\n>\n> > How are we to see a box full of photons?\n> >\n> > We can increase the mass of the box by shoving in more photons.\n> > But from what you have said, the implication is that photons do\nnot\n> > interact with the higgs field, hence they are massless. How do we\nget\n> > mass being created by adding massless photons to the box?\n>\n> This problem is due to two (or many) different uses of the word\n> "mass". The mass that the Higgs is supposed to produce is the mass\n> appearing in the Lagrangian and in the wave equation, which is a\n> totally invariant numerical quantity which we can call m. For a\n> physical (detectable) particle we have m^2 = E^2 - p.p (forgetting\n> about factors of c). For a single particle, this turns out to be the\n> same as what relativists call the "invariant mass".\n>\n> Now, there is the "mass" that is talked about by relativists as the\n> 0-component of the energy-momentum 4-vector. That quantity p^0 will\n> change as a massive (m/=0) particle is accelerated and is non-zero\n> even for particles with zero mass (m=0).\n>\n> And there is the so-called "invariant mass" for systems of many\n> particles, which is just equal to the sum of p^0 in the\ncentre-of-mass\n> frame (I think).\n\nNo. The invariant mass of a system of particles is the energy of the\nsystem in that system of coordinates in which the resultant momentum\nis zero.\n\n[snip]\n\nFranz\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>"Thomas Dent" <tdent@auth.gr> wrote in message
news:cb504c2c.0407260634.31c5406f@posting.google.com...
>
>
>
>
> Oz <oz@farmeroz.port995.com> wrote
>
> > How are we to see a box full of photons?
> >
> > We can increase the mass of the box by shoving in more photons.
> > But from what you have said, the implication is that photons do
not
> > interact with the higgs field, hence they are massless. How do we
get
> > mass being created by adding massless photons to the box?
>
> This problem is due to two (or many) different uses of the word
> "mass". The mass that the Higgs is supposed to produce is the mass
> appearing in the Lagrangian and in the wave equation, which is a
> totally invariant numerical quantity which we can call m. For a
> physical (detectable) particle we have m^2 = E^2 - p.p (forgetting
> about factors of c). For a single particle, this turns out to be the
> same as what relativists call the "invariant mass".
>
> Now, there is the "mass" that is talked about by relativists as the
> 0-component of the energy-momentum 4-vector. That quantity p^0 will
> change as a massive (m/=0) particle is accelerated and is non-zero
> even for particles with zero mass (m=0).
>
> And there is the so-called "invariant mass" for systems of many
> particles, which is just equal to the sum of p^0 in the
centre-of-mass
> frame (I think).
No. The invariant mass of a system of particles is the energy of the
system in that system of coordinates in which the resultant momentum
is zero.
[snip]
Franz
Franz Heymann
Jul27-04, 01:50 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>\n\n\n"Oz" <oz@farmeroz.port995.com> wrote in message\nnews:YXv5dgTxiQBBFw1a@farmeroz.port995.com...\n\n[snip]\n\n> I\'m perfectly happy about a boxfull of photons having a (rest) mass.\n> I\'m unclear about how the higgs field generates this mass though.\n\nSince it is possible to calculate the invariant mass of a box of\nphotons without making use of the Higgs mechanism, it is clear that\nthat mass is not generated by that mechanism.\n\n[snip]\n\nFranz\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>"Oz" <oz@farmeroz.port995.com> wrote in message
news:YXv5dgTxiQBBFw1a@farmeroz.port995.com...
[snip]
> I'm perfectly happy about a boxfull of photons having a (rest) mass.
> I'm unclear about how the higgs field generates this mass though.
Since it is possible to calculate the invariant mass of a box of
photons without making use of the Higgs mechanism, it is clear that
that mass is not generated by that mechanism.
[snip]
Franz
Oz
Jul27-04, 01:50 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>\n\n\nThomas Dent <tdent@auth.gr> writes\n\n>This problem is due to two (or many) different uses of the word\n>"mass". The mass that the Higgs is supposed to produce is the mass\n>appearing in the Lagrangian and in the wave equation, which is a\n>totally invariant numerical quantity which we can call m. For a\n>physical (detectable) particle we have m^2 = E^2 - p.p (forgetting\n>about factors of c). For a single particle, this turns out to be the\n>same as what relativists call the "invariant mass".\n\nI think \'rest mass\' is a better description (ie p=0).\n\n>Now, there is the "mass" that is talked about by relativists as the\n>0-component of the energy-momentum 4-vector. That quantity p^0 will\n>change as a massive (m/=0) particle is accelerated and is non-zero\n>even for particles with zero mass (m=0).\n\nMomentum in the time direction.\nI think \'relativistic mass\' would be a good description for that.\n\n>And there is the so-called "invariant mass" for systems of many\n>particles, which is just equal to the sum of p^0 in the centre-of-mass\n>frame (I think).\n\nHmmm....\nThat\'s really the \'particles in a box mass\' about which we are\ndiscussing.\n\n>The only thing the Higgs does for you is give you the mass m of\n>elementary particles. The other "masses" are a combination of m and\n>other kinematic variables.\n\nI am uncomfortable about this right now. I really don\'t want two\nseparate mechanisms that generate mass. I\'m pretty happy with mass as\ngenerated kinematically, and one might plausibly argue that all mass is\ngenerated this way. Its unclear to me how a coupling to a field should\ngenerate mass, hence my stumbling efforts to clarify this.\n\n>Composite particles such as the proton can have a large mass even\n>though they may be made of massless (m=0) constituents (gluons and\n>quarks). (The quarks in question do have small masses m_q, but much\n>smaller than the proton mass scale.) Then the mass of the proton is\n>made up of the energy of the constituent fields which can be thought\n>of as constituents moving rapidly inside the proton. Roughly, the\n>proton mass can be thought of as the "invariant mass" for all the\n>gluons and quarks inside it. (It is more complicated than that because\n>there is also a quark condensate in the vacuum which gives the proton\n>more energy even though it cannot be seen as the kinetic energy of\n>anything...)\n\nThat\'s fine. I have no problems with massless particles generating mass.\nThis is simply another way of saying \'all mass is kinematic mass\'.\n\n==============\nAlfred Einstead <whopkins@csd.uwm.edu> wrote\n>At the semi-classical level, the Dirac fermion would be coiling\n>in a helical worldline of about this radius (for electrons) at\n>light speed.\n>If you plug the velocity operator v = alpha c into the Heisenberg\n>equation of motion\n> i h-har d()/dt = [(), H]\n>using the Dirac Hamiltonian\n> H = alpha.pc + beta mc^2\n>the picture that emerges is of a worldline with the following\n>properties:\n\n> (1) It\'s helical with mean free motion at a velocity\n> v = p c^2/E (in a momentum eigenstate), parallel\n> to p (opposite to p for negative energy states).\n> (2) The velocity, counting the helical motion is just\n> light speed.\n> (3) The radius of the circular part of the motion is\n> r = L (1 - (v/c)^2)\n> where L is the Compton wavelength.\n>As it goes faster, the helical motion uncoils and more of the\n>motion goes into the linear part. Essentially, the fermion\n>pictured is like a flywheel, which stores its motion into the\n>circular part and releases more of it when it needs\n>to go faster.\n>This gets to the root of the relation between the Poincare\' and\n>Galilei irreducible reps, where you see the role played by m\n>in the Galilei reps now being taken over by the E/c^2 in the\n>Poincare\' rep.\n>Essentially, what it\'s saying is that the mass is dynamic in\n>origin, arising from the stored energy of the flywheel. The\n>particle energy E is related to the frequency f of the helical\n>motion as:\n> E = 1/2 h f = spin x frequency.\n====================\n\nWhich I think is beautiful.\n\nHowever its unclear to me how I would figure a higgs field into this\nmodel.\n\n--\nOz\nThis post is worth absolutely nothing and is probably fallacious.\n\nBTOPENWORLD address about to cease. DEMON address no longer in use.\n>>Use oz@farmeroz.port995.com<<\nozacoohdb@despammed.com still functions.\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>Thomas Dent <tdent@auth.gr> writes
>This problem is due to two (or many) different uses of the word
>"mass". The mass that the Higgs is supposed to produce is the mass
>appearing in the Lagrangian and in the wave equation, which is a
>totally invariant numerical quantity which we can call m. For a
>physical (detectable) particle we have m^2 = E^2 - p.p (forgetting
>about factors of c). For a single particle, this turns out to be the
>same as what relativists call the "invariant mass".
I think 'rest mass' is a better description (ie p=0).
>Now, there is the "mass" that is talked about by relativists as the
>0-component of the energy-momentum 4-vector. That quantity p^0 will
>change as a massive (m/=0) particle is accelerated and is non-zero
>even for particles with zero mass (m=0).
Momentum in the time direction.
I think 'relativistic mass' would be a good description for that.
>And there is the so-called "invariant mass" for systems of many
>particles, which is just equal to the sum of p^0 in the centre-of-mass
>frame (I think).
Hmmm....
That's really the 'particles in a box mass' about which we are
discussing.
>The only thing the Higgs does for you is give you the mass m of
>elementary particles. The other "masses" are a combination of m and
>other kinematic variables.
I am uncomfortable about this right now. I really don't want two
separate mechanisms that generate mass. I'm pretty happy with mass as
generated kinematically, and one might plausibly argue that all mass is
generated this way. Its unclear to me how a coupling to a field should
generate mass, hence my stumbling efforts to clarify this.
>Composite particles such as the proton can have a large mass even
>though they may be made of massless (m=0) constituents (gluons and
>quarks). (The quarks in question do have small masses m_q, but much
>smaller than the proton mass scale.) Then the mass of the proton is
>made up of the energy of the constituent fields which can be thought
>of as constituents moving rapidly inside the proton. Roughly, the
>proton mass can be thought of as the "invariant mass" for all the
>gluons and quarks inside it. (It is more complicated than that because
>there is also a quark condensate in the vacuum which gives the proton
>more energy even though it cannot be seen as the kinetic energy of
>anything...)
That's fine. I have no problems with massless particles generating mass.
This is simply another way of saying 'all mass is kinematic mass'.
==============
Alfred Einstead <whopkins@csd.uwm.edu> wrote
>At the semi-classical level, the Dirac fermion would be coiling
>in a helical worldline of about this radius (for electrons) at
>light speed.
>If you plug the velocity operator v = \alpha c into the Heisenberg
>equation of motion
> i h-har d()/dt = [(), H]
>using the Dirac Hamiltonian
> H = \alpha.pc + \beta mc^2
>the picture that emerges is of a worldline with the following
>properties:
> (1) It's helical with mean free motion at a velocity
> v = p c^2/E (in a momentum eigenstate), parallel
> to p (opposite to p for negative energy states).
> (2) The velocity, counting the helical motion is just
> light speed.
> (3) The radius of the circular part of the motion is
> r = L (1 - (v/c)^2)
> where L is the Compton wavelength.
>As it goes faster, the helical motion uncoils and more of the
>motion goes into the linear part. Essentially, the fermion
>pictured is like a flywheel, which stores its motion into the
>circular part and releases more of it when it needs
>to go faster.
>This gets to the root of the relation between the Poincare' and
>Galilei irreducible reps, where you see the role played by m
>in the Galilei reps now being taken over by the E/c^2 in the
>Poincare' rep.
>Essentially, what it's saying is that the mass is dynamic in
>origin, arising from the stored energy of the flywheel. The
>particle energy E is related to the frequency f of the helical
>motion as:
> E = 1/2 h f = spin x frequency.
====================
Which I think is beautiful.
However its unclear to me how I would figure a higgs field into this
model.
--
Oz
This post is worth absolutely nothing and is probably fallacious.
BTOPENWORLD address about to cease. DEMON address no longer in use.
>>Use oz@farmeroz.port995.com<<
ozacoohdb@despammed.com still functions.
Creighton Hogg
Jul27-04, 01:50 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>\n\n\n\n\n\nOn 27 Jul 2004, Oz wrote:\n\n> Creighton Hogg <wchogg@hep.wisc.edu> writes\n>\n> >Nope. It\'s a different type of coupling. The massive electroweak bosons\n> >gain an effective mass because of their coupling to the Higgs, which then\n> >acquires a VEV. This supplies a mass term without making the theory\n> >non-renormalizable.\n>\n> OK, so what you are saying is that these three massless particles gain\n> mass by coupling to the higgs field. Nothing else.\n>\n> So it doesn\'t explain (say) masses of electrons or quarks.\n\nWell, I\'ve never seen it said explicitly, but often it seems to be assumed\nthat the Higgs couples to the charged leptons and the quarks as well to\nprovide their mass. I\'m a little fuzzy as to *why* this needs to be true,\nespecially since most neutrino mass models I\'m familiar with rely upon a\nsee-saw mechanism, radiative corrections, or majorana mass to generate\nthe mass of the neutrino and thus I\'m not clear why the neutrino is\nspecial and doesn\'t play be the same rules as the charged lepton...is\nthere a phenomenologist in the house?\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>On 27 Jul 2004, Oz wrote:
> Creighton Hogg <wchogg@hep.wisc.edu> writes
>
> >Nope. It's a different type of coupling. The massive electroweak bosons
> >gain an effective mass because of their coupling to the Higgs, which then
> >acquires a VEV. This supplies a mass term without making the theory
> >non-renormalizable.
>
> OK, so what you are saying is that these three massless particles gain
> mass by coupling to the higgs field. Nothing else.
>
> So it doesn't explain (say) masses of electrons or quarks.
Well, I've never seen it said explicitly, but often it seems to be assumed
that the Higgs couples to the charged leptons and the quarks as well to
provide their mass. I'm a little fuzzy as to *why* this needs to be true,
especially since most neutrino mass models I'm familiar with rely upon a
see-saw mechanism, radiative corrections, or majorana mass to generate
the mass of the neutrino and thus I'm not clear why the neutrino is
special and doesn't play be the same rules as the charged lepton...is
there a phenomenologist in the house?
Hendrik van Hees
Jul28-04, 04:58 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\nFranz Heymann wrote:\n\n\n> Since it is possible to calculate the invariant mass of a box of\n> photons without making use of the Higgs mechanism, it is clear that\n> that mass is not generated by that mechanism.\n\nActually very little of the mass of the matter surrounding us is\ngenerated by the Higgs mechanism.\n\nThis matter is indeed made of the lightest quarks (up and down) which\nhave a nearly negligible bare mass. One can even use this fact to\nestablish effective hadronic actions to describe the hadrons due to the\nchiral symmetry associated with massless fermions (and the strong\ninteraction does not break it). The bare (or current) quark masses are\nonly a tiny correction which can be treated as perturbation around the\nexactly chirally symmetric theory (that\'s what is known as chiral\nperturbation theory).\n\nThe much bigger effect on the hadron masses is the quark condensate\nwhich leads to a spontaneous breaking of the approximate chiral\nsymmetry.\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</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>Franz Heymann wrote:
> Since it is possible to calculate the invariant mass of a box of
> photons without making use of the Higgs mechanism, it is clear that
> that mass is not generated by that mechanism.
Actually very little of the mass of the matter surrounding us is
generated by the Higgs mechanism.
This matter is indeed made of the lightest quarks (up and down) which
have a nearly negligible bare mass. One can even use this fact to
establish effective hadronic actions to describe the hadrons due to the
chiral symmetry associated with massless fermions (and the strong
interaction does not break it). The bare (or current) quark masses are
only a tiny correction which can be treated as perturbation around the
exactly chirally symmetric theory (that's what is known as chiral
perturbation theory).
The much bigger effect on the hadron masses is the quark condensate
which leads to a spontaneous breaking of the approximate chiral
symmetry.
--
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
Oz
Aug7-04, 06:05 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>Greg Egan <gregegan@netspace.zebra.net.au> writes\n\n>Each photon\'s 4-momentum individually has a magnitude of zero, but:\n>\n> p_e + p_p = p_total = p_{photon 1} + p_{photon 2}\n>\n>i.e. conservation of total 4-momentum is required.\n>\n>And in general, the total 4-momentum of any large number of photons won\'t\n>be a null vector. (One obvious exception are the photons in a flash of\n>light that are all moving in the same direction.)\n\nAs ever, a clear and lucid reply.\n\nNow let me consider a photon with a spherical probability function.\n\nNaively we might consider this as a superposition of a near infinity of\nphotons spanning all possible directions. We can sum them up in opposing\npairs and find we have (some probability of) a 4-momentum in any\ndirection. Hmm, I suppose that\'s OK, if somewhat counterintuitive.\n\nIn the unlikely event the above is right (give or take some spherical\nintegration) we can do the same for a wavefront. If what I have been\ntold is correct, we can consider a wavefront as a superposition of a\nbunch of photons (keeping tabs on the phase) each with spherical\nsymmetry. These will interfere to propagate yet another wavefront and so\non.\n\nThis will mimic a single photon with a defined definite direction and\nthus a 4-magnitude of zero. Ngggth,\n\nHey, this is quite fun....\n\nBut I\'m missing something ....\n\n--\nOz\nThis post is worth absolutely nothing and is probably fallacious.\n\nBTOPENWORLD address about to cease. DEMON address no longer in use.\n>>Use oz@farmeroz.port995.com<<\nozacoohdb@despammed.com still functions.\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>Greg Egan <gregegan@netspace.zebra.net.au> writes
>Each photon's 4-momentum individually has a magnitude of zero, but:
>
> p_e + p_p = p_{total} = p_{photon 1} + p_{photon 2}
>
>i.e. conservation of total 4-momentum is required.
>
>And in general, the total 4-momentum of any large number of photons won't
>be a null vector. (One obvious exception are the photons in a flash of
>light that are all moving in the same direction.)
As ever, a clear and lucid reply.
Now let me consider a photon with a spherical probability function.
Naively we might consider this as a superposition of a near infinity of
photons spanning all possible directions. We can sum them up in opposing
pairs and find we have (some probability of) a 4-momentum in any
direction. Hmm, I suppose that's OK, if somewhat counterintuitive.
In the unlikely event the above is right (give or take some spherical
integration) we can do the same for a wavefront. If what I have been
told is correct, we can consider a wavefront as a superposition of a
bunch of photons (keeping tabs on the phase) each with spherical
symmetry. These will interfere to propagate yet another wavefront and so
on.
This will mimic a single photon with a defined definite direction and
thus a 4-magnitude of zero. Ngggth,
Hey, this is quite fun....
But I'm missing something ....
--
Oz
This post is worth absolutely nothing and is probably fallacious.
BTOPENWORLD address about to cease. DEMON address no longer in use.
>>Use oz@farmeroz.port995.com<<
ozacoohdb@despammed.com still functions.
alistair
Aug31-04, 04:55 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\nZefram C said:\n\n>All particles, except the\n>photon and gluons, are assumed to interact with the Higgs field, and\n>they develop masses proportional to the strength of this coupling,\n>which for the fermions is arbitrary and independent of other constants\n>(it is not so for the W and Z). The electron\'s coupling to the Higgs is\n>smaller than the proton\'s, resulting in a smaller apparent mass for the\n>electron.\n\n\nHave there been any attempts by theorists to predict\nthe fermion coupling constants that are needed to give\nthe masses of protons and electrons?\nOr is this unchartered territory?\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>Zefram C said:
>All particles, except the
>photon and gluons, are assumed to interact with the Higgs field, and
>they develop masses proportional to the strength of this coupling,
>which for the fermions is arbitrary and independent of other constants
>(it is not so for the W and Z). The electron's coupling to the Higgs is
>smaller than the proton's, resulting in a smaller apparent mass for the
>electron.
Have there been any attempts by theorists to predict
the fermion coupling constants that are needed to give
the masses of protons and electrons?
Or is this unchartered territory?
alistair
Aug31-04, 03:39 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>Can the Higgs field be associated with some kind of Higgs charge on particles\nin analogy to electric charges and the electric field?\nAnd is there a theory behind the size of the couplings\nof a proton and electron with the Higgs field?\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Can the Higgs field be associated with some kind of Higgs charge on particles
in analogy to electric charges and the electric field?
And is there a theory behind the size of the couplings
of a proton and electron with the Higgs field?
Cl.Massé
Sep4-04, 03:07 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>"alistair" <alistair@goforit64.fsnet.co.uk> a écrit dans le message de\nnews:861c1b21.0408301149.3d549dbd@posting.google.com...\n\n> Can the Higgs field be associated with some kind of Higgs charge on\n> particles in analogy to electric charges and the electric field?\n> And is there a theory behind the size of the couplings\n> of a proton and electron with the Higgs field?\n\nThe Higgs field is like the pre-relativity aether. It seems necessary\nand granted, all the formula to describe it are ther, but it isn\'t\nunderstood and resist any attempt of being seen.\n\n--\n~~~~ clmasse on free dot F-country\nLiberty, Equality, Profitability.\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>"alistair" <alistair@goforit64.fsnet.co.uk> a écrit dans le message de
news:861c1b21.0408301149.3d549dbd@posting.google.com...
> Can the Higgs field be associated with some kind of Higgs charge on
> particles in analogy to electric charges and the electric field?
> And is there a theory behind the size of the couplings
> of a proton and electron with the Higgs field?
The Higgs field is like the pre-relativity aether. It seems necessary
and granted, all the formula to describe it are ther, but it isn't
understood and resist any attempt of being seen.
--
~~~~ clmasse on free dot F-country
Liberty, Equality, Profitability.
arivero
Sep9-04, 04:02 PM
Somewhere in a FAQ it should be a distinction between a Higgs field and a Higgs particle.
Very basically, Higgs bosons (or Higgs particles) are the remain after that the Higgs field has mixed with the gauge fields, giving mass to them in the process. Thus generically the number of Higgs bosons will depend both of the number of Higgs fields and the number of gauge fields getting mass.
In the standard model, a Higgs doublet is the input, and a Higgs boson remain. In SUSY, two Higgs doublets are input, and five Higgs bosons remain.