View Full Version : mass of bound system
Pedro Tamirez
Oct20-04, 10:48 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>\nHello,\n\nI seem to miss a quite fundamental point concerning the mass of a\nbound system. When I ask people what makes the mass of the proton, I\nalways get vague answers as: E=mc2, mass is equivalent to energy, ...\nand I always have the feeling that they haven\'t understood it either.\nDoes anybody have a good way of explaining why a bound system of\nmassless quarks can give a very massive particle, and why increasing\nthe kinetic energy (or orbital angular momentum) of the system will\nincrease its mass?\n\nThanks a lot in advance!\nPedro\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Hello,
I seem to miss a quite fundamental point concerning the mass of a
bound system. When I ask people what makes the mass of the proton, I
always get vague answers as: E=mc2, mass is equivalent to energy, ...
and I always have the feeling that they haven't understood it either.
Does anybody have a good way of explaining why a bound system of
massless quarks can give a very massive particle, and why increasing
the kinetic energy (or orbital angular momentum) of the system will
increase its mass?
Thanks a lot in advance!
Pedro
Arnold Neumaier
Oct21-04, 02:47 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\nPedro Tamirez wrote:\n>\n> I seem to miss a quite fundamental point concerning the mass of a\n> bound system. When I ask people what makes the mass of the proton, I\n> always get vague answers as: E=mc2, mass is equivalent to energy, ...\n> and I always have the feeling that they haven\'t understood it either.\n> Does anybody have a good way of explaining why a bound system of\n> massless quarks can give a very massive particle, and why increasing\n> the kinetic energy (or orbital angular momentum) of the system will\n> increase its mass?\n\nA system has a well-defined mass if it is in an eigenstate of p^2,\nwhere p is the total momentum operator (whatever this is;\nrelativistically, bound states are very poorly understood).\n\nSo to understand, view it from a nonrelativistic perspective.\nBecause of E=mc^2, the mass shows up as energy, i.e., as eigenstate\nof the Hamiltonian.\n\nNow a bound state at rest defines the rest energy, and by giving\nit uniform motion you can increase the energy by an arbitrary amount\nof kinetic energy. The rest energy (and hence the rest mass), on the\nother hand, is determined by the discrete spectrum of the Hamiltonian\nin reduced coordinates, i.e., with center of mass motion separated out.\n\nFor forces that decay with distance, a bound state necessarily has\na mass that is less than the sum of the masses of the constituents.\nFor particles involving quarks, this does not apply since the strong\nforce increases with distance. Hence the rest mass of a bound state of\nquarks could be anything.\n\nOf course, the kinetic energy again increases the mass if you put\nthe bound state in uniform motion.\n\n\nArnold Neumaier\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>Pedro Tamirez wrote:
>
> I seem to miss a quite fundamental point concerning the mass of a
> bound system. When I ask people what makes the mass of the proton, I
> always get vague answers as: E=mc2, mass is equivalent to energy, ...
> and I always have the feeling that they haven't understood it either.
> Does anybody have a good way of explaining why a bound system of
> massless quarks can give a very massive particle, and why increasing
> the kinetic energy (or orbital angular momentum) of the system will
> increase its mass?
A system has a well-defined mass if it is in an eigenstate of p^2,
where p is the total momentum operator (whatever this is;
relativistically, bound states are very poorly understood).
So to understand, view it from a nonrelativistic perspective.
Because of E=mc^2, the mass shows up as energy, i.e., as eigenstate
of the Hamiltonian.
Now a bound state at rest defines the rest energy, and by giving
it uniform motion you can increase the energy by an arbitrary amount
of kinetic energy. The rest energy (and hence the rest mass), on the
other hand, is determined by the discrete spectrum of the Hamiltonian
in reduced coordinates, i.e., with center of mass motion separated out.
For forces that decay with distance, a bound state necessarily has
a mass that is less than the sum of the masses of the constituents.
For particles involving quarks, this does not apply since the strong
force increases with distance. Hence the rest mass of a bound state of
quarks could be anything.
Of course, the kinetic energy again increases the mass if you put
the bound state in uniform motion.
Arnold Neumaier
mathman
Oct22-04, 12:42 PM
This is not my field of expertise. However quarks do have mass, although not enough to account for proton or neutron mass. The bulk is a result of the energy associated with the gluons holding the quarks together to form nucleons.
J. J. Lodder
Oct24-04, 09:04 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\nmathman <mathnucl@optonline.net> wrote:\n\n> This is not my field of expertise. However quarks do have mass,\n> although not enough to account for proton or neutron mass. The bulk is\n> a result of the energy associated with the gluons holding the quarks\n> together to form nucleons.\n\nDo not forget the kinetic energy of the quarks.\nYou can estimate it immediately from Heisenberg,\napplied to a light particle confined in a box of nuclear size.\nThe precise mass doesn\'t matter,\nfor they are relativistic anyway.\n\nBest,\n\nJan\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>mathman <mathnucl@optonline.net> wrote:
> This is not my field of expertise. However quarks do have mass,
> although not enough to account for proton or neutron mass. The bulk is
> a result of the energy associated with the gluons holding the quarks
> together to form nucleons.
Do not forget the kinetic energy of the quarks.
You can estimate it immediately from Heisenberg,
applied to a light particle confined in a box of nuclear size.
The precise mass doesn't matter,
for they are relativistic anyway.
Best,
Jan
josefmatz
Oct25-04, 08:10 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>\nAs far as i know free quarks are existing and they have mass. Or is my\ninformation on the measurements in the collider rings wrong ? Only one\nparticle is missed so far and because of this unobservable particle the mass\nof the quarks only can be measured with certain uncertainties.\n\nOr am i wrong ?\n\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>As far as i know free quarks are existing and they have mass. Or is my
information on the measurements in the collider rings wrong ? Only one
particle is missed so far and because of this unobservable particle the mass
of the quarks only can be measured with certain uncertainties.
Or am i wrong ?
<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\nJ. J. Lodder <nospam@de-ster.demon.nl> writes\n>Do not forget the kinetic energy of the quarks.\n>You can estimate it immediately from Heisenberg,\n>applied to a light particle confined in a box of nuclear size.\n>The precise mass doesn\'t matter,\n>for they are relativistic anyway.\n\nThat\'s a remarkably good point and I\'m quite surprised I haven\'t seen it\nmentioned before. However I\'ve also heard it said that within the\nnucleus the quarks are \'unbound\' because they are so close to the other\nquarks. Offhand I don\'t see how to rationalise both these points. This\nbeing so I would expect to see excited nuclear states \'depending on how\nmany wavelengths fit in the box\', and I don\'t think we generally see\nthis. Mind you, the energy gap would likely be \'rather large\' and might\nhave a very short (and probably destructive) halflife.\n\n--\nOz\nThis post is worth absolutely nothing and is probably fallacious.\n\nUse oz@farmeroz.port995.com [ozacoohdb@despammed.com functions].\nBTOPENWORLD address has ceased. DEMON address has ceased.\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>J. J. Lodder <nospam@de-ster.demon.nl> writes
>Do not forget the kinetic energy of the quarks.
>You can estimate it immediately from Heisenberg,
>applied to a light particle confined in a box of nuclear size.
>The precise mass doesn't matter,
>for they are relativistic anyway.
That's a remarkably good point and I'm quite surprised I haven't seen it
mentioned before. However I've also heard it said that within the
nucleus the quarks are 'unbound' because they are so close to the other
quarks. Offhand I don't see how to rationalise both these points. This
being so I would expect to see excited nuclear states 'depending on how
many wavelengths fit in the box', and I don't think we generally see
this. Mind you, the energy gap would likely be 'rather large' and might
have a very short (and probably destructive) halflife.
--
Oz
This post is worth absolutely nothing and is probably fallacious.
Use oz@farmeroz.port995.com [ozacoohdb@despammed.com functions].
BTOPENWORLD address has ceased. DEMON address has ceased.
Paul Draper
Oct27-04, 10:56 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"josefmatz" <josefmatz@arcor.de> wrote in message news:<4177b764_1@news.arcor-ip.de>...\n> As far as i know free quarks are existing and they have mass. Or is my\n> information on the measurements in the collider rings wrong ? Only one\n> particle is missed so far and because of this unobservable particle the mass\n> of the quarks only can be measured with certain uncertainties.\n>\n> Or am i wrong ?\n\nYou are wrong. There are no free quarks observed, only jets of\nparticles that come from quarks or the decay products of quarks. In\nthe latter case, especially for the heavy quarks like b and t quarks,\nwe can estimate the mass from the invariant mass of the decay\nproducts. Even then, it\'s not easy.\n\nYour "missing particle" I\'m presuming is the Higgs. However, it is not\nunobservable, it just hasn\'t been seen yet above the noise. Nor does\nits absence so far cause uncertainty in the quark masses.\n\nPD\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>"josefmatz" <josefmatz@arcor.de> wrote in message news:<4177b764_1@news.arcor-ip.de>...
> As far as i know free quarks are existing and they have mass. Or is my
> information on the measurements in the collider rings wrong ? Only one
> particle is missed so far and because of this unobservable particle the mass
> of the quarks only can be measured with certain uncertainties.
>
> Or am i wrong ?
You are wrong. There are no free quarks observed, only jets of
particles that come from quarks or the decay products of quarks. In
the latter case, especially for the heavy quarks like b and t quarks,
we can estimate the mass from the invariant mass of the decay
products. Even then, it's not easy.
Your "missing particle" I'm presuming is the Higgs. However, it is not
unobservable, it just hasn't been seen yet above the noise. Nor does
its absence so far cause uncertainty in the quark masses.
PD
Richard Saam
Oct29-04, 02:08 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\nArnold Neumaier wrote:\n\n>\n>A system has a well-defined mass if it is in an eigenstate of p^2,\n>where p is the total momentum operator (whatever this is;\n>relativistically, bound states are very poorly understood).\n>\n>So to understand, view it from a nonrelativistic perspective.\n>Because of E=mc^2, the mass shows up as energy, i.e., as eigenstate\n>of the Hamiltonian.\n>\n>Now a bound state at rest defines the rest energy, and by giving\n>it uniform motion you can increase the energy by an arbitrary amount\n>of kinetic energy. The rest energy (and hence the rest mass), on the\n>other hand, is determined by the discrete spectrum of the Hamiltonian\n>in reduced coordinates, i.e., with center of mass motion separated out.\n>\n>For forces that decay with distance, a bound state necessarily has\n>a mass that is less than the sum of the masses of the constituents.\n>For particles involving quarks, this does not apply since the strong\n>force increases with distance. Hence the rest mass of a bound state of\n>quarks could be anything.\n>\n>Of course, the kinetic energy again increases the mass if you put\n>the bound state in uniform motion.\n>\n>\n>Arnold Neumaier\n>\n>\nJust wondering if a simple approach may have application here.\n\nAssume basic physics of elasticity in that there is conservation of\nenergy(E) and momentum(p) such that:\n\np1 + p2 = p3 + p4\n\nE1 + E2 = E3 + E4\n\nnow E = p^2 / m\n\np1^2 / m1 + p1^2 / m2 = p1^2 / m3 + p1^2 / m4\n\nNow for system perfect elasticity (infinitely stable bound system) to\nexist there has to be some kind of mass differential in\n\nm1, m2, m3, m4.\n\nRight??\n\nSystem could be generalized 1, 2, 3, 4, ...... infinity.\n\nRichard Saam\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Arnold Neumaier wrote:
>
>A system has a well-defined mass if it is in an eigenstate of p^2,
>where p is the total momentum operator (whatever this is;
>relativistically, bound states are very poorly understood).
>
>So to understand, view it from a nonrelativistic perspective.
>Because of E=mc^2, the mass shows up as energy, i.e., as eigenstate
>of the Hamiltonian.
>
>Now a bound state at rest defines the rest energy, and by giving
>it uniform motion you can increase the energy by an arbitrary amount
>of kinetic energy. The rest energy (and hence the rest mass), on the
>other hand, is determined by the discrete spectrum of the Hamiltonian
>in reduced coordinates, i.e., with center of mass motion separated out.
>
>For forces that decay with distance, a bound state necessarily has
>a mass that is less than the sum of the masses of the constituents.
>For particles involving quarks, this does not apply since the strong
>force increases with distance. Hence the rest mass of a bound state of
>quarks could be anything.
>
>Of course, the kinetic energy again increases the mass if you put
>the bound state in uniform motion.
>
>
>Arnold Neumaier
>
>
Just wondering if a simple approach may have application here.
Assume basic physics of elasticity in that there is conservation of
energy(E) and momentum(p) such that:
p1 + p2 = p3 + p4E1 + E2 = E3 + E4
now E = p^2 / m
p1^2 / m1 + p1^2 / m2 = p1^2 / m3 + p1^2 / m4
Now for system perfect elasticity (infinitely stable bound system) to
exist there has to be some kind of mass differential in
m1, m2, m3, m4.
Right??
System could be generalized 1, 2, 3, 4, ...... infinity.
Richard Saam
Arnold Neumaier
Nov2-04, 12:38 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\nRichard Saam wrote:\n> Arnold Neumaier wrote:\n>\n>>For forces that decay with distance, a bound state necessarily has\n>>a mass that is less than the sum of the masses of the constituents.\n>>For particles involving quarks, this does not apply since the strong\n>>force increases with distance. Hence the rest mass of a bound state of\n>>quarks could be anything.\n>\n> Just wondering if a simple approach may have application here.\n>\n> Assume basic physics of elasticity in that there is conservation of\n> energy(E) and momentum(p) such that:\n> p1 + p2 = p3 + p4\n> E1 + E2 = E3 + E4\n> now E = p^2 / m\n> p1^2 / m1 + p1^2 / m2 = p1^2 / m3 + p1^2 / m4\n\nNo, but:\np1^2 / m1 + p1^2 / m2 = p3^2 / m3 + p4^2 / m4\nThis can be solved in terms of Mandelshtam variables, as is done in\nrelativistic scattering calculations. It has nothing to do with\nwhether bound states can have larger mass than the sum of the masses\nof their constituents.\n\n\nArnold Neumaier\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Richard Saam wrote:
> Arnold Neumaier wrote:
>
>>For forces that decay with distance, a bound state necessarily has
>>a mass that is less than the sum of the masses of the constituents.
>>For particles involving quarks, this does not apply since the strong
>>force increases with distance. Hence the rest mass of a bound state of
>>quarks could be anything.
>
> Just wondering if a simple approach may have application here.
>
> Assume basic physics of elasticity in that there is conservation of
> energy(E) and momentum(p) such that:
> p1 + p2 = p3 + p4
> E1 + E2 = E3 + E4
> now E = p^2 / m
> p1^2 / m1 + p1^2 / m2 = p1^2 / m3 + p1^2 / m4
No, but:
p1^2 / m1 + p1^2 / m2 = p3^2 / m3 + p4^2 / m4
This can be solved in terms of Mandelshtam variables, as is done in
relativistic scattering calculations. It has nothing to do with
whether bound states can have larger mass than the sum of the masses
of their constituents.
Arnold Neumaier
Richard Saam
Nov4-04, 03:40 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>Arnold Neumaier wrote:\n\n>Richard Saam wrote:\n>\n>\n>>Arnold Neumaier wrote:\n>>\n>>\n>>\n>>>For forces that decay with distance, a bound state necessarily has\n>>>a mass that is less than the sum of the masses of the constituents.\n>>>For particles involving quarks, this does not apply since the strong\n>>>force increases with distance. Hence the rest mass of a bound state of\n>>>quarks could be anything.\n>>>\n>>>\n>>Just wondering if a simple approach may have application here.\n>>\n>>Assume basic physics of elasticity in that there is conservation of\n>>energy(E) and momentum(p) such that:\n>>p1 + p2 = p3 + p4\n>>E1 + E2 = E3 + E4\n>>now E = p^2 / m\n>>p1^2 / m1 + p1^2 / m2 = p1^2 / m3 + p1^2 / m4\n>>\n>>\n>\n>No, but:\n> p1^2 / m1 + p1^2 / m2 = p3^2 / m3 + p4^2 / m4\n>This can be solved in terms of Mandelshtam variables, as is done in\n>relativistic scattering calculations. It has nothing to do with\n>whether bound states can have larger mass than the sum of the masses\n>of their constituents.\n>\n>\n>Arnold Neumaier\n>\n>\n\nThanks but let me restate the proposition with correct notation.\nAssume basic physics of elasticity in that there is conservation of\nenergy(E) and momentum(p) such that:\np1 + p2 = p3 + p4\nE1 + E2 = E3 + E4\nnow E = p^2/m\np1^2/m1 + p2^2/m2 = p3^2/m3 + p4^2/m4\n\nNow assume that a bound system defined by states 1,2,3,4\n\nresonates between\n\nstates 1&2 and 3&4\n\nI have done some calculations on a particular bound system wherein\n\nm1=m2=m3=m4\n"nothing to do with\nwhether bound states can have larger mass than the sum of the masses\nof their constituents"\n\nand\n\n(p1 + p2)*k = p3 + p4\n(p1^2 + p2^2)/k = p3^2 + p4^2\n\nwhere k is a constant.\n\nWithin this dimensionless constant k, there is a larger mass than for the unbound system.\n\nIs this a necessary condition for a bound system?\n\nRichard Saam\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Arnold Neumaier wrote:
>Richard Saam wrote:
>
>
>>Arnold Neumaier wrote:
>>
>>
>>
>>>For forces that decay with distance, a bound state necessarily has
>>>a mass that is less than the sum of the masses of the constituents.
>>>For particles involving quarks, this does not apply since the strong
>>>force increases with distance. Hence the rest mass of a bound state of
>>>quarks could be anything.
>>>
>>>
>>Just wondering if a simple approach may have application here.
>>
>>Assume basic physics of elasticity in that there is conservation of
>>energy(E) and momentum(p) such that:
>>p1 + p2 = p3 + p4>>E1 + E2 = E3 + E4
>>now E = p^2 / m>>p1^2 / m1 + p1^2 / m2 = p1^2 / m3 + p1^2 / m4
>>
>>
>
>No, but:
> p1^2 / m1 + p1^2 / m2 = p3^2 / m3 + p4^2 / m4
>This can be solved in terms of Mandelshtam variables, as is done in
>relativistic scattering calculations. It has nothing to do with
>whether bound states can have larger mass than the sum of the masses
>of their constituents.
>
>
>Arnold Neumaier
>
>
Thanks but let me restate the proposition with correct notation.
Assume basic physics of elasticity in that there is conservation of
energy(E) and momentum(p) such that:
p1 + p2 = p3 + p4E1 + E2 = E3 + E4
now E = p^2/mp1^2/m1 + p2^2/m2 = p3^2/m3 + p4^2/m4
Now assume that a bound system defined by states 1,2,3,4
resonates between
states 1&2 and 3&4
I have done some calculations on a particular bound system wherein
m1=m2=m3=m4[/itex]
"nothing to do with
whether bound states can have larger mass than the sum of the masses
of their constituents"
and
[itex](p1 + p2)*k = p3 + p4(p1^2 + p2^2)/k = p3^2 + p4^2
where k is a constant.
Within this dimensionless constant k, there is a larger mass than for the unbound system.
Is this a necessary condition for a bound system?
Richard Saam
Arnold Neumaier
Nov14-04, 07:08 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>Kefka G wrote:\n> Arnold Neumaier wrote:\n>\n>>What you are describing is elastic scattering of two particles;\n>>this has nothing to do with bound systems. For a bound system, you\'d\n>>have instead somnething like\n>> E_p + E_e = E_H + binding energy.\n>>For forces which vanish at infinity, the binding energy is always\n>>positive; for quarks it need not be.\n>\n> I\'ve not seen the rest of this thread, so perhaps I shouldn\'t comment on it,\n> but I\'m a bit confused by the binding energy > 0 claim...\n>\n> Suppose we had a potential which vanishes at infinity, comes in and reaches a\n> local maximum at distance xa, then dips down to a local minimum at xb < xa. In\n> such a circumstance, we could in principle have the energy at separation xb be\n> anything that we want it to be - I\'ll grant you that if the binding energy were\n> negative we\'d have tunnelling through the barrier, but certainly there are\n> situations where the bound state would have a long enough lifetime to consider\n> it as such, right? Or by "bound state" do you mean strictly a state that\n> remains bound forever?\n\nOf course. That\'s in the definition of a bound state.\nStates that can decay by tunnneling are unstable, hence not\nbound states but \'resonances\'.\n\n\nArnold Neumaier\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Kefka G wrote:
> Arnold Neumaier wrote:
>
>>What you are describing is elastic scattering of two particles;
>>this has nothing to do with bound systems. For a bound system, you'd
>>have instead somnething like
>> E_p + E_e = E_H + binding energy.
>>For forces which vanish at infinity, the binding energy is always
>>positive; for quarks it need not be.
>
> I've not seen the rest of this thread, so perhaps I shouldn't comment on it,
> but I'm a bit confused by the binding energy > claim...
>
> Suppose we had a potential which vanishes at infinity, comes in and reaches a
> local maximum at distance xa, then dips down to a local minimum at xb < xa. In
> such a circumstance, we could in principle have the energy at separation xb be
> anything that we want it to be - I'll grant you that if the binding energy were
> negative we'd have tunnelling through the barrier, but certainly there are
> situations where the bound state would have a long enough lifetime to consider
> it as such, right? Or by "bound state" do you mean strictly a state that
> remains bound forever?
Of course. That's in the definition of a bound state.
States that can decay by tunnneling are unstable, hence not
bound states but 'resonances'.
Arnold Neumaier
Richard Saam
Nov16-04, 02:56 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\nArnold Neumaier wrote:\n\n> *************\n>\n>What you are describing is elastic scattering of two particles;\n>this has nothing to do with bound systems. For a bound system, you\'d\n>have instead somnething like\n> E_p + E_e = E_H + binding energy.\n>For forces which vanish at infinity, the binding energy is always\n>positive; for quarks it need not be.\n>\n>\n>Arnold Neumaier\n>\n>\nI understand the description of bound systems as exampled by hydrogen\nabove but would like to describe another system of particles that are\n"bound" in a "coordinated scattering" sense.\n\nAssume a rectangular grid x,y of charged particles\n\nAll particles move in x direction with velocity vx\nAll particles move in y direction with velocity -vy with respect to\nadjacent y direction particle and do not move past any adjacent y\ndirection particle. In other words, particles elastically oscillate in\ny direction against adjacent y direction particles as they move at\nconstant velocity in x direction.\n\nNow overlaying this particle grid is a charged lattice to which the\nparticles elastically interact keeping them on track.\n\nNow without going into geometric details, assume four states that define\nthis bound assembly of particles\n\nstates 1, 2, 3, 4\nwith resonance between\nstates 1&2 and 3&4\n\nand test the hypothesis\n\np1 + p2 = p3 + p4\nE1 + E2 = E3 + E4\nwhere E = p2/m\nand therefore\np1^2/m1 + p2^2/m2 = p3^2/m3 + p4^2/m4\n\n\nI have done some calculations on a particular system wherein\n\nm1=m2=m3=m4\n\na constant k must be introduced to get the equality to establish perfect system elasticity\n\n(p1 + p2)*k = p3 + p4\n(p1^2 + p2^2)/k = p3^2 + p4^2\n\nI am wondering if this is a general type of conclusion? There may be an infinite number of states 1,2,3,4, that would describe the particle system above, but do they all have in common a factor k to establish elasticity?\n\nThis dimensionless constant k could be assumed to contain a larger mass (in some type of ratio) than for the "unbound" system.\n\n\nRichard Saam\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Arnold Neumaier wrote:
> *************
>
>What you are describing is elastic scattering of two particles;
>this has nothing to do with bound systems. For a bound system, you'd
>have instead somnething like
> E_p + E_e = E_H + binding energy.
>For forces which vanish at infinity, the binding energy is always
>positive; for quarks it need not be.
>
>
>Arnold Neumaier
>
>
I understand the description of bound systems as exampled by hydrogen
above but would like to describe another system of particles that are
"bound" in a "coordinated scattering" sense.
Assume a rectangular grid x,y of charged particles
All particles move in x direction with velocity vx
All particles move in y direction with velocity -vy with respect to
adjacent y direction particle and do not move past any adjacent y
direction particle. In other words, particles elastically oscillate in
y direction against adjacent y direction particles as they move at
constant velocity in x direction.
Now overlaying this particle grid is a charged lattice to which the
particles elastically interact keeping them on track.
Now without going into geometric details, assume four states that define
this bound assembly of particles
states 1, 2, 3, 4
with resonance between
states 1&2 and 3&4
and test the hypothesis
p1 + p2 = p3 + p4E1 + E2 = E3 + E4[/itex]
where E = p2/m
and therefore
[itex]p1^2/m1 + p2^2/m2 = p3^2/m3 + p4^2/m4
I have done some calculations on a particular system wherein
m1=m2=m3=m4
a constant k must be introduced to get the equality to establish perfect system elasticity
(p1 + p2)*k = p3 + p4(p1^2 + p2^2)/k = p3^2 + p4^2
I am wondering if this is a general type of conclusion? There may be an infinite number of states 1,2,3,4, that would describe the particle system above, but do they all have in common a factor k to establish elasticity?
This dimensionless constant k could be assumed to contain a larger mass (in some type of ratio) than for the "unbound" system.
Richard Saam
A. T. Wilson
Nov22-04, 05:14 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\nHello there,\n\nThere is another way to think about this, due, I believe, to Kaluza\nand Klein. Imagine some extra space dimensions that are wrapped up\nsmall, so that a massless (i.e. rest mass zero) particle, which must\nafter all travel at the speed of light, moves mostly around a wrapped\nup dimension.\n\nConsider E^2 = m^2 + (p_x)^2 + (p_y)^2 + (p_z)^2. Then rest mass may\nsimply be more momentum, trapped in some wrapped-up dimension or\ndimensions...\n\nATW\n\n\nptamirez@yahoo.co.uk (Pedro Tamirez) wrote in message news:<11d563b8.0410200643.5f16692d@posting.google. com>...\n> Hello,\n>\n> I seem to miss a quite fundamental point concerning the mass of a\n> bound system. When I ask people what makes the mass of the proton, I\n> always get vague answers as: E=mc2, mass is equivalent to energy, ...\n> and I always have the feeling that they haven\'t understood it either.\n> Does anybody have a good way of explaining why a bound system of\n\n> massless quarks can give a very massive particle, and why increasing\n> the kinetic energy (or orbital angular momentum) of the system will\n> increase its mass?\n>\n> Thanks a lot in advance!\n> Pedro\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Hello there,
There is another way to think about this, due, I believe, to Kaluza
and Klein. Imagine some extra space dimensions that are wrapped up
small, so that a massless (i.e. rest mass zero) particle, which must
after all travel at the speed of light, moves mostly around a wrapped
up dimension.
Consider E^2 = m^2 + (p_x)^2 + (p_y)^2 + (p_z)^2. Then rest mass may
simply be more momentum, trapped in some wrapped-up dimension or
dimensions...
ATW
ptamirez@yahoo.co.uk (Pedro Tamirez) wrote in message news:<11d563b8.0410200643.5f16692d@posting.google.com>...
> Hello,
>
> I seem to miss a quite fundamental point concerning the mass of a
> bound system. When I ask people what makes the mass of the proton, I
> always get vague answers as: E=mc2, mass is equivalent to energy, ...
> and I always have the feeling that they haven't understood it either.
> Does anybody have a good way of explaining why a bound system of
> massless quarks can give a very massive particle, and why increasing
> the kinetic energy (or orbital angular momentum) of the system will
> increase its mass?
>
> Thanks a lot in advance!
> Pedro
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