<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no,location=no,scrollbars=yes,resizable=yes,status=no,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Hi everyone\n\nI have the impression that the chemical potential mu for particles\nobeying Bose-Einstein statistics is always less than zero. What about\nfor fermions? Are they always positive, or can they take any value?\n\nAs an example I calculate the number density for ultrarelativistic fermions:\n\nn_f\n= (g/2Pi^2)Integral[p^2/(exp[beta(p-mu)] + 1),{p,0,Infinity}]\n= -(g/2Pi^2) T^3 PolyLog[3,-exp[mu/T]]\n\nAnd for relativistic bosons we have\n\nn_b\n= (g/2Pi^2)Integral[p^2/(exp[beta(p-mu)] - 1),{p,0,Infinity}]\n= (g/2Pi^2) T^3 PolyLog[3,exp[mu/T]]\n\nwhere g is the number of spin degrees of freedom and p is the momentum.\n\nFor the case of bosons, we use the fact that the PolyLog[p,x] - if we\n_define_ it as the sum of x^q/(q+1)^p from q = 0 to infinity - diverges\nfor |x|>1, i.e. for exp[mu/T]>1 or mu/T > 0. Then we say therefore mu/T\n< 0 always. Is this the correct reason why bosons have negative mu?\n\nWhat about for fermions? It seems by the same argument above mu/T < 0\nalways too. But I realize PolyLog can be "analytic continued" for x < -1\non the real line, so this seems to allow mu/T to take any value. So are\nthere any restrictions on the chemical potential of fermions?\n\nOne of the reasons I ask this is because it seems in\nantiparticle-particle reactions, say e- + e+ <-> gamma + gamma if we\nassume there is some kind of equilibrium (such as in the very early\nuniverse) then I suppose we can say something like mu_e+ + mu_e- =\nmu_gamma + mu_gamma = 0, because mu_gamma = 0 always. And therefore in\nchemical equilibrium the chemical potentials of antiparticles are\nnegative of that of their particle counterparts. But what about bosons,\nsince mu_bosons/T < 0 always?\n\nThanks for helping me understand this.\n\nYi-Zen\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Hi everyone
I have the impression that the chemical potential \mu for particles
obeying Bose-Einstein statistics is always less than zero. What about
for fermions? Are they always positive, or can they take any value?
As an example I calculate the number density for ultrarelativistic fermions:
where g is the number of spin degrees of freedom and p is the momentum.
For the case of bosons, we use the fact that the PolyLog[p,x] - if we
_define_ it as the sum of x^q/(q+1)^p from q = to infinity - diverges
for |x|>1, i.e. for \exp[\mu/T]>1 or \mu/T > . Then we say therefore \mu/T
< always. Is this the correct reason why bosons have negative \mu?
What about for fermions? It seems by the same argument above \mu/T <
always too. But I realize PolyLog can be "analytic continued" for x < -1
on the real line, so this seems to allow \mu/T to take any value. So are
there any restrictions on the chemical potential of fermions?
One of the reasons I ask this is because it seems in
antiparticle-particle reactions, say e- + e+ <-> \gamma + \gamma if we
assume there is some kind of equilibrium (such as in the very early
universe) then I suppose we can say something like \mu_e+ + \mu_e- =\mu_gamma + \mu_gamma = 0, because \mu_gamma = always. And therefore in
chemical equilibrium the chemical potentials of antiparticles are
negative of that of their particle counterparts. But what about bosons,
since \mu_bosons/T < always?
Thanks for helping me understand this.
Yi-Zen
Yi-Zen Chu; Yiren Qu
Jul4-04, 08:39 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>\nYi-Zen Chu; Yiren Qu wrote:\n>\n> For the case of bosons, we use the fact that the PolyLog[p,x] - if we\n> _define_ it as the sum of x^q/(q+1)^p from q = 0 to infinity\n\nI think the qth term should actually be x^(q+1)/(q+1)^p.\n\n> assume there is some kind of equilibrium (such as in the very early\n> universe) then I suppose we can say something like mu_e+ + mu_e- =\n> mu_gamma + mu_gamma = 0, because mu_gamma = 0 always.\n\nI wondered why is it in equilibrium, given a reaction A + B <-> C + D,\nthe sum of the chemical potentials on both sides are equal - mu_A + mu_B\n= mu_C + mu_D? The reason given by Landau (Statistical Physics) is that\nin equilibrium since by definition the particle number is constant the\nfree energy F = T ln[Z] is minimized with respect to the variation of\nparticle number - that leads directly to the above condition if one\nrecalls that the chemical potential of a given particle species is the\nderivative of the free energy with respect to particle number of the same.\n\nYi-Zen\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>Yi-Zen Chu; Yiren Qu wrote:
>
> For the case of bosons, we use the fact that the PolyLog[p,x] - if we
> _define_ it as the sum of x^q/(q+1)^p from q = to infinity
I think the qth term should actually be x^(q+1)/(q+1)^p.
> assume there is some kind of equilibrium (such as in the very early
> universe) then I suppose we can say something like \mu_e+ + \mu_e- =
> \mu_gamma + \mu_gamma = 0, because \mu_gamma = always.
I wondered why is it in equilibrium, given a reaction A + B <-> C + D,
the sum of the chemical potentials on both sides are equal - \mu_A + \mu_B= \mu_C + \mu_D? The reason given by Landau (Statistical Physics) is that
in equilibrium since by definition the particle number is constant the
free energy F = T ln[Z] is minimized with respect to the variation of
particle number - that leads directly to the above condition if one
recalls that the chemical potential of a given particle species is the
derivative of the free energy with respect to particle number of the same.
Yi-Zen
Yi-Zen Chu; Yiren Qu
Jul4-04, 08:39 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>\nYi-Zen Chu; Yiren Qu wrote:\n>\n> One of the reasons I ask this is because it seems in\n> antiparticle-particle reactions, say e- + e+ <-> gamma + gamma if we\n> assume there is some kind of equilibrium (such as in the very early\n> universe) then I suppose we can say something like mu_e+ + mu_e- =\n> mu_gamma + mu_gamma\n\nI want to understand this too, actually: why is it that if I have a\nreaction A + B <-> C + D, and if this is in some kind of chemical\nequilibrium - by that I assume what is meant is that the number of\nparticles of A, B, C and D remain constant - then mu_A + mu_B = mu_C +\nmu_D? That is why is the sum of the chemical potentials equal?\n\nYi-Zen\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>Yi-Zen Chu; Yiren Qu wrote:
>
> One of the reasons I ask this is because it seems in
> antiparticle-particle reactions, say e- + e+ <-> \gamma + \gamma if we
> assume there is some kind of equilibrium (such as in the very early
> universe) then I suppose we can say something like \mu_e+ + \mu_e- =
> \mu_gamma + \mu_gamma
I want to understand this too, actually: why is it that if I have a
reaction A + B <-> C + D, and if this is in some kind of chemical
equilibrium - by that I assume what is meant is that the number of
particles of A, B, C and D remain constant - then \mu_A + \mu_B = \mu_C +\mu_D? That is why is the sum of the chemical potentials equal?
Yi-Zen
MP
Jul4-04, 08:40 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>\n\n"Yi-Zen Chu; Yiren Qu" <y#i#-#z#e#n#.#c#h#u#@#y#a#l#e#.#e#d#u> wrote in\nmessage news:cc48fp\\$977\\$1@news.wss.yale.edu...\n> Hi everyone\n>\n> I have the impression that the chemical potential mu for particles\n> obeying Bose-Einstein statistics is always less than zero. What about\n> for fermions? Are they always positive, or can they take any value?\n>\n> As an example I calculate the number density for ultrarelativistic\nfermions:\n>\n> n_f\n> = (g/2Pi^2)Integral[p^2/(exp[beta(p-mu)] + 1),{p,0,Infinity}]\n> = -(g/2Pi^2) T^3 PolyLog[3,-exp[mu/T]]\n>\n> And for relativistic bosons we have\n>\n> n_b\n> = (g/2Pi^2)Integral[p^2/(exp[beta(p-mu)] - 1),{p,0,Infinity}]\n> = (g/2Pi^2) T^3 PolyLog[3,exp[mu/T]]\n>\n> where g is the number of spin degrees of freedom and p is the momentum.\n>\n> For the case of bosons, we use the fact that the PolyLog[p,x] - if we\n> _define_ it as the sum of x^q/(q+1)^p from q = 0 to infinity - diverges\n> for |x|>1, i.e. for exp[mu/T]>1 or mu/T > 0. Then we say therefore mu/T\n> < 0 always. Is this the correct reason why bosons have negative mu?\n>\n\nDear Yi-Zen,\n\nthe calculations you are doing are about ultra-relativistic particles (you\nhave set E=p) in the thermodynamic expressions. So the results you\nhave found only apply to the situation T >> m (or m=0)\n\nHowever, it is correct that in the case of *ultra-relativistic* bosons\nthe PolyLog-function becomes imaginary whenever \\mu_Bosons > 0.\nTherefore ultra-relativistic bosons cannot have a chemical potential\nwhich is positive. I.e. simply from microscopic statistical thermodynamics\nwe have the condition \\mu_Boson <= 0 (for ultra-relativistic bosons).\n\nNote that I will be assuming T>0 throughout the whole discussion. For\nthe thermodynamics of an ultra-relativistic gas of fermions and bosons\nonly the *ratio* /mu / T is important (it is the quantity that appears in\nthe polylog function and determines whether polylog is real, or becomes\nimaginary - which is not desirable).\n\nNow at ultra-relativistic energies you will have particle-antiparticle\npair production in abundance. On the other hand, the chemical potentials\nof particle and anti-particle are necessarily opposite to each other:\n\n\\mu_particle + \\mu_antiparticle = 0.\n\nCombined with the thermodynamic result \\mu_Boson <=0 this only leaves\n\\mu_boson = 0 for ultra-relativistic bosons. In fact, this is the chemical\npotential of the photon (which quite definitely is an ultra-relativistic\nboson, as m=0), but the result is more general: *Any* ultra-relativistic\n(or zero rest mass) boson *must* have a chemical potential of *zero*,\nat least if thermodynamic reasoning is still appropriate at\nultra-relativistic\nenergies.\n\nHowever, this result is only true at ultra-relativistic energies. At *lower*\nenergies, massive bosons can even have a chemical potential that is\npositive, as long as \\mu < m. (See the appendix of gr-qc/0405010 for the\nthermodynamics of an ideal gas of fermions and bosons with m arbitrary -\nyou can probably find the same results elsewhere, however, when I needed\nthem I was too lazy to look and therefore decided to derive the necessary\nresults myself - if you need to *understand* what you are doing, doing the\ncalculations oneself is usually quicker than looking them up ;-)\n\n> What about for fermions? It seems by the same argument above mu/T < 0\n> always too. But I realize PolyLog can be "analytic continued" for x < -1\n> on the real line, so this seems to allow mu/T to take any value. So are\n> there any restrictions on the chemical potential of fermions?\n\nNo, there aren\'t. For fermions the polylog-function is well defined for\n*any* value of the chemical potential. The chemical potential of\nultra-relativistic fermions is not restricted. However, one still has the\nrestriction that fermions and anti-fermions have opposite chemical\npotentials. Therefore, whenever an ultra-relativistic fermion species has\na *non-zero* chemical potential, we can distinguish between particles\nand anti-particles in a thermodynamic sense.\nThe particle in the particle-antiparticle pair with *positive* chemical\npotential would be called "normal matter", whereas its antiparticle\n(with negative chemical potential) would be called "antimatter".\n\n> One of the reasons I ask this is because it seems in\n> antiparticle-particle reactions, say e- + e+ <-> gamma + gamma if we\n> assume there is some kind of equilibrium (such as in the very early\n> universe) then I suppose we can say something like mu_e+ + mu_e- =\n> mu_gamma + mu_gamma = 0, because mu_gamma = 0 always. And therefore in\n> chemical equilibrium the chemical potentials of antiparticles are\n> negative of that of their particle counterparts. But what about bosons,\n> since mu_bosons/T < 0 always?\n\nI would think, that you *always* can say \\mu_particle + \\mu_antiparticle =0,\nindependent of the nature of the particle. At least this is what I thought I\nhave learned from QFT. Or does anyone in this group disagree? This\nwould in fact be interesting.\n\nBest wishes, M\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>"Yi-Zen Chu; Yiren Qu" <y#i#-#z#e#n#.#c#h#u#@#y#a#l#e#.#e#d#u> wrote in
message news:cc48fp$977$1@news.wss.yale.edu...
> Hi everyone
>
> I have the impression that the chemical potential \mu for particles
> obeying Bose-Einstein statistics is always less than zero. What about
> for fermions? Are they always positive, or can they take any value?
>
> As an example I calculate the number density for ultrarelativistic
fermions:
>
> n_f
> = (g/2Pi^2)Integral[p^2/(\exp[\beta(p-\mu)] + 1),{p,0,Infinity}]
> = -(g/2Pi^2) T^3 PolyLog[3,-\exp[\mu/T]]
>
> And for relativistic bosons we have
>
> n_b
> = (g/2Pi^2)Integral[p^2/(\exp[\beta(p-\mu)] - 1),{p,0,Infinity}]
> = (g/2Pi^2) T^3 PolyLog[3,\exp[\mu/T]]
>
> where g is the number of spin degrees of freedom and p is the momentum.
>
> For the case of bosons, we use the fact that the PolyLog[p,x] - if we
> _define_ it as the sum of x^q/(q+1)^p from q = to infinity - diverges
> for |x|>1, i.e. for \exp[\mu/T]>1 or \mu/T > . Then we say therefore \mu/T
> < always. Is this the correct reason why bosons have negative \mu?
>
Dear Yi-Zen,
the calculations you are doing are about ultra-relativistic particles (you
have set E=p) in the thermodynamic expressions. So the results you
have found only apply to the situation T >> m (or m=0)
However, it is correct that in the case of *ultra-relativistic* bosons
the PolyLog-function becomes imaginary whenever \mu_Bosons > .
Therefore ultra-relativistic bosons cannot have a chemical potential
which is positive. I.e. simply from microscopic statistical thermodynamics
we have the condition \mu_Boson <= (for ultra-relativistic bosons).
Note that I will be assuming T>0 throughout the whole discussion. For
the thermodynamics of an ultra-relativistic gas of fermions and bosons
only the *ratio* /\mu / T is important (it is the quantity that appears in
the polylog function and determines whether polylog is real, or becomes
imaginary - which is not desirable).
Now at ultra-relativistic energies you will have particle-antiparticle
pair production in abundance. On the other hand, the chemical potentials
of particle and anti-particle are necessarily opposite to each other:
\mu_particle + \mu_antiparticle = .
Combined with the thermodynamic result \mu_Boson <=0 this only leaves
\mu_boson = for ultra-relativistic bosons. In fact, this is the chemical
potential of the photon (which quite definitely is an ultra-relativistic
boson, as m=0), but the result is more general: *Any* ultra-relativistic
(or zero rest mass) boson *must* have a chemical potential of *zero*,
at least if thermodynamic reasoning is still appropriate at
ultra-relativistic
energies.
However, this result is only true at ultra-relativistic energies. At *lower*
energies, massive bosons can even have a chemical potential that is
positive, as long as \mu < m. (See the appendix of http://www.arxiv.org/abs/gr-qc/0405010 for the
thermodynamics of an ideal gas of fermions and bosons with m arbitrary -
you can probably find the same results elsewhere, however, when I needed
them I was too lazy to look and therefore decided to derive the necessary
results myself - if you need to *understand* what you are doing, doing the
calculations oneself is usually quicker than looking them up ;-)
> What about for fermions? It seems by the same argument above \mu/T <
> always too. But I realize PolyLog can be "analytic continued" for x < -1
> on the real line, so this seems to allow \mu/T to take any value. So are
> there any restrictions on the chemical potential of fermions?
No, there aren't. For fermions the polylog-function is well defined for
*any* value of the chemical potential. The chemical potential of
ultra-relativistic fermions is not restricted. However, one still has the
restriction that fermions and anti-fermions have opposite chemical
potentials. Therefore, whenever an ultra-relativistic fermion species has
a *non-zero* chemical potential, we can distinguish between particles
and anti-particles in a thermodynamic sense.
The particle in the particle-antiparticle pair with *positive* chemical
potential would be called "normal matter", whereas its antiparticle
(with negative chemical potential) would be called "antimatter".
> One of the reasons I ask this is because it seems in
> antiparticle-particle reactions, say e- + e+ <-> \gamma + \gamma if we
> assume there is some kind of equilibrium (such as in the very early
> universe) then I suppose we can say something like \mu_e+ + \mu_e- =
> \mu_gamma + \mu_gamma = 0, because \mu_gamma = always. And therefore in
> chemical equilibrium the chemical potentials of antiparticles are
> negative of that of their particle counterparts. But what about bosons,
> since \mu_bosons/T < always?
I would think, that you *always* can say \mu_particle + \mu_antiparticle =0,
independent of the nature of the particle. At least this is what I thought I
have learned from QFT. Or does anyone in this group disagree? This
would in fact be interesting.