View Full Version : Maxwell energy versus speed distribution
A. Carkaci
Jun12-04, 07:21 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>"According to Maxwell energy distribution the probability that a\nparticle is in energy state E = 1/A exp(-E/kT)".\n\nHow can we interpret the above equation ?\n\nDoes It mean, for a given Temprature T, most of the particles are in\nthe lower energy states ?\n\nIf yes, does it make a contradiction with the maxwell SPEED\ndistribution ?\n\nAlso, please look the energy versus number-of-molecules graph at the\nweb address http://www.webchem.net/notes/how_far/kinetics/rate_factors.htm.\n\nDoes anyone carify the situation ?\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>"According to Maxwell energy distribution the probability that a
particle is in energy state E = 1/A \exp(-E/kT)".
How can we interpret the above equation ?
Does It mean, for a given Temprature T, most of the particles are in
the lower energy states ?
If yes, does it make a contradiction with the maxwell SPEED
distribution ?
Also, please look the energy versus number-of-molecules graph at the
web address http://www.webchem.net/notes/how_far/kinetics/rate_factors.htm.
Does anyone carify the situation ?
Peter Tobias
Jun14-04, 03:06 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>A. Carkaci:\n> "According to Maxwell energy distribution the probability that a\n> particle is in energy state E = 1/A exp(-E/kT)".\n> How can we interpret the above equation ?\n\nI will call the probability P, and I assume that your prefactor 1/A is\nthere to make the integral over all energy states equal to 1:\nP(E) = 1/A * exp(-E/(k*T))\n\nThen P for a certain energy level E is the ratio of the occupied states\nto all the states at this level. To get the number of particles, N, at\nan energy level, E,you have to multiply P with the number of states, M,\nat this energy level:\nN(E) = M(E) * 1/A * exp(-E/(k*T))\n\nM(E) may also depend on E, for example in a three dimensional gas or\nliquid, M(E) is a linear function of E. You can find an explanation for\nthis in quantum mechanics: Momentum space is discrete with a constant\ndensity, and we can describe the number of states close to a certain\nsize of momentum by the area of a sphere around the origin in momentum\nspace. The area depends linearly on p^2, therefore also linearly on E.\n\n> Does It mean, for a given Temprature T, most of the particles are in\n> the lower energy states ?\n\nNo, not in general. If M(E) becomes zero for E = 0, then P(E) will have\na maximum at an E larger than 0.\n\n> If yes, does it make a contradiction with the maxwell SPEED\n> distribution ?\n\nBecause M(E) depends linearly on E, no contradiction is found. On the\ncontrary, the speed distribution is one of the most famous example of an\nexperimentally confirmed Maxwell-Boltzmann distribution.\n\nRegards,\n\nPeter\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>A. Carkaci:
> "According to Maxwell energy distribution the probability that a
> particle is in energy state E = 1/A \exp(-E/kT)".
> How can we interpret the above equation ?
I will call the probability P, and I assume that your prefactor 1/A is
there to make the integral over all energy states equal to 1:
P(E) = 1/A * \exp(-E/(k*T))
Then P for a certain energy level E is the ratio of the occupied states
to all the states at this level. To get the number of particles, N, at
an energy level, E,you have to multiply P with the number of states, M,
at this energy level:
N(E) = M(E) * 1/A * \exp(-E/(k*T))
M(E) may also depend on E, for example in a three dimensional gas or
liquid, M(E) is a linear function of E. You can find an explanation for
this in quantum mechanics: Momentum space is discrete with a constant
density, and we can describe the number of states close to a certain
size of momentum by the area of a sphere around the origin in momentum
space. The area depends linearly on p^2, therefore also linearly on E.
> Does It mean, for a given Temprature T, most of the particles are in
> the lower energy states ?
No, not in general. If M(E) becomes zero for E = 0, then P(E) will have
a maximum at an E larger than .
> If yes, does it make a contradiction with the maxwell SPEED
> distribution ?
Because M(E) depends linearly on E, no contradiction is found. On the
contrary, the speed distribution is one of the most famous example of an
experimentally confirmed Maxwell-Boltzmann distribution.
Regards,
Peter
Igor Khavkine
Jun14-04, 03:07 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>On Sat, 12 Jun 2004 12:21:00 +0000, A. Carkaci wrote:\n\n> "According to Maxwell energy distribution the probability that a particle\n> is in energy state E = 1/A exp(-E/kT)".\n>\n> How can we interpret the above equation ?\n\nThe probability that a particular state is occupied is proportional to\nexp(-E/kT) where E is the energy of this state.\n\n> Does It mean, for a given Temperature T, most of the particles are in\n> the lower energy states ?\n\nNot necessarily, this depends on the number of accessible states for\neach energy. For example, given two energy states E1 < E2 with respective\nnumber of available states 1 and 1000, even if exp(-E1/kT) ~ .5 and\nexp(-E2/kT) ~ 0.05, on average you would find ~ .5 particles at energy E\nand ~ 5 particles at energy E2 (these numbers should be scaled by\nsomething like the total number of particles, but only their ratios matter).\n\n> If yes, does it make a contradiction with the maxwell SPEED distribution ?\n\nNo. The Maxwell velocity distribution is obtained by a simple change of\nvariables from the Maxwell energy distribution. For a classical ideal gas,\nthe states are labeled by points in phase space (x,v), x being the\nposition and v the velocity. The energy of a particle in a classical ideal\ngas is E(x,v) = mv^2/2. So the energy distribution says that the\nprobability of being in this state is\n\nP ~ exp(-mv^2/2kT) dV,\n\nwhere dV = dv_x dv_y dv_z. If we change from Cartesian to spherical\ncoordinates the volume element changes to dV = v^2 dv dOmega, where\ndOmega is the element of solid angle. Now we can rewrite the probability\nas\n\nP ~ v^2 exp(-mv^2/2kT) dv dOmega.\n\nAfter integrating out the angular degrees of freedom and normalizing,\nwe obtain the Maxwell velocity distribution.\n\nThe lesson to learn here is that the number of accessible states per, say,\nenergy and per, say, velocity are not the same. This leads to a\nmodification of the probability distribution when expressed in different\nvariables.\n\n> Also, please look the energy versus number-of-molecules graph at the web\n> address http://www.webchem.net/notes/how_far/kinetics/rate_factors.htm.\n>\n> Does anyone carify the situation ?\n\nIf you plot v^2 exp(-mv^2/2kT) you get a similar graph. Does that clarify\nthe situation?\n\nHope this helps.\n\nIgor\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 Sat, 12 Jun 2004 12:21:00 +0000, A. Carkaci wrote:
> "According to Maxwell energy distribution the probability that a particle
> is in energy state E = 1/A \exp(-E/kT)".
>
> How can we interpret the above equation ?
The probability that a particular state is occupied is proportional to
\exp(-E/kT) where E is the energy of this state.
> Does It mean, for a given Temperature T, most of the particles are in
> the lower energy states ?
Not necessarily, this depends on the number of accessible states for
each energy. For example, given two energy states E1 < E2 with respective
number of available states 1 and 1000, even if \exp(-E1/kT) ~ .5 and
\exp(-E2/kT) ~ .05, on average you would find ~ .5 particles at energy E
and ~ 5 particles at energy E2 (these numbers should be scaled by
something like the total number of particles, but only their ratios matter).
> If yes, does it make a contradiction with the maxwell SPEED distribution ?
No. The Maxwell velocity distribution is obtained by a simple change of
variables from the Maxwell energy distribution. For a classical ideal gas,
the states are labeled by points in phase space (x,v), x being the
position and v the velocity. The energy of a particle in a classical ideal
gas is E(x,v) = mv^2/2. So the energy distribution says that the
probability of being in this state is
P ~ \exp(-mv^2/2kT) dV,
where dV = dv_x dv_y dv_z. If we change from Cartesian to spherical
coordinates the volume element changes to dV = v^2 dv dOmega, where
dOmega is the element of solid angle. Now we can rewrite the probability
as
P ~ v^2 \exp(-mv^2/2kT) dv dOmega.
After integrating out the angular degrees of freedom and normalizing,
we obtain the Maxwell velocity distribution.
The lesson to learn here is that the number of accessible states per, say,
energy and per, say, velocity are not the same. This leads to a
modification of the probability distribution when expressed in different
variables.
> Also, please look the energy versus number-of-molecules graph at the web
> address http://www.webchem.net/notes/how_far/kinetics/rate_factors.htm.
>
> Does anyone carify the situation ?
If you plot v^2 \exp(-mv^2/2kT) you get a similar graph. Does that clarify
the situation?
Hope this helps.
Igor
A. Carkaci
Jun16-04, 05:27 PM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Peter Tobias <tobiasep@comcast.net> wrote in message news:<c6OdnRx9uby3j1bdRVn-jw@comcast.com>...\n> A. Carkaci:\n> > "According to Maxwell energy distribution the probability that a\n> > particle is in energy state E = 1/A exp(-E/kT)".\n> > How can we interpret the above equation ?\n>\n> I will call the probability P, and I assume that your prefactor 1/A is\n> there to make the integral over all energy states equal to 1:\n> P(E) = 1/A * exp(-E/(k*T))\n>\n> Then P for a certain energy level E is the ratio of the occupied states\n> to all the states at this level. To get the number of particles, N, at\n> an energy level, E,you have to multiply P with the number of states, M,\n> at this energy level:\n> N(E) = M(E) * 1/A * exp(-E/(k*T))\n>\n> M(E) may also depend on E, for example in a three dimensional gas or\n> liquid, M(E) is a linear function of E. You can find an explanation for\n> this in quantum mechanics: Momentum space is discrete with a constant\n> density, and we can describe the number of states close to a certain\n> size of momentum by the area of a sphere around the origin in momentum\n> space. The area depends linearly on p^2, therefore also linearly on E.\n>\n> > Does It mean, for a given Temprature T, most of the particles are in\n> > the lower energy states ?\n>\n> No, not in general. If M(E) becomes zero for E = 0, then P(E) will have\n> a maximum at an E larger than 0.\n>\n> > If yes, does it make a contradiction with the maxwell SPEED\n> > distribution ?\n>\n> Because M(E) depends linearly on E, no contradiction is found. On the\n> contrary, the speed distribution is one of the most famous example of an\n> experimentally confirmed Maxwell-Boltzmann distribution.\n>\n> Regards,\n>\n> Peter\n\nThank you so much for the function N(E) = M(E) * 1/A * exp(-E/(k*T)).\nEverything is now clear.\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>Peter Tobias <tobiasep@comcast.net> wrote in message news:<c6OdnRx9uby3j1bdRVn-jw@comcast.com>...
> A. Carkaci:
> > "According to Maxwell energy distribution the probability that a
> > particle is in energy state E = 1/A \exp(-E/kT)".
> > How can we interpret the above equation ?
>
> I will call the probability P, and I assume that your prefactor 1/A is
> there to make the integral over all energy states equal to 1:
> P(E) = 1/A * \exp(-E/(k*T))
>
> Then P for a certain energy level E is the ratio of the occupied states
> to all the states at this level. To get the number of particles, N, at
> an energy level, E,you have to multiply P with the number of states, M,
> at this energy level:
> N(E) = M(E) * 1/A * \exp(-E/(k*T))
>
> M(E) may also depend on E, for example in a three dimensional gas or
> liquid, M(E) is a linear function of E. You can find an explanation for
> this in quantum mechanics: Momentum space is discrete with a constant
> density, and we can describe the number of states close to a certain
> size of momentum by the area of a sphere around the origin in momentum
> space. The area depends linearly on p^2, therefore also linearly on E.
>
> > Does It mean, for a given Temprature T, most of the particles are in
> > the lower energy states ?
>
> No, not in general. If M(E) becomes zero for E = 0, then P(E) will have
> a maximum at an E larger than .
>
> > If yes, does it make a contradiction with the maxwell SPEED
> > distribution ?
>
> Because M(E) depends linearly on E, no contradiction is found. On the
> contrary, the speed distribution is one of the most famous example of an
> experimentally confirmed Maxwell-Boltzmann distribution.
>
> Regards,
>
> Peter
Thank you so much for the function N(E) = M(E) * 1/A * \exp(-E/(k*T)).
Everything is now clear.
carkaci
Dec22-04, 06:00 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>Peter Tobias wrote:\n> A. Carkaci:\n> > "According to Maxwell energy distribution the probability that a\n> > particle is in energy state E = 1/A exp(-E/kT)".\n> > How can we interpret the above equation ?\n>\n> I will call the probability P, and I assume that your prefactor 1/A is\n> there to make the integral over all energy states equal to 1:\n> P(E) = 1/A * exp(-E/(k*T))\n>\n> Then P for a certain energy level E is the ratio of the occupied states\n> to all the states at this level. To get the number of particles, N, at\n> an energy level, E,you have to multiply P with the number of states, M,\n> at this energy level:\n> N(E) = M(E) * 1/A * exp(-E/(k*T))\n>\n> M(E) may also depend on E, for example in a three dimensional gas or\n> liquid, M(E) is a linear function of E. You can find an explanation for\n> this in quantum mechanics: Momentum space is discrete with a constant\n\n> density, and we can describe the number of states close to a certain\n> size of momentum by the area of a sphere around the origin in momentum\n> space. The area depends linearly on p^2, therefore also linearly on E.\n>\n> > Does It mean, for a given Temprature T, most of the particles are in\n> > the lower energy states ?\n>\n> No, not in general. If M(E) becomes zero for E = 0, then P(E) will have\n> a maximum at an E larger than 0.\n>\n> > If yes, does it make a contradiction with the maxwell SPEED\n> > distribution ?\n>\n> Because M(E) depends linearly on E, no contradiction is found. On the\n> contrary, the speed distribution is one of the most famous example of an\n> experimentally confirmed Maxwell-Boltzmann distribution.\n>\n> Regards,\n>\n> Peter\n\nSuppose, that M(E) depends linearly on E, what is the most probable\nenergy state in terms of T ?\n\nThanks,\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Peter Tobias wrote:
> A. Carkaci:
> > "According to Maxwell energy distribution the probability that a
> > particle is in energy state E = 1/A \exp(-E/kT)".
> > How can we interpret the above equation ?
>
> I will call the probability P, and I assume that your prefactor 1/A is
> there to make the integral over all energy states equal to 1:
> P(E) = 1/A * \exp(-E/(k*T))
>
> Then P for a certain energy level E is the ratio of the occupied states
> to all the states at this level. To get the number of particles, N, at
> an energy level, E,you have to multiply P with the number of states, M,
> at this energy level:
> N(E) = M(E) * 1/A * \exp(-E/(k*T))
>
> M(E) may also depend on E, for example in a three dimensional gas or
> liquid, M(E) is a linear function of E. You can find an explanation for
> this in quantum mechanics: Momentum space is discrete with a constant
> density, and we can describe the number of states close to a certain
> size of momentum by the area of a sphere around the origin in momentum
> space. The area depends linearly on p^2, therefore also linearly on E.
>
> > Does It mean, for a given Temprature T, most of the particles are in
> > the lower energy states ?
>
> No, not in general. If M(E) becomes zero for E = 0, then P(E) will have
> a maximum at an E larger than .
>
> > If yes, does it make a contradiction with the maxwell SPEED
> > distribution ?
>
> Because M(E) depends linearly on E, no contradiction is found. On the
> contrary, the speed distribution is one of the most famous example of an
> experimentally confirmed Maxwell-Boltzmann distribution.
>
> Regards,
>
> Peter
Suppose, that M(E) depends linearly on E, what is the most probable
energy state in terms of T ?
Thanks,
Peter Tobias
Jan8-05, 04:00 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>carkaci:\n> Suppose, that M(E) depends linearly on E, what is the most probable\n> energy state in terms of T ?\n\nYou can find the maximum in the usual way, by taking the derivative of\nP(E)=M(E)*exp(-E/kT)/A with respect to E. With a linear M(E), we have\nP(E) = C * E * exp(-E / (k*T)) / A\nand\ndP/dE = C / A * ( exp(...) - E / (k*T) * exp(...))\n\ndP/dE is zero, if and only if (1-E/kT)=0. The (only) maximum of P(E)\nis therefore at E = kT.\nRegards,\n\nPeter Tobias\n\nPS\nNo guarantee against errors in calculation.\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>carkaci:
> Suppose, that M(E) depends linearly on E, what is the most probable
> energy state in terms of T ?
You can find the maximum in the usual way, by taking the derivative of
P(E)=M(E)*\exp(-E/kT)/A with respect to E. With a linear M(E), we have
P(E) = C * E * \exp(-E / (k*T)) / A
and
dP/dE = C / A * ( \exp(...) - E / (k*T) * \exp(...))dP/dE is zero, if and only if (1-E/kT)=0. The (only) maximum of P(E)
is therefore at E = kT.
Regards,
Peter Tobias
PS
No guarantee against errors in calculation.
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