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Statistical mechanics and Planck constant universality
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In questions that ask about Planck's constant entering into statistical mechanics, a common and accepted answer is that Planck's constant is an arbitrary normalization that falls out when calculating experimentally measurable quantities.
Specifically, it's said in the above questions that $h$ enters to de-dimensionalize the product $dp dq$ or to normalize $dp dq$ to count states, and the choice for such a constant with units of action is nearly arbitrary.
To the contrary, consider how one calculates quantities in the grand canonical ensemble with a variable number of particles - mentioned in this question.
In the grand canonical ensemble, the key potential is the grand potential, which is $$Phi = U - TS-mu N .$$ The link to statistical mechanics comes from
$$Phi = -kT ln(mathcalZ), $$
where $$mathcalZ = sum_N e^beta mu N Z(V,N,T).$$
For simplicity, consider a classical partition function $Z$ for N non-interacting particles. We have then that
$$Z = frac1h^3N N! Big(int d^3p d^3q , e^-beta E(p, q)Big)^N$$
This yields in turn that
$$mathcalZ = e^frac1h^3 Big(int d^3p d^3q , e^-beta E(p, q)Big)e^beta mu$$
and in turn that $Phi$ is proportional to $frac1h^3$:
$$Phi = -kT frac1h^3 Big(int d^3p d^3q , e^- fracE(p, q)kTBig)e^fracmukT. $$
For example, for non-interacting massless particles with $E = cp$ in a box with two internal degrees of freedom with $mu = 0$ (i.e. a naive picture of light, without any $h$ put into the energy or whatnot), we have
$$Phi = -frac16 pi k^4h^3 c^3 T^4 V.$$ This leads to a pressure
$$P = frac16 pi k^4h^3 c^3 T^4$$ from the thermodynamic relation above.
As you can see, $h$ enters into an experimentally measurable quantity! Thus one could experimentally measure Planck's constant via the pressure of such a gas. I'll leave it to you to check that $h$ also enters into the pressure of a non-relativistic gas at a fixed $mu$.
I will take from the above arguments that $h$ is not simply an arbitrary constant for de-dimensionalizing the product $dp dq$. Given this point of view that the de-dimensionalization of $dp dq$ has experimental ramifications, can we show that for every system in thermodynamic equilibrium that the de-dimensionalizing constant $h$ must be a universal constant? That is, since the de-dimensionalization is not arbitrary, can we show the de-dimensionalizing constant must be the same for all equilibrium systems in classical statistical mechanics?
quantum-mechanics thermodynamics statistical-mechanics physical-constants
asked Sep 8 at 20:33
user196574
514
add a comment |Â
up vote
5
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In questions that ask about Planck's constant entering into statistical mechanics, a common and accepted answer is that Planck's constant is an arbitrary normalization that falls out when calculating experimentally measurable quantities.
Specifically, it's said in the above questions that $h$ enters to de-dimensionalize the product $dp dq$ or to normalize $dp dq$ to count states, and the choice for such a constant with units of action is nearly arbitrary.
To the contrary, consider how one calculates quantities in the grand canonical ensemble with a variable number of particles - mentioned in this question.
In the grand canonical ensemble, the key potential is the grand potential, which is $$Phi = U - TS-mu N .$$ The link to statistical mechanics comes from
$$Phi = -kT ln(mathcalZ), $$
where $$mathcalZ = sum_N e^beta mu N Z(V,N,T).$$
For simplicity, consider a classical partition function $Z$ for N non-interacting particles. We have then that
$$Z = frac1h^3N N! Big(int d^3p d^3q , e^-beta E(p, q)Big)^N$$
This yields in turn that
$$mathcalZ = e^frac1h^3 Big(int d^3p d^3q , e^-beta E(p, q)Big)e^beta mu$$
and in turn that $Phi$ is proportional to $frac1h^3$:
$$Phi = -kT frac1h^3 Big(int d^3p d^3q , e^- fracE(p, q)kTBig)e^fracmukT. $$
For example, for non-interacting massless particles with $E = cp$ in a box with two internal degrees of freedom with $mu = 0$ (i.e. a naive picture of light, without any $h$ put into the energy or whatnot), we have
$$Phi = -frac16 pi k^4h^3 c^3 T^4 V.$$ This leads to a pressure
$$P = frac16 pi k^4h^3 c^3 T^4$$ from the thermodynamic relation above.
As you can see, $h$ enters into an experimentally measurable quantity! Thus one could experimentally measure Planck's constant via the pressure of such a gas. I'll leave it to you to check that $h$ also enters into the pressure of a non-relativistic gas at a fixed $mu$.
I will take from the above arguments that $h$ is not simply an arbitrary constant for de-dimensionalizing the product $dp dq$. Given this point of view that the de-dimensionalization of $dp dq$ has experimental ramifications, can we show that for every system in thermodynamic equilibrium that the de-dimensionalizing constant $h$ must be a universal constant? That is, since the de-dimensionalization is not arbitrary, can we show the de-dimensionalizing constant must be the same for all equilibrium systems in classical statistical mechanics?
quantum-mechanics thermodynamics statistical-mechanics physical-constants
asked Sep 8 at 20:33
user196574
514
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
In questions that ask about Planck's constant entering into statistical mechanics, a common and accepted answer is that Planck's constant is an arbitrary normalization that falls out when calculating experimentally measurable quantities.
Specifically, it's said in the above questions that $h$ enters to de-dimensionalize the product $dp dq$ or to normalize $dp dq$ to count states, and the choice for such a constant with units of action is nearly arbitrary.
To the contrary, consider how one calculates quantities in the grand canonical ensemble with a variable number of particles - mentioned in this question.
In the grand canonical ensemble, the key potential is the grand potential, which is $$Phi = U - TS-mu N .$$ The link to statistical mechanics comes from
$$Phi = -kT ln(mathcalZ), $$
where $$mathcalZ = sum_N e^beta mu N Z(V,N,T).$$
For simplicity, consider a classical partition function $Z$ for N non-interacting particles. We have then that
$$Z = frac1h^3N N! Big(int d^3p d^3q , e^-beta E(p, q)Big)^N$$
This yields in turn that
$$mathcalZ = e^frac1h^3 Big(int d^3p d^3q , e^-beta E(p, q)Big)e^beta mu$$
and in turn that $Phi$ is proportional to $frac1h^3$:
$$Phi = -kT frac1h^3 Big(int d^3p d^3q , e^- fracE(p, q)kTBig)e^fracmukT. $$
For example, for non-interacting massless particles with $E = cp$ in a box with two internal degrees of freedom with $mu = 0$ (i.e. a naive picture of light, without any $h$ put into the energy or whatnot), we have
$$Phi = -frac16 pi k^4h^3 c^3 T^4 V.$$ This leads to a pressure
$$P = frac16 pi k^4h^3 c^3 T^4$$ from the thermodynamic relation above.
As you can see, $h$ enters into an experimentally measurable quantity! Thus one could experimentally measure Planck's constant via the pressure of such a gas. I'll leave it to you to check that $h$ also enters into the pressure of a non-relativistic gas at a fixed $mu$.
I will take from the above arguments that $h$ is not simply an arbitrary constant for de-dimensionalizing the product $dp dq$. Given this point of view that the de-dimensionalization of $dp dq$ has experimental ramifications, can we show that for every system in thermodynamic equilibrium that the de-dimensionalizing constant $h$ must be a universal constant? That is, since the de-dimensionalization is not arbitrary, can we show the de-dimensionalizing constant must be the same for all equilibrium systems in classical statistical mechanics?
quantum-mechanics thermodynamics statistical-mechanics physical-constants
asked Sep 8 at 20:33
user196574
514
In questions that ask about Planck's constant entering into statistical mechanics, a common and accepted answer is that Planck's constant is an arbitrary normalization that falls out when calculating experimentally measurable quantities.
Specifically, it's said in the above questions that $h$ enters to de-dimensionalize the product $dp dq$ or to normalize $dp dq$ to count states, and the choice for such a constant with units of action is nearly arbitrary.
To the contrary, consider how one calculates quantities in the grand canonical ensemble with a variable number of particles - mentioned in this question.
In the grand canonical ensemble, the key potential is the grand potential, which is $$Phi = U - TS-mu N .$$ The link to statistical mechanics comes from
$$Phi = -kT ln(mathcalZ), $$
where $$mathcalZ = sum_N e^beta mu N Z(V,N,T).$$
For simplicity, consider a classical partition function $Z$ for N non-interacting particles. We have then that
$$Z = frac1h^3N N! Big(int d^3p d^3q , e^-beta E(p, q)Big)^N$$
This yields in turn that
$$mathcalZ = e^frac1h^3 Big(int d^3p d^3q , e^-beta E(p, q)Big)e^beta mu$$
and in turn that $Phi$ is proportional to $frac1h^3$:
$$Phi = -kT frac1h^3 Big(int d^3p d^3q , e^- fracE(p, q)kTBig)e^fracmukT. $$
For example, for non-interacting massless particles with $E = cp$ in a box with two internal degrees of freedom with $mu = 0$ (i.e. a naive picture of light, without any $h$ put into the energy or whatnot), we have
$$Phi = -frac16 pi k^4h^3 c^3 T^4 V.$$ This leads to a pressure
$$P = frac16 pi k^4h^3 c^3 T^4$$ from the thermodynamic relation above.
As you can see, $h$ enters into an experimentally measurable quantity! Thus one could experimentally measure Planck's constant via the pressure of such a gas. I'll leave it to you to check that $h$ also enters into the pressure of a non-relativistic gas at a fixed $mu$.
I will take from the above arguments that $h$ is not simply an arbitrary constant for de-dimensionalizing the product $dp dq$. Given this point of view that the de-dimensionalization of $dp dq$ has experimental ramifications, can we show that for every system in thermodynamic equilibrium that the de-dimensionalizing constant $h$ must be a universal constant? That is, since the de-dimensionalization is not arbitrary, can we show the de-dimensionalizing constant must be the same for all equilibrium systems in classical statistical mechanics?
quantum-mechanics thermodynamics statistical-mechanics physical-constants
quantum-mechanics thermodynamics statistical-mechanics physical-constants
asked Sep 8 at 20:33
user196574
514
asked Sep 8 at 20:33
user196574
514
asked Sep 8 at 20:33
user196574
514
asked Sep 8 at 20:33
user196574
514
asked Sep 8 at 20:33
user196574
514
514
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add a comment |Â
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I do not think that the value of the normalization constant that de-dimensionalizes the phase space integral in the partition function is arbitrary. Already in 1912, Sackur and and Tetrode used vapor pressure data of mercury to determine its numerical value, and found that it was the same as the constant Planck discovered when studying blackbody radiation.
I do not know if at that time there was a fundamental reason to assume that this constant is universal, i.e., independent of the used material. Heisenberg's uncertainty principle, discovered 15 years later, provided that reason.
I highly recommend this paper for an insightful discussion and historical perspective of Sackur and Tetrode's work.
answered Sep 8 at 22:06
user8153
33016
1
(h/t to Semiclassical who helped me figure this out) The Sackur and Tetrode equation only yields an experimentally measurable $h$ from vapor pressure if you have a good model for the heat capacity of the solid as a function of temperature. They both use variants of the Einstein model for this [see the bottom of page 10 in the paper you link], which is a quantum model, so this is not establishing the claim that $h$ is measurable purely in classical statistical physics.
– ACuriousMind♦
yesterday
1
@ACuriousMind Thank you for looking into this. Two comments: (1) As you point out, the cited paper used an Einstein-ish model to compute the heat capacity of solid mercury. But is that strictly necessary, or due to lack of experimental data? I would assume that at least in principle the heat capcaity could be measured, which would make the Sackur-Tetrode approach a viable way to determine $h$ wihout an underlying quantum-mechanical model.
– user8153
yesterday
1
@ACuriousMind (2) Perhaps more importantly, I believe it is incorrect to say that the value of the quantity $h$ that appears in the classical partition function is "arbitrary", which to me sounds like "it doesn't matter which value you use, the results will always be the same". It seems to me that there is only one correct valuer (6.626E-34 m^2 kg / s), but this value is not predicted by classical theory, and must be obtained from experiment instead.
– user8153
yesterday
1
@user8153 Maybe the following is a better way to say what $h$ is: In pure classical statistical mechanics, the value of $h$ is irrelevant - you can express all physically measurable quantities in terms that do not involve it. Not only is there no classical prediction, there is also no purely classical measurement prescription. Classical statistical mechanics is agnostic as to whether $h$ is a fundamental feature of the universe or a trick of the theorist. Our knowledge of quantum mechanics nowadays fixes its role firmly to be the former.
– ACuriousMind♦
yesterday
1
Note also that a measurement of the heat capacity at discrete points is not enough - the ST equation contains an integral over the heat capacity at constant pressure all the way from absolute zero. As good as we may be in figuring out experiments, I doubt that there is an experiment that allows you to collect the data needed for a good approximation of this integral.
– ACuriousMind♦
yesterday
 |Â
show 5 more comments
up vote
0
down vote
The derivation in the question is misleading because it prematurely sets the chemical potential to zero. Even in systems where it is zero, you must carry it with you until the final evaluation of expressions, since it is a free parameter of a macrostate in the grand canonical ensemble.
In particular, the chemical potential contains a $ln(h^3/textstuff)$ factor (e.g. the classical ideal non-relativistic gas has $mu propto lnleft(fraclambda^3 NVright)$, with $lambda$ the thermal de Broglie wavelength proportional to $h$), that, since it is the fugacity that enters into physical quantities, cancels against the $h^3$ in the denominator.
Therefore, your approach does not necessitate a universal classical value for $h$, since the results can be re-expressed in terms of quantities that do not involve $h$. Note, in particular, that you cannot experimentally directly measure the chemical potential $mu$ - you need to compute it from indirect measurements, with formulae that in this case also will involve $h$.
answered yesterday
ACuriousMind♦
68k17118289
Thanks for the response. In the grand canonical ensemble, the fixed parameters are T, V, and $mu$. The pressure and number of particles fluctuate because they are not fixed. I don't believe that $mu$ is a free parameter.
– user196574
yesterday
Also, there are cases where nature "fixes" $mu$ and not number of particles, the classic example being a photon gas. However, we can even fix $mu$ and not the number of particles by affixing our system to a "particle number" bath. This is how I think about it, please let me hear your thoughts: A heat bath can give heat without noticeably changing its temperature until an adjoined smaller system is at the same temperature. A particle number bath can give particles without noticeably changing its chemical potential until an adjoined smaller system is at the same chemical potential.
– user196574
yesterday
@user196574 $mu$ is fixed for a specific macrostate in the grand canonical example. It's a free parameter in the sense that derivations that are supposed to hold for a system must hold for all possible values of this parameter, just like they must hold for all values of $T$ and $V$ (that obey the equation of state). You can't just say "but $T = 1mathrmK$ in my system". That's not a property of the system, but of a state of the system. In any experimental test, you would have to establish that indeed $mu = 0$ for the state you're measuring, thus losing the ability to measure $h$.
– ACuriousMind♦
yesterday
I think I will need to refresh my knowledge on the chemical potential before returning to the role of $h$ in the grand canonical ensemble. Specifically, whether it is necessary to have some theoretical expression for $mu$ (that depends on $h$) in order to be able to set $mu$, or whether they may be ways to experimentally determine a chemical potential without knowledge of $h$, just as there are ways to experimentally determine a volume without knowledge of $h$.
– user196574
yesterday
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2 Answers
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2 Answers
2
active
oldest
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active
oldest
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active
oldest
votes
up vote
5
down vote
I do not think that the value of the normalization constant that de-dimensionalizes the phase space integral in the partition function is arbitrary. Already in 1912, Sackur and and Tetrode used vapor pressure data of mercury to determine its numerical value, and found that it was the same as the constant Planck discovered when studying blackbody radiation.
I do not know if at that time there was a fundamental reason to assume that this constant is universal, i.e., independent of the used material. Heisenberg's uncertainty principle, discovered 15 years later, provided that reason.
I highly recommend this paper for an insightful discussion and historical perspective of Sackur and Tetrode's work.
answered Sep 8 at 22:06
user8153
33016
1
(h/t to Semiclassical who helped me figure this out) The Sackur and Tetrode equation only yields an experimentally measurable $h$ from vapor pressure if you have a good model for the heat capacity of the solid as a function of temperature. They both use variants of the Einstein model for this [see the bottom of page 10 in the paper you link], which is a quantum model, so this is not establishing the claim that $h$ is measurable purely in classical statistical physics.
– ACuriousMind♦
yesterday
1
@ACuriousMind Thank you for looking into this. Two comments: (1) As you point out, the cited paper used an Einstein-ish model to compute the heat capacity of solid mercury. But is that strictly necessary, or due to lack of experimental data? I would assume that at least in principle the heat capcaity could be measured, which would make the Sackur-Tetrode approach a viable way to determine $h$ wihout an underlying quantum-mechanical model.
– user8153
yesterday
1
@ACuriousMind (2) Perhaps more importantly, I believe it is incorrect to say that the value of the quantity $h$ that appears in the classical partition function is "arbitrary", which to me sounds like "it doesn't matter which value you use, the results will always be the same". It seems to me that there is only one correct valuer (6.626E-34 m^2 kg / s), but this value is not predicted by classical theory, and must be obtained from experiment instead.
– user8153
yesterday
1
@user8153 Maybe the following is a better way to say what $h$ is: In pure classical statistical mechanics, the value of $h$ is irrelevant - you can express all physically measurable quantities in terms that do not involve it. Not only is there no classical prediction, there is also no purely classical measurement prescription. Classical statistical mechanics is agnostic as to whether $h$ is a fundamental feature of the universe or a trick of the theorist. Our knowledge of quantum mechanics nowadays fixes its role firmly to be the former.
– ACuriousMind♦
yesterday
1
Note also that a measurement of the heat capacity at discrete points is not enough - the ST equation contains an integral over the heat capacity at constant pressure all the way from absolute zero. As good as we may be in figuring out experiments, I doubt that there is an experiment that allows you to collect the data needed for a good approximation of this integral.
– ACuriousMind♦
yesterday
 |Â
show 5 more comments
up vote
5
down vote
I do not think that the value of the normalization constant that de-dimensionalizes the phase space integral in the partition function is arbitrary. Already in 1912, Sackur and and Tetrode used vapor pressure data of mercury to determine its numerical value, and found that it was the same as the constant Planck discovered when studying blackbody radiation.
I do not know if at that time there was a fundamental reason to assume that this constant is universal, i.e., independent of the used material. Heisenberg's uncertainty principle, discovered 15 years later, provided that reason.
I highly recommend this paper for an insightful discussion and historical perspective of Sackur and Tetrode's work.
answered Sep 8 at 22:06
user8153
33016
1
(h/t to Semiclassical who helped me figure this out) The Sackur and Tetrode equation only yields an experimentally measurable $h$ from vapor pressure if you have a good model for the heat capacity of the solid as a function of temperature. They both use variants of the Einstein model for this [see the bottom of page 10 in the paper you link], which is a quantum model, so this is not establishing the claim that $h$ is measurable purely in classical statistical physics.
– ACuriousMind♦
yesterday
1
@ACuriousMind Thank you for looking into this. Two comments: (1) As you point out, the cited paper used an Einstein-ish model to compute the heat capacity of solid mercury. But is that strictly necessary, or due to lack of experimental data? I would assume that at least in principle the heat capcaity could be measured, which would make the Sackur-Tetrode approach a viable way to determine $h$ wihout an underlying quantum-mechanical model.
– user8153
yesterday
1
@ACuriousMind (2) Perhaps more importantly, I believe it is incorrect to say that the value of the quantity $h$ that appears in the classical partition function is "arbitrary", which to me sounds like "it doesn't matter which value you use, the results will always be the same". It seems to me that there is only one correct valuer (6.626E-34 m^2 kg / s), but this value is not predicted by classical theory, and must be obtained from experiment instead.
– user8153
yesterday
1
@user8153 Maybe the following is a better way to say what $h$ is: In pure classical statistical mechanics, the value of $h$ is irrelevant - you can express all physically measurable quantities in terms that do not involve it. Not only is there no classical prediction, there is also no purely classical measurement prescription. Classical statistical mechanics is agnostic as to whether $h$ is a fundamental feature of the universe or a trick of the theorist. Our knowledge of quantum mechanics nowadays fixes its role firmly to be the former.
– ACuriousMind♦
yesterday
1
Note also that a measurement of the heat capacity at discrete points is not enough - the ST equation contains an integral over the heat capacity at constant pressure all the way from absolute zero. As good as we may be in figuring out experiments, I doubt that there is an experiment that allows you to collect the data needed for a good approximation of this integral.
– ACuriousMind♦
yesterday
 |Â
show 5 more comments
up vote
5
down vote
up vote
5
down vote
I do not think that the value of the normalization constant that de-dimensionalizes the phase space integral in the partition function is arbitrary. Already in 1912, Sackur and and Tetrode used vapor pressure data of mercury to determine its numerical value, and found that it was the same as the constant Planck discovered when studying blackbody radiation.
I do not know if at that time there was a fundamental reason to assume that this constant is universal, i.e., independent of the used material. Heisenberg's uncertainty principle, discovered 15 years later, provided that reason.
I highly recommend this paper for an insightful discussion and historical perspective of Sackur and Tetrode's work.
answered Sep 8 at 22:06
user8153
33016
I do not think that the value of the normalization constant that de-dimensionalizes the phase space integral in the partition function is arbitrary. Already in 1912, Sackur and and Tetrode used vapor pressure data of mercury to determine its numerical value, and found that it was the same as the constant Planck discovered when studying blackbody radiation.
I do not know if at that time there was a fundamental reason to assume that this constant is universal, i.e., independent of the used material. Heisenberg's uncertainty principle, discovered 15 years later, provided that reason.
I highly recommend this paper for an insightful discussion and historical perspective of Sackur and Tetrode's work.
answered Sep 8 at 22:06
user8153
33016
answered Sep 8 at 22:06
user8153
33016
answered Sep 8 at 22:06
user8153
33016
answered Sep 8 at 22:06
user8153
33016
33016
1
(h/t to Semiclassical who helped me figure this out) The Sackur and Tetrode equation only yields an experimentally measurable $h$ from vapor pressure if you have a good model for the heat capacity of the solid as a function of temperature. They both use variants of the Einstein model for this [see the bottom of page 10 in the paper you link], which is a quantum model, so this is not establishing the claim that $h$ is measurable purely in classical statistical physics.
– ACuriousMind♦
yesterday
1
@ACuriousMind Thank you for looking into this. Two comments: (1) As you point out, the cited paper used an Einstein-ish model to compute the heat capacity of solid mercury. But is that strictly necessary, or due to lack of experimental data? I would assume that at least in principle the heat capcaity could be measured, which would make the Sackur-Tetrode approach a viable way to determine $h$ wihout an underlying quantum-mechanical model.
– user8153
yesterday
1
@ACuriousMind (2) Perhaps more importantly, I believe it is incorrect to say that the value of the quantity $h$ that appears in the classical partition function is "arbitrary", which to me sounds like "it doesn't matter which value you use, the results will always be the same". It seems to me that there is only one correct valuer (6.626E-34 m^2 kg / s), but this value is not predicted by classical theory, and must be obtained from experiment instead.
– user8153
yesterday
1
@user8153 Maybe the following is a better way to say what $h$ is: In pure classical statistical mechanics, the value of $h$ is irrelevant - you can express all physically measurable quantities in terms that do not involve it. Not only is there no classical prediction, there is also no purely classical measurement prescription. Classical statistical mechanics is agnostic as to whether $h$ is a fundamental feature of the universe or a trick of the theorist. Our knowledge of quantum mechanics nowadays fixes its role firmly to be the former.
– ACuriousMind♦
yesterday
1
Note also that a measurement of the heat capacity at discrete points is not enough - the ST equation contains an integral over the heat capacity at constant pressure all the way from absolute zero. As good as we may be in figuring out experiments, I doubt that there is an experiment that allows you to collect the data needed for a good approximation of this integral.
– ACuriousMind♦
yesterday
 |Â
show 5 more comments
1
(h/t to Semiclassical who helped me figure this out) The Sackur and Tetrode equation only yields an experimentally measurable $h$ from vapor pressure if you have a good model for the heat capacity of the solid as a function of temperature. They both use variants of the Einstein model for this [see the bottom of page 10 in the paper you link], which is a quantum model, so this is not establishing the claim that $h$ is measurable purely in classical statistical physics.
– ACuriousMind♦
yesterday
1
@ACuriousMind Thank you for looking into this. Two comments: (1) As you point out, the cited paper used an Einstein-ish model to compute the heat capacity of solid mercury. But is that strictly necessary, or due to lack of experimental data? I would assume that at least in principle the heat capcaity could be measured, which would make the Sackur-Tetrode approach a viable way to determine $h$ wihout an underlying quantum-mechanical model.
– user8153
yesterday
1
@ACuriousMind (2) Perhaps more importantly, I believe it is incorrect to say that the value of the quantity $h$ that appears in the classical partition function is "arbitrary", which to me sounds like "it doesn't matter which value you use, the results will always be the same". It seems to me that there is only one correct valuer (6.626E-34 m^2 kg / s), but this value is not predicted by classical theory, and must be obtained from experiment instead.
– user8153
yesterday
1
@user8153 Maybe the following is a better way to say what $h$ is: In pure classical statistical mechanics, the value of $h$ is irrelevant - you can express all physically measurable quantities in terms that do not involve it. Not only is there no classical prediction, there is also no purely classical measurement prescription. Classical statistical mechanics is agnostic as to whether $h$ is a fundamental feature of the universe or a trick of the theorist. Our knowledge of quantum mechanics nowadays fixes its role firmly to be the former.
– ACuriousMind♦
yesterday
1
Note also that a measurement of the heat capacity at discrete points is not enough - the ST equation contains an integral over the heat capacity at constant pressure all the way from absolute zero. As good as we may be in figuring out experiments, I doubt that there is an experiment that allows you to collect the data needed for a good approximation of this integral.
– ACuriousMind♦
yesterday
1
1
(h/t to Semiclassical who helped me figure this out) The Sackur and Tetrode equation only yields an experimentally measurable $h$ from vapor pressure if you have a good model for the heat capacity of the solid as a function of temperature. They both use variants of the Einstein model for this [see the bottom of page 10 in the paper you link], which is a quantum model, so this is not establishing the claim that $h$ is measurable purely in classical statistical physics.
– ACuriousMind♦
yesterday
(h/t to Semiclassical who helped me figure this out) The Sackur and Tetrode equation only yields an experimentally measurable $h$ from vapor pressure if you have a good model for the heat capacity of the solid as a function of temperature. They both use variants of the Einstein model for this [see the bottom of page 10 in the paper you link], which is a quantum model, so this is not establishing the claim that $h$ is measurable purely in classical statistical physics.
– ACuriousMind♦
yesterday
1
1
@ACuriousMind Thank you for looking into this. Two comments: (1) As you point out, the cited paper used an Einstein-ish model to compute the heat capacity of solid mercury. But is that strictly necessary, or due to lack of experimental data? I would assume that at least in principle the heat capcaity could be measured, which would make the Sackur-Tetrode approach a viable way to determine $h$ wihout an underlying quantum-mechanical model.
– user8153
yesterday
@ACuriousMind Thank you for looking into this. Two comments: (1) As you point out, the cited paper used an Einstein-ish model to compute the heat capacity of solid mercury. But is that strictly necessary, or due to lack of experimental data? I would assume that at least in principle the heat capcaity could be measured, which would make the Sackur-Tetrode approach a viable way to determine $h$ wihout an underlying quantum-mechanical model.
– user8153
yesterday
1
1
@ACuriousMind (2) Perhaps more importantly, I believe it is incorrect to say that the value of the quantity $h$ that appears in the classical partition function is "arbitrary", which to me sounds like "it doesn't matter which value you use, the results will always be the same". It seems to me that there is only one correct valuer (6.626E-34 m^2 kg / s), but this value is not predicted by classical theory, and must be obtained from experiment instead.
– user8153
yesterday
@ACuriousMind (2) Perhaps more importantly, I believe it is incorrect to say that the value of the quantity $h$ that appears in the classical partition function is "arbitrary", which to me sounds like "it doesn't matter which value you use, the results will always be the same". It seems to me that there is only one correct valuer (6.626E-34 m^2 kg / s), but this value is not predicted by classical theory, and must be obtained from experiment instead.
– user8153
yesterday
1
1
@user8153 Maybe the following is a better way to say what $h$ is: In pure classical statistical mechanics, the value of $h$ is irrelevant - you can express all physically measurable quantities in terms that do not involve it. Not only is there no classical prediction, there is also no purely classical measurement prescription. Classical statistical mechanics is agnostic as to whether $h$ is a fundamental feature of the universe or a trick of the theorist. Our knowledge of quantum mechanics nowadays fixes its role firmly to be the former.
– ACuriousMind♦
yesterday
@user8153 Maybe the following is a better way to say what $h$ is: In pure classical statistical mechanics, the value of $h$ is irrelevant - you can express all physically measurable quantities in terms that do not involve it. Not only is there no classical prediction, there is also no purely classical measurement prescription. Classical statistical mechanics is agnostic as to whether $h$ is a fundamental feature of the universe or a trick of the theorist. Our knowledge of quantum mechanics nowadays fixes its role firmly to be the former.
– ACuriousMind♦
yesterday
1
1
Note also that a measurement of the heat capacity at discrete points is not enough - the ST equation contains an integral over the heat capacity at constant pressure all the way from absolute zero. As good as we may be in figuring out experiments, I doubt that there is an experiment that allows you to collect the data needed for a good approximation of this integral.
– ACuriousMind♦
yesterday
Note also that a measurement of the heat capacity at discrete points is not enough - the ST equation contains an integral over the heat capacity at constant pressure all the way from absolute zero. As good as we may be in figuring out experiments, I doubt that there is an experiment that allows you to collect the data needed for a good approximation of this integral.
– ACuriousMind♦
yesterday
 |Â
show 5 more comments
up vote
0
down vote
The derivation in the question is misleading because it prematurely sets the chemical potential to zero. Even in systems where it is zero, you must carry it with you until the final evaluation of expressions, since it is a free parameter of a macrostate in the grand canonical ensemble.
In particular, the chemical potential contains a $ln(h^3/textstuff)$ factor (e.g. the classical ideal non-relativistic gas has $mu propto lnleft(fraclambda^3 NVright)$, with $lambda$ the thermal de Broglie wavelength proportional to $h$), that, since it is the fugacity that enters into physical quantities, cancels against the $h^3$ in the denominator.
Therefore, your approach does not necessitate a universal classical value for $h$, since the results can be re-expressed in terms of quantities that do not involve $h$. Note, in particular, that you cannot experimentally directly measure the chemical potential $mu$ - you need to compute it from indirect measurements, with formulae that in this case also will involve $h$.
answered yesterday
ACuriousMind♦
68k17118289
Thanks for the response. In the grand canonical ensemble, the fixed parameters are T, V, and $mu$. The pressure and number of particles fluctuate because they are not fixed. I don't believe that $mu$ is a free parameter.
– user196574
yesterday
Also, there are cases where nature "fixes" $mu$ and not number of particles, the classic example being a photon gas. However, we can even fix $mu$ and not the number of particles by affixing our system to a "particle number" bath. This is how I think about it, please let me hear your thoughts: A heat bath can give heat without noticeably changing its temperature until an adjoined smaller system is at the same temperature. A particle number bath can give particles without noticeably changing its chemical potential until an adjoined smaller system is at the same chemical potential.
– user196574
yesterday
@user196574 $mu$ is fixed for a specific macrostate in the grand canonical example. It's a free parameter in the sense that derivations that are supposed to hold for a system must hold for all possible values of this parameter, just like they must hold for all values of $T$ and $V$ (that obey the equation of state). You can't just say "but $T = 1mathrmK$ in my system". That's not a property of the system, but of a state of the system. In any experimental test, you would have to establish that indeed $mu = 0$ for the state you're measuring, thus losing the ability to measure $h$.
– ACuriousMind♦
yesterday
I think I will need to refresh my knowledge on the chemical potential before returning to the role of $h$ in the grand canonical ensemble. Specifically, whether it is necessary to have some theoretical expression for $mu$ (that depends on $h$) in order to be able to set $mu$, or whether they may be ways to experimentally determine a chemical potential without knowledge of $h$, just as there are ways to experimentally determine a volume without knowledge of $h$.
– user196574
yesterday
add a comment |Â
up vote
0
down vote
The derivation in the question is misleading because it prematurely sets the chemical potential to zero. Even in systems where it is zero, you must carry it with you until the final evaluation of expressions, since it is a free parameter of a macrostate in the grand canonical ensemble.
In particular, the chemical potential contains a $ln(h^3/textstuff)$ factor (e.g. the classical ideal non-relativistic gas has $mu propto lnleft(fraclambda^3 NVright)$, with $lambda$ the thermal de Broglie wavelength proportional to $h$), that, since it is the fugacity that enters into physical quantities, cancels against the $h^3$ in the denominator.
Therefore, your approach does not necessitate a universal classical value for $h$, since the results can be re-expressed in terms of quantities that do not involve $h$. Note, in particular, that you cannot experimentally directly measure the chemical potential $mu$ - you need to compute it from indirect measurements, with formulae that in this case also will involve $h$.
answered yesterday
ACuriousMind♦
68k17118289
Thanks for the response. In the grand canonical ensemble, the fixed parameters are T, V, and $mu$. The pressure and number of particles fluctuate because they are not fixed. I don't believe that $mu$ is a free parameter.
– user196574
yesterday
Also, there are cases where nature "fixes" $mu$ and not number of particles, the classic example being a photon gas. However, we can even fix $mu$ and not the number of particles by affixing our system to a "particle number" bath. This is how I think about it, please let me hear your thoughts: A heat bath can give heat without noticeably changing its temperature until an adjoined smaller system is at the same temperature. A particle number bath can give particles without noticeably changing its chemical potential until an adjoined smaller system is at the same chemical potential.
– user196574
yesterday
@user196574 $mu$ is fixed for a specific macrostate in the grand canonical example. It's a free parameter in the sense that derivations that are supposed to hold for a system must hold for all possible values of this parameter, just like they must hold for all values of $T$ and $V$ (that obey the equation of state). You can't just say "but $T = 1mathrmK$ in my system". That's not a property of the system, but of a state of the system. In any experimental test, you would have to establish that indeed $mu = 0$ for the state you're measuring, thus losing the ability to measure $h$.
– ACuriousMind♦
yesterday
I think I will need to refresh my knowledge on the chemical potential before returning to the role of $h$ in the grand canonical ensemble. Specifically, whether it is necessary to have some theoretical expression for $mu$ (that depends on $h$) in order to be able to set $mu$, or whether they may be ways to experimentally determine a chemical potential without knowledge of $h$, just as there are ways to experimentally determine a volume without knowledge of $h$.
– user196574
yesterday
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The derivation in the question is misleading because it prematurely sets the chemical potential to zero. Even in systems where it is zero, you must carry it with you until the final evaluation of expressions, since it is a free parameter of a macrostate in the grand canonical ensemble.
In particular, the chemical potential contains a $ln(h^3/textstuff)$ factor (e.g. the classical ideal non-relativistic gas has $mu propto lnleft(fraclambda^3 NVright)$, with $lambda$ the thermal de Broglie wavelength proportional to $h$), that, since it is the fugacity that enters into physical quantities, cancels against the $h^3$ in the denominator.
Therefore, your approach does not necessitate a universal classical value for $h$, since the results can be re-expressed in terms of quantities that do not involve $h$. Note, in particular, that you cannot experimentally directly measure the chemical potential $mu$ - you need to compute it from indirect measurements, with formulae that in this case also will involve $h$.
answered yesterday
ACuriousMind♦
68k17118289
The derivation in the question is misleading because it prematurely sets the chemical potential to zero. Even in systems where it is zero, you must carry it with you until the final evaluation of expressions, since it is a free parameter of a macrostate in the grand canonical ensemble.
In particular, the chemical potential contains a $ln(h^3/textstuff)$ factor (e.g. the classical ideal non-relativistic gas has $mu propto lnleft(fraclambda^3 NVright)$, with $lambda$ the thermal de Broglie wavelength proportional to $h$), that, since it is the fugacity that enters into physical quantities, cancels against the $h^3$ in the denominator.
Therefore, your approach does not necessitate a universal classical value for $h$, since the results can be re-expressed in terms of quantities that do not involve $h$. Note, in particular, that you cannot experimentally directly measure the chemical potential $mu$ - you need to compute it from indirect measurements, with formulae that in this case also will involve $h$.
answered yesterday
ACuriousMind♦
68k17118289
answered yesterday
ACuriousMind♦
68k17118289
answered yesterday
ACuriousMind♦
68k17118289
answered yesterday
ACuriousMind♦
68k17118289
68k17118289
Thanks for the response. In the grand canonical ensemble, the fixed parameters are T, V, and $mu$. The pressure and number of particles fluctuate because they are not fixed. I don't believe that $mu$ is a free parameter.
– user196574
yesterday
Also, there are cases where nature "fixes" $mu$ and not number of particles, the classic example being a photon gas. However, we can even fix $mu$ and not the number of particles by affixing our system to a "particle number" bath. This is how I think about it, please let me hear your thoughts: A heat bath can give heat without noticeably changing its temperature until an adjoined smaller system is at the same temperature. A particle number bath can give particles without noticeably changing its chemical potential until an adjoined smaller system is at the same chemical potential.
– user196574
yesterday
@user196574 $mu$ is fixed for a specific macrostate in the grand canonical example. It's a free parameter in the sense that derivations that are supposed to hold for a system must hold for all possible values of this parameter, just like they must hold for all values of $T$ and $V$ (that obey the equation of state). You can't just say "but $T = 1mathrmK$ in my system". That's not a property of the system, but of a state of the system. In any experimental test, you would have to establish that indeed $mu = 0$ for the state you're measuring, thus losing the ability to measure $h$.
– ACuriousMind♦
yesterday
I think I will need to refresh my knowledge on the chemical potential before returning to the role of $h$ in the grand canonical ensemble. Specifically, whether it is necessary to have some theoretical expression for $mu$ (that depends on $h$) in order to be able to set $mu$, or whether they may be ways to experimentally determine a chemical potential without knowledge of $h$, just as there are ways to experimentally determine a volume without knowledge of $h$.
– user196574
yesterday
add a comment |Â
Thanks for the response. In the grand canonical ensemble, the fixed parameters are T, V, and $mu$. The pressure and number of particles fluctuate because they are not fixed. I don't believe that $mu$ is a free parameter.
– user196574
yesterday
Also, there are cases where nature "fixes" $mu$ and not number of particles, the classic example being a photon gas. However, we can even fix $mu$ and not the number of particles by affixing our system to a "particle number" bath. This is how I think about it, please let me hear your thoughts: A heat bath can give heat without noticeably changing its temperature until an adjoined smaller system is at the same temperature. A particle number bath can give particles without noticeably changing its chemical potential until an adjoined smaller system is at the same chemical potential.
– user196574
yesterday
@user196574 $mu$ is fixed for a specific macrostate in the grand canonical example. It's a free parameter in the sense that derivations that are supposed to hold for a system must hold for all possible values of this parameter, just like they must hold for all values of $T$ and $V$ (that obey the equation of state). You can't just say "but $T = 1mathrmK$ in my system". That's not a property of the system, but of a state of the system. In any experimental test, you would have to establish that indeed $mu = 0$ for the state you're measuring, thus losing the ability to measure $h$.
– ACuriousMind♦
yesterday
I think I will need to refresh my knowledge on the chemical potential before returning to the role of $h$ in the grand canonical ensemble. Specifically, whether it is necessary to have some theoretical expression for $mu$ (that depends on $h$) in order to be able to set $mu$, or whether they may be ways to experimentally determine a chemical potential without knowledge of $h$, just as there are ways to experimentally determine a volume without knowledge of $h$.
– user196574
yesterday
Thanks for the response. In the grand canonical ensemble, the fixed parameters are T, V, and $mu$. The pressure and number of particles fluctuate because they are not fixed. I don't believe that $mu$ is a free parameter.
– user196574
yesterday
Thanks for the response. In the grand canonical ensemble, the fixed parameters are T, V, and $mu$. The pressure and number of particles fluctuate because they are not fixed. I don't believe that $mu$ is a free parameter.
– user196574
yesterday
Also, there are cases where nature "fixes" $mu$ and not number of particles, the classic example being a photon gas. However, we can even fix $mu$ and not the number of particles by affixing our system to a "particle number" bath. This is how I think about it, please let me hear your thoughts: A heat bath can give heat without noticeably changing its temperature until an adjoined smaller system is at the same temperature. A particle number bath can give particles without noticeably changing its chemical potential until an adjoined smaller system is at the same chemical potential.
– user196574
yesterday
Also, there are cases where nature "fixes" $mu$ and not number of particles, the classic example being a photon gas. However, we can even fix $mu$ and not the number of particles by affixing our system to a "particle number" bath. This is how I think about it, please let me hear your thoughts: A heat bath can give heat without noticeably changing its temperature until an adjoined smaller system is at the same temperature. A particle number bath can give particles without noticeably changing its chemical potential until an adjoined smaller system is at the same chemical potential.
– user196574
yesterday
@user196574 $mu$ is fixed for a specific macrostate in the grand canonical example. It's a free parameter in the sense that derivations that are supposed to hold for a system must hold for all possible values of this parameter, just like they must hold for all values of $T$ and $V$ (that obey the equation of state). You can't just say "but $T = 1mathrmK$ in my system". That's not a property of the system, but of a state of the system. In any experimental test, you would have to establish that indeed $mu = 0$ for the state you're measuring, thus losing the ability to measure $h$.
– ACuriousMind♦
yesterday
@user196574 $mu$ is fixed for a specific macrostate in the grand canonical example. It's a free parameter in the sense that derivations that are supposed to hold for a system must hold for all possible values of this parameter, just like they must hold for all values of $T$ and $V$ (that obey the equation of state). You can't just say "but $T = 1mathrmK$ in my system". That's not a property of the system, but of a state of the system. In any experimental test, you would have to establish that indeed $mu = 0$ for the state you're measuring, thus losing the ability to measure $h$.
– ACuriousMind♦
yesterday
I think I will need to refresh my knowledge on the chemical potential before returning to the role of $h$ in the grand canonical ensemble. Specifically, whether it is necessary to have some theoretical expression for $mu$ (that depends on $h$) in order to be able to set $mu$, or whether they may be ways to experimentally determine a chemical potential without knowledge of $h$, just as there are ways to experimentally determine a volume without knowledge of $h$.
– user196574
yesterday
I think I will need to refresh my knowledge on the chemical potential before returning to the role of $h$ in the grand canonical ensemble. Specifically, whether it is necessary to have some theoretical expression for $mu$ (that depends on $h$) in order to be able to set $mu$, or whether they may be ways to experimentally determine a chemical potential without knowledge of $h$, just as there are ways to experimentally determine a volume without knowledge of $h$.
– user196574
yesterday
add a comment |Â
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