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Entropy as a function of temperature: is temperature well defined?
How do we define temperature?Relation between entropy and internal energyProving that the Boltzmann entropy is equal to the thermodynamic entropyWhy is Boltzmann entropy equation used for configurational entropy?How to avoid temperature being infinite at entropy maximumEntropy definition, additivity, laws in different ensembleshow to increase entropy without increasing temperature?Is entropy zero at temperature zero for ideal gas?Entropy and temperature of small systemsTemperature from the Gibbs entropy
$begingroup$
Considering volume and particle number constant, the internal energy $U$ is a function of the entropy: $U=U(S)$. The temperature is then defined as $T=dU(S)/dS$. From here, the temperature is a function of the entropy: $T=T(S)$.
One can now define the free energy $F=F(T)$ as
$$F(T) = U-TS = U(S(T)) - TS(T)$$
For this, however, $T(S)$ has to be invertable. My question is, what happens if $U(S)$ is some wavy function such that its derivative is the same at many distinct $S$ values? In other words, what happens if $T^*=T(S^*)$ has multiple solutions for $S^*$? Is it possible not to have a well defined temperature at equilibrium?
My wild guess is that this shouldn't happen in the thermodynamic limit, but can't see this. Also, small, even single particle systems have well defined temperature so is there some constraint on the shape of $U(S)$ I'm missing here?
thermodynamics statistical-mechanics temperature entropy
asked Apr 29 at 18:11
BotondBotond
26710
$endgroup$
add a comment |
$begingroup$
Considering volume and particle number constant, the internal energy $U$ is a function of the entropy: $U=U(S)$. The temperature is then defined as $T=dU(S)/dS$. From here, the temperature is a function of the entropy: $T=T(S)$.
One can now define the free energy $F=F(T)$ as
$$F(T) = U-TS = U(S(T)) - TS(T)$$
For this, however, $T(S)$ has to be invertable. My question is, what happens if $U(S)$ is some wavy function such that its derivative is the same at many distinct $S$ values? In other words, what happens if $T^*=T(S^*)$ has multiple solutions for $S^*$? Is it possible not to have a well defined temperature at equilibrium?
My wild guess is that this shouldn't happen in the thermodynamic limit, but can't see this. Also, small, even single particle systems have well defined temperature so is there some constraint on the shape of $U(S)$ I'm missing here?
thermodynamics statistical-mechanics temperature entropy
asked Apr 29 at 18:11
BotondBotond
26710
$endgroup$
1
$begingroup$
You can prove $U(S)$ is always a convex function, so that puts some limits on what can go wrong here.
$endgroup$
– Jahan Claes
Apr 29 at 19:30
$begingroup$
@JahanClaes Could you sketch or refer to a proof that $U(S)$ is always convex?
$endgroup$
– Botond
Apr 29 at 21:54
1
$begingroup$
If you want a book on thermodynamics which explicitly addresses the convex nature of the entropy function, and what it implies, including for phase transitions, then I have written one; publisher is Oxford University Press. I hope this comment does not break self-advertising rules on the site. I think it is fair because it is directly addressed to the question asked.
$endgroup$
– Andrew Steane
yesterday
add a comment |
$begingroup$
Considering volume and particle number constant, the internal energy $U$ is a function of the entropy: $U=U(S)$. The temperature is then defined as $T=dU(S)/dS$. From here, the temperature is a function of the entropy: $T=T(S)$.
One can now define the free energy $F=F(T)$ as
$$F(T) = U-TS = U(S(T)) - TS(T)$$
For this, however, $T(S)$ has to be invertable. My question is, what happens if $U(S)$ is some wavy function such that its derivative is the same at many distinct $S$ values? In other words, what happens if $T^*=T(S^*)$ has multiple solutions for $S^*$? Is it possible not to have a well defined temperature at equilibrium?
My wild guess is that this shouldn't happen in the thermodynamic limit, but can't see this. Also, small, even single particle systems have well defined temperature so is there some constraint on the shape of $U(S)$ I'm missing here?
thermodynamics statistical-mechanics temperature entropy
asked Apr 29 at 18:11
BotondBotond
26710
$endgroup$
Considering volume and particle number constant, the internal energy $U$ is a function of the entropy: $U=U(S)$. The temperature is then defined as $T=dU(S)/dS$. From here, the temperature is a function of the entropy: $T=T(S)$.
One can now define the free energy $F=F(T)$ as
$$F(T) = U-TS = U(S(T)) - TS(T)$$
For this, however, $T(S)$ has to be invertable. My question is, what happens if $U(S)$ is some wavy function such that its derivative is the same at many distinct $S$ values? In other words, what happens if $T^*=T(S^*)$ has multiple solutions for $S^*$? Is it possible not to have a well defined temperature at equilibrium?
My wild guess is that this shouldn't happen in the thermodynamic limit, but can't see this. Also, small, even single particle systems have well defined temperature so is there some constraint on the shape of $U(S)$ I'm missing here?
thermodynamics statistical-mechanics temperature entropy
thermodynamics statistical-mechanics temperature entropy
asked Apr 29 at 18:11
BotondBotond
26710
asked Apr 29 at 18:11
BotondBotond
26710
asked Apr 29 at 18:11
BotondBotond
26710
asked Apr 29 at 18:11
BotondBotond
26710
asked Apr 29 at 18:11
BotondBotond
26710
26710
1
$begingroup$
You can prove $U(S)$ is always a convex function, so that puts some limits on what can go wrong here.
$endgroup$
– Jahan Claes
Apr 29 at 19:30
$begingroup$
@JahanClaes Could you sketch or refer to a proof that $U(S)$ is always convex?
$endgroup$
– Botond
Apr 29 at 21:54
1
$begingroup$
If you want a book on thermodynamics which explicitly addresses the convex nature of the entropy function, and what it implies, including for phase transitions, then I have written one; publisher is Oxford University Press. I hope this comment does not break self-advertising rules on the site. I think it is fair because it is directly addressed to the question asked.
$endgroup$
– Andrew Steane
yesterday
add a comment |
1
$begingroup$
You can prove $U(S)$ is always a convex function, so that puts some limits on what can go wrong here.
$endgroup$
– Jahan Claes
Apr 29 at 19:30
$begingroup$
@JahanClaes Could you sketch or refer to a proof that $U(S)$ is always convex?
$endgroup$
– Botond
Apr 29 at 21:54
1
$begingroup$
If you want a book on thermodynamics which explicitly addresses the convex nature of the entropy function, and what it implies, including for phase transitions, then I have written one; publisher is Oxford University Press. I hope this comment does not break self-advertising rules on the site. I think it is fair because it is directly addressed to the question asked.
$endgroup$
– Andrew Steane
yesterday
1
1
$begingroup$
You can prove $U(S)$ is always a convex function, so that puts some limits on what can go wrong here.
$endgroup$
– Jahan Claes
Apr 29 at 19:30
$begingroup$
You can prove $U(S)$ is always a convex function, so that puts some limits on what can go wrong here.
$endgroup$
– Jahan Claes
Apr 29 at 19:30
$begingroup$
@JahanClaes Could you sketch or refer to a proof that $U(S)$ is always convex?
$endgroup$
– Botond
Apr 29 at 21:54
$begingroup$
@JahanClaes Could you sketch or refer to a proof that $U(S)$ is always convex?
$endgroup$
– Botond
Apr 29 at 21:54
1
1
$begingroup$
If you want a book on thermodynamics which explicitly addresses the convex nature of the entropy function, and what it implies, including for phase transitions, then I have written one; publisher is Oxford University Press. I hope this comment does not break self-advertising rules on the site. I think it is fair because it is directly addressed to the question asked.
$endgroup$
– Andrew Steane
yesterday
$begingroup$
If you want a book on thermodynamics which explicitly addresses the convex nature of the entropy function, and what it implies, including for phase transitions, then I have written one; publisher is Oxford University Press. I hope this comment does not break self-advertising rules on the site. I think it is fair because it is directly addressed to the question asked.
$endgroup$
– Andrew Steane
yesterday
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Your question requires an answer on three different levels:
- mathematical;
- pure thermodynamics;
- statistical mechanics.
1. mathematics
The definition of the Helmholtz free energy you refer to is nothing but the Legendre transform of the fundamental equation $U(S,V,N)$ with respect to its first variable $S$ in term of the conjugate variable $T=left( frac partialUpartialS right)_V,N$. The original Legendre transform would be easily done by requiring $U$ to be a twice differentiable function of $S$ with a positively defined hessian matrix. However, such a request is too strong for real thermodynamic systems. It is well known that a useful extension of the Legendre transform is the so-called Legendre-Fenchel transform (LF) (or convex-conjugate).
In the case of thermodynamics, the definition of LF transform is slightly different from the most common mathematical definition. In the case of the Helmholtz free energy, it would be written as
$$
F(T,V,N) = inf_S( U(S,V,N) - TS )
$$
Such a definition reduces to the usual Legendre transform in the part of the domain of $U(S,V,N)$ where the function is strictly convex (and twice differentiable with respect to $S$). Where the function is convex but not strictly convex (i.e. what mathematicians call an affine function, i.e. a linear function), the LF transform maps the entire affine interval into a single point where left and right derivatives differ.
Since fundamental equations must be convex (o cancave) but not strictly convex (or strictly concave), it turns out that the LF transform is the proper mathematical tool for a change of variable $S leftrightarrow T$.
2. Pure thermodynamics
Affine regions of $U(S,V,N)$ are to be expected, due to the phenomenon of phase coexistence. In such regions, the thermodynamic potential must be a linear function of its extensive variables since it corresponds to an equilibrium condition of an inhomogeneous system made by more coexisting phases. At coexistence, $T(S,V,N)$ is a constant as a function of $S$. But this is physically consistent with the presence of a latent heat at the first order phase transition.
3. Statistical mechanics
Statistial mechanics is deemed to give access to Thermodynamics, by starting with a model for the Hamiltonian of the system. However, such a program in general requires the so-called thermodynamic limit (TL). TL is required for different reasons. Summarizing, these are:
- only TL can introduce the non-analiticity required to recover phase transitions;
- only at TL (if it exists) it is possible to recover the extensiveness
- only at TL (if it exists) it is possible to recover the convexity properties.
Without TL many properties, which are considered typical for thermodynamic systems, would not be valid. On the other hand, working with a finite number of degrees of freedom, although unavoidable from the numerical point of view, in general introduces non-convex (unphysical) regions. Therefore, TL is required, but at TL the $T(S,V,N)$ is not invertible for $S$ in the whole coexistence region. Nevertheless, LF transform can cope without problem with the situation.
answered Apr 29 at 21:16
GiorgioPGiorgioP
5,2472833
$endgroup$
$begingroup$
Thank you for the detailed answer. Still digesting. Can you please explain what you mean by "fundamental equations must be convex" and why this is true? Also, can you please elaborate on your definition of the free energy as the infimum over S? Does this just cuts the whole domain of $S$ to chunks where $U(S)$ is convex/concave?
$endgroup$
– Botond
Apr 29 at 21:57
$begingroup$
Convexity/concavity of fundamental equations (thermodynamic potentials in the present case) is a basic consequence of the minimum pincipe for internal energy. You'll find an argument for concavity of entropy in the Callen's textbook. Definition of free energy as inf is just the application of LF definition (according to thermodynamicists) to the case of S <->T change of variables. Finally, $U(S)$ must be convex everywhere in its domain. Concave regions are forbidden in a stable thermodynamic system. This is the short answer. If you need the long answer I could do but not immediately.
$endgroup$
– GiorgioP
Apr 29 at 22:39
add a comment |
$begingroup$
Function $T(S)=(partial U/partial S)_V$ does not have to be strictly increasing with $S$. For example, when heat is added to a mix of ice and water at 0 degrees Celsius, entropy increases but temperature stays the same until all ice is melted. So it is not, in general, possible to express entropy as function of $T,V,N$.
Free (Helmholtz) energy is usually defined as
$$
F(T,V,N) = U - TS
$$
but in general neither $U$ nor $S$ can be expressed as functions of $T,V,N$.
answered Apr 29 at 18:45
Ján LalinskýJán Lalinský
16.2k1441
$endgroup$
$begingroup$
I'm confused. I thought that if you externally impose a fixed $T,V,N$ to the system, that defines your $F(T,V,N)$ free energy unambiguously. You can write a free energy as a function of the molar concentrations, but the equilibrium free energy will be the one that minimizes $F(T,V,N)$. Thus the equilibrium molar concentrations are also functions of $T,V,N$.
$endgroup$
– Botond
Apr 29 at 19:23
$begingroup$
My mistake, the part about molar numbers was wrong. I changed my answer.
$endgroup$
– Ján Lalinský
Apr 29 at 21:07
$begingroup$
Thank you. What still bugs me is that if neither $U$ nor $S$ can be expressed as a function of $T,V,N$, how would you possibly know that $U-TS$ will only depend on $T,V,N$ and not on $U$ and/or $S$?
$endgroup$
– Botond
Apr 29 at 21:47
$begingroup$
It follows from the definition of $F$. We have $dU = TdS - pdV$, so from the differentials we see $U$ is a function of $S,V$. Then since $F=U-TS$, we have $dF=dU - TdS - SdT$ so $dF = -SdT -pdV$ so again from differentials $F$ must be function of $T,V$.
$endgroup$
– Ján Lalinský
Apr 29 at 22:33
1
$begingroup$
What the differential tells me is that $S=partial F(T,V,N) / partial T$ thus $S=S(T,V,N)$
$endgroup$
– Botond
2 days ago
|
show 2 more comments
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2 Answers
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2 Answers
2
active
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active
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active
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$begingroup$
Your question requires an answer on three different levels:
- mathematical;
- pure thermodynamics;
- statistical mechanics.
1. mathematics
The definition of the Helmholtz free energy you refer to is nothing but the Legendre transform of the fundamental equation $U(S,V,N)$ with respect to its first variable $S$ in term of the conjugate variable $T=left( frac partialUpartialS right)_V,N$. The original Legendre transform would be easily done by requiring $U$ to be a twice differentiable function of $S$ with a positively defined hessian matrix. However, such a request is too strong for real thermodynamic systems. It is well known that a useful extension of the Legendre transform is the so-called Legendre-Fenchel transform (LF) (or convex-conjugate).
In the case of thermodynamics, the definition of LF transform is slightly different from the most common mathematical definition. In the case of the Helmholtz free energy, it would be written as
$$
F(T,V,N) = inf_S( U(S,V,N) - TS )
$$
Such a definition reduces to the usual Legendre transform in the part of the domain of $U(S,V,N)$ where the function is strictly convex (and twice differentiable with respect to $S$). Where the function is convex but not strictly convex (i.e. what mathematicians call an affine function, i.e. a linear function), the LF transform maps the entire affine interval into a single point where left and right derivatives differ.
Since fundamental equations must be convex (o cancave) but not strictly convex (or strictly concave), it turns out that the LF transform is the proper mathematical tool for a change of variable $S leftrightarrow T$.
2. Pure thermodynamics
Affine regions of $U(S,V,N)$ are to be expected, due to the phenomenon of phase coexistence. In such regions, the thermodynamic potential must be a linear function of its extensive variables since it corresponds to an equilibrium condition of an inhomogeneous system made by more coexisting phases. At coexistence, $T(S,V,N)$ is a constant as a function of $S$. But this is physically consistent with the presence of a latent heat at the first order phase transition.
3. Statistical mechanics
Statistial mechanics is deemed to give access to Thermodynamics, by starting with a model for the Hamiltonian of the system. However, such a program in general requires the so-called thermodynamic limit (TL). TL is required for different reasons. Summarizing, these are:
- only TL can introduce the non-analiticity required to recover phase transitions;
- only at TL (if it exists) it is possible to recover the extensiveness
- only at TL (if it exists) it is possible to recover the convexity properties.
Without TL many properties, which are considered typical for thermodynamic systems, would not be valid. On the other hand, working with a finite number of degrees of freedom, although unavoidable from the numerical point of view, in general introduces non-convex (unphysical) regions. Therefore, TL is required, but at TL the $T(S,V,N)$ is not invertible for $S$ in the whole coexistence region. Nevertheless, LF transform can cope without problem with the situation.
answered Apr 29 at 21:16
GiorgioPGiorgioP
5,2472833
$endgroup$
$begingroup$
Thank you for the detailed answer. Still digesting. Can you please explain what you mean by "fundamental equations must be convex" and why this is true? Also, can you please elaborate on your definition of the free energy as the infimum over S? Does this just cuts the whole domain of $S$ to chunks where $U(S)$ is convex/concave?
$endgroup$
– Botond
Apr 29 at 21:57
$begingroup$
Convexity/concavity of fundamental equations (thermodynamic potentials in the present case) is a basic consequence of the minimum pincipe for internal energy. You'll find an argument for concavity of entropy in the Callen's textbook. Definition of free energy as inf is just the application of LF definition (according to thermodynamicists) to the case of S <->T change of variables. Finally, $U(S)$ must be convex everywhere in its domain. Concave regions are forbidden in a stable thermodynamic system. This is the short answer. If you need the long answer I could do but not immediately.
$endgroup$
– GiorgioP
Apr 29 at 22:39
add a comment |
$begingroup$
Your question requires an answer on three different levels:
- mathematical;
- pure thermodynamics;
- statistical mechanics.
1. mathematics
The definition of the Helmholtz free energy you refer to is nothing but the Legendre transform of the fundamental equation $U(S,V,N)$ with respect to its first variable $S$ in term of the conjugate variable $T=left( frac partialUpartialS right)_V,N$. The original Legendre transform would be easily done by requiring $U$ to be a twice differentiable function of $S$ with a positively defined hessian matrix. However, such a request is too strong for real thermodynamic systems. It is well known that a useful extension of the Legendre transform is the so-called Legendre-Fenchel transform (LF) (or convex-conjugate).
In the case of thermodynamics, the definition of LF transform is slightly different from the most common mathematical definition. In the case of the Helmholtz free energy, it would be written as
$$
F(T,V,N) = inf_S( U(S,V,N) - TS )
$$
Such a definition reduces to the usual Legendre transform in the part of the domain of $U(S,V,N)$ where the function is strictly convex (and twice differentiable with respect to $S$). Where the function is convex but not strictly convex (i.e. what mathematicians call an affine function, i.e. a linear function), the LF transform maps the entire affine interval into a single point where left and right derivatives differ.
Since fundamental equations must be convex (o cancave) but not strictly convex (or strictly concave), it turns out that the LF transform is the proper mathematical tool for a change of variable $S leftrightarrow T$.
2. Pure thermodynamics
Affine regions of $U(S,V,N)$ are to be expected, due to the phenomenon of phase coexistence. In such regions, the thermodynamic potential must be a linear function of its extensive variables since it corresponds to an equilibrium condition of an inhomogeneous system made by more coexisting phases. At coexistence, $T(S,V,N)$ is a constant as a function of $S$. But this is physically consistent with the presence of a latent heat at the first order phase transition.
3. Statistical mechanics
Statistial mechanics is deemed to give access to Thermodynamics, by starting with a model for the Hamiltonian of the system. However, such a program in general requires the so-called thermodynamic limit (TL). TL is required for different reasons. Summarizing, these are:
- only TL can introduce the non-analiticity required to recover phase transitions;
- only at TL (if it exists) it is possible to recover the extensiveness
- only at TL (if it exists) it is possible to recover the convexity properties.
Without TL many properties, which are considered typical for thermodynamic systems, would not be valid. On the other hand, working with a finite number of degrees of freedom, although unavoidable from the numerical point of view, in general introduces non-convex (unphysical) regions. Therefore, TL is required, but at TL the $T(S,V,N)$ is not invertible for $S$ in the whole coexistence region. Nevertheless, LF transform can cope without problem with the situation.
answered Apr 29 at 21:16
GiorgioPGiorgioP
5,2472833
$endgroup$
$begingroup$
Thank you for the detailed answer. Still digesting. Can you please explain what you mean by "fundamental equations must be convex" and why this is true? Also, can you please elaborate on your definition of the free energy as the infimum over S? Does this just cuts the whole domain of $S$ to chunks where $U(S)$ is convex/concave?
$endgroup$
– Botond
Apr 29 at 21:57
$begingroup$
Convexity/concavity of fundamental equations (thermodynamic potentials in the present case) is a basic consequence of the minimum pincipe for internal energy. You'll find an argument for concavity of entropy in the Callen's textbook. Definition of free energy as inf is just the application of LF definition (according to thermodynamicists) to the case of S <->T change of variables. Finally, $U(S)$ must be convex everywhere in its domain. Concave regions are forbidden in a stable thermodynamic system. This is the short answer. If you need the long answer I could do but not immediately.
$endgroup$
– GiorgioP
Apr 29 at 22:39
add a comment |
$begingroup$
Your question requires an answer on three different levels:
- mathematical;
- pure thermodynamics;
- statistical mechanics.
1. mathematics
The definition of the Helmholtz free energy you refer to is nothing but the Legendre transform of the fundamental equation $U(S,V,N)$ with respect to its first variable $S$ in term of the conjugate variable $T=left( frac partialUpartialS right)_V,N$. The original Legendre transform would be easily done by requiring $U$ to be a twice differentiable function of $S$ with a positively defined hessian matrix. However, such a request is too strong for real thermodynamic systems. It is well known that a useful extension of the Legendre transform is the so-called Legendre-Fenchel transform (LF) (or convex-conjugate).
In the case of thermodynamics, the definition of LF transform is slightly different from the most common mathematical definition. In the case of the Helmholtz free energy, it would be written as
$$
F(T,V,N) = inf_S( U(S,V,N) - TS )
$$
Such a definition reduces to the usual Legendre transform in the part of the domain of $U(S,V,N)$ where the function is strictly convex (and twice differentiable with respect to $S$). Where the function is convex but not strictly convex (i.e. what mathematicians call an affine function, i.e. a linear function), the LF transform maps the entire affine interval into a single point where left and right derivatives differ.
Since fundamental equations must be convex (o cancave) but not strictly convex (or strictly concave), it turns out that the LF transform is the proper mathematical tool for a change of variable $S leftrightarrow T$.
2. Pure thermodynamics
Affine regions of $U(S,V,N)$ are to be expected, due to the phenomenon of phase coexistence. In such regions, the thermodynamic potential must be a linear function of its extensive variables since it corresponds to an equilibrium condition of an inhomogeneous system made by more coexisting phases. At coexistence, $T(S,V,N)$ is a constant as a function of $S$. But this is physically consistent with the presence of a latent heat at the first order phase transition.
3. Statistical mechanics
Statistial mechanics is deemed to give access to Thermodynamics, by starting with a model for the Hamiltonian of the system. However, such a program in general requires the so-called thermodynamic limit (TL). TL is required for different reasons. Summarizing, these are:
- only TL can introduce the non-analiticity required to recover phase transitions;
- only at TL (if it exists) it is possible to recover the extensiveness
- only at TL (if it exists) it is possible to recover the convexity properties.
Without TL many properties, which are considered typical for thermodynamic systems, would not be valid. On the other hand, working with a finite number of degrees of freedom, although unavoidable from the numerical point of view, in general introduces non-convex (unphysical) regions. Therefore, TL is required, but at TL the $T(S,V,N)$ is not invertible for $S$ in the whole coexistence region. Nevertheless, LF transform can cope without problem with the situation.
answered Apr 29 at 21:16
GiorgioPGiorgioP
5,2472833
$endgroup$
Your question requires an answer on three different levels:
- mathematical;
- pure thermodynamics;
- statistical mechanics.
1. mathematics
The definition of the Helmholtz free energy you refer to is nothing but the Legendre transform of the fundamental equation $U(S,V,N)$ with respect to its first variable $S$ in term of the conjugate variable $T=left( frac partialUpartialS right)_V,N$. The original Legendre transform would be easily done by requiring $U$ to be a twice differentiable function of $S$ with a positively defined hessian matrix. However, such a request is too strong for real thermodynamic systems. It is well known that a useful extension of the Legendre transform is the so-called Legendre-Fenchel transform (LF) (or convex-conjugate).
In the case of thermodynamics, the definition of LF transform is slightly different from the most common mathematical definition. In the case of the Helmholtz free energy, it would be written as
$$
F(T,V,N) = inf_S( U(S,V,N) - TS )
$$
Such a definition reduces to the usual Legendre transform in the part of the domain of $U(S,V,N)$ where the function is strictly convex (and twice differentiable with respect to $S$). Where the function is convex but not strictly convex (i.e. what mathematicians call an affine function, i.e. a linear function), the LF transform maps the entire affine interval into a single point where left and right derivatives differ.
Since fundamental equations must be convex (o cancave) but not strictly convex (or strictly concave), it turns out that the LF transform is the proper mathematical tool for a change of variable $S leftrightarrow T$.
2. Pure thermodynamics
Affine regions of $U(S,V,N)$ are to be expected, due to the phenomenon of phase coexistence. In such regions, the thermodynamic potential must be a linear function of its extensive variables since it corresponds to an equilibrium condition of an inhomogeneous system made by more coexisting phases. At coexistence, $T(S,V,N)$ is a constant as a function of $S$. But this is physically consistent with the presence of a latent heat at the first order phase transition.
3. Statistical mechanics
Statistial mechanics is deemed to give access to Thermodynamics, by starting with a model for the Hamiltonian of the system. However, such a program in general requires the so-called thermodynamic limit (TL). TL is required for different reasons. Summarizing, these are:
- only TL can introduce the non-analiticity required to recover phase transitions;
- only at TL (if it exists) it is possible to recover the extensiveness
- only at TL (if it exists) it is possible to recover the convexity properties.
Without TL many properties, which are considered typical for thermodynamic systems, would not be valid. On the other hand, working with a finite number of degrees of freedom, although unavoidable from the numerical point of view, in general introduces non-convex (unphysical) regions. Therefore, TL is required, but at TL the $T(S,V,N)$ is not invertible for $S$ in the whole coexistence region. Nevertheless, LF transform can cope without problem with the situation.
answered Apr 29 at 21:16
GiorgioPGiorgioP
5,2472833
answered Apr 29 at 21:16
GiorgioPGiorgioP
5,2472833
answered Apr 29 at 21:16
GiorgioPGiorgioP
5,2472833
answered Apr 29 at 21:16
GiorgioPGiorgioP
5,2472833
5,2472833
$begingroup$
Thank you for the detailed answer. Still digesting. Can you please explain what you mean by "fundamental equations must be convex" and why this is true? Also, can you please elaborate on your definition of the free energy as the infimum over S? Does this just cuts the whole domain of $S$ to chunks where $U(S)$ is convex/concave?
$endgroup$
– Botond
Apr 29 at 21:57
$begingroup$
Convexity/concavity of fundamental equations (thermodynamic potentials in the present case) is a basic consequence of the minimum pincipe for internal energy. You'll find an argument for concavity of entropy in the Callen's textbook. Definition of free energy as inf is just the application of LF definition (according to thermodynamicists) to the case of S <->T change of variables. Finally, $U(S)$ must be convex everywhere in its domain. Concave regions are forbidden in a stable thermodynamic system. This is the short answer. If you need the long answer I could do but not immediately.
$endgroup$
– GiorgioP
Apr 29 at 22:39
add a comment |
$begingroup$
Thank you for the detailed answer. Still digesting. Can you please explain what you mean by "fundamental equations must be convex" and why this is true? Also, can you please elaborate on your definition of the free energy as the infimum over S? Does this just cuts the whole domain of $S$ to chunks where $U(S)$ is convex/concave?
$endgroup$
– Botond
Apr 29 at 21:57
$begingroup$
Convexity/concavity of fundamental equations (thermodynamic potentials in the present case) is a basic consequence of the minimum pincipe for internal energy. You'll find an argument for concavity of entropy in the Callen's textbook. Definition of free energy as inf is just the application of LF definition (according to thermodynamicists) to the case of S <->T change of variables. Finally, $U(S)$ must be convex everywhere in its domain. Concave regions are forbidden in a stable thermodynamic system. This is the short answer. If you need the long answer I could do but not immediately.
$endgroup$
– GiorgioP
Apr 29 at 22:39
$begingroup$
Thank you for the detailed answer. Still digesting. Can you please explain what you mean by "fundamental equations must be convex" and why this is true? Also, can you please elaborate on your definition of the free energy as the infimum over S? Does this just cuts the whole domain of $S$ to chunks where $U(S)$ is convex/concave?
$endgroup$
– Botond
Apr 29 at 21:57
$begingroup$
Thank you for the detailed answer. Still digesting. Can you please explain what you mean by "fundamental equations must be convex" and why this is true? Also, can you please elaborate on your definition of the free energy as the infimum over S? Does this just cuts the whole domain of $S$ to chunks where $U(S)$ is convex/concave?
$endgroup$
– Botond
Apr 29 at 21:57
$begingroup$
Convexity/concavity of fundamental equations (thermodynamic potentials in the present case) is a basic consequence of the minimum pincipe for internal energy. You'll find an argument for concavity of entropy in the Callen's textbook. Definition of free energy as inf is just the application of LF definition (according to thermodynamicists) to the case of S <->T change of variables. Finally, $U(S)$ must be convex everywhere in its domain. Concave regions are forbidden in a stable thermodynamic system. This is the short answer. If you need the long answer I could do but not immediately.
$endgroup$
– GiorgioP
Apr 29 at 22:39
$begingroup$
Convexity/concavity of fundamental equations (thermodynamic potentials in the present case) is a basic consequence of the minimum pincipe for internal energy. You'll find an argument for concavity of entropy in the Callen's textbook. Definition of free energy as inf is just the application of LF definition (according to thermodynamicists) to the case of S <->T change of variables. Finally, $U(S)$ must be convex everywhere in its domain. Concave regions are forbidden in a stable thermodynamic system. This is the short answer. If you need the long answer I could do but not immediately.
$endgroup$
– GiorgioP
Apr 29 at 22:39
add a comment |
$begingroup$
Function $T(S)=(partial U/partial S)_V$ does not have to be strictly increasing with $S$. For example, when heat is added to a mix of ice and water at 0 degrees Celsius, entropy increases but temperature stays the same until all ice is melted. So it is not, in general, possible to express entropy as function of $T,V,N$.
Free (Helmholtz) energy is usually defined as
$$
F(T,V,N) = U - TS
$$
but in general neither $U$ nor $S$ can be expressed as functions of $T,V,N$.
answered Apr 29 at 18:45
Ján LalinskýJán Lalinský
16.2k1441
$endgroup$
$begingroup$
I'm confused. I thought that if you externally impose a fixed $T,V,N$ to the system, that defines your $F(T,V,N)$ free energy unambiguously. You can write a free energy as a function of the molar concentrations, but the equilibrium free energy will be the one that minimizes $F(T,V,N)$. Thus the equilibrium molar concentrations are also functions of $T,V,N$.
$endgroup$
– Botond
Apr 29 at 19:23
$begingroup$
My mistake, the part about molar numbers was wrong. I changed my answer.
$endgroup$
– Ján Lalinský
Apr 29 at 21:07
$begingroup$
Thank you. What still bugs me is that if neither $U$ nor $S$ can be expressed as a function of $T,V,N$, how would you possibly know that $U-TS$ will only depend on $T,V,N$ and not on $U$ and/or $S$?
$endgroup$
– Botond
Apr 29 at 21:47
$begingroup$
It follows from the definition of $F$. We have $dU = TdS - pdV$, so from the differentials we see $U$ is a function of $S,V$. Then since $F=U-TS$, we have $dF=dU - TdS - SdT$ so $dF = -SdT -pdV$ so again from differentials $F$ must be function of $T,V$.
$endgroup$
– Ján Lalinský
Apr 29 at 22:33
1
$begingroup$
What the differential tells me is that $S=partial F(T,V,N) / partial T$ thus $S=S(T,V,N)$
$endgroup$
– Botond
2 days ago
|
show 2 more comments
$begingroup$
Function $T(S)=(partial U/partial S)_V$ does not have to be strictly increasing with $S$. For example, when heat is added to a mix of ice and water at 0 degrees Celsius, entropy increases but temperature stays the same until all ice is melted. So it is not, in general, possible to express entropy as function of $T,V,N$.
Free (Helmholtz) energy is usually defined as
$$
F(T,V,N) = U - TS
$$
but in general neither $U$ nor $S$ can be expressed as functions of $T,V,N$.
answered Apr 29 at 18:45
Ján LalinskýJán Lalinský
16.2k1441
$endgroup$
$begingroup$
I'm confused. I thought that if you externally impose a fixed $T,V,N$ to the system, that defines your $F(T,V,N)$ free energy unambiguously. You can write a free energy as a function of the molar concentrations, but the equilibrium free energy will be the one that minimizes $F(T,V,N)$. Thus the equilibrium molar concentrations are also functions of $T,V,N$.
$endgroup$
– Botond
Apr 29 at 19:23
$begingroup$
My mistake, the part about molar numbers was wrong. I changed my answer.
$endgroup$
– Ján Lalinský
Apr 29 at 21:07
$begingroup$
Thank you. What still bugs me is that if neither $U$ nor $S$ can be expressed as a function of $T,V,N$, how would you possibly know that $U-TS$ will only depend on $T,V,N$ and not on $U$ and/or $S$?
$endgroup$
– Botond
Apr 29 at 21:47
$begingroup$
It follows from the definition of $F$. We have $dU = TdS - pdV$, so from the differentials we see $U$ is a function of $S,V$. Then since $F=U-TS$, we have $dF=dU - TdS - SdT$ so $dF = -SdT -pdV$ so again from differentials $F$ must be function of $T,V$.
$endgroup$
– Ján Lalinský
Apr 29 at 22:33
1
$begingroup$
What the differential tells me is that $S=partial F(T,V,N) / partial T$ thus $S=S(T,V,N)$
$endgroup$
– Botond
2 days ago
|
show 2 more comments
$begingroup$
Function $T(S)=(partial U/partial S)_V$ does not have to be strictly increasing with $S$. For example, when heat is added to a mix of ice and water at 0 degrees Celsius, entropy increases but temperature stays the same until all ice is melted. So it is not, in general, possible to express entropy as function of $T,V,N$.
Free (Helmholtz) energy is usually defined as
$$
F(T,V,N) = U - TS
$$
but in general neither $U$ nor $S$ can be expressed as functions of $T,V,N$.
answered Apr 29 at 18:45
Ján LalinskýJán Lalinský
16.2k1441
$endgroup$
Function $T(S)=(partial U/partial S)_V$ does not have to be strictly increasing with $S$. For example, when heat is added to a mix of ice and water at 0 degrees Celsius, entropy increases but temperature stays the same until all ice is melted. So it is not, in general, possible to express entropy as function of $T,V,N$.
Free (Helmholtz) energy is usually defined as
$$
F(T,V,N) = U - TS
$$
but in general neither $U$ nor $S$ can be expressed as functions of $T,V,N$.
answered Apr 29 at 18:45
Ján LalinskýJán Lalinský
16.2k1441
edited yesterday
answered Apr 29 at 18:45
Ján LalinskýJán Lalinský
16.2k1441
answered Apr 29 at 18:45
Ján LalinskýJán Lalinský
16.2k1441
answered Apr 29 at 18:45
Ján LalinskýJán Lalinský
16.2k1441
16.2k1441
$begingroup$
I'm confused. I thought that if you externally impose a fixed $T,V,N$ to the system, that defines your $F(T,V,N)$ free energy unambiguously. You can write a free energy as a function of the molar concentrations, but the equilibrium free energy will be the one that minimizes $F(T,V,N)$. Thus the equilibrium molar concentrations are also functions of $T,V,N$.
$endgroup$
– Botond
Apr 29 at 19:23
$begingroup$
My mistake, the part about molar numbers was wrong. I changed my answer.
$endgroup$
– Ján Lalinský
Apr 29 at 21:07
$begingroup$
Thank you. What still bugs me is that if neither $U$ nor $S$ can be expressed as a function of $T,V,N$, how would you possibly know that $U-TS$ will only depend on $T,V,N$ and not on $U$ and/or $S$?
$endgroup$
– Botond
Apr 29 at 21:47
$begingroup$
It follows from the definition of $F$. We have $dU = TdS - pdV$, so from the differentials we see $U$ is a function of $S,V$. Then since $F=U-TS$, we have $dF=dU - TdS - SdT$ so $dF = -SdT -pdV$ so again from differentials $F$ must be function of $T,V$.
$endgroup$
– Ján Lalinský
Apr 29 at 22:33
1
$begingroup$
What the differential tells me is that $S=partial F(T,V,N) / partial T$ thus $S=S(T,V,N)$
$endgroup$
– Botond
2 days ago
|
show 2 more comments
$begingroup$
I'm confused. I thought that if you externally impose a fixed $T,V,N$ to the system, that defines your $F(T,V,N)$ free energy unambiguously. You can write a free energy as a function of the molar concentrations, but the equilibrium free energy will be the one that minimizes $F(T,V,N)$. Thus the equilibrium molar concentrations are also functions of $T,V,N$.
$endgroup$
– Botond
Apr 29 at 19:23
$begingroup$
My mistake, the part about molar numbers was wrong. I changed my answer.
$endgroup$
– Ján Lalinský
Apr 29 at 21:07
$begingroup$
Thank you. What still bugs me is that if neither $U$ nor $S$ can be expressed as a function of $T,V,N$, how would you possibly know that $U-TS$ will only depend on $T,V,N$ and not on $U$ and/or $S$?
$endgroup$
– Botond
Apr 29 at 21:47
$begingroup$
It follows from the definition of $F$. We have $dU = TdS - pdV$, so from the differentials we see $U$ is a function of $S,V$. Then since $F=U-TS$, we have $dF=dU - TdS - SdT$ so $dF = -SdT -pdV$ so again from differentials $F$ must be function of $T,V$.
$endgroup$
– Ján Lalinský
Apr 29 at 22:33
1
$begingroup$
What the differential tells me is that $S=partial F(T,V,N) / partial T$ thus $S=S(T,V,N)$
$endgroup$
– Botond
2 days ago
$begingroup$
I'm confused. I thought that if you externally impose a fixed $T,V,N$ to the system, that defines your $F(T,V,N)$ free energy unambiguously. You can write a free energy as a function of the molar concentrations, but the equilibrium free energy will be the one that minimizes $F(T,V,N)$. Thus the equilibrium molar concentrations are also functions of $T,V,N$.
$endgroup$
– Botond
Apr 29 at 19:23
$begingroup$
I'm confused. I thought that if you externally impose a fixed $T,V,N$ to the system, that defines your $F(T,V,N)$ free energy unambiguously. You can write a free energy as a function of the molar concentrations, but the equilibrium free energy will be the one that minimizes $F(T,V,N)$. Thus the equilibrium molar concentrations are also functions of $T,V,N$.
$endgroup$
– Botond
Apr 29 at 19:23
$begingroup$
My mistake, the part about molar numbers was wrong. I changed my answer.
$endgroup$
– Ján Lalinský
Apr 29 at 21:07
$begingroup$
My mistake, the part about molar numbers was wrong. I changed my answer.
$endgroup$
– Ján Lalinský
Apr 29 at 21:07
$begingroup$
Thank you. What still bugs me is that if neither $U$ nor $S$ can be expressed as a function of $T,V,N$, how would you possibly know that $U-TS$ will only depend on $T,V,N$ and not on $U$ and/or $S$?
$endgroup$
– Botond
Apr 29 at 21:47
$begingroup$
Thank you. What still bugs me is that if neither $U$ nor $S$ can be expressed as a function of $T,V,N$, how would you possibly know that $U-TS$ will only depend on $T,V,N$ and not on $U$ and/or $S$?
$endgroup$
– Botond
Apr 29 at 21:47
$begingroup$
It follows from the definition of $F$. We have $dU = TdS - pdV$, so from the differentials we see $U$ is a function of $S,V$. Then since $F=U-TS$, we have $dF=dU - TdS - SdT$ so $dF = -SdT -pdV$ so again from differentials $F$ must be function of $T,V$.
$endgroup$
– Ján Lalinský
Apr 29 at 22:33
$begingroup$
It follows from the definition of $F$. We have $dU = TdS - pdV$, so from the differentials we see $U$ is a function of $S,V$. Then since $F=U-TS$, we have $dF=dU - TdS - SdT$ so $dF = -SdT -pdV$ so again from differentials $F$ must be function of $T,V$.
$endgroup$
– Ján Lalinský
Apr 29 at 22:33
1
1
$begingroup$
What the differential tells me is that $S=partial F(T,V,N) / partial T$ thus $S=S(T,V,N)$
$endgroup$
– Botond
2 days ago
$begingroup$
What the differential tells me is that $S=partial F(T,V,N) / partial T$ thus $S=S(T,V,N)$
$endgroup$
– Botond
2 days ago
|
show 2 more comments
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$begingroup$
You can prove $U(S)$ is always a convex function, so that puts some limits on what can go wrong here.
$endgroup$
– Jahan Claes
Apr 29 at 19:30
$begingroup$
@JahanClaes Could you sketch or refer to a proof that $U(S)$ is always convex?
$endgroup$
– Botond
Apr 29 at 21:54
1
$begingroup$
If you want a book on thermodynamics which explicitly addresses the convex nature of the entropy function, and what it implies, including for phase transitions, then I have written one; publisher is Oxford University Press. I hope this comment does not break self-advertising rules on the site. I think it is fair because it is directly addressed to the question asked.
$endgroup$
– Andrew Steane
yesterday