Maximum entropy at equilibrium for closed system: Local maximum or global maximum?
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For a closed system at equilibrium the entropy is maximum. Is this a local maximum or is it a global maximum?
I am an undergraduate physics student and it seems that the possibility of entropy having local maximums was not discussed. It was always assumed it was a global maximum. Is this true in all cases?
thermodynamics statistical-mechanics entropy equilibrium
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up vote
8
down vote
favorite
For a closed system at equilibrium the entropy is maximum. Is this a local maximum or is it a global maximum?
I am an undergraduate physics student and it seems that the possibility of entropy having local maximums was not discussed. It was always assumed it was a global maximum. Is this true in all cases?
thermodynamics statistical-mechanics entropy equilibrium
Entropy is a convex function - at least of it's natural variables. Hence it only has a global maximum.
– Nephente
7 hours ago
add a comment |
up vote
8
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favorite
up vote
8
down vote
favorite
For a closed system at equilibrium the entropy is maximum. Is this a local maximum or is it a global maximum?
I am an undergraduate physics student and it seems that the possibility of entropy having local maximums was not discussed. It was always assumed it was a global maximum. Is this true in all cases?
thermodynamics statistical-mechanics entropy equilibrium
For a closed system at equilibrium the entropy is maximum. Is this a local maximum or is it a global maximum?
I am an undergraduate physics student and it seems that the possibility of entropy having local maximums was not discussed. It was always assumed it was a global maximum. Is this true in all cases?
thermodynamics statistical-mechanics entropy equilibrium
thermodynamics statistical-mechanics entropy equilibrium
edited 5 hours ago
Qmechanic♦
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100k121811133
asked 8 hours ago
TaeNyFan
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Entropy is a convex function - at least of it's natural variables. Hence it only has a global maximum.
– Nephente
7 hours ago
add a comment |
Entropy is a convex function - at least of it's natural variables. Hence it only has a global maximum.
– Nephente
7 hours ago
Entropy is a convex function - at least of it's natural variables. Hence it only has a global maximum.
– Nephente
7 hours ago
Entropy is a convex function - at least of it's natural variables. Hence it only has a global maximum.
– Nephente
7 hours ago
add a comment |
3 Answers
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up vote
4
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For a closed system at equilibrium the entropy is maximum. Is this a local maximum or is it a global maximum?
To allow a meaningful answer, it would be necessary to qualify the maximum. Maximum with respect to which variable? In thermodynamics the correct (and meaningful) statement is "maximum with respect to the variables (different from the thermodynamic state variables describing the isolated system) which represent the entropy dependence on all possible internal constraints" (i.e. constraints in the isolated system).
From this principle, i.e. from this sentence which condensate a long series of experiences, it is possible to obtain many consequences, like the equilibrium conditions or even the condition of concavity of the entropy as function of the variables which describe the macroscopic state of the isolated system, which, I stress, are not the same which describe the constraints.
So, from the maximum principle, one can get the concavity of entropy with respect to the state variables, by carefully choosing the kind of constraint.
However, such concavity as function of the state variables, does not imply strict concavity, or even concavity of entropy with respect to any possible internal constraint. For example, one could think of a constraint forcing an atomic system to stay only in two ordered crystalline structures (maybe not easy in a lab but not complicate in a computer simulation). For such constrained system one could have local maxima, with the highest being the true stable state and the remaining one, being a metastable system.
Probably the most interesting question could be: if we remove all the internal constraints, how can we know if there is a unique final equilibrium state?
And maybe this was the intended original question. Well, at the best of my knowledge, there is no definite answer. And there is a good reason for that. It is possible to imagine systems which do not reach equilibrium at all (non ergodic systems). Thus, I would consider the request of a unique maximum value of the entropy as an additional request for thermodynamic systems.
add a comment |
up vote
3
down vote
Good question! At a fundamental level, entropy depends on the probability distribution $p(x)$ of the microscopic states $xin Omega$ of a system, which is given by the following equation:
$$H(p) = -sum_{xin Omega}p(x)log p(x)$$
Just specifying a few macroscopic variables of a system (e.g. $U$, $V$, and $N$) isn't enough to determine a unique probability distribution over the microscopic states, but the principle of maximum entropy says that the equilibrium distribution over microscopic states satisfying these macroscopic constraints is the one that has the greatest entropy.
Mathematically, $H(p)$ is a (strictly) concave function of the probability distributions, which means that it can only increase when we average over probability distributions:
$$H(lambda p_1 + (1-lambda) p_2) geq lambda H(p_1) + (1-lambda)H(p_2)$$
An incredibly useful property of (strictly) concave functions is that they can only have one local maximum point, which is then guaranteed to be the global maximum. This is the reason why people ignore the possibility of multiple local maxima of the entropy, because the concave nature of the entropy guarantees that you'll only ever have one (see these notes, for example).
This of course isn't the whole story, because in practice you can get things like metastable states, where a system gets stuck in a non-equilibrium state for a long time. But at least on paper, that's why we only ever talk about "the" maximum entropy state.
New contributor
Does metastable state correspond to entropy's local maximum?
– Gec
6 hours ago
4
Uniqueness of the maximum follows from the strict (without parentheses) concavity. Unfortunately, thermodynamic entropy is just concave, not strict concave. It is strictly concave almost everywhere, but in the presence of a first order phase transition, the physical coexistence of phases implies a non strict concavity. Therefore it is not possible to claim uniqueness of the maximum, in general.
– GiorgioP
5 hours ago
add a comment |
up vote
2
down vote
To quote H.B. Callen Thermodynamics book, his second postulate about the formal development of Thermodynamics is:
Postulate II - There exists a function (called the entropy $S$) of the extensive parameters of any composite system, defined for all equilibrium states and having the following property. The values assumed by the extensive parameters in the absence of an internal constraint are those that maximize the entropy over the manifold of constrained equilibrium states.
As an example: assume $S$ is a function of $U,V,N$ and suppose your system can only exchange heat, so $V$ and $N$ are constants. Out of all the possible values the unconstrained parameter $U$ may take, the system at equilibrium will assume the value of $U$ such that $S$ is a maximum. So $S$ will be a global maximum in respect to $U$, but not necessarily in respect to $V$ and $N$ in a particular problem. However, if you also allow your system to expand and to exchange matter, by this postulate the values assumed by $U,V,N$ (which are now all unconstrained) will be such that $S$ is a global maximum.
Edit. I was forgetting about phase transitions. When near a phase transition there will be states such that the Gibbs Potential $G$ is a local minimum (so $S$ is a local maximum). These states are, however, metastable and your thermodynamic system will typically prefer more stable states that correspond to a global minimum of $G$ (and consequently, to a global maximum of $S$).
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
For a closed system at equilibrium the entropy is maximum. Is this a local maximum or is it a global maximum?
To allow a meaningful answer, it would be necessary to qualify the maximum. Maximum with respect to which variable? In thermodynamics the correct (and meaningful) statement is "maximum with respect to the variables (different from the thermodynamic state variables describing the isolated system) which represent the entropy dependence on all possible internal constraints" (i.e. constraints in the isolated system).
From this principle, i.e. from this sentence which condensate a long series of experiences, it is possible to obtain many consequences, like the equilibrium conditions or even the condition of concavity of the entropy as function of the variables which describe the macroscopic state of the isolated system, which, I stress, are not the same which describe the constraints.
So, from the maximum principle, one can get the concavity of entropy with respect to the state variables, by carefully choosing the kind of constraint.
However, such concavity as function of the state variables, does not imply strict concavity, or even concavity of entropy with respect to any possible internal constraint. For example, one could think of a constraint forcing an atomic system to stay only in two ordered crystalline structures (maybe not easy in a lab but not complicate in a computer simulation). For such constrained system one could have local maxima, with the highest being the true stable state and the remaining one, being a metastable system.
Probably the most interesting question could be: if we remove all the internal constraints, how can we know if there is a unique final equilibrium state?
And maybe this was the intended original question. Well, at the best of my knowledge, there is no definite answer. And there is a good reason for that. It is possible to imagine systems which do not reach equilibrium at all (non ergodic systems). Thus, I would consider the request of a unique maximum value of the entropy as an additional request for thermodynamic systems.
add a comment |
up vote
4
down vote
For a closed system at equilibrium the entropy is maximum. Is this a local maximum or is it a global maximum?
To allow a meaningful answer, it would be necessary to qualify the maximum. Maximum with respect to which variable? In thermodynamics the correct (and meaningful) statement is "maximum with respect to the variables (different from the thermodynamic state variables describing the isolated system) which represent the entropy dependence on all possible internal constraints" (i.e. constraints in the isolated system).
From this principle, i.e. from this sentence which condensate a long series of experiences, it is possible to obtain many consequences, like the equilibrium conditions or even the condition of concavity of the entropy as function of the variables which describe the macroscopic state of the isolated system, which, I stress, are not the same which describe the constraints.
So, from the maximum principle, one can get the concavity of entropy with respect to the state variables, by carefully choosing the kind of constraint.
However, such concavity as function of the state variables, does not imply strict concavity, or even concavity of entropy with respect to any possible internal constraint. For example, one could think of a constraint forcing an atomic system to stay only in two ordered crystalline structures (maybe not easy in a lab but not complicate in a computer simulation). For such constrained system one could have local maxima, with the highest being the true stable state and the remaining one, being a metastable system.
Probably the most interesting question could be: if we remove all the internal constraints, how can we know if there is a unique final equilibrium state?
And maybe this was the intended original question. Well, at the best of my knowledge, there is no definite answer. And there is a good reason for that. It is possible to imagine systems which do not reach equilibrium at all (non ergodic systems). Thus, I would consider the request of a unique maximum value of the entropy as an additional request for thermodynamic systems.
add a comment |
up vote
4
down vote
up vote
4
down vote
For a closed system at equilibrium the entropy is maximum. Is this a local maximum or is it a global maximum?
To allow a meaningful answer, it would be necessary to qualify the maximum. Maximum with respect to which variable? In thermodynamics the correct (and meaningful) statement is "maximum with respect to the variables (different from the thermodynamic state variables describing the isolated system) which represent the entropy dependence on all possible internal constraints" (i.e. constraints in the isolated system).
From this principle, i.e. from this sentence which condensate a long series of experiences, it is possible to obtain many consequences, like the equilibrium conditions or even the condition of concavity of the entropy as function of the variables which describe the macroscopic state of the isolated system, which, I stress, are not the same which describe the constraints.
So, from the maximum principle, one can get the concavity of entropy with respect to the state variables, by carefully choosing the kind of constraint.
However, such concavity as function of the state variables, does not imply strict concavity, or even concavity of entropy with respect to any possible internal constraint. For example, one could think of a constraint forcing an atomic system to stay only in two ordered crystalline structures (maybe not easy in a lab but not complicate in a computer simulation). For such constrained system one could have local maxima, with the highest being the true stable state and the remaining one, being a metastable system.
Probably the most interesting question could be: if we remove all the internal constraints, how can we know if there is a unique final equilibrium state?
And maybe this was the intended original question. Well, at the best of my knowledge, there is no definite answer. And there is a good reason for that. It is possible to imagine systems which do not reach equilibrium at all (non ergodic systems). Thus, I would consider the request of a unique maximum value of the entropy as an additional request for thermodynamic systems.
For a closed system at equilibrium the entropy is maximum. Is this a local maximum or is it a global maximum?
To allow a meaningful answer, it would be necessary to qualify the maximum. Maximum with respect to which variable? In thermodynamics the correct (and meaningful) statement is "maximum with respect to the variables (different from the thermodynamic state variables describing the isolated system) which represent the entropy dependence on all possible internal constraints" (i.e. constraints in the isolated system).
From this principle, i.e. from this sentence which condensate a long series of experiences, it is possible to obtain many consequences, like the equilibrium conditions or even the condition of concavity of the entropy as function of the variables which describe the macroscopic state of the isolated system, which, I stress, are not the same which describe the constraints.
So, from the maximum principle, one can get the concavity of entropy with respect to the state variables, by carefully choosing the kind of constraint.
However, such concavity as function of the state variables, does not imply strict concavity, or even concavity of entropy with respect to any possible internal constraint. For example, one could think of a constraint forcing an atomic system to stay only in two ordered crystalline structures (maybe not easy in a lab but not complicate in a computer simulation). For such constrained system one could have local maxima, with the highest being the true stable state and the remaining one, being a metastable system.
Probably the most interesting question could be: if we remove all the internal constraints, how can we know if there is a unique final equilibrium state?
And maybe this was the intended original question. Well, at the best of my knowledge, there is no definite answer. And there is a good reason for that. It is possible to imagine systems which do not reach equilibrium at all (non ergodic systems). Thus, I would consider the request of a unique maximum value of the entropy as an additional request for thermodynamic systems.
answered 4 hours ago
GiorgioP
1,099212
1,099212
add a comment |
add a comment |
up vote
3
down vote
Good question! At a fundamental level, entropy depends on the probability distribution $p(x)$ of the microscopic states $xin Omega$ of a system, which is given by the following equation:
$$H(p) = -sum_{xin Omega}p(x)log p(x)$$
Just specifying a few macroscopic variables of a system (e.g. $U$, $V$, and $N$) isn't enough to determine a unique probability distribution over the microscopic states, but the principle of maximum entropy says that the equilibrium distribution over microscopic states satisfying these macroscopic constraints is the one that has the greatest entropy.
Mathematically, $H(p)$ is a (strictly) concave function of the probability distributions, which means that it can only increase when we average over probability distributions:
$$H(lambda p_1 + (1-lambda) p_2) geq lambda H(p_1) + (1-lambda)H(p_2)$$
An incredibly useful property of (strictly) concave functions is that they can only have one local maximum point, which is then guaranteed to be the global maximum. This is the reason why people ignore the possibility of multiple local maxima of the entropy, because the concave nature of the entropy guarantees that you'll only ever have one (see these notes, for example).
This of course isn't the whole story, because in practice you can get things like metastable states, where a system gets stuck in a non-equilibrium state for a long time. But at least on paper, that's why we only ever talk about "the" maximum entropy state.
New contributor
Does metastable state correspond to entropy's local maximum?
– Gec
6 hours ago
4
Uniqueness of the maximum follows from the strict (without parentheses) concavity. Unfortunately, thermodynamic entropy is just concave, not strict concave. It is strictly concave almost everywhere, but in the presence of a first order phase transition, the physical coexistence of phases implies a non strict concavity. Therefore it is not possible to claim uniqueness of the maximum, in general.
– GiorgioP
5 hours ago
add a comment |
up vote
3
down vote
Good question! At a fundamental level, entropy depends on the probability distribution $p(x)$ of the microscopic states $xin Omega$ of a system, which is given by the following equation:
$$H(p) = -sum_{xin Omega}p(x)log p(x)$$
Just specifying a few macroscopic variables of a system (e.g. $U$, $V$, and $N$) isn't enough to determine a unique probability distribution over the microscopic states, but the principle of maximum entropy says that the equilibrium distribution over microscopic states satisfying these macroscopic constraints is the one that has the greatest entropy.
Mathematically, $H(p)$ is a (strictly) concave function of the probability distributions, which means that it can only increase when we average over probability distributions:
$$H(lambda p_1 + (1-lambda) p_2) geq lambda H(p_1) + (1-lambda)H(p_2)$$
An incredibly useful property of (strictly) concave functions is that they can only have one local maximum point, which is then guaranteed to be the global maximum. This is the reason why people ignore the possibility of multiple local maxima of the entropy, because the concave nature of the entropy guarantees that you'll only ever have one (see these notes, for example).
This of course isn't the whole story, because in practice you can get things like metastable states, where a system gets stuck in a non-equilibrium state for a long time. But at least on paper, that's why we only ever talk about "the" maximum entropy state.
New contributor
Does metastable state correspond to entropy's local maximum?
– Gec
6 hours ago
4
Uniqueness of the maximum follows from the strict (without parentheses) concavity. Unfortunately, thermodynamic entropy is just concave, not strict concave. It is strictly concave almost everywhere, but in the presence of a first order phase transition, the physical coexistence of phases implies a non strict concavity. Therefore it is not possible to claim uniqueness of the maximum, in general.
– GiorgioP
5 hours ago
add a comment |
up vote
3
down vote
up vote
3
down vote
Good question! At a fundamental level, entropy depends on the probability distribution $p(x)$ of the microscopic states $xin Omega$ of a system, which is given by the following equation:
$$H(p) = -sum_{xin Omega}p(x)log p(x)$$
Just specifying a few macroscopic variables of a system (e.g. $U$, $V$, and $N$) isn't enough to determine a unique probability distribution over the microscopic states, but the principle of maximum entropy says that the equilibrium distribution over microscopic states satisfying these macroscopic constraints is the one that has the greatest entropy.
Mathematically, $H(p)$ is a (strictly) concave function of the probability distributions, which means that it can only increase when we average over probability distributions:
$$H(lambda p_1 + (1-lambda) p_2) geq lambda H(p_1) + (1-lambda)H(p_2)$$
An incredibly useful property of (strictly) concave functions is that they can only have one local maximum point, which is then guaranteed to be the global maximum. This is the reason why people ignore the possibility of multiple local maxima of the entropy, because the concave nature of the entropy guarantees that you'll only ever have one (see these notes, for example).
This of course isn't the whole story, because in practice you can get things like metastable states, where a system gets stuck in a non-equilibrium state for a long time. But at least on paper, that's why we only ever talk about "the" maximum entropy state.
New contributor
Good question! At a fundamental level, entropy depends on the probability distribution $p(x)$ of the microscopic states $xin Omega$ of a system, which is given by the following equation:
$$H(p) = -sum_{xin Omega}p(x)log p(x)$$
Just specifying a few macroscopic variables of a system (e.g. $U$, $V$, and $N$) isn't enough to determine a unique probability distribution over the microscopic states, but the principle of maximum entropy says that the equilibrium distribution over microscopic states satisfying these macroscopic constraints is the one that has the greatest entropy.
Mathematically, $H(p)$ is a (strictly) concave function of the probability distributions, which means that it can only increase when we average over probability distributions:
$$H(lambda p_1 + (1-lambda) p_2) geq lambda H(p_1) + (1-lambda)H(p_2)$$
An incredibly useful property of (strictly) concave functions is that they can only have one local maximum point, which is then guaranteed to be the global maximum. This is the reason why people ignore the possibility of multiple local maxima of the entropy, because the concave nature of the entropy guarantees that you'll only ever have one (see these notes, for example).
This of course isn't the whole story, because in practice you can get things like metastable states, where a system gets stuck in a non-equilibrium state for a long time. But at least on paper, that's why we only ever talk about "the" maximum entropy state.
New contributor
New contributor
answered 7 hours ago
jemisjoky
612
612
New contributor
New contributor
Does metastable state correspond to entropy's local maximum?
– Gec
6 hours ago
4
Uniqueness of the maximum follows from the strict (without parentheses) concavity. Unfortunately, thermodynamic entropy is just concave, not strict concave. It is strictly concave almost everywhere, but in the presence of a first order phase transition, the physical coexistence of phases implies a non strict concavity. Therefore it is not possible to claim uniqueness of the maximum, in general.
– GiorgioP
5 hours ago
add a comment |
Does metastable state correspond to entropy's local maximum?
– Gec
6 hours ago
4
Uniqueness of the maximum follows from the strict (without parentheses) concavity. Unfortunately, thermodynamic entropy is just concave, not strict concave. It is strictly concave almost everywhere, but in the presence of a first order phase transition, the physical coexistence of phases implies a non strict concavity. Therefore it is not possible to claim uniqueness of the maximum, in general.
– GiorgioP
5 hours ago
Does metastable state correspond to entropy's local maximum?
– Gec
6 hours ago
Does metastable state correspond to entropy's local maximum?
– Gec
6 hours ago
4
4
Uniqueness of the maximum follows from the strict (without parentheses) concavity. Unfortunately, thermodynamic entropy is just concave, not strict concave. It is strictly concave almost everywhere, but in the presence of a first order phase transition, the physical coexistence of phases implies a non strict concavity. Therefore it is not possible to claim uniqueness of the maximum, in general.
– GiorgioP
5 hours ago
Uniqueness of the maximum follows from the strict (without parentheses) concavity. Unfortunately, thermodynamic entropy is just concave, not strict concave. It is strictly concave almost everywhere, but in the presence of a first order phase transition, the physical coexistence of phases implies a non strict concavity. Therefore it is not possible to claim uniqueness of the maximum, in general.
– GiorgioP
5 hours ago
add a comment |
up vote
2
down vote
To quote H.B. Callen Thermodynamics book, his second postulate about the formal development of Thermodynamics is:
Postulate II - There exists a function (called the entropy $S$) of the extensive parameters of any composite system, defined for all equilibrium states and having the following property. The values assumed by the extensive parameters in the absence of an internal constraint are those that maximize the entropy over the manifold of constrained equilibrium states.
As an example: assume $S$ is a function of $U,V,N$ and suppose your system can only exchange heat, so $V$ and $N$ are constants. Out of all the possible values the unconstrained parameter $U$ may take, the system at equilibrium will assume the value of $U$ such that $S$ is a maximum. So $S$ will be a global maximum in respect to $U$, but not necessarily in respect to $V$ and $N$ in a particular problem. However, if you also allow your system to expand and to exchange matter, by this postulate the values assumed by $U,V,N$ (which are now all unconstrained) will be such that $S$ is a global maximum.
Edit. I was forgetting about phase transitions. When near a phase transition there will be states such that the Gibbs Potential $G$ is a local minimum (so $S$ is a local maximum). These states are, however, metastable and your thermodynamic system will typically prefer more stable states that correspond to a global minimum of $G$ (and consequently, to a global maximum of $S$).
add a comment |
up vote
2
down vote
To quote H.B. Callen Thermodynamics book, his second postulate about the formal development of Thermodynamics is:
Postulate II - There exists a function (called the entropy $S$) of the extensive parameters of any composite system, defined for all equilibrium states and having the following property. The values assumed by the extensive parameters in the absence of an internal constraint are those that maximize the entropy over the manifold of constrained equilibrium states.
As an example: assume $S$ is a function of $U,V,N$ and suppose your system can only exchange heat, so $V$ and $N$ are constants. Out of all the possible values the unconstrained parameter $U$ may take, the system at equilibrium will assume the value of $U$ such that $S$ is a maximum. So $S$ will be a global maximum in respect to $U$, but not necessarily in respect to $V$ and $N$ in a particular problem. However, if you also allow your system to expand and to exchange matter, by this postulate the values assumed by $U,V,N$ (which are now all unconstrained) will be such that $S$ is a global maximum.
Edit. I was forgetting about phase transitions. When near a phase transition there will be states such that the Gibbs Potential $G$ is a local minimum (so $S$ is a local maximum). These states are, however, metastable and your thermodynamic system will typically prefer more stable states that correspond to a global minimum of $G$ (and consequently, to a global maximum of $S$).
add a comment |
up vote
2
down vote
up vote
2
down vote
To quote H.B. Callen Thermodynamics book, his second postulate about the formal development of Thermodynamics is:
Postulate II - There exists a function (called the entropy $S$) of the extensive parameters of any composite system, defined for all equilibrium states and having the following property. The values assumed by the extensive parameters in the absence of an internal constraint are those that maximize the entropy over the manifold of constrained equilibrium states.
As an example: assume $S$ is a function of $U,V,N$ and suppose your system can only exchange heat, so $V$ and $N$ are constants. Out of all the possible values the unconstrained parameter $U$ may take, the system at equilibrium will assume the value of $U$ such that $S$ is a maximum. So $S$ will be a global maximum in respect to $U$, but not necessarily in respect to $V$ and $N$ in a particular problem. However, if you also allow your system to expand and to exchange matter, by this postulate the values assumed by $U,V,N$ (which are now all unconstrained) will be such that $S$ is a global maximum.
Edit. I was forgetting about phase transitions. When near a phase transition there will be states such that the Gibbs Potential $G$ is a local minimum (so $S$ is a local maximum). These states are, however, metastable and your thermodynamic system will typically prefer more stable states that correspond to a global minimum of $G$ (and consequently, to a global maximum of $S$).
To quote H.B. Callen Thermodynamics book, his second postulate about the formal development of Thermodynamics is:
Postulate II - There exists a function (called the entropy $S$) of the extensive parameters of any composite system, defined for all equilibrium states and having the following property. The values assumed by the extensive parameters in the absence of an internal constraint are those that maximize the entropy over the manifold of constrained equilibrium states.
As an example: assume $S$ is a function of $U,V,N$ and suppose your system can only exchange heat, so $V$ and $N$ are constants. Out of all the possible values the unconstrained parameter $U$ may take, the system at equilibrium will assume the value of $U$ such that $S$ is a maximum. So $S$ will be a global maximum in respect to $U$, but not necessarily in respect to $V$ and $N$ in a particular problem. However, if you also allow your system to expand and to exchange matter, by this postulate the values assumed by $U,V,N$ (which are now all unconstrained) will be such that $S$ is a global maximum.
Edit. I was forgetting about phase transitions. When near a phase transition there will be states such that the Gibbs Potential $G$ is a local minimum (so $S$ is a local maximum). These states are, however, metastable and your thermodynamic system will typically prefer more stable states that correspond to a global minimum of $G$ (and consequently, to a global maximum of $S$).
edited 7 hours ago
answered 7 hours ago
ErickShock
465
465
add a comment |
add a comment |
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Entropy is a convex function - at least of it's natural variables. Hence it only has a global maximum.
– Nephente
7 hours ago