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?










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    up vote
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    down vote

    favorite
    1












    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?










    share|cite|improve this question


























      up vote
      9
      down vote

      favorite
      1









      up vote
      9
      down vote

      favorite
      1






      1





      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?










      share|cite|improve this question















      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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      edited 18 hours ago









      Qmechanic

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      TaeNyFan

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          3 Answers
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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.






          share|cite|improve this answer




























            up vote
            4
            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.






            share|cite|improve this answer








            New contributor




            jemisjoky is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.


















            • Does metastable state correspond to entropy's local maximum?
              – Gec
              20 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
              19 hours ago


















            up vote
            3
            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 $S$).






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              up vote
              5
              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.






              share|cite|improve this answer

























                up vote
                5
                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.






                share|cite|improve this answer























                  up vote
                  5
                  down vote










                  up vote
                  5
                  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.






                  share|cite|improve this answer













                  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.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 18 hours ago









                  GiorgioP

                  1,109212




                  1,109212






















                      up vote
                      4
                      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.






                      share|cite|improve this answer








                      New contributor




                      jemisjoky is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                      Check out our Code of Conduct.


















                      • Does metastable state correspond to entropy's local maximum?
                        – Gec
                        20 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
                        19 hours ago















                      up vote
                      4
                      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.






                      share|cite|improve this answer








                      New contributor




                      jemisjoky is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                      Check out our Code of Conduct.


















                      • Does metastable state correspond to entropy's local maximum?
                        – Gec
                        20 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
                        19 hours ago













                      up vote
                      4
                      down vote










                      up vote
                      4
                      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.






                      share|cite|improve this answer








                      New contributor




                      jemisjoky is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                      Check out our Code of Conduct.









                      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.







                      share|cite|improve this answer








                      New contributor




                      jemisjoky is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                      Check out our Code of Conduct.









                      share|cite|improve this answer



                      share|cite|improve this answer






                      New contributor




                      jemisjoky is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                      Check out our Code of Conduct.









                      answered 20 hours ago









                      jemisjoky

                      712




                      712




                      New contributor




                      jemisjoky is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                      Check out our Code of Conduct.





                      New contributor





                      jemisjoky is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                      Check out our Code of Conduct.






                      jemisjoky is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                      Check out our Code of Conduct.












                      • Does metastable state correspond to entropy's local maximum?
                        – Gec
                        20 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
                        19 hours ago


















                      • Does metastable state correspond to entropy's local maximum?
                        – Gec
                        20 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
                        19 hours ago
















                      Does metastable state correspond to entropy's local maximum?
                      – Gec
                      20 hours ago






                      Does metastable state correspond to entropy's local maximum?
                      – Gec
                      20 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
                      19 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
                      19 hours ago










                      up vote
                      3
                      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 $S$).






                      share|cite|improve this answer



























                        up vote
                        3
                        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 $S$).






                        share|cite|improve this answer

























                          up vote
                          3
                          down vote










                          up vote
                          3
                          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 $S$).






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                          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 $S$).







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited 21 hours ago

























                          answered 21 hours ago









                          ErickShock

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