Ambiguity in the definition of entropy

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The entropy $S$ of a system is defined as $$S = kln Omega.$$ What precisely is $Omega$? It refers to "the number of microstates" of the system, but is this the number of all accessible microstates or just the number of microstates corresponding to the systems current macrostate? Or is it something else that eludes me?










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    $begingroup$


    The entropy $S$ of a system is defined as $$S = kln Omega.$$ What precisely is $Omega$? It refers to "the number of microstates" of the system, but is this the number of all accessible microstates or just the number of microstates corresponding to the systems current macrostate? Or is it something else that eludes me?










    share|cite|improve this question











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      2





      $begingroup$


      The entropy $S$ of a system is defined as $$S = kln Omega.$$ What precisely is $Omega$? It refers to "the number of microstates" of the system, but is this the number of all accessible microstates or just the number of microstates corresponding to the systems current macrostate? Or is it something else that eludes me?










      share|cite|improve this question











      $endgroup$




      The entropy $S$ of a system is defined as $$S = kln Omega.$$ What precisely is $Omega$? It refers to "the number of microstates" of the system, but is this the number of all accessible microstates or just the number of microstates corresponding to the systems current macrostate? Or is it something else that eludes me?







      statistical-mechanics entropy






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          $begingroup$

          Entropy is a property of a macrostate, not a system. So $Omega$ is the number of microstates that correspond to the macrostate in question.



          Since it is almost always the change in entropy, not the absolute entropy, that is considered, and we're taking the log of $Omega$, it actually doesn't matter if the definition of S is ambiguous up to a constant multiplicative factor, as that will cancel out when we take dS.






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            Corresponding to the current macrostate. The principle of entropy is that a system seeks out the macro state that has the most microstates in it: in other words, our uncertainty about the underlying state of the system keeps multiplying and multiplying, until, with certain assumptions, we cannot do much better than just choosing a microstate uniformly at random.






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            • $begingroup$
              I would say corresponding to the macrostate whose entropy is being evaluated, simply because you can compute subjunctive entropies.
              $endgroup$
              – dmckee
              14 mins ago












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            2 Answers
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            2 Answers
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            7












            $begingroup$

            Entropy is a property of a macrostate, not a system. So $Omega$ is the number of microstates that correspond to the macrostate in question.



            Since it is almost always the change in entropy, not the absolute entropy, that is considered, and we're taking the log of $Omega$, it actually doesn't matter if the definition of S is ambiguous up to a constant multiplicative factor, as that will cancel out when we take dS.






            share|cite|improve this answer









            $endgroup$


















              7












              $begingroup$

              Entropy is a property of a macrostate, not a system. So $Omega$ is the number of microstates that correspond to the macrostate in question.



              Since it is almost always the change in entropy, not the absolute entropy, that is considered, and we're taking the log of $Omega$, it actually doesn't matter if the definition of S is ambiguous up to a constant multiplicative factor, as that will cancel out when we take dS.






              share|cite|improve this answer









              $endgroup$
















                7












                7








                7





                $begingroup$

                Entropy is a property of a macrostate, not a system. So $Omega$ is the number of microstates that correspond to the macrostate in question.



                Since it is almost always the change in entropy, not the absolute entropy, that is considered, and we're taking the log of $Omega$, it actually doesn't matter if the definition of S is ambiguous up to a constant multiplicative factor, as that will cancel out when we take dS.






                share|cite|improve this answer









                $endgroup$



                Entropy is a property of a macrostate, not a system. So $Omega$ is the number of microstates that correspond to the macrostate in question.



                Since it is almost always the change in entropy, not the absolute entropy, that is considered, and we're taking the log of $Omega$, it actually doesn't matter if the definition of S is ambiguous up to a constant multiplicative factor, as that will cancel out when we take dS.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 1 hour ago









                AcccumulationAcccumulation

                2,794312




                2,794312























                    5












                    $begingroup$

                    Corresponding to the current macrostate. The principle of entropy is that a system seeks out the macro state that has the most microstates in it: in other words, our uncertainty about the underlying state of the system keeps multiplying and multiplying, until, with certain assumptions, we cannot do much better than just choosing a microstate uniformly at random.






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      I would say corresponding to the macrostate whose entropy is being evaluated, simply because you can compute subjunctive entropies.
                      $endgroup$
                      – dmckee
                      14 mins ago
















                    5












                    $begingroup$

                    Corresponding to the current macrostate. The principle of entropy is that a system seeks out the macro state that has the most microstates in it: in other words, our uncertainty about the underlying state of the system keeps multiplying and multiplying, until, with certain assumptions, we cannot do much better than just choosing a microstate uniformly at random.






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      I would say corresponding to the macrostate whose entropy is being evaluated, simply because you can compute subjunctive entropies.
                      $endgroup$
                      – dmckee
                      14 mins ago














                    5












                    5








                    5





                    $begingroup$

                    Corresponding to the current macrostate. The principle of entropy is that a system seeks out the macro state that has the most microstates in it: in other words, our uncertainty about the underlying state of the system keeps multiplying and multiplying, until, with certain assumptions, we cannot do much better than just choosing a microstate uniformly at random.






                    share|cite|improve this answer









                    $endgroup$



                    Corresponding to the current macrostate. The principle of entropy is that a system seeks out the macro state that has the most microstates in it: in other words, our uncertainty about the underlying state of the system keeps multiplying and multiplying, until, with certain assumptions, we cannot do much better than just choosing a microstate uniformly at random.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 1 hour ago









                    CR DrostCR Drost

                    22.5k11961




                    22.5k11961












                    • $begingroup$
                      I would say corresponding to the macrostate whose entropy is being evaluated, simply because you can compute subjunctive entropies.
                      $endgroup$
                      – dmckee
                      14 mins ago


















                    • $begingroup$
                      I would say corresponding to the macrostate whose entropy is being evaluated, simply because you can compute subjunctive entropies.
                      $endgroup$
                      – dmckee
                      14 mins ago
















                    $begingroup$
                    I would say corresponding to the macrostate whose entropy is being evaluated, simply because you can compute subjunctive entropies.
                    $endgroup$
                    – dmckee
                    14 mins ago




                    $begingroup$
                    I would say corresponding to the macrostate whose entropy is being evaluated, simply because you can compute subjunctive entropies.
                    $endgroup$
                    – dmckee
                    14 mins ago


















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