Ambiguity in the definition of entropyHow are possible microstates discerned in Gibb's entropy formula?Statistical interpretation of EntropyEntropy as an arrow of timeWhat precisely does the 2nd law of thermo state, considering that entropy depends on how we define macrostate?The statistical interpretation of EntropyWhat is the cause for the inclusion of 'thermal equilibrium' in the statement of Ergodic hypothesis?Do the results of statistical mechanics depend upon the choice of macrostates?Entropy definition, additivity, laws in different ensemblesDefinition of entropy and other StatMech variablesWhat is the definition of entropy in microcanonical ensemble?
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Ambiguity in the definition of entropy
How are possible microstates discerned in Gibb's entropy formula?Statistical interpretation of EntropyEntropy as an arrow of timeWhat precisely does the 2nd law of thermo state, considering that entropy depends on how we define macrostate?The statistical interpretation of EntropyWhat is the cause for the inclusion of 'thermal equilibrium' in the statement of Ergodic hypothesis?Do the results of statistical mechanics depend upon the choice of macrostates?Entropy definition, additivity, laws in different ensemblesDefinition of entropy and other StatMech variablesWhat is the definition of entropy in microcanonical ensemble?
$begingroup$
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?
statistical-mechanics entropy
asked 1 hour ago
PiKindOfGuyPiKindOfGuy
596622
$endgroup$
add a comment |
$begingroup$
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?
statistical-mechanics entropy
asked 1 hour ago
PiKindOfGuyPiKindOfGuy
596622
$endgroup$
add a comment |
$begingroup$
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?
statistical-mechanics entropy
asked 1 hour ago
PiKindOfGuyPiKindOfGuy
596622
$endgroup$
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
statistical-mechanics entropy
asked 1 hour ago
PiKindOfGuyPiKindOfGuy
596622
asked 1 hour ago
PiKindOfGuyPiKindOfGuy
596622
edited 30 mins ago
PiKindOfGuy
asked 1 hour ago
PiKindOfGuyPiKindOfGuy
596622
asked 1 hour ago
PiKindOfGuyPiKindOfGuy
596622
asked 1 hour ago
PiKindOfGuyPiKindOfGuy
596622
596622
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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.
answered 40 mins ago
AcccumulationAcccumulation
2,774312
$endgroup$
add a comment |
$begingroup$
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.
answered 39 mins ago
CR DrostCR Drost
22.5k11961
$endgroup$
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
votes
active
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active
oldest
votes
$begingroup$
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.
answered 40 mins ago
AcccumulationAcccumulation
2,774312
$endgroup$
add a comment |
$begingroup$
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.
answered 40 mins ago
AcccumulationAcccumulation
2,774312
$endgroup$
add a comment |
$begingroup$
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.
answered 40 mins ago
AcccumulationAcccumulation
2,774312
$endgroup$
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.
answered 40 mins ago
AcccumulationAcccumulation
2,774312
answered 40 mins ago
AcccumulationAcccumulation
2,774312
answered 40 mins ago
AcccumulationAcccumulation
2,774312
answered 40 mins ago
AcccumulationAcccumulation
2,774312
2,774312
add a comment |
add a comment |
$begingroup$
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.
answered 39 mins ago
CR DrostCR Drost
22.5k11961
$endgroup$
add a comment |
$begingroup$
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.
answered 39 mins ago
CR DrostCR Drost
22.5k11961
$endgroup$
add a comment |
$begingroup$
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.
answered 39 mins ago
CR DrostCR Drost
22.5k11961
$endgroup$
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.
answered 39 mins ago
CR DrostCR Drost
22.5k11961
answered 39 mins ago
CR DrostCR Drost
22.5k11961
answered 39 mins ago
CR DrostCR Drost
22.5k11961
answered 39 mins ago
CR DrostCR Drost
22.5k11961
22.5k11961
add a comment |
add a comment |
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