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Ambiguity in the definition of entropy
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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?
statistical-mechanics entropy
asked 2 hours ago
PiKindOfGuyPiKindOfGuy
601622
$endgroup$
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 2 hours ago
PiKindOfGuyPiKindOfGuy
601622
$endgroup$
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 2 hours ago
PiKindOfGuyPiKindOfGuy
601622
$endgroup$
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 2 hours ago
PiKindOfGuyPiKindOfGuy
601622
asked 2 hours ago
PiKindOfGuyPiKindOfGuy
601622
edited 1 hour ago
PiKindOfGuy
asked 2 hours ago
PiKindOfGuyPiKindOfGuy
601622
asked 2 hours ago
PiKindOfGuyPiKindOfGuy
601622
asked 2 hours ago
PiKindOfGuyPiKindOfGuy
601622
601622
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 2 hours ago
AcccumulationAcccumulation
2,794312
$endgroup$
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 2 hours ago
CR DrostCR Drost
22.5k11961
$endgroup$
$begingroup$
I would say corresponding to the macrostate whose entropy is being evaluated, simply because you can compute subjunctive entropies.
$endgroup$
– dmckee♦
44 mins ago
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
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active
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active
oldest
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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.
answered 2 hours ago
AcccumulationAcccumulation
2,794312
$endgroup$
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 2 hours ago
AcccumulationAcccumulation
2,794312
$endgroup$
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 2 hours ago
AcccumulationAcccumulation
2,794312
$endgroup$
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 2 hours ago
AcccumulationAcccumulation
2,794312
answered 2 hours ago
AcccumulationAcccumulation
2,794312
answered 2 hours ago
AcccumulationAcccumulation
2,794312
answered 2 hours ago
AcccumulationAcccumulation
2,794312
2,794312
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 2 hours ago
CR DrostCR Drost
22.5k11961
$endgroup$
$begingroup$
I would say corresponding to the macrostate whose entropy is being evaluated, simply because you can compute subjunctive entropies.
$endgroup$
– dmckee♦
44 mins ago
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 2 hours ago
CR DrostCR Drost
22.5k11961
$endgroup$
$begingroup$
I would say corresponding to the macrostate whose entropy is being evaluated, simply because you can compute subjunctive entropies.
$endgroup$
– dmckee♦
44 mins ago
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 2 hours 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 2 hours ago
CR DrostCR Drost
22.5k11961
answered 2 hours ago
CR DrostCR Drost
22.5k11961
answered 2 hours ago
CR DrostCR Drost
22.5k11961
answered 2 hours 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♦
44 mins ago
add a comment |
$begingroup$
I would say corresponding to the macrostate whose entropy is being evaluated, simply because you can compute subjunctive entropies.
$endgroup$
– dmckee♦
44 mins ago
$begingroup$
I would say corresponding to the macrostate whose entropy is being evaluated, simply because you can compute subjunctive entropies.
$endgroup$
– dmckee♦
44 mins ago
$begingroup$
I would say corresponding to the macrostate whose entropy is being evaluated, simply because you can compute subjunctive entropies.
$endgroup$
– dmckee♦
44 mins ago
add a comment |
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