Ambiguity in the definition of entropy
$begingroup$
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 35 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 45 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 44 mins ago
CR DrostCR Drost
22.5k11961
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "151"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f470202%2fambiguity-in-the-definition-of-entropy%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
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 45 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 45 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 45 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 45 mins ago
AcccumulationAcccumulation
2,774312
answered 45 mins ago
AcccumulationAcccumulation
2,774312
answered 45 mins ago
AcccumulationAcccumulation
2,774312
answered 45 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 44 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 44 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 44 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 44 mins ago
CR DrostCR Drost
22.5k11961
answered 44 mins ago
CR DrostCR Drost
22.5k11961
answered 44 mins ago
CR DrostCR Drost
22.5k11961
answered 44 mins ago
CR DrostCR Drost
22.5k11961
22.5k11961
add a comment |
add a comment |
Thanks for contributing an answer to Physics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f470202%2fambiguity-in-the-definition-of-entropy%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
