Ambiguity in the definition of entropyHow are possible microstates discerned in Gibb's entropy...
What are the G forces leaving Earth orbit?
What's the meaning of "Sollensaussagen"?
Can someone clarify Hamming's notion of important problems in relation to modern academia?
Implication of namely
How to install cross-compiler on Ubuntu 18.04?
Avoiding the "not like other girls" trope?
OP Amp not amplifying audio signal
How to coordinate airplane tickets?
How can a day be of 24 hours?
One verb to replace 'be a member of' a club
Is there a hemisphere-neutral way of specifying a season?
How to Prove P(a) → ∀x(P(x) ∨ ¬(x = a)) using Natural Deduction
What historical events would have to change in order to make 19th century "steampunk" technology possible?
Car headlights in a world without electricity
Mathematica command that allows it to read my intentions
How can I deal with my CEO asking me to hire someone with a higher salary than me, a co-founder?
files created then deleted at every second in tmp directory
Ambiguity in the definition of entropy
Why are UK visa biometrics appointments suspended at USCIS Application Support Centers?
Can compressed videos be decoded back to their uncompresed original format?
Why is the sentence "Das ist eine Nase" correct?
How does a dynamic QR code work?
Where would I need my direct neural interface to be implanted?
In the UK, is it possible to get a referendum by a court decision?
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
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 1 hour 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 1 hour 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 1 hour ago
PiKindOfGuyPiKindOfGuy
601622
asked 1 hour ago
PiKindOfGuyPiKindOfGuy
601622
edited 56 mins ago
PiKindOfGuy
asked 1 hour ago
PiKindOfGuyPiKindOfGuy
601622
asked 1 hour ago
PiKindOfGuyPiKindOfGuy
601622
asked 1 hour 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 1 hour ago
AcccumulationAcccumulation
2,784312
$endgroup$
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 1 hour 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 1 hour ago
AcccumulationAcccumulation
2,784312
$endgroup$
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 1 hour ago
AcccumulationAcccumulation
2,784312
$endgroup$
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 1 hour ago
AcccumulationAcccumulation
2,784312
$endgroup$
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 1 hour ago
AcccumulationAcccumulation
2,784312
answered 1 hour ago
AcccumulationAcccumulation
2,784312
answered 1 hour ago
AcccumulationAcccumulation
2,784312
answered 1 hour ago
AcccumulationAcccumulation
2,784312
2,784312
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 1 hour 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 1 hour 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 1 hour 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 1 hour ago
CR DrostCR Drost
22.5k11961
answered 1 hour ago
CR DrostCR Drost
22.5k11961
answered 1 hour ago
CR DrostCR Drost
22.5k11961
answered 1 hour 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
