Interpreting heat using information entropy
$begingroup$
In an answer to the post about Microscopic Definition of Heat and Work, Ronan says,
$$<dE> = sum epsilon_idp_i + p_idepsilon_i$$
We can see that the change in average energy is partly due to a change
in the distribution of probability of occurrence of microscopic state
$epsilon_i$ and partly due to a change in the eigen values $epsilon_i$ of the
N-particles microscopic eigen states.
and the first term corresponds to heat:
$$TdS = sum epsilon_idp_i.$$
I am having a hard time imagining what it means.
If the following is true:
- probabilities $p_i$ can not change the shape of Gibbs distribution if we assume that the new state is in equilibrium
- $sum p_i = 1$
then the temperature must have risen (without any work done)! Is that correct, or can anything else happen?
thermodynamics statistical-mechanics entropy
asked yesterday
thehorseisbrownthehorseisbrown
6017
$endgroup$
add a comment |
$begingroup$
In an answer to the post about Microscopic Definition of Heat and Work, Ronan says,
$$<dE> = sum epsilon_idp_i + p_idepsilon_i$$
We can see that the change in average energy is partly due to a change
in the distribution of probability of occurrence of microscopic state
$epsilon_i$ and partly due to a change in the eigen values $epsilon_i$ of the
N-particles microscopic eigen states.
and the first term corresponds to heat:
$$TdS = sum epsilon_idp_i.$$
I am having a hard time imagining what it means.
If the following is true:
- probabilities $p_i$ can not change the shape of Gibbs distribution if we assume that the new state is in equilibrium
- $sum p_i = 1$
then the temperature must have risen (without any work done)! Is that correct, or can anything else happen?
thermodynamics statistical-mechanics entropy
asked yesterday
thehorseisbrownthehorseisbrown
6017
$endgroup$
add a comment |
$begingroup$
In an answer to the post about Microscopic Definition of Heat and Work, Ronan says,
$$<dE> = sum epsilon_idp_i + p_idepsilon_i$$
We can see that the change in average energy is partly due to a change
in the distribution of probability of occurrence of microscopic state
$epsilon_i$ and partly due to a change in the eigen values $epsilon_i$ of the
N-particles microscopic eigen states.
and the first term corresponds to heat:
$$TdS = sum epsilon_idp_i.$$
I am having a hard time imagining what it means.
If the following is true:
- probabilities $p_i$ can not change the shape of Gibbs distribution if we assume that the new state is in equilibrium
- $sum p_i = 1$
then the temperature must have risen (without any work done)! Is that correct, or can anything else happen?
thermodynamics statistical-mechanics entropy
asked yesterday
thehorseisbrownthehorseisbrown
6017
$endgroup$
In an answer to the post about Microscopic Definition of Heat and Work, Ronan says,
$$<dE> = sum epsilon_idp_i + p_idepsilon_i$$
We can see that the change in average energy is partly due to a change
in the distribution of probability of occurrence of microscopic state
$epsilon_i$ and partly due to a change in the eigen values $epsilon_i$ of the
N-particles microscopic eigen states.
and the first term corresponds to heat:
$$TdS = sum epsilon_idp_i.$$
I am having a hard time imagining what it means.
If the following is true:
- probabilities $p_i$ can not change the shape of Gibbs distribution if we assume that the new state is in equilibrium
- $sum p_i = 1$
then the temperature must have risen (without any work done)! Is that correct, or can anything else happen?
thermodynamics statistical-mechanics entropy
thermodynamics statistical-mechanics entropy
asked yesterday
thehorseisbrownthehorseisbrown
6017
asked yesterday
thehorseisbrownthehorseisbrown
6017
edited 17 hours ago
thehorseisbrown
asked yesterday
thehorseisbrownthehorseisbrown
6017
asked yesterday
thehorseisbrownthehorseisbrown
6017
asked yesterday
thehorseisbrownthehorseisbrown
6017
6017
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
When you heat a single-phase system without performing any work, then yes, the temperature must increase, which is interpreted as a change in the dispersion of energy levels.
Some alternatives are:
A two-phase system, described by expanding the fundamental relation $dU=T,dS-P,dV+Sigma_imu_i,dN_i$ to include the final term, where $mu$ is the chemical potential and $N$ is the amount of stuff and where the temperature stays constant (in a first-order transition) even though the amount of each phase changes.
Doing work on a system, in which the energy levels might vary even though their dispersion is unchanged. An example is accelerating a mass, in which the speed of each constitute particle changes by a constant amount even though its speed relative to the center of mass remains the same, as does the temperature.
answered yesterday
ChemomechanicsChemomechanics
4,91531023
$endgroup$
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%2f459083%2finterpreting-heat-using-information-entropy%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
When you heat a single-phase system without performing any work, then yes, the temperature must increase, which is interpreted as a change in the dispersion of energy levels.
Some alternatives are:
A two-phase system, described by expanding the fundamental relation $dU=T,dS-P,dV+Sigma_imu_i,dN_i$ to include the final term, where $mu$ is the chemical potential and $N$ is the amount of stuff and where the temperature stays constant (in a first-order transition) even though the amount of each phase changes.
Doing work on a system, in which the energy levels might vary even though their dispersion is unchanged. An example is accelerating a mass, in which the speed of each constitute particle changes by a constant amount even though its speed relative to the center of mass remains the same, as does the temperature.
answered yesterday
ChemomechanicsChemomechanics
4,91531023
$endgroup$
add a comment |
$begingroup$
When you heat a single-phase system without performing any work, then yes, the temperature must increase, which is interpreted as a change in the dispersion of energy levels.
Some alternatives are:
A two-phase system, described by expanding the fundamental relation $dU=T,dS-P,dV+Sigma_imu_i,dN_i$ to include the final term, where $mu$ is the chemical potential and $N$ is the amount of stuff and where the temperature stays constant (in a first-order transition) even though the amount of each phase changes.
Doing work on a system, in which the energy levels might vary even though their dispersion is unchanged. An example is accelerating a mass, in which the speed of each constitute particle changes by a constant amount even though its speed relative to the center of mass remains the same, as does the temperature.
answered yesterday
ChemomechanicsChemomechanics
4,91531023
$endgroup$
add a comment |
$begingroup$
When you heat a single-phase system without performing any work, then yes, the temperature must increase, which is interpreted as a change in the dispersion of energy levels.
Some alternatives are:
A two-phase system, described by expanding the fundamental relation $dU=T,dS-P,dV+Sigma_imu_i,dN_i$ to include the final term, where $mu$ is the chemical potential and $N$ is the amount of stuff and where the temperature stays constant (in a first-order transition) even though the amount of each phase changes.
Doing work on a system, in which the energy levels might vary even though their dispersion is unchanged. An example is accelerating a mass, in which the speed of each constitute particle changes by a constant amount even though its speed relative to the center of mass remains the same, as does the temperature.
answered yesterday
ChemomechanicsChemomechanics
4,91531023
$endgroup$
When you heat a single-phase system without performing any work, then yes, the temperature must increase, which is interpreted as a change in the dispersion of energy levels.
Some alternatives are:
A two-phase system, described by expanding the fundamental relation $dU=T,dS-P,dV+Sigma_imu_i,dN_i$ to include the final term, where $mu$ is the chemical potential and $N$ is the amount of stuff and where the temperature stays constant (in a first-order transition) even though the amount of each phase changes.
Doing work on a system, in which the energy levels might vary even though their dispersion is unchanged. An example is accelerating a mass, in which the speed of each constitute particle changes by a constant amount even though its speed relative to the center of mass remains the same, as does the temperature.
answered yesterday
ChemomechanicsChemomechanics
4,91531023
answered yesterday
ChemomechanicsChemomechanics
4,91531023
answered yesterday
ChemomechanicsChemomechanics
4,91531023
answered yesterday
ChemomechanicsChemomechanics
4,91531023
4,91531023
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%2f459083%2finterpreting-heat-using-information-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
