Are the net microstates of the universe increasing?
$begingroup$
In physics and chemistry we learn that entropy is given by
$$S=klnOmega$$
where $S$ is entropy, $k$ is Boltzmann's constant, and $Omega$ is the number of microstates in the system. We also learn by the 2nd Law of Thermodynamics, that the entropy of the universe always increases. So if we apply both ideas, can we say that the number of microstates in the universe is always increasing? Or is this a naïve approach to the subject?
thermodynamics entropy
edited Feb 8 at 2:04
Chris
9,42372942
asked Feb 8 at 1:59
StobyStoby
768
$endgroup$
add a comment |
$begingroup$
In physics and chemistry we learn that entropy is given by
$$S=klnOmega$$
where $S$ is entropy, $k$ is Boltzmann's constant, and $Omega$ is the number of microstates in the system. We also learn by the 2nd Law of Thermodynamics, that the entropy of the universe always increases. So if we apply both ideas, can we say that the number of microstates in the universe is always increasing? Or is this a naïve approach to the subject?
thermodynamics entropy
edited Feb 8 at 2:04
Chris
9,42372942
asked Feb 8 at 1:59
StobyStoby
768
$endgroup$
1
$begingroup$
Formally, this expression of entropy is only valid for an equilibrium macrostate.
$endgroup$
– Cham
Feb 8 at 2:16
1
$begingroup$
For one, $Omega$ should be something more like "the number of microstates consistent with the observed macrostate." This can increase without the total number of microstates itself changing; the usual classroom problem of removing a partition between two gases and allowing them to mix provides an example of this.
$endgroup$
– Rococo
Feb 8 at 2:28
add a comment |
$begingroup$
In physics and chemistry we learn that entropy is given by
$$S=klnOmega$$
where $S$ is entropy, $k$ is Boltzmann's constant, and $Omega$ is the number of microstates in the system. We also learn by the 2nd Law of Thermodynamics, that the entropy of the universe always increases. So if we apply both ideas, can we say that the number of microstates in the universe is always increasing? Or is this a naïve approach to the subject?
thermodynamics entropy
edited Feb 8 at 2:04
Chris
9,42372942
asked Feb 8 at 1:59
StobyStoby
768
$endgroup$
In physics and chemistry we learn that entropy is given by
$$S=klnOmega$$
where $S$ is entropy, $k$ is Boltzmann's constant, and $Omega$ is the number of microstates in the system. We also learn by the 2nd Law of Thermodynamics, that the entropy of the universe always increases. So if we apply both ideas, can we say that the number of microstates in the universe is always increasing? Or is this a naïve approach to the subject?
thermodynamics entropy
thermodynamics entropy
edited Feb 8 at 2:04
Chris
9,42372942
asked Feb 8 at 1:59
StobyStoby
768
edited Feb 8 at 2:04
Chris
9,42372942
asked Feb 8 at 1:59
StobyStoby
768
edited Feb 8 at 2:04
Chris
9,42372942
edited Feb 8 at 2:04
Chris
9,42372942
edited Feb 8 at 2:04
Chris
9,42372942
9,42372942
asked Feb 8 at 1:59
StobyStoby
768
asked Feb 8 at 1:59
StobyStoby
768
asked Feb 8 at 1:59
StobyStoby
768
768
1
$begingroup$
Formally, this expression of entropy is only valid for an equilibrium macrostate.
$endgroup$
– Cham
Feb 8 at 2:16
1
$begingroup$
For one, $Omega$ should be something more like "the number of microstates consistent with the observed macrostate." This can increase without the total number of microstates itself changing; the usual classroom problem of removing a partition between two gases and allowing them to mix provides an example of this.
$endgroup$
– Rococo
Feb 8 at 2:28
add a comment |
1
$begingroup$
Formally, this expression of entropy is only valid for an equilibrium macrostate.
$endgroup$
– Cham
Feb 8 at 2:16
1
$begingroup$
For one, $Omega$ should be something more like "the number of microstates consistent with the observed macrostate." This can increase without the total number of microstates itself changing; the usual classroom problem of removing a partition between two gases and allowing them to mix provides an example of this.
$endgroup$
– Rococo
Feb 8 at 2:28
1
1
$begingroup$
Formally, this expression of entropy is only valid for an equilibrium macrostate.
$endgroup$
– Cham
Feb 8 at 2:16
$begingroup$
Formally, this expression of entropy is only valid for an equilibrium macrostate.
$endgroup$
– Cham
Feb 8 at 2:16
1
1
$begingroup$
For one, $Omega$ should be something more like "the number of microstates consistent with the observed macrostate." This can increase without the total number of microstates itself changing; the usual classroom problem of removing a partition between two gases and allowing them to mix provides an example of this.
$endgroup$
– Rococo
Feb 8 at 2:28
$begingroup$
For one, $Omega$ should be something more like "the number of microstates consistent with the observed macrostate." This can increase without the total number of microstates itself changing; the usual classroom problem of removing a partition between two gases and allowing them to mix provides an example of this.
$endgroup$
– Rococo
Feb 8 at 2:28
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
In statistical mechanics, the word entropy is used for $klnOmega$ where $Omega$ is the number of microstates that are compatible with specified conditions, such as a given total energy and/or a given total volume. If no conditions are specified, then it's the total number of possible microstates. Therefore, whenever we talk about (or read about) entropy in statistical mechanics, it is important to specify what, if any, conditions are being imposed.
Here's an example. Suppose we start with an airtight box full of gas sitting inside a bigger airtight box with no gas. If the smaller box breaks open, so that the gas molecules are free to move into the larger space, then the number of possible microstates of the gas has suddenly increased. In other words, if we define $S$ in terms of the number of microstates compatible with the given constraints, then $S$ has suddenly increased, at least if we pretend that the change in the available volume is the only information we have.
(By the way, in quantum theory, we count mutually orthogonal microstates.)
Now: is the entropy of the universe always increasing?
Although there are many possible microstates, the universe is in only one of them. We have only partial information about which one we're in, so many (maybe infinitely many) microstates are consistent with the partial information we have. We can define $S=klnOmega$ where $Omega$ is the number of microstates that are consistent with the information we have, and this $S$ may change with time, if it is finite at all. I don't think our understanding of the laws of nature are complete enough yet to quantify this, at least not for the whole universe, as noted in this post.
Some related thoughts can be found here:
Explain the second principle of thermodynamics without the notion of entropy
answered Feb 8 at 2:47
Dan YandDan Yand
11.6k21540
$endgroup$
1
$begingroup$
Are you assuming the volume of the gas increases as soon as the small box breaks, or are you assuming the new system comes to equilibrium first?
$endgroup$
– Aaron Stevens
Feb 8 at 3:07
1
$begingroup$
@AaronStevens That's an excellent question/point. I assumed for simplicity that we don't know anything else about what happens to the gas after the inner box breaks, as though it could instantly be in any one of the possible microstates compatible with that larger $V$. In reality, of course, we do have more info. We know that real gas molecules move at finite speeds, etc, and accounting for that knowledge would give an $S$ that increases smoothly. Or, as you said, we could get the same answer by waiting until equilibrium is reached: "equilibrium" = we've lost track of everything except $V$.
$endgroup$
– Dan Yand
Feb 8 at 3:20
1
$begingroup$
In what sense can the number of microstates increase if the volume of the "universe" is infinite, or is the increase of the statistical entropy an indication that the eventual volume any gas expands must be finite?
$endgroup$
– hyportnex
Feb 8 at 15:57
$begingroup$
@hyportnex I don't know how to apply the statistical-mechanics concept of "entropy" to the whole universe. That's why I wrote "I don't think our understanding of the laws of nature are complete enough yet to quantify this, at least not for the whole universe, as noted in this post." However, according to arxiv.org/abs/hep-th/0007146, an asymptotically de Sitter spacetime (like ours) has a finite entropy; but this idea involves quantum gravity, so I didn't go there. Practical applic'ns of the 2nd law involve finite systems.
$endgroup$
– Dan Yand
Feb 8 at 17:33
1
$begingroup$
Thanks. My question was actually hinting at this statement by Clausius because I have always been a bit surprised by its implication from a historic point of view: "The energy of the universe is constant. The entropy of the universe tends to a maximum." I cannot make any sense out of this unless I assume that energy/entropy/size, etc., are all finite for the whole universe, whatever that may mean in classical physics but in classical physics with "Eucledian 3D space" must be infinite... what am I missing?
$endgroup$
– hyportnex
Feb 8 at 18:02
|
show 1 more 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%2f459496%2fare-the-net-microstates-of-the-universe-increasing%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
$begingroup$
In statistical mechanics, the word entropy is used for $klnOmega$ where $Omega$ is the number of microstates that are compatible with specified conditions, such as a given total energy and/or a given total volume. If no conditions are specified, then it's the total number of possible microstates. Therefore, whenever we talk about (or read about) entropy in statistical mechanics, it is important to specify what, if any, conditions are being imposed.
Here's an example. Suppose we start with an airtight box full of gas sitting inside a bigger airtight box with no gas. If the smaller box breaks open, so that the gas molecules are free to move into the larger space, then the number of possible microstates of the gas has suddenly increased. In other words, if we define $S$ in terms of the number of microstates compatible with the given constraints, then $S$ has suddenly increased, at least if we pretend that the change in the available volume is the only information we have.
(By the way, in quantum theory, we count mutually orthogonal microstates.)
Now: is the entropy of the universe always increasing?
Although there are many possible microstates, the universe is in only one of them. We have only partial information about which one we're in, so many (maybe infinitely many) microstates are consistent with the partial information we have. We can define $S=klnOmega$ where $Omega$ is the number of microstates that are consistent with the information we have, and this $S$ may change with time, if it is finite at all. I don't think our understanding of the laws of nature are complete enough yet to quantify this, at least not for the whole universe, as noted in this post.
Some related thoughts can be found here:
Explain the second principle of thermodynamics without the notion of entropy
answered Feb 8 at 2:47
Dan YandDan Yand
11.6k21540
$endgroup$
1
$begingroup$
Are you assuming the volume of the gas increases as soon as the small box breaks, or are you assuming the new system comes to equilibrium first?
$endgroup$
– Aaron Stevens
Feb 8 at 3:07
1
$begingroup$
@AaronStevens That's an excellent question/point. I assumed for simplicity that we don't know anything else about what happens to the gas after the inner box breaks, as though it could instantly be in any one of the possible microstates compatible with that larger $V$. In reality, of course, we do have more info. We know that real gas molecules move at finite speeds, etc, and accounting for that knowledge would give an $S$ that increases smoothly. Or, as you said, we could get the same answer by waiting until equilibrium is reached: "equilibrium" = we've lost track of everything except $V$.
$endgroup$
– Dan Yand
Feb 8 at 3:20
1
$begingroup$
In what sense can the number of microstates increase if the volume of the "universe" is infinite, or is the increase of the statistical entropy an indication that the eventual volume any gas expands must be finite?
$endgroup$
– hyportnex
Feb 8 at 15:57
$begingroup$
@hyportnex I don't know how to apply the statistical-mechanics concept of "entropy" to the whole universe. That's why I wrote "I don't think our understanding of the laws of nature are complete enough yet to quantify this, at least not for the whole universe, as noted in this post." However, according to arxiv.org/abs/hep-th/0007146, an asymptotically de Sitter spacetime (like ours) has a finite entropy; but this idea involves quantum gravity, so I didn't go there. Practical applic'ns of the 2nd law involve finite systems.
$endgroup$
– Dan Yand
Feb 8 at 17:33
1
$begingroup$
Thanks. My question was actually hinting at this statement by Clausius because I have always been a bit surprised by its implication from a historic point of view: "The energy of the universe is constant. The entropy of the universe tends to a maximum." I cannot make any sense out of this unless I assume that energy/entropy/size, etc., are all finite for the whole universe, whatever that may mean in classical physics but in classical physics with "Eucledian 3D space" must be infinite... what am I missing?
$endgroup$
– hyportnex
Feb 8 at 18:02
|
show 1 more comment
$begingroup$
In statistical mechanics, the word entropy is used for $klnOmega$ where $Omega$ is the number of microstates that are compatible with specified conditions, such as a given total energy and/or a given total volume. If no conditions are specified, then it's the total number of possible microstates. Therefore, whenever we talk about (or read about) entropy in statistical mechanics, it is important to specify what, if any, conditions are being imposed.
Here's an example. Suppose we start with an airtight box full of gas sitting inside a bigger airtight box with no gas. If the smaller box breaks open, so that the gas molecules are free to move into the larger space, then the number of possible microstates of the gas has suddenly increased. In other words, if we define $S$ in terms of the number of microstates compatible with the given constraints, then $S$ has suddenly increased, at least if we pretend that the change in the available volume is the only information we have.
(By the way, in quantum theory, we count mutually orthogonal microstates.)
Now: is the entropy of the universe always increasing?
Although there are many possible microstates, the universe is in only one of them. We have only partial information about which one we're in, so many (maybe infinitely many) microstates are consistent with the partial information we have. We can define $S=klnOmega$ where $Omega$ is the number of microstates that are consistent with the information we have, and this $S$ may change with time, if it is finite at all. I don't think our understanding of the laws of nature are complete enough yet to quantify this, at least not for the whole universe, as noted in this post.
Some related thoughts can be found here:
Explain the second principle of thermodynamics without the notion of entropy
answered Feb 8 at 2:47
Dan YandDan Yand
11.6k21540
$endgroup$
1
$begingroup$
Are you assuming the volume of the gas increases as soon as the small box breaks, or are you assuming the new system comes to equilibrium first?
$endgroup$
– Aaron Stevens
Feb 8 at 3:07
1
$begingroup$
@AaronStevens That's an excellent question/point. I assumed for simplicity that we don't know anything else about what happens to the gas after the inner box breaks, as though it could instantly be in any one of the possible microstates compatible with that larger $V$. In reality, of course, we do have more info. We know that real gas molecules move at finite speeds, etc, and accounting for that knowledge would give an $S$ that increases smoothly. Or, as you said, we could get the same answer by waiting until equilibrium is reached: "equilibrium" = we've lost track of everything except $V$.
$endgroup$
– Dan Yand
Feb 8 at 3:20
1
$begingroup$
In what sense can the number of microstates increase if the volume of the "universe" is infinite, or is the increase of the statistical entropy an indication that the eventual volume any gas expands must be finite?
$endgroup$
– hyportnex
Feb 8 at 15:57
$begingroup$
@hyportnex I don't know how to apply the statistical-mechanics concept of "entropy" to the whole universe. That's why I wrote "I don't think our understanding of the laws of nature are complete enough yet to quantify this, at least not for the whole universe, as noted in this post." However, according to arxiv.org/abs/hep-th/0007146, an asymptotically de Sitter spacetime (like ours) has a finite entropy; but this idea involves quantum gravity, so I didn't go there. Practical applic'ns of the 2nd law involve finite systems.
$endgroup$
– Dan Yand
Feb 8 at 17:33
1
$begingroup$
Thanks. My question was actually hinting at this statement by Clausius because I have always been a bit surprised by its implication from a historic point of view: "The energy of the universe is constant. The entropy of the universe tends to a maximum." I cannot make any sense out of this unless I assume that energy/entropy/size, etc., are all finite for the whole universe, whatever that may mean in classical physics but in classical physics with "Eucledian 3D space" must be infinite... what am I missing?
$endgroup$
– hyportnex
Feb 8 at 18:02
|
show 1 more comment
$begingroup$
In statistical mechanics, the word entropy is used for $klnOmega$ where $Omega$ is the number of microstates that are compatible with specified conditions, such as a given total energy and/or a given total volume. If no conditions are specified, then it's the total number of possible microstates. Therefore, whenever we talk about (or read about) entropy in statistical mechanics, it is important to specify what, if any, conditions are being imposed.
Here's an example. Suppose we start with an airtight box full of gas sitting inside a bigger airtight box with no gas. If the smaller box breaks open, so that the gas molecules are free to move into the larger space, then the number of possible microstates of the gas has suddenly increased. In other words, if we define $S$ in terms of the number of microstates compatible with the given constraints, then $S$ has suddenly increased, at least if we pretend that the change in the available volume is the only information we have.
(By the way, in quantum theory, we count mutually orthogonal microstates.)
Now: is the entropy of the universe always increasing?
Although there are many possible microstates, the universe is in only one of them. We have only partial information about which one we're in, so many (maybe infinitely many) microstates are consistent with the partial information we have. We can define $S=klnOmega$ where $Omega$ is the number of microstates that are consistent with the information we have, and this $S$ may change with time, if it is finite at all. I don't think our understanding of the laws of nature are complete enough yet to quantify this, at least not for the whole universe, as noted in this post.
Some related thoughts can be found here:
Explain the second principle of thermodynamics without the notion of entropy
answered Feb 8 at 2:47
Dan YandDan Yand
11.6k21540
$endgroup$
In statistical mechanics, the word entropy is used for $klnOmega$ where $Omega$ is the number of microstates that are compatible with specified conditions, such as a given total energy and/or a given total volume. If no conditions are specified, then it's the total number of possible microstates. Therefore, whenever we talk about (or read about) entropy in statistical mechanics, it is important to specify what, if any, conditions are being imposed.
Here's an example. Suppose we start with an airtight box full of gas sitting inside a bigger airtight box with no gas. If the smaller box breaks open, so that the gas molecules are free to move into the larger space, then the number of possible microstates of the gas has suddenly increased. In other words, if we define $S$ in terms of the number of microstates compatible with the given constraints, then $S$ has suddenly increased, at least if we pretend that the change in the available volume is the only information we have.
(By the way, in quantum theory, we count mutually orthogonal microstates.)
Now: is the entropy of the universe always increasing?
Although there are many possible microstates, the universe is in only one of them. We have only partial information about which one we're in, so many (maybe infinitely many) microstates are consistent with the partial information we have. We can define $S=klnOmega$ where $Omega$ is the number of microstates that are consistent with the information we have, and this $S$ may change with time, if it is finite at all. I don't think our understanding of the laws of nature are complete enough yet to quantify this, at least not for the whole universe, as noted in this post.
Some related thoughts can be found here:
Explain the second principle of thermodynamics without the notion of entropy
answered Feb 8 at 2:47
Dan YandDan Yand
11.6k21540
edited Feb 8 at 2:55
answered Feb 8 at 2:47
Dan YandDan Yand
11.6k21540
answered Feb 8 at 2:47
Dan YandDan Yand
11.6k21540
answered Feb 8 at 2:47
Dan YandDan Yand
11.6k21540
11.6k21540
1
$begingroup$
Are you assuming the volume of the gas increases as soon as the small box breaks, or are you assuming the new system comes to equilibrium first?
$endgroup$
– Aaron Stevens
Feb 8 at 3:07
1
$begingroup$
@AaronStevens That's an excellent question/point. I assumed for simplicity that we don't know anything else about what happens to the gas after the inner box breaks, as though it could instantly be in any one of the possible microstates compatible with that larger $V$. In reality, of course, we do have more info. We know that real gas molecules move at finite speeds, etc, and accounting for that knowledge would give an $S$ that increases smoothly. Or, as you said, we could get the same answer by waiting until equilibrium is reached: "equilibrium" = we've lost track of everything except $V$.
$endgroup$
– Dan Yand
Feb 8 at 3:20
1
$begingroup$
In what sense can the number of microstates increase if the volume of the "universe" is infinite, or is the increase of the statistical entropy an indication that the eventual volume any gas expands must be finite?
$endgroup$
– hyportnex
Feb 8 at 15:57
$begingroup$
@hyportnex I don't know how to apply the statistical-mechanics concept of "entropy" to the whole universe. That's why I wrote "I don't think our understanding of the laws of nature are complete enough yet to quantify this, at least not for the whole universe, as noted in this post." However, according to arxiv.org/abs/hep-th/0007146, an asymptotically de Sitter spacetime (like ours) has a finite entropy; but this idea involves quantum gravity, so I didn't go there. Practical applic'ns of the 2nd law involve finite systems.
$endgroup$
– Dan Yand
Feb 8 at 17:33
1
$begingroup$
Thanks. My question was actually hinting at this statement by Clausius because I have always been a bit surprised by its implication from a historic point of view: "The energy of the universe is constant. The entropy of the universe tends to a maximum." I cannot make any sense out of this unless I assume that energy/entropy/size, etc., are all finite for the whole universe, whatever that may mean in classical physics but in classical physics with "Eucledian 3D space" must be infinite... what am I missing?
$endgroup$
– hyportnex
Feb 8 at 18:02
|
show 1 more comment
1
$begingroup$
Are you assuming the volume of the gas increases as soon as the small box breaks, or are you assuming the new system comes to equilibrium first?
$endgroup$
– Aaron Stevens
Feb 8 at 3:07
1
$begingroup$
@AaronStevens That's an excellent question/point. I assumed for simplicity that we don't know anything else about what happens to the gas after the inner box breaks, as though it could instantly be in any one of the possible microstates compatible with that larger $V$. In reality, of course, we do have more info. We know that real gas molecules move at finite speeds, etc, and accounting for that knowledge would give an $S$ that increases smoothly. Or, as you said, we could get the same answer by waiting until equilibrium is reached: "equilibrium" = we've lost track of everything except $V$.
$endgroup$
– Dan Yand
Feb 8 at 3:20
1
$begingroup$
In what sense can the number of microstates increase if the volume of the "universe" is infinite, or is the increase of the statistical entropy an indication that the eventual volume any gas expands must be finite?
$endgroup$
– hyportnex
Feb 8 at 15:57
$begingroup$
@hyportnex I don't know how to apply the statistical-mechanics concept of "entropy" to the whole universe. That's why I wrote "I don't think our understanding of the laws of nature are complete enough yet to quantify this, at least not for the whole universe, as noted in this post." However, according to arxiv.org/abs/hep-th/0007146, an asymptotically de Sitter spacetime (like ours) has a finite entropy; but this idea involves quantum gravity, so I didn't go there. Practical applic'ns of the 2nd law involve finite systems.
$endgroup$
– Dan Yand
Feb 8 at 17:33
1
$begingroup$
Thanks. My question was actually hinting at this statement by Clausius because I have always been a bit surprised by its implication from a historic point of view: "The energy of the universe is constant. The entropy of the universe tends to a maximum." I cannot make any sense out of this unless I assume that energy/entropy/size, etc., are all finite for the whole universe, whatever that may mean in classical physics but in classical physics with "Eucledian 3D space" must be infinite... what am I missing?
$endgroup$
– hyportnex
Feb 8 at 18:02
1
1
$begingroup$
Are you assuming the volume of the gas increases as soon as the small box breaks, or are you assuming the new system comes to equilibrium first?
$endgroup$
– Aaron Stevens
Feb 8 at 3:07
$begingroup$
Are you assuming the volume of the gas increases as soon as the small box breaks, or are you assuming the new system comes to equilibrium first?
$endgroup$
– Aaron Stevens
Feb 8 at 3:07
1
1
$begingroup$
@AaronStevens That's an excellent question/point. I assumed for simplicity that we don't know anything else about what happens to the gas after the inner box breaks, as though it could instantly be in any one of the possible microstates compatible with that larger $V$. In reality, of course, we do have more info. We know that real gas molecules move at finite speeds, etc, and accounting for that knowledge would give an $S$ that increases smoothly. Or, as you said, we could get the same answer by waiting until equilibrium is reached: "equilibrium" = we've lost track of everything except $V$.
$endgroup$
– Dan Yand
Feb 8 at 3:20
$begingroup$
@AaronStevens That's an excellent question/point. I assumed for simplicity that we don't know anything else about what happens to the gas after the inner box breaks, as though it could instantly be in any one of the possible microstates compatible with that larger $V$. In reality, of course, we do have more info. We know that real gas molecules move at finite speeds, etc, and accounting for that knowledge would give an $S$ that increases smoothly. Or, as you said, we could get the same answer by waiting until equilibrium is reached: "equilibrium" = we've lost track of everything except $V$.
$endgroup$
– Dan Yand
Feb 8 at 3:20
1
1
$begingroup$
In what sense can the number of microstates increase if the volume of the "universe" is infinite, or is the increase of the statistical entropy an indication that the eventual volume any gas expands must be finite?
$endgroup$
– hyportnex
Feb 8 at 15:57
$begingroup$
In what sense can the number of microstates increase if the volume of the "universe" is infinite, or is the increase of the statistical entropy an indication that the eventual volume any gas expands must be finite?
$endgroup$
– hyportnex
Feb 8 at 15:57
$begingroup$
@hyportnex I don't know how to apply the statistical-mechanics concept of "entropy" to the whole universe. That's why I wrote "I don't think our understanding of the laws of nature are complete enough yet to quantify this, at least not for the whole universe, as noted in this post." However, according to arxiv.org/abs/hep-th/0007146, an asymptotically de Sitter spacetime (like ours) has a finite entropy; but this idea involves quantum gravity, so I didn't go there. Practical applic'ns of the 2nd law involve finite systems.
$endgroup$
– Dan Yand
Feb 8 at 17:33
$begingroup$
@hyportnex I don't know how to apply the statistical-mechanics concept of "entropy" to the whole universe. That's why I wrote "I don't think our understanding of the laws of nature are complete enough yet to quantify this, at least not for the whole universe, as noted in this post." However, according to arxiv.org/abs/hep-th/0007146, an asymptotically de Sitter spacetime (like ours) has a finite entropy; but this idea involves quantum gravity, so I didn't go there. Practical applic'ns of the 2nd law involve finite systems.
$endgroup$
– Dan Yand
Feb 8 at 17:33
1
1
$begingroup$
Thanks. My question was actually hinting at this statement by Clausius because I have always been a bit surprised by its implication from a historic point of view: "The energy of the universe is constant. The entropy of the universe tends to a maximum." I cannot make any sense out of this unless I assume that energy/entropy/size, etc., are all finite for the whole universe, whatever that may mean in classical physics but in classical physics with "Eucledian 3D space" must be infinite... what am I missing?
$endgroup$
– hyportnex
Feb 8 at 18:02
$begingroup$
Thanks. My question was actually hinting at this statement by Clausius because I have always been a bit surprised by its implication from a historic point of view: "The energy of the universe is constant. The entropy of the universe tends to a maximum." I cannot make any sense out of this unless I assume that energy/entropy/size, etc., are all finite for the whole universe, whatever that may mean in classical physics but in classical physics with "Eucledian 3D space" must be infinite... what am I missing?
$endgroup$
– hyportnex
Feb 8 at 18:02
|
show 1 more 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%2f459496%2fare-the-net-microstates-of-the-universe-increasing%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
1
$begingroup$
Formally, this expression of entropy is only valid for an equilibrium macrostate.
$endgroup$
– Cham
Feb 8 at 2:16
1
$begingroup$
For one, $Omega$ should be something more like "the number of microstates consistent with the observed macrostate." This can increase without the total number of microstates itself changing; the usual classroom problem of removing a partition between two gases and allowing them to mix provides an example of this.
$endgroup$
– Rococo
Feb 8 at 2:28