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 4 hours ago
Chris
9,29772842
asked 5 hours ago
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 4 hours ago
Chris
9,29772842
asked 5 hours ago
StobyStoby
768
$endgroup$
1
$begingroup$
Formally, this expression of entropy is only valid for an equilibrium macrostate.
$endgroup$
– Cham
4 hours ago
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
4 hours ago
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 4 hours ago
Chris
9,29772842
asked 5 hours ago
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 4 hours ago
Chris
9,29772842
asked 5 hours ago
StobyStoby
768
edited 4 hours ago
Chris
9,29772842
asked 5 hours ago
StobyStoby
768
edited 4 hours ago
Chris
9,29772842
edited 4 hours ago
Chris
9,29772842
edited 4 hours ago
Chris
9,29772842
9,29772842
asked 5 hours ago
StobyStoby
768
asked 5 hours ago
StobyStoby
768
asked 5 hours ago
StobyStoby
768
768
1
$begingroup$
Formally, this expression of entropy is only valid for an equilibrium macrostate.
$endgroup$
– Cham
4 hours ago
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
4 hours ago
add a comment |
1
$begingroup$
Formally, this expression of entropy is only valid for an equilibrium macrostate.
$endgroup$
– Cham
4 hours ago
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
4 hours ago
1
1
$begingroup$
Formally, this expression of entropy is only valid for an equilibrium macrostate.
$endgroup$
– Cham
4 hours ago
$begingroup$
Formally, this expression of entropy is only valid for an equilibrium macrostate.
$endgroup$
– Cham
4 hours ago
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
4 hours ago
$begingroup$
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
4 hours ago
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 4 hours ago
Dan YandDan Yand
10.1k21538
$endgroup$
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
3 hours ago
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
3 hours ago
$begingroup$
Yeah, that's what I was getting after. Thanks!
$endgroup$
– Aaron Stevens
3 hours ago
add a comment |
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1 Answer
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$begingroup$
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 4 hours ago
Dan YandDan Yand
10.1k21538
$endgroup$
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
3 hours ago
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
3 hours ago
$begingroup$
Yeah, that's what I was getting after. Thanks!
$endgroup$
– Aaron Stevens
3 hours ago
add a 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 4 hours ago
Dan YandDan Yand
10.1k21538
$endgroup$
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
3 hours ago
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
3 hours ago
$begingroup$
Yeah, that's what I was getting after. Thanks!
$endgroup$
– Aaron Stevens
3 hours ago
add a 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 4 hours ago
Dan YandDan Yand
10.1k21538
$endgroup$
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 4 hours ago
Dan YandDan Yand
10.1k21538
edited 4 hours ago
answered 4 hours ago
Dan YandDan Yand
10.1k21538
answered 4 hours ago
Dan YandDan Yand
10.1k21538
answered 4 hours ago
Dan YandDan Yand
10.1k21538
10.1k21538
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
3 hours ago
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
3 hours ago
$begingroup$
Yeah, that's what I was getting after. Thanks!
$endgroup$
– Aaron Stevens
3 hours ago
add a 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
3 hours ago
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
3 hours ago
$begingroup$
Yeah, that's what I was getting after. Thanks!
$endgroup$
– Aaron Stevens
3 hours ago
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
3 hours ago
$begingroup$
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
3 hours ago
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
3 hours ago
$begingroup$
@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
3 hours ago
$begingroup$
Yeah, that's what I was getting after. Thanks!
$endgroup$
– Aaron Stevens
3 hours ago
$begingroup$
Yeah, that's what I was getting after. Thanks!
$endgroup$
– Aaron Stevens
3 hours ago
add a comment |
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1
$begingroup$
Formally, this expression of entropy is only valid for an equilibrium macrostate.
$endgroup$
– Cham
4 hours ago
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
4 hours ago