Ambiguity in the definition of entropyHow can temperature be defined as it is if entropy isn't a function of energy?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?Axioms behind entropy!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?
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Ambiguity in the definition of entropy
How can temperature be defined as it is if entropy isn't a function of energy?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?Axioms behind entropy!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 definition
edited Apr 3 at 16:35
Qmechanic♦
107k121991239
asked Apr 3 at 0:46
PiKindOfGuyPiKindOfGuy
626624
$endgroup$
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 definition
edited Apr 3 at 16:35
Qmechanic♦
107k121991239
asked Apr 3 at 0:46
PiKindOfGuyPiKindOfGuy
626624
$endgroup$
$begingroup$
Uh, ambiguity is the definition of entropy.
$endgroup$
– Hot Licks
2 days ago
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 definition
edited Apr 3 at 16:35
Qmechanic♦
107k121991239
asked Apr 3 at 0:46
PiKindOfGuyPiKindOfGuy
626624
$endgroup$
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 definition
statistical-mechanics entropy definition
edited Apr 3 at 16:35
Qmechanic♦
107k121991239
asked Apr 3 at 0:46
PiKindOfGuyPiKindOfGuy
626624
edited Apr 3 at 16:35
Qmechanic♦
107k121991239
asked Apr 3 at 0:46
PiKindOfGuyPiKindOfGuy
626624
edited Apr 3 at 16:35
Qmechanic♦
107k121991239
edited Apr 3 at 16:35
Qmechanic♦
107k121991239
edited Apr 3 at 16:35
Qmechanic♦
107k121991239
107k121991239
asked Apr 3 at 0:46
PiKindOfGuyPiKindOfGuy
626624
asked Apr 3 at 0:46
PiKindOfGuyPiKindOfGuy
626624
asked Apr 3 at 0:46
PiKindOfGuyPiKindOfGuy
626624
626624
$begingroup$
Uh, ambiguity is the definition of entropy.
$endgroup$
– Hot Licks
2 days ago
add a comment |
$begingroup$
Uh, ambiguity is the definition of entropy.
$endgroup$
– Hot Licks
2 days ago
$begingroup$
Uh, ambiguity is the definition of entropy.
$endgroup$
– Hot Licks
2 days ago
$begingroup$
Uh, ambiguity is the definition of entropy.
$endgroup$
– Hot Licks
2 days ago
add a comment |
3 Answers
3
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.
Putting aside quantization, it might appear that there are an infinite number of microstates, and thus the entropy is infinite, but for any level of resolution, the number is finite. And changing the level of resolution simply multiplies the number of microstates by a constant amount. 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. So with a little hand waving (aka "normalization"), we can ignore the apparent infinity of entropy.
answered Apr 3 at 1:22
AcccumulationAcccumulation
3,014514
$endgroup$
$begingroup$
+1; You may add for completeness that $Omega$ is actually the volume occupied by all the possible microstates corresponding to the given macrostate, in phase space. I Always found this expression clearer, since both positions and momentum are continuous variables, the number of microstates available would be infinite.
$endgroup$
– Run like hell
Apr 3 at 14:59
1
$begingroup$
@Runlikehell I was trying to get at the infinity problem with my last paragraph, but I think I'll make that a bit more clear.
$endgroup$
– Acccumulation
Apr 3 at 15:31
$begingroup$
+1. "Entropy is a property of a macrostate, not a system." is completely clear, but I have never heard that before. It clarified my thinking a lot.
$endgroup$
– M. Winter
2 days ago
$begingroup$
@M. Winter, in physics, the macrostate " in question " is the state in which system exists or more accurately spend most of the time, so we (or the system) must choose the macrostate with the highest stat. weight, that is, highest $ Omega$. Analogy: the extremum of the function (entropy) is a property of the function (system) and not a particular function value of particular argument value
$endgroup$
– Aleksey Druggist
2 days ago
add a comment |
$begingroup$
Entropy logarithmically measure of the number of microscopic states corresponding to some specific macroscopically-observable state, not the system as a whole. Put another way: systems that have not yet found their equilibrium state, when left alone, increase their entropy. This would not be possible if the system had the same entropy for all macrostates.
Indeed, the driving principle of entropy in modern stat-mech says that we have some uncertainty about the underlying microscopic state of the system and that from a certain perspective (basically, the one where every macroscopic quantity we can determine is conserved) we can treat nature as simply choosing a microstate uniformly at random. (We have to tread carefully about what exactly uniformly means here but an “obvious” choice seems to replicate certain nice features, like that metals will have specific heats that look like $3R$ where $R$ is the gas constant—a result that I want to say is due to Einstein but I am not 100% sure.)
As a result of this principle of nature picking microstates at random, our equilibrium state is the macrostate which contains the most microstates, and our regression to equilibrium is a process of macrostates getting larger and larger.
answered Apr 3 at 1:23
CR DrostCR Drost
22.7k11962
$endgroup$
add a comment |
$begingroup$
Entropy is a matter of perspective.
You pick a way to describe a system at large scales. This effectively subdivides the system into macrostates, or "macroscopic states".
Each of these macroscopic states corresponds to a number of "microstates"; different configurations of the system that are clumped together in one macrostate.
If, for each macrostate, you take the log of the number of microstates in it, the principle of Entropy is that whatever macrostate it is in, it will move towards macrostates with a higher value almost certainly.
Now you can move to a lower Entropy value only by increasing the Entropy of another system. This basically consists of merging the two systems into one and applying the first rule.
The number of microstates multiply when they are combined; if we have two systems A and B, and they have macrostates A_0 and B_0 with 7 and 10 microstates apiece, the system A+B with macrostate A_0+B_0 has 70 microstates (7*10).
Taking the log of the number of microstates simply allows us to use addition instead of multiplication; the entropy of $log(7)$ and $log(10)$ add to $log(7)+log(10)$ = $log(7*10)$.
Any function that has the property that $f(a*b)=f(a)+f(b)$ will do just as well, which is why we don't care what the base of our logarithm is.
The fun part is that this applies regardless of how you clump the microstates into macrostates so long as you do the clumping before the experiment. So we go and pick sensible macrostates that correspond to things we care about, and the result holds. Crazy choice of macrostates don't actually help us; the vast majority of the possible configuration space of any system is completely useless chaos, only a ridiculously small fraction of the system configuration space is going to be "useful", and no matter how we label it that space is going to have very few microstates in it.
answered Apr 3 at 14:17
YakkYakk
3,0511714
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
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.
Putting aside quantization, it might appear that there are an infinite number of microstates, and thus the entropy is infinite, but for any level of resolution, the number is finite. And changing the level of resolution simply multiplies the number of microstates by a constant amount. 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. So with a little hand waving (aka "normalization"), we can ignore the apparent infinity of entropy.
answered Apr 3 at 1:22
AcccumulationAcccumulation
3,014514
$endgroup$
$begingroup$
+1; You may add for completeness that $Omega$ is actually the volume occupied by all the possible microstates corresponding to the given macrostate, in phase space. I Always found this expression clearer, since both positions and momentum are continuous variables, the number of microstates available would be infinite.
$endgroup$
– Run like hell
Apr 3 at 14:59
1
$begingroup$
@Runlikehell I was trying to get at the infinity problem with my last paragraph, but I think I'll make that a bit more clear.
$endgroup$
– Acccumulation
Apr 3 at 15:31
$begingroup$
+1. "Entropy is a property of a macrostate, not a system." is completely clear, but I have never heard that before. It clarified my thinking a lot.
$endgroup$
– M. Winter
2 days ago
$begingroup$
@M. Winter, in physics, the macrostate " in question " is the state in which system exists or more accurately spend most of the time, so we (or the system) must choose the macrostate with the highest stat. weight, that is, highest $ Omega$. Analogy: the extremum of the function (entropy) is a property of the function (system) and not a particular function value of particular argument value
$endgroup$
– Aleksey Druggist
2 days ago
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.
Putting aside quantization, it might appear that there are an infinite number of microstates, and thus the entropy is infinite, but for any level of resolution, the number is finite. And changing the level of resolution simply multiplies the number of microstates by a constant amount. 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. So with a little hand waving (aka "normalization"), we can ignore the apparent infinity of entropy.
answered Apr 3 at 1:22
AcccumulationAcccumulation
3,014514
$endgroup$
$begingroup$
+1; You may add for completeness that $Omega$ is actually the volume occupied by all the possible microstates corresponding to the given macrostate, in phase space. I Always found this expression clearer, since both positions and momentum are continuous variables, the number of microstates available would be infinite.
$endgroup$
– Run like hell
Apr 3 at 14:59
1
$begingroup$
@Runlikehell I was trying to get at the infinity problem with my last paragraph, but I think I'll make that a bit more clear.
$endgroup$
– Acccumulation
Apr 3 at 15:31
$begingroup$
+1. "Entropy is a property of a macrostate, not a system." is completely clear, but I have never heard that before. It clarified my thinking a lot.
$endgroup$
– M. Winter
2 days ago
$begingroup$
@M. Winter, in physics, the macrostate " in question " is the state in which system exists or more accurately spend most of the time, so we (or the system) must choose the macrostate with the highest stat. weight, that is, highest $ Omega$. Analogy: the extremum of the function (entropy) is a property of the function (system) and not a particular function value of particular argument value
$endgroup$
– Aleksey Druggist
2 days ago
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.
Putting aside quantization, it might appear that there are an infinite number of microstates, and thus the entropy is infinite, but for any level of resolution, the number is finite. And changing the level of resolution simply multiplies the number of microstates by a constant amount. 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. So with a little hand waving (aka "normalization"), we can ignore the apparent infinity of entropy.
answered Apr 3 at 1:22
AcccumulationAcccumulation
3,014514
$endgroup$
Entropy is a property of a macrostate, not a system. So $Omega$ is the number of microstates that correspond to the macrostate in question.
Putting aside quantization, it might appear that there are an infinite number of microstates, and thus the entropy is infinite, but for any level of resolution, the number is finite. And changing the level of resolution simply multiplies the number of microstates by a constant amount. 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. So with a little hand waving (aka "normalization"), we can ignore the apparent infinity of entropy.
answered Apr 3 at 1:22
AcccumulationAcccumulation
3,014514
edited Apr 3 at 15:35
answered Apr 3 at 1:22
AcccumulationAcccumulation
3,014514
answered Apr 3 at 1:22
AcccumulationAcccumulation
3,014514
answered Apr 3 at 1:22
AcccumulationAcccumulation
3,014514
3,014514
$begingroup$
+1; You may add for completeness that $Omega$ is actually the volume occupied by all the possible microstates corresponding to the given macrostate, in phase space. I Always found this expression clearer, since both positions and momentum are continuous variables, the number of microstates available would be infinite.
$endgroup$
– Run like hell
Apr 3 at 14:59
1
$begingroup$
@Runlikehell I was trying to get at the infinity problem with my last paragraph, but I think I'll make that a bit more clear.
$endgroup$
– Acccumulation
Apr 3 at 15:31
$begingroup$
+1. "Entropy is a property of a macrostate, not a system." is completely clear, but I have never heard that before. It clarified my thinking a lot.
$endgroup$
– M. Winter
2 days ago
$begingroup$
@M. Winter, in physics, the macrostate " in question " is the state in which system exists or more accurately spend most of the time, so we (or the system) must choose the macrostate with the highest stat. weight, that is, highest $ Omega$. Analogy: the extremum of the function (entropy) is a property of the function (system) and not a particular function value of particular argument value
$endgroup$
– Aleksey Druggist
2 days ago
add a comment |
$begingroup$
+1; You may add for completeness that $Omega$ is actually the volume occupied by all the possible microstates corresponding to the given macrostate, in phase space. I Always found this expression clearer, since both positions and momentum are continuous variables, the number of microstates available would be infinite.
$endgroup$
– Run like hell
Apr 3 at 14:59
1
$begingroup$
@Runlikehell I was trying to get at the infinity problem with my last paragraph, but I think I'll make that a bit more clear.
$endgroup$
– Acccumulation
Apr 3 at 15:31
$begingroup$
+1. "Entropy is a property of a macrostate, not a system." is completely clear, but I have never heard that before. It clarified my thinking a lot.
$endgroup$
– M. Winter
2 days ago
$begingroup$
@M. Winter, in physics, the macrostate " in question " is the state in which system exists or more accurately spend most of the time, so we (or the system) must choose the macrostate with the highest stat. weight, that is, highest $ Omega$. Analogy: the extremum of the function (entropy) is a property of the function (system) and not a particular function value of particular argument value
$endgroup$
– Aleksey Druggist
2 days ago
$begingroup$
+1; You may add for completeness that $Omega$ is actually the volume occupied by all the possible microstates corresponding to the given macrostate, in phase space. I Always found this expression clearer, since both positions and momentum are continuous variables, the number of microstates available would be infinite.
$endgroup$
– Run like hell
Apr 3 at 14:59
$begingroup$
+1; You may add for completeness that $Omega$ is actually the volume occupied by all the possible microstates corresponding to the given macrostate, in phase space. I Always found this expression clearer, since both positions and momentum are continuous variables, the number of microstates available would be infinite.
$endgroup$
– Run like hell
Apr 3 at 14:59
1
1
$begingroup$
@Runlikehell I was trying to get at the infinity problem with my last paragraph, but I think I'll make that a bit more clear.
$endgroup$
– Acccumulation
Apr 3 at 15:31
$begingroup$
@Runlikehell I was trying to get at the infinity problem with my last paragraph, but I think I'll make that a bit more clear.
$endgroup$
– Acccumulation
Apr 3 at 15:31
$begingroup$
+1. "Entropy is a property of a macrostate, not a system." is completely clear, but I have never heard that before. It clarified my thinking a lot.
$endgroup$
– M. Winter
2 days ago
$begingroup$
+1. "Entropy is a property of a macrostate, not a system." is completely clear, but I have never heard that before. It clarified my thinking a lot.
$endgroup$
– M. Winter
2 days ago
$begingroup$
@M. Winter, in physics, the macrostate " in question " is the state in which system exists or more accurately spend most of the time, so we (or the system) must choose the macrostate with the highest stat. weight, that is, highest $ Omega$. Analogy: the extremum of the function (entropy) is a property of the function (system) and not a particular function value of particular argument value
$endgroup$
– Aleksey Druggist
2 days ago
$begingroup$
@M. Winter, in physics, the macrostate " in question " is the state in which system exists or more accurately spend most of the time, so we (or the system) must choose the macrostate with the highest stat. weight, that is, highest $ Omega$. Analogy: the extremum of the function (entropy) is a property of the function (system) and not a particular function value of particular argument value
$endgroup$
– Aleksey Druggist
2 days ago
add a comment |
$begingroup$
Entropy logarithmically measure of the number of microscopic states corresponding to some specific macroscopically-observable state, not the system as a whole. Put another way: systems that have not yet found their equilibrium state, when left alone, increase their entropy. This would not be possible if the system had the same entropy for all macrostates.
Indeed, the driving principle of entropy in modern stat-mech says that we have some uncertainty about the underlying microscopic state of the system and that from a certain perspective (basically, the one where every macroscopic quantity we can determine is conserved) we can treat nature as simply choosing a microstate uniformly at random. (We have to tread carefully about what exactly uniformly means here but an “obvious” choice seems to replicate certain nice features, like that metals will have specific heats that look like $3R$ where $R$ is the gas constant—a result that I want to say is due to Einstein but I am not 100% sure.)
As a result of this principle of nature picking microstates at random, our equilibrium state is the macrostate which contains the most microstates, and our regression to equilibrium is a process of macrostates getting larger and larger.
answered Apr 3 at 1:23
CR DrostCR Drost
22.7k11962
$endgroup$
add a comment |
$begingroup$
Entropy logarithmically measure of the number of microscopic states corresponding to some specific macroscopically-observable state, not the system as a whole. Put another way: systems that have not yet found their equilibrium state, when left alone, increase their entropy. This would not be possible if the system had the same entropy for all macrostates.
Indeed, the driving principle of entropy in modern stat-mech says that we have some uncertainty about the underlying microscopic state of the system and that from a certain perspective (basically, the one where every macroscopic quantity we can determine is conserved) we can treat nature as simply choosing a microstate uniformly at random. (We have to tread carefully about what exactly uniformly means here but an “obvious” choice seems to replicate certain nice features, like that metals will have specific heats that look like $3R$ where $R$ is the gas constant—a result that I want to say is due to Einstein but I am not 100% sure.)
As a result of this principle of nature picking microstates at random, our equilibrium state is the macrostate which contains the most microstates, and our regression to equilibrium is a process of macrostates getting larger and larger.
answered Apr 3 at 1:23
CR DrostCR Drost
22.7k11962
$endgroup$
add a comment |
$begingroup$
Entropy logarithmically measure of the number of microscopic states corresponding to some specific macroscopically-observable state, not the system as a whole. Put another way: systems that have not yet found their equilibrium state, when left alone, increase their entropy. This would not be possible if the system had the same entropy for all macrostates.
Indeed, the driving principle of entropy in modern stat-mech says that we have some uncertainty about the underlying microscopic state of the system and that from a certain perspective (basically, the one where every macroscopic quantity we can determine is conserved) we can treat nature as simply choosing a microstate uniformly at random. (We have to tread carefully about what exactly uniformly means here but an “obvious” choice seems to replicate certain nice features, like that metals will have specific heats that look like $3R$ where $R$ is the gas constant—a result that I want to say is due to Einstein but I am not 100% sure.)
As a result of this principle of nature picking microstates at random, our equilibrium state is the macrostate which contains the most microstates, and our regression to equilibrium is a process of macrostates getting larger and larger.
answered Apr 3 at 1:23
CR DrostCR Drost
22.7k11962
$endgroup$
Entropy logarithmically measure of the number of microscopic states corresponding to some specific macroscopically-observable state, not the system as a whole. Put another way: systems that have not yet found their equilibrium state, when left alone, increase their entropy. This would not be possible if the system had the same entropy for all macrostates.
Indeed, the driving principle of entropy in modern stat-mech says that we have some uncertainty about the underlying microscopic state of the system and that from a certain perspective (basically, the one where every macroscopic quantity we can determine is conserved) we can treat nature as simply choosing a microstate uniformly at random. (We have to tread carefully about what exactly uniformly means here but an “obvious” choice seems to replicate certain nice features, like that metals will have specific heats that look like $3R$ where $R$ is the gas constant—a result that I want to say is due to Einstein but I am not 100% sure.)
As a result of this principle of nature picking microstates at random, our equilibrium state is the macrostate which contains the most microstates, and our regression to equilibrium is a process of macrostates getting larger and larger.
answered Apr 3 at 1:23
CR DrostCR Drost
22.7k11962
edited Apr 3 at 14:55
answered Apr 3 at 1:23
CR DrostCR Drost
22.7k11962
answered Apr 3 at 1:23
CR DrostCR Drost
22.7k11962
answered Apr 3 at 1:23
CR DrostCR Drost
22.7k11962
22.7k11962
add a comment |
add a comment |
$begingroup$
Entropy is a matter of perspective.
You pick a way to describe a system at large scales. This effectively subdivides the system into macrostates, or "macroscopic states".
Each of these macroscopic states corresponds to a number of "microstates"; different configurations of the system that are clumped together in one macrostate.
If, for each macrostate, you take the log of the number of microstates in it, the principle of Entropy is that whatever macrostate it is in, it will move towards macrostates with a higher value almost certainly.
Now you can move to a lower Entropy value only by increasing the Entropy of another system. This basically consists of merging the two systems into one and applying the first rule.
The number of microstates multiply when they are combined; if we have two systems A and B, and they have macrostates A_0 and B_0 with 7 and 10 microstates apiece, the system A+B with macrostate A_0+B_0 has 70 microstates (7*10).
Taking the log of the number of microstates simply allows us to use addition instead of multiplication; the entropy of $log(7)$ and $log(10)$ add to $log(7)+log(10)$ = $log(7*10)$.
Any function that has the property that $f(a*b)=f(a)+f(b)$ will do just as well, which is why we don't care what the base of our logarithm is.
The fun part is that this applies regardless of how you clump the microstates into macrostates so long as you do the clumping before the experiment. So we go and pick sensible macrostates that correspond to things we care about, and the result holds. Crazy choice of macrostates don't actually help us; the vast majority of the possible configuration space of any system is completely useless chaos, only a ridiculously small fraction of the system configuration space is going to be "useful", and no matter how we label it that space is going to have very few microstates in it.
answered Apr 3 at 14:17
YakkYakk
3,0511714
$endgroup$
add a comment |
$begingroup$
Entropy is a matter of perspective.
You pick a way to describe a system at large scales. This effectively subdivides the system into macrostates, or "macroscopic states".
Each of these macroscopic states corresponds to a number of "microstates"; different configurations of the system that are clumped together in one macrostate.
If, for each macrostate, you take the log of the number of microstates in it, the principle of Entropy is that whatever macrostate it is in, it will move towards macrostates with a higher value almost certainly.
Now you can move to a lower Entropy value only by increasing the Entropy of another system. This basically consists of merging the two systems into one and applying the first rule.
The number of microstates multiply when they are combined; if we have two systems A and B, and they have macrostates A_0 and B_0 with 7 and 10 microstates apiece, the system A+B with macrostate A_0+B_0 has 70 microstates (7*10).
Taking the log of the number of microstates simply allows us to use addition instead of multiplication; the entropy of $log(7)$ and $log(10)$ add to $log(7)+log(10)$ = $log(7*10)$.
Any function that has the property that $f(a*b)=f(a)+f(b)$ will do just as well, which is why we don't care what the base of our logarithm is.
The fun part is that this applies regardless of how you clump the microstates into macrostates so long as you do the clumping before the experiment. So we go and pick sensible macrostates that correspond to things we care about, and the result holds. Crazy choice of macrostates don't actually help us; the vast majority of the possible configuration space of any system is completely useless chaos, only a ridiculously small fraction of the system configuration space is going to be "useful", and no matter how we label it that space is going to have very few microstates in it.
answered Apr 3 at 14:17
YakkYakk
3,0511714
$endgroup$
add a comment |
$begingroup$
Entropy is a matter of perspective.
You pick a way to describe a system at large scales. This effectively subdivides the system into macrostates, or "macroscopic states".
Each of these macroscopic states corresponds to a number of "microstates"; different configurations of the system that are clumped together in one macrostate.
If, for each macrostate, you take the log of the number of microstates in it, the principle of Entropy is that whatever macrostate it is in, it will move towards macrostates with a higher value almost certainly.
Now you can move to a lower Entropy value only by increasing the Entropy of another system. This basically consists of merging the two systems into one and applying the first rule.
The number of microstates multiply when they are combined; if we have two systems A and B, and they have macrostates A_0 and B_0 with 7 and 10 microstates apiece, the system A+B with macrostate A_0+B_0 has 70 microstates (7*10).
Taking the log of the number of microstates simply allows us to use addition instead of multiplication; the entropy of $log(7)$ and $log(10)$ add to $log(7)+log(10)$ = $log(7*10)$.
Any function that has the property that $f(a*b)=f(a)+f(b)$ will do just as well, which is why we don't care what the base of our logarithm is.
The fun part is that this applies regardless of how you clump the microstates into macrostates so long as you do the clumping before the experiment. So we go and pick sensible macrostates that correspond to things we care about, and the result holds. Crazy choice of macrostates don't actually help us; the vast majority of the possible configuration space of any system is completely useless chaos, only a ridiculously small fraction of the system configuration space is going to be "useful", and no matter how we label it that space is going to have very few microstates in it.
answered Apr 3 at 14:17
YakkYakk
3,0511714
$endgroup$
Entropy is a matter of perspective.
You pick a way to describe a system at large scales. This effectively subdivides the system into macrostates, or "macroscopic states".
Each of these macroscopic states corresponds to a number of "microstates"; different configurations of the system that are clumped together in one macrostate.
If, for each macrostate, you take the log of the number of microstates in it, the principle of Entropy is that whatever macrostate it is in, it will move towards macrostates with a higher value almost certainly.
Now you can move to a lower Entropy value only by increasing the Entropy of another system. This basically consists of merging the two systems into one and applying the first rule.
The number of microstates multiply when they are combined; if we have two systems A and B, and they have macrostates A_0 and B_0 with 7 and 10 microstates apiece, the system A+B with macrostate A_0+B_0 has 70 microstates (7*10).
Taking the log of the number of microstates simply allows us to use addition instead of multiplication; the entropy of $log(7)$ and $log(10)$ add to $log(7)+log(10)$ = $log(7*10)$.
Any function that has the property that $f(a*b)=f(a)+f(b)$ will do just as well, which is why we don't care what the base of our logarithm is.
The fun part is that this applies regardless of how you clump the microstates into macrostates so long as you do the clumping before the experiment. So we go and pick sensible macrostates that correspond to things we care about, and the result holds. Crazy choice of macrostates don't actually help us; the vast majority of the possible configuration space of any system is completely useless chaos, only a ridiculously small fraction of the system configuration space is going to be "useful", and no matter how we label it that space is going to have very few microstates in it.
answered Apr 3 at 14:17
YakkYakk
3,0511714
answered Apr 3 at 14:17
YakkYakk
3,0511714
answered Apr 3 at 14:17
YakkYakk
3,0511714
answered Apr 3 at 14:17
YakkYakk
3,0511714
3,0511714
add a comment |
add a comment |
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$begingroup$
Uh, ambiguity is the definition of entropy.
$endgroup$
– Hot Licks
2 days ago