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 Feb 5 at 22:52
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 Feb 5 at 22:52
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 Feb 5 at 22:52
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 Feb 5 at 22:52
thehorseisbrownthehorseisbrown
6017
asked Feb 5 at 22:52
thehorseisbrownthehorseisbrown
6017
edited Feb 6 at 10:38
thehorseisbrown
asked Feb 5 at 22:52
thehorseisbrownthehorseisbrown
6017
asked Feb 5 at 22:52
thehorseisbrownthehorseisbrown
6017
asked Feb 5 at 22:52
thehorseisbrownthehorseisbrown
6017
6017
add a comment |
add a comment |
1 Answer
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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 Feb 5 at 23:10
ChemomechanicsChemomechanics
5,14231124
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1 Answer
1
active
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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 Feb 5 at 23:10
ChemomechanicsChemomechanics
5,14231124
$endgroup$
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 Feb 5 at 23:10
ChemomechanicsChemomechanics
5,14231124
$endgroup$
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 Feb 5 at 23:10
ChemomechanicsChemomechanics
5,14231124
$endgroup$
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 Feb 5 at 23:10
ChemomechanicsChemomechanics
5,14231124
answered Feb 5 at 23:10
ChemomechanicsChemomechanics
5,14231124
answered Feb 5 at 23:10
ChemomechanicsChemomechanics
5,14231124
answered Feb 5 at 23:10
ChemomechanicsChemomechanics
5,14231124
5,14231124
add a comment |
add a comment |
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