Interpretation of the Boltzmann factor and partition function
Clash Royale CLAN TAG#URR8PPP
$begingroup$
$$p_i = frac expleft(-fracepsilon _ik_BT right)Z $$
$$ Z= sum_i expleft(-fracepsilon _ik_BT right)$$
A) Is $p_i$ the probability of the system having an energy equal to $epsilon_i$? (Probability to be in any of the many microstates that have energy $epsilon_i$).
B) Or is $p_i$ the probability of the system being in one particular microstate which happens to have energy $epsilon_i$? (This microstate is not the only microstate with the same energy).
If A) is correct then:
$$ Z= sum_epsilon_i expleft(-fracepsilon _ik_BT right)$$
If B) is correct then:
$$ Z= sum_epsilon_i Omega_iexpleft(-fracepsilon _ik_BT right),$$
where $Omega_i$ is the multiplicity of the macrostate of energy $epsilon_i$.
From the derivation of the Boltzmann distribution I am inclined to understand it as B). But I have never seen the multiplicity in the partition function.
What is the correct interpretation of the Boltzmann distribution?
statistical-mechanics probability partition-function
edited Jan 7 at 22:42
Qmechanic♦
103k121851175
asked Jan 7 at 19:52
Daniel DuqueDaniel Duque
14010
$endgroup$
add a comment |
$begingroup$
$$p_i = frac expleft(-fracepsilon _ik_BT right)Z $$
$$ Z= sum_i expleft(-fracepsilon _ik_BT right)$$
A) Is $p_i$ the probability of the system having an energy equal to $epsilon_i$? (Probability to be in any of the many microstates that have energy $epsilon_i$).
B) Or is $p_i$ the probability of the system being in one particular microstate which happens to have energy $epsilon_i$? (This microstate is not the only microstate with the same energy).
If A) is correct then:
$$ Z= sum_epsilon_i expleft(-fracepsilon _ik_BT right)$$
If B) is correct then:
$$ Z= sum_epsilon_i Omega_iexpleft(-fracepsilon _ik_BT right),$$
where $Omega_i$ is the multiplicity of the macrostate of energy $epsilon_i$.
From the derivation of the Boltzmann distribution I am inclined to understand it as B). But I have never seen the multiplicity in the partition function.
What is the correct interpretation of the Boltzmann distribution?
statistical-mechanics probability partition-function
edited Jan 7 at 22:42
Qmechanic♦
103k121851175
asked Jan 7 at 19:52
Daniel DuqueDaniel Duque
14010
$endgroup$
add a comment |
$begingroup$
$$p_i = frac expleft(-fracepsilon _ik_BT right)Z $$
$$ Z= sum_i expleft(-fracepsilon _ik_BT right)$$
A) Is $p_i$ the probability of the system having an energy equal to $epsilon_i$? (Probability to be in any of the many microstates that have energy $epsilon_i$).
B) Or is $p_i$ the probability of the system being in one particular microstate which happens to have energy $epsilon_i$? (This microstate is not the only microstate with the same energy).
If A) is correct then:
$$ Z= sum_epsilon_i expleft(-fracepsilon _ik_BT right)$$
If B) is correct then:
$$ Z= sum_epsilon_i Omega_iexpleft(-fracepsilon _ik_BT right),$$
where $Omega_i$ is the multiplicity of the macrostate of energy $epsilon_i$.
From the derivation of the Boltzmann distribution I am inclined to understand it as B). But I have never seen the multiplicity in the partition function.
What is the correct interpretation of the Boltzmann distribution?
statistical-mechanics probability partition-function
edited Jan 7 at 22:42
Qmechanic♦
103k121851175
asked Jan 7 at 19:52
Daniel DuqueDaniel Duque
14010
$endgroup$
$$p_i = frac expleft(-fracepsilon _ik_BT right)Z $$
$$ Z= sum_i expleft(-fracepsilon _ik_BT right)$$
A) Is $p_i$ the probability of the system having an energy equal to $epsilon_i$? (Probability to be in any of the many microstates that have energy $epsilon_i$).
B) Or is $p_i$ the probability of the system being in one particular microstate which happens to have energy $epsilon_i$? (This microstate is not the only microstate with the same energy).
If A) is correct then:
$$ Z= sum_epsilon_i expleft(-fracepsilon _ik_BT right)$$
If B) is correct then:
$$ Z= sum_epsilon_i Omega_iexpleft(-fracepsilon _ik_BT right),$$
where $Omega_i$ is the multiplicity of the macrostate of energy $epsilon_i$.
From the derivation of the Boltzmann distribution I am inclined to understand it as B). But I have never seen the multiplicity in the partition function.
What is the correct interpretation of the Boltzmann distribution?
statistical-mechanics probability partition-function
statistical-mechanics probability partition-function
edited Jan 7 at 22:42
Qmechanic♦
103k121851175
asked Jan 7 at 19:52
Daniel DuqueDaniel Duque
14010
edited Jan 7 at 22:42
Qmechanic♦
103k121851175
asked Jan 7 at 19:52
Daniel DuqueDaniel Duque
14010
edited Jan 7 at 22:42
Qmechanic♦
103k121851175
edited Jan 7 at 22:42
Qmechanic♦
103k121851175
edited Jan 7 at 22:42
Qmechanic♦
103k121851175
103k121851175
asked Jan 7 at 19:52
Daniel DuqueDaniel Duque
14010
asked Jan 7 at 19:52
Daniel DuqueDaniel Duque
14010
asked Jan 7 at 19:52
Daniel DuqueDaniel Duque
14010
14010
add a comment |
add a comment |
1 Answer
1
active
oldest
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$begingroup$
To the first question, the answer is B: $p_i$ is the probability of being in the $i$-th microstate, which happens to have an energy $varepsilon_i$. However, microstates other than the $i$-th one may also have an energy $varepsilon_i$.
The reason you never see the multiplicity in the partition function is because you are probably looking at summations done over the microstates: $$Z=sum_i e^-fracvarepsilon_ik_BT$$ instead of over the internal energies as you’ve written above.
answered Jan 7 at 20:04
Riley Scott JacobRiley Scott Jacob
1,007110
$endgroup$
$begingroup$
I would not say that it is bizarre to write the partition function a a sum over energies. I have certainly seen it done. It is common, for example to do this as a precursor to moving representation of the partition function as an integral over energy levels, in which the multiplicity $g_i$ becomes a contribution to the density of states $g(epsilon)$ states.
$endgroup$
– By Symmetry
Jan 7 at 21:09
$begingroup$
I do apologize - I am only an undergraduate and have never seen it as a sum over the energies in my limited experience. I will amend my statement.
$endgroup$
– Riley Scott Jacob
Jan 7 at 21:10
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
To the first question, the answer is B: $p_i$ is the probability of being in the $i$-th microstate, which happens to have an energy $varepsilon_i$. However, microstates other than the $i$-th one may also have an energy $varepsilon_i$.
The reason you never see the multiplicity in the partition function is because you are probably looking at summations done over the microstates: $$Z=sum_i e^-fracvarepsilon_ik_BT$$ instead of over the internal energies as you’ve written above.
answered Jan 7 at 20:04
Riley Scott JacobRiley Scott Jacob
1,007110
$endgroup$
$begingroup$
I would not say that it is bizarre to write the partition function a a sum over energies. I have certainly seen it done. It is common, for example to do this as a precursor to moving representation of the partition function as an integral over energy levels, in which the multiplicity $g_i$ becomes a contribution to the density of states $g(epsilon)$ states.
$endgroup$
– By Symmetry
Jan 7 at 21:09
$begingroup$
I do apologize - I am only an undergraduate and have never seen it as a sum over the energies in my limited experience. I will amend my statement.
$endgroup$
– Riley Scott Jacob
Jan 7 at 21:10
add a comment |
$begingroup$
To the first question, the answer is B: $p_i$ is the probability of being in the $i$-th microstate, which happens to have an energy $varepsilon_i$. However, microstates other than the $i$-th one may also have an energy $varepsilon_i$.
The reason you never see the multiplicity in the partition function is because you are probably looking at summations done over the microstates: $$Z=sum_i e^-fracvarepsilon_ik_BT$$ instead of over the internal energies as you’ve written above.
answered Jan 7 at 20:04
Riley Scott JacobRiley Scott Jacob
1,007110
$endgroup$
$begingroup$
I would not say that it is bizarre to write the partition function a a sum over energies. I have certainly seen it done. It is common, for example to do this as a precursor to moving representation of the partition function as an integral over energy levels, in which the multiplicity $g_i$ becomes a contribution to the density of states $g(epsilon)$ states.
$endgroup$
– By Symmetry
Jan 7 at 21:09
$begingroup$
I do apologize - I am only an undergraduate and have never seen it as a sum over the energies in my limited experience. I will amend my statement.
$endgroup$
– Riley Scott Jacob
Jan 7 at 21:10
add a comment |
$begingroup$
To the first question, the answer is B: $p_i$ is the probability of being in the $i$-th microstate, which happens to have an energy $varepsilon_i$. However, microstates other than the $i$-th one may also have an energy $varepsilon_i$.
The reason you never see the multiplicity in the partition function is because you are probably looking at summations done over the microstates: $$Z=sum_i e^-fracvarepsilon_ik_BT$$ instead of over the internal energies as you’ve written above.
answered Jan 7 at 20:04
Riley Scott JacobRiley Scott Jacob
1,007110
$endgroup$
To the first question, the answer is B: $p_i$ is the probability of being in the $i$-th microstate, which happens to have an energy $varepsilon_i$. However, microstates other than the $i$-th one may also have an energy $varepsilon_i$.
The reason you never see the multiplicity in the partition function is because you are probably looking at summations done over the microstates: $$Z=sum_i e^-fracvarepsilon_ik_BT$$ instead of over the internal energies as you’ve written above.
answered Jan 7 at 20:04
Riley Scott JacobRiley Scott Jacob
1,007110
edited Jan 7 at 21:13
answered Jan 7 at 20:04
Riley Scott JacobRiley Scott Jacob
1,007110
answered Jan 7 at 20:04
Riley Scott JacobRiley Scott Jacob
1,007110
answered Jan 7 at 20:04
Riley Scott JacobRiley Scott Jacob
1,007110
1,007110
$begingroup$
I would not say that it is bizarre to write the partition function a a sum over energies. I have certainly seen it done. It is common, for example to do this as a precursor to moving representation of the partition function as an integral over energy levels, in which the multiplicity $g_i$ becomes a contribution to the density of states $g(epsilon)$ states.
$endgroup$
– By Symmetry
Jan 7 at 21:09
$begingroup$
I do apologize - I am only an undergraduate and have never seen it as a sum over the energies in my limited experience. I will amend my statement.
$endgroup$
– Riley Scott Jacob
Jan 7 at 21:10
add a comment |
$begingroup$
I would not say that it is bizarre to write the partition function a a sum over energies. I have certainly seen it done. It is common, for example to do this as a precursor to moving representation of the partition function as an integral over energy levels, in which the multiplicity $g_i$ becomes a contribution to the density of states $g(epsilon)$ states.
$endgroup$
– By Symmetry
Jan 7 at 21:09
$begingroup$
I do apologize - I am only an undergraduate and have never seen it as a sum over the energies in my limited experience. I will amend my statement.
$endgroup$
– Riley Scott Jacob
Jan 7 at 21:10
$begingroup$
I would not say that it is bizarre to write the partition function a a sum over energies. I have certainly seen it done. It is common, for example to do this as a precursor to moving representation of the partition function as an integral over energy levels, in which the multiplicity $g_i$ becomes a contribution to the density of states $g(epsilon)$ states.
$endgroup$
– By Symmetry
Jan 7 at 21:09
$begingroup$
I would not say that it is bizarre to write the partition function a a sum over energies. I have certainly seen it done. It is common, for example to do this as a precursor to moving representation of the partition function as an integral over energy levels, in which the multiplicity $g_i$ becomes a contribution to the density of states $g(epsilon)$ states.
$endgroup$
– By Symmetry
Jan 7 at 21:09
$begingroup$
I do apologize - I am only an undergraduate and have never seen it as a sum over the energies in my limited experience. I will amend my statement.
$endgroup$
– Riley Scott Jacob
Jan 7 at 21:10
$begingroup$
I do apologize - I am only an undergraduate and have never seen it as a sum over the energies in my limited experience. I will amend my statement.
$endgroup$
– Riley Scott Jacob
Jan 7 at 21:10
add a comment |
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