Interpretation of the Boltzmann factor and partition function












4














$$p_i = frac{ expleft(-frac{epsilon _i}{k_BT} right)}{Z} $$
$$ Z= sum_{i} expleft(-frac{epsilon _i}{k_BT} right)$$



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:
$$ Z= sum_{epsilon_i} expleft(-frac{epsilon _i}{k_BT} right)$$



If B) is correct then:
$$ Z= sum_{epsilon_i} Omega_iexpleft(-frac{epsilon _i}{k_BT} right),$$
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







share|cite|improve this question





























    4














    $$p_i = frac{ expleft(-frac{epsilon _i}{k_BT} right)}{Z} $$
    $$ Z= sum_{i} expleft(-frac{epsilon _i}{k_BT} right)$$



    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:
    $$ Z= sum_{epsilon_i} expleft(-frac{epsilon _i}{k_BT} right)$$



    If B) is correct then:
    $$ Z= sum_{epsilon_i} Omega_iexpleft(-frac{epsilon _i}{k_BT} right),$$
    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







    share|cite|improve this question



























      4












      4








      4







      $$p_i = frac{ expleft(-frac{epsilon _i}{k_BT} right)}{Z} $$
      $$ Z= sum_{i} expleft(-frac{epsilon _i}{k_BT} right)$$



      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:
      $$ Z= sum_{epsilon_i} expleft(-frac{epsilon _i}{k_BT} right)$$



      If B) is correct then:
      $$ Z= sum_{epsilon_i} Omega_iexpleft(-frac{epsilon _i}{k_BT} right),$$
      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







      share|cite|improve this question















      $$p_i = frac{ expleft(-frac{epsilon _i}{k_BT} right)}{Z} $$
      $$ Z= sum_{i} expleft(-frac{epsilon _i}{k_BT} right)$$



      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:
      $$ Z= sum_{epsilon_i} expleft(-frac{epsilon _i}{k_BT} right)$$



      If B) is correct then:
      $$ Z= sum_{epsilon_i} Omega_iexpleft(-frac{epsilon _i}{k_BT} right),$$
      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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited yesterday









      Qmechanic♦

      102k121831160




      102k121831160










      asked yesterday









      Daniel DuqueDaniel Duque

      13510




      13510






















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          9














          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: $$Z=sum_i e^{-frac{varepsilon_i}{k_BT}}$$ instead of over the internal energies as you’ve written above.






          share|cite|improve this answer























          • 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.
            – By Symmetry
            yesterday










          • 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.
            – Riley Scott Jacob
            yesterday











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          1 Answer
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          oldest

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          1 Answer
          1






          active

          oldest

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          active

          oldest

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          active

          oldest

          votes









          9














          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: $$Z=sum_i e^{-frac{varepsilon_i}{k_BT}}$$ instead of over the internal energies as you’ve written above.






          share|cite|improve this answer























          • 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.
            – By Symmetry
            yesterday










          • 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.
            – Riley Scott Jacob
            yesterday
















          9














          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: $$Z=sum_i e^{-frac{varepsilon_i}{k_BT}}$$ instead of over the internal energies as you’ve written above.






          share|cite|improve this answer























          • 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.
            – By Symmetry
            yesterday










          • 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.
            – Riley Scott Jacob
            yesterday














          9












          9








          9






          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: $$Z=sum_i e^{-frac{varepsilon_i}{k_BT}}$$ instead of over the internal energies as you’ve written above.






          share|cite|improve this answer














          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: $$Z=sum_i e^{-frac{varepsilon_i}{k_BT}}$$ instead of over the internal energies as you’ve written above.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited yesterday

























          answered yesterday









          Riley Scott JacobRiley Scott Jacob

          2937




          2937












          • 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.
            – By Symmetry
            yesterday










          • 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.
            – Riley Scott Jacob
            yesterday


















          • 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.
            – By Symmetry
            yesterday










          • 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.
            – Riley Scott Jacob
            yesterday
















          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.
          – By Symmetry
          yesterday




          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.
          – By Symmetry
          yesterday












          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.
          – Riley Scott Jacob
          yesterday




          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.
          – Riley Scott Jacob
          yesterday


















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