Interpretation of the Boltzmann factor and partition function

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4












$begingroup$


$$p_i = frac expleft(-fracepsilon _ik_BT right)Z $$
$$ Z= sum_i expleft(-fracepsilon _ik_BT right)$$



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?










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$endgroup$
















    4












    $begingroup$


    $$p_i = frac expleft(-fracepsilon _ik_BT right)Z $$
    $$ Z= sum_i expleft(-fracepsilon _ik_BT right)$$



    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?










    share|cite|improve this question











    $endgroup$














      4












      4








      4





      $begingroup$


      $$p_i = frac expleft(-fracepsilon _ik_BT right)Z $$
      $$ Z= sum_i expleft(-fracepsilon _ik_BT right)$$



      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?










      share|cite|improve this question











      $endgroup$




      $$p_i = frac expleft(-fracepsilon _ik_BT right)Z $$
      $$ Z= sum_i expleft(-fracepsilon _ik_BT right)$$



      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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 7 at 22:42









      Qmechanic

      103k121851175




      103k121851175










      asked Jan 7 at 19:52









      Daniel DuqueDaniel Duque

      14010




      14010




















          1 Answer
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          9












          $begingroup$

          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.






          share|cite|improve this answer











          $endgroup$












          • $begingroup$
            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










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          1 Answer
          1






          active

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          9












          $begingroup$

          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.






          share|cite|improve this answer











          $endgroup$












          • $begingroup$
            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















          9












          $begingroup$

          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.






          share|cite|improve this answer











          $endgroup$












          • $begingroup$
            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













          9












          9








          9





          $begingroup$

          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.






          share|cite|improve this answer











          $endgroup$



          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.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 7 at 21:13

























          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
















          • $begingroup$
            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

















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