Interpreting heat using information entropy












3












$begingroup$


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.




then 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:




  1. probabilities $p_i$ can not change the shape of Gibbs distribution if we assume that the new state is in equilibrium

  2. $sum p_i = 1$


then the temperature must have risen (without any work done)! Is that correct, or can anything else happen?






thermodynamics statistical-mechanics entropy







share|cite|improve this question









$endgroup$

















    3












    $begingroup$


    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.




    then 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:




    1. probabilities $p_i$ can not change the shape of Gibbs distribution if we assume that the new state is in equilibrium

    2. $sum p_i = 1$


    then the temperature must have risen (without any work done)! Is that correct, or can anything else happen?






    thermodynamics statistical-mechanics entropy







    share|cite|improve this question









    $endgroup$















      3












      3








      3





      $begingroup$


      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.




      then 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:




      1. probabilities $p_i$ can not change the shape of Gibbs distribution if we assume that the new state is in equilibrium

      2. $sum p_i = 1$


      then the temperature must have risen (without any work done)! Is that correct, or can anything else happen?






      thermodynamics statistical-mechanics entropy







      share|cite|improve this question









      $endgroup$




      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.




      then 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:




      1. probabilities $p_i$ can not change the shape of Gibbs distribution if we assume that the new state is in equilibrium

      2. $sum p_i = 1$


      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






      share|cite|improve this question













      share|cite|improve this question











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      asked 3 hours ago









      thehorseisbrownthehorseisbrown

      5517




      5517






















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          $begingroup$

          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.







          share|cite|improve this answer









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            1 Answer
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            active

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            $begingroup$

            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.







            share|cite|improve this answer









            $endgroup$


















              4












              $begingroup$

              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.







              share|cite|improve this answer









              $endgroup$
















                4












                4








                4





                $begingroup$

                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.







                share|cite|improve this answer









                $endgroup$



                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.








                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 3 hours ago









                ChemomechanicsChemomechanics

                4,91531023




                4,91531023






























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