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Dario56
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Canonical ensemble can be used to derive probability distribution for the internal energy of the closed system at constant volume ##V## and number of particles ##N## in thermal contact with the reservoir.
Also, it is stated that the temperature of both system and reservoir is the same, i.e. they are in the thermal equilibrium.
In statistical thermodynamics, the thermal equilibrium arises naturally from following consideration:
If we consider closed system of interest in thermal contact with the reservoir such that these two form an isolated system than total energy ##E## is defined as: ##E = E_S + E_R##
Total number of microstates of the isolated system with internal energy ##E## is given by: $$ \Omega(E) = \sum_{E_S}\Omega_S(E_S)\Omega_R(E - E_S) = \sum_{E_S}f(E_S) $$
If we wanted to find the internal energy of the system ##E_S## for which the number of microstates of the closed system ##f(E_S)## has a maximum value, we do the following: $$\frac {df(E_S)} {dE_S} = 0$$
This leads us to: $$ \Omega_R\frac {d\Omega_S} {dE_S} + \Omega_S \frac {d\Omega_R} {dE_S} = 0 $$
Because the total system is isolated: $$dE_S = - dE_R$$
Which leads us towards: $$ \frac {dln\Omega_S} {dE_S} = \frac {dln\Omega_R} {dE_R} $$
Taking into account Boltzmann definition of entropy, we know that the previous expression defines temperature equality of the closed system and the reservoir or thermal equilibrium condition.
Important thing to note is that thermal equilibrium is defined when the closed system of interest has the internal energy which corresponds to the maximum number of microstates of the isolated system.
Such state is the most probable and in the thermodynamic limit we are basically certain to find the system's internal energy very close to the value which maximizes the number of microstates.
However, canonical ensemble can be used, as it is mentioned earlier, to derive the probability distribution of the internal energy of the system in the more general case when the thermodynamic limit isn't supposed.
It is from the canonical ensemble that we know that the variance of internal energy for the closed system tends to zero when the number of particles tends to infinity.
This means that if the thermodynamic limit isn't assumed, thermal equilibrium condition losses its significance as system can be found in many different energy states with considerable probability not only in the energy state which maximizes the number of microstates as it is the case in the thermodynamic limit.
This is what doesn't really make sense because the canonical ensemble should be applicable in more general case than thermodynamic limit and yet it uses the concept of thermal equilibrium as its condition which only has sense in the thermodynamic limit.
I think solution to this question may be that the thermal equilibrium isn't the general requirement for the canonical ensemble and that it is only valid in the thermodynamic limit which covers almost all cases in reality.
Although, if you look on the Wikipedia: https://en.m.wikipedia.org/wiki/Canonical_ensemble
you'll find that thermal equilibrium is included in the definition of the canonical ensemble.
Also, it is stated that the temperature of both system and reservoir is the same, i.e. they are in the thermal equilibrium.
In statistical thermodynamics, the thermal equilibrium arises naturally from following consideration:
If we consider closed system of interest in thermal contact with the reservoir such that these two form an isolated system than total energy ##E## is defined as: ##E = E_S + E_R##
Total number of microstates of the isolated system with internal energy ##E## is given by: $$ \Omega(E) = \sum_{E_S}\Omega_S(E_S)\Omega_R(E - E_S) = \sum_{E_S}f(E_S) $$
If we wanted to find the internal energy of the system ##E_S## for which the number of microstates of the closed system ##f(E_S)## has a maximum value, we do the following: $$\frac {df(E_S)} {dE_S} = 0$$
This leads us to: $$ \Omega_R\frac {d\Omega_S} {dE_S} + \Omega_S \frac {d\Omega_R} {dE_S} = 0 $$
Because the total system is isolated: $$dE_S = - dE_R$$
Which leads us towards: $$ \frac {dln\Omega_S} {dE_S} = \frac {dln\Omega_R} {dE_R} $$
Taking into account Boltzmann definition of entropy, we know that the previous expression defines temperature equality of the closed system and the reservoir or thermal equilibrium condition.
Important thing to note is that thermal equilibrium is defined when the closed system of interest has the internal energy which corresponds to the maximum number of microstates of the isolated system.
Such state is the most probable and in the thermodynamic limit we are basically certain to find the system's internal energy very close to the value which maximizes the number of microstates.
However, canonical ensemble can be used, as it is mentioned earlier, to derive the probability distribution of the internal energy of the system in the more general case when the thermodynamic limit isn't supposed.
It is from the canonical ensemble that we know that the variance of internal energy for the closed system tends to zero when the number of particles tends to infinity.
This means that if the thermodynamic limit isn't assumed, thermal equilibrium condition losses its significance as system can be found in many different energy states with considerable probability not only in the energy state which maximizes the number of microstates as it is the case in the thermodynamic limit.
This is what doesn't really make sense because the canonical ensemble should be applicable in more general case than thermodynamic limit and yet it uses the concept of thermal equilibrium as its condition which only has sense in the thermodynamic limit.
I think solution to this question may be that the thermal equilibrium isn't the general requirement for the canonical ensemble and that it is only valid in the thermodynamic limit which covers almost all cases in reality.
Although, if you look on the Wikipedia: https://en.m.wikipedia.org/wiki/Canonical_ensemble
you'll find that thermal equilibrium is included in the definition of the canonical ensemble.
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