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I fully understand the concept of a microstate, but I'm having some trouble reconciling two seemingly contradictory definitions of a macrostate.

Here are the two definitions:

  1. A macrostate is defined by a set of definite values of some observable quantities, such as temperature and pressure. Every microstate has an associated macrostate, and so it is possible to calculate the value of temperature and pressure etc for a single microstate. This definition is illustrated by the image below.

Or alternatively

  1. A macrostate is defined by a probability distribution $\{p_i\}$. Thermodynamic quantities can be calculated by taking expectation values in this distribution e.g. $P=-\sum_i p_i \frac{\partial E_i}{\partial V} $, where $E_i$ is the energy of the $i^{\text{th}}$ state. It is not possible to calculate the temperature or pressure of a single microstate.

I think what I really don't understand is why the second definition doesn't allow the pressure of a single microstate to be defined. If you were to consider a single microstate of a gas, surely it would still be possible to calculate the pressure by considering the distribution of momenta etc? Why is it necessary to take an ensemble average?

enter image description here

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – ACuriousMind
    Commented Mar 23, 2019 at 13:23

1 Answer 1

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I have to say that both the definitions you cite are unsatisfactory. I am curious about their origin.

Def. n.1:

A macrostate is defined by a set of definite values of some observable quantities, such as temperature and pressure. Every microstate has an associated macrostate, and so it is possible to calculate the value of temperature and pressure etc for a single microstate. This definition is illustrated by the image below.

In the word macrostate prefix macro conveys the concept of a state of the macroscopic system. For system at equilibrium, it is a result of classical thermodynamics that a few thermodynamic quantities are sufficient to provide the relevant information about the possible evolution of the system if external conditions are changed and enough time is allowed to reach a new equilibrium state. Such thermodynamic quantities are clearly observable quantities, but are the macroscopic quantities. Only in very special cases there is a microscopic observable coinciding with them. The part of the definition i find misleading is the next sentence.

There are many microstates compatible with one macrostate. However, it is not a unique correspondence. The same microstate may be part of the sets of microscopic states corresponding to more than one macrostate. The difference is in the probability of the microstate in correspondence to each possible choice of the macrostate. The immediate consequence of this fact is that it is in general impossible to evaluate a macroscopic thermodynamic quantity by the knowledge of a single microstate.

There is an exception to such impossibility and that is the case of the so-called self-averaging quantities, i.e. quantities such that the ensemble average can be replaced by an average over a sample of increasing size. However, not all thermodynamic quantities can be obtained as statistical averages (noticeable examples are entropy, free energy and in general all thermodynamic potentials). Moreover, if the sample is not large enough, self-averaging can only provide the macroscopic parameters within a (sometimes sizable) error bar.

Def. n.2:

A macrostate is defined by a probability distribution $p_i$. Thermodynamic quantities can be calculated by taking expectation values in this distribution e.g. $P=-\sum_i p_i \frac{\partial{}E_i}{\partial{V}} $, where $E_i$ is the energy of the 𝑖th state. It is not possible to calculate the temperature or pressure of a single microstate.

This definition of macrostate is even worse. The probability distribution of the microstates is not a macroscopic object but it would require the knowledge of microscopic states. Moreover, this definition emphasize the character of average of macroscopic quantities, which is not true for all the possible thermodynamic state functions.

Thus, what could be an acceptable definition of macrostate and microstate?

As stated before, when referring to equilibrium states, a macrostate can be described by any set of macroscopic observable which completely characterize the thermodynamic state of the system.

A microstate, is characterized by a full description of the microscopic dynamical state of the (many) elementary degrees of freedom.

Relations between the two concepts go as follows;

Many microstates correspond in general to one macrostate. The correspondence is not unique. In the ensemble approach to statistical mechanics, fixing the macrostate induces, for finite systems, a probability measure over the set of microstates.

It is possible to define some observables, functions of an individual microscopic configuration, whose ensemble average corresponds to one of the macroscopic parameters (for instance temperature). However such "single configuration quantities" are in general subject to fluctuations and their usage as proxies of the macroscopic parameters can be done only within the size of the unavoidable fluctuations in finite size systems.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – ACuriousMind
    Commented Mar 23, 2019 at 19:30
  • $\begingroup$ So according to the last paragraph we can specify an isolated system (assuming thermodynamic limit) by giving $(E, V, N)$ but we can not specify its state by averaged quantities like $T$ or $P$. In general the specification of the macrostate depends on the ensemble under consideration? $\endgroup$ Commented Feb 19, 2022 at 19:18
  • $\begingroup$ @AntoniosSarikas My point is not exactly like that. The last paragraph refers to the possibility that some observable exists, whose average is equal to a quantity fixed by the external conditions. If the system is large enough to be well approximated by the thermodynamic limit the whole machinery f thermodynamics can be exploited and every set of thermodynamic quantities can be used. However, each ensemble is defined in terms of a specific set of fixed thermodynamic variables and those variables do not fluctuate (by definition). $\endgroup$ Commented Feb 19, 2022 at 23:48
  • $\begingroup$ @GiorgioP So for an isolated system although $(E, V, N)$ remain constant there are some quantities like $T$ that fluctuate. Whereas $\langle T \rangle$ remains constant, $T$ will not. Only in the limit of very large number of particles can we use without problem any macroscopic property to completely characterize the system? $\endgroup$ Commented Feb 20, 2022 at 12:46
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    $\begingroup$ @AntoniosSarikas Yes, that's the main content of the statement that relative fluctuations of observables go to zero at the thermodynamic limit. And it is at the root of the ensemble equivalence for obtaining thermodynamic. $\endgroup$ Commented Feb 20, 2022 at 12:50

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