Temperature in an isolated system is defined as: $$\frac{1}{T} = -\frac{\partial{S(E,V,N)}}{\partial{E}} $$ But I wonder how one can generalize this to a random system. Or for instance to a point in a system. Because in these books about statistical physics they talk often about "temperature gradients in a system", but for these to exist, temperature has first to be defined in every point (although I can't find a general definition).

I hope someone can help me out.

  • $\begingroup$ Related: physics.stackexchange.com/q/43858/2451 $\endgroup$ – Qmechanic May 20 '13 at 14:44
  • $\begingroup$ Statistical physics is about random systems, this definition apply only in case of thermal equilibrium, it can be generalized by identifying the logarithm of the number of possible states with the entropy, if you can express this number in terms of the energy of the system you consider. $\endgroup$ – Ikiperu May 20 '13 at 17:14
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    $\begingroup$ You can't define the temperature in a point, since it depends on the average velocity of the atoms/molecule. Therefore temperature is a system feature at a macro scale. $\endgroup$ – fibonatic May 21 '13 at 8:54

Length scales are not accounted for properly in your question. When you have a system at local equilibrium where a temperature gradient can be defined then each "point" in this description contains say $10^{10}$ molecules and can be seen as a thermostatistical system at equilibrium. We call that "local" equilibrium because intensive quantities such as temperature and chemical potential might not be uniform throughout the whole system i.e. they may vary from one "point" to another. There are evolution equations of these mesoscopic quantities that deal with such local equilibrium situations. The simplest are the Fourier (for temperature) and the Fick (for particle density) equations but they can be derived from more general equations with a collision kernel such as e.g. the Boltzmann equation.

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  • $\begingroup$ what you said is true, but I think we can define temperature even for a 1 particle systems, consider for example a box that produce a gaussian random with a particle $m$ inside, we can define an effective temperature for this system with the aid of the definition, of the order of the mean energy given to the system time the boltzmann factor, that could be used for an approximating the probability distribution of the particle's height using Gibbs-Boltzmann law, the difference between the case in which you have lots of particles is that the variance of this probability distribution is very high $\endgroup$ – Ikiperu May 20 '13 at 20:04
  • $\begingroup$ @Ikiperu: You are perfectly right but I answered the part about temperature gradients in a system where the temperature would have to be defined at every "point". I simply specified that in this specific case a point is not actually a point. I do not think this contradicts what you say. Having a big number of particles at every mesoscopic point is however important to a) get some "inertia" to changes in temperature and concentration and b) have vanishing fluctuations once the equilibrium is reached. $\endgroup$ – gatsu May 20 '13 at 20:21

A "point" in a macroscopic system is not a geometrical point. It is a volume element that is small on a macroscopic scale and yet has a large number of molecules for entropy and internal energy to be defined. Your temperature probe does not measure its value at a geometrical point but for a small volume of the system in whose contact it is put.

The local thermal equilibrium pointed in the previous answer (by @gatsu) means that $S$ and $E$ are uniform in a volume element but may vary from one element to another leading to a gradient.

The definition of $T$ in terms of $S$ and $E$ is applicable to any macroscopic system for which $S$ and $E$ are defined.

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The maximum entropy probability distribution for a system of fixed expected total energy is the Boltzmann distribution. This means that Boltzmann distribution is appropriate for systems where we know the total energy but little else. The distribution is given by $p(E) = Z^{-1} e^{-E/kT}$, where $Z$ is the normalization constant (making sure probabilities add to 1), $k$ is Boltzmann's constant and $T$ is another parameter that we name temperature.

When you talk about localized temperature at a specific location and time then at that time we assume that we have a fixed and well defined expected energy distribution. This assumption allows you to conclude that Boltzmann distribution is valid at every point of the system of interest and therefore you can define a location dependent temperature (ie a particle at location $\vec{r}$ has the probability of having energy $E$ of $p(E, \vec{r}) = Z(\vec{r}) e^{-E/kT(\vec{r})}$).

If you have detailed information about the particles in your system then temperature may still be a useful concept (it can be extended somewhat by adding chemical potentials), but some information is in general likely to be lost through its use.

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  • $\begingroup$ A slight correction: the maximum entropy probability distribution for a system of fixed expected total energy is the Boltzmann distribution. $\endgroup$ – Nathaniel May 21 '13 at 12:35
  • $\begingroup$ Another little thing: you can use temperature when you have information about the particles, without losing that information. In that case you just get a distribution that looks like $p(E,\{N_i\}) = Z^{-1}e^{(-E + \sum_i \mu_i N_i)/kT}$ (the maximum entropy distribution for fixed expected energy and particle numbers, with $\{\mu_i\}$ being parameters we call chemical potential), so temperature is still well defined and $1/T$ still turns out to be equal to $\partial S(E,\{N_i\})/\partial E$. $\endgroup$ – Nathaniel May 21 '13 at 12:43
  • $\begingroup$ Yes, but what if for example you have exact information about all the particles in your system (ie full momentum-position information), then temperature will cease to be useful. If you are lucky enough for the particles to be in the ground state then you can set it to zero, otherwise not sure how you can continue this representation. $\endgroup$ – SMeznaric May 21 '13 at 12:57
  • $\begingroup$ Ok fair enough, I see what you meant now. $\endgroup$ – Nathaniel May 21 '13 at 15:05

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