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When analyzing the probability of finding the system in a particular state, we observe that the equilibrium condition ($\beta$) is an integral part of the distribution's structure. Furthermore, it becomes evident that the function that best models the system is the logarithm of the number of states, which is associated with what we will term entropy.

>Model

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Image

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When we multiply the number of cases, we obtain a function with a very pronounced peak.

The system is more likely to be found at the energy where the peak of the probability curve occurs.

ID:(11543, 0)

Equation

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To study the behavior of the number of states function, we can expand it around the equilibrium energy value $\bar{E}$. If we do this expansion in the logarithm of the number of states, we obtain

$\ln\Omega(E)=\ln\Omega(\bar{E})+\displaystyle\frac{\partial\ln\Omega}{\partial E}\eta+\displaystyle\frac{1}{2}\displaystyle\frac{\partial^2\ln\Omega}{\partial E^2}\eta^2\ldots$

where $\eta=E-\bar{E}$. Using for the definition of

and

$\lambda\equiv-\displaystyle\frac{\partial^2\ln\Omega}{\partial E^2}=-\displaystyle\frac{\partial\ln\beta}{\partial E}$

we obtain the expression with :

$\ln\Omega(E)=\ln\Omega(\bar{E})+\beta\eta-\displaystyle\frac{1}{2}\lambda\eta^2\ldots$

Conditions

Variables

$\beta$

Beta del sistema

$1/J$

8823

$\eta$

Desviación de la energía

$J$

8829

$\ln(\Omega(E))$

Logaritmo del numero de estados del sistema con la energía $E$

$-$

8833

$\ln(\Omega(\bar{E}))$

Logaritmo del numero de estados del sistema con la energía media $\bar{E}$

$-$

8834

$\lambda$

Medida del ancho de la distribución de probabilidad

$1/J^2$

8830

Relation

ID:(3443, 0)

Equation

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If we consider the expansion of the number of states with beta del sistema $1/J$, desviación de la energía $J$, logaritmo del numero de estados del sistema con la energía $E$ $-$, logaritmo del numero de estados del sistema con la energía media $\bar{E}$ $-$ and medida del ancho de la distribución de probabilidad $1/J^2$:

we can estimate the logarithm of the probability:

$\ln P = \ln[\Omega(E)\Omega'(E')] = \ln\Omega(E) + \ln\Omega'(E') = \ln\Omega(\bar{E}) + \ln\Omega'(\bar{E'}) + (\beta - \beta')\eta - \frac{1}{2}(\lambda + \lambda')\eta^2\ldots$

For the case of equilibrium, both betas are equal, and the probability that one of the systems has an energy $E$ is reduced to a Gaussian distribution, which is expressed with beta del sistema $1/J$, desviación de la energía $J$, logaritmo del numero de estados del sistema con la energía $E$ $-$, logaritmo del numero de estados del sistema con la energía media $\bar{E}$ $-$ and medida del ancho de la distribución de probabilidad $1/J^2$ as:

where

$\lambda_0 = \lambda + \lambda'$

ID:(3444, 0)

Equation

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The factor that defines the width of the probability curve is the quadratic factor in the Taylor series expansion, which is expressed with as follows:

It can be shown that the introduced number $\lambda$ is always positive. An indication of this comes from the function of the number of states we have already calculated for the case of free particles. In that case, as the number of states is proportional to the energy raised to the power of the number of degrees of freedom $f$, we find that with

$\Omega\sim E^f$

we have

$\lambda_0\equiv-\displaystyle\frac{\partial^2\ln\Omega}{\partial E^2}=-f\displaystyle\frac{\partial^2\ln E}{\partial E^2}=\displaystyle\frac{f}{E^2}$

.

ID:(11549, 0)

Equation

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The key parameter in the study of equilibrium is given by the logarithm of the number of states, which with is expressed as:

The natural logarithm of the number of states multiplied by the Boltzmann constant $k_B$ is defined as the system's entropy, which is expressed with as follows:

ID:(3439, 0)

Equation

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The definition of $\beta$ is found in

that of temperature is in

and that of entropy is in constante de Boltzmann $J/K$, entropia del sistema $J/K$ and numero de estados del sistema con la energía $E$ $-$

These definitions lead us to a thermodynamic relationship that indicates how temperature $T$ is related to constante de Boltzmann $J/K$, entropia del sistema $J/K$ and numero de estados del sistema con la energía $E$ $-$:

$\displaystyle\frac{1}{T}=\displaystyle\frac{\partial S}{\partial E}$

Conditions

Variables

$E$

Energía del sistema

$J$

8719

$S$

Entropia del sistema

$J/K$

8831

$T$

Temperatura del sistema

$K$

8825

Relation

ID:(3442, 0)

Equation

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With the definition of entropy as constante de Boltzmann $J/K$, entropia del sistema $J/K$ and numero de estados del sistema con la energía $E$ $-$:

and considering that in equilibrium it holds with :

we conclude that in equilibrium, the energy $E$ must always be maximum with :

ID:(3440, 0)

0

Video

Video: Distribution and Entropy