Druyd:Knowledge

Distribution and Entropy

Storyboard

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

ID:(437, 0)

Comparing the number of state curves

Definition

ID:(11543, 0)

Distribution and Entropy

Description

Variables

Symbol

Text

Variable

Value

Units

Calculate

MKS Value

MKS Units

$\beta$

beta

Beta del sistema

1/J

$k_B$

k_B

Constante de Boltzmann

J/K

$\eta$

eta

Desviación de la energía

$E_2$

E_2

Energía del reservorio

$E$

Energía del sistema

$\bar{E}$

Energía media del sistema

$S$

Entropia del sistema

J/K

$S_{max}$

S_max

Entropia máxima

J/K

$\ln(\Omega(E))$

ln_Omega_E

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

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

ln_Omega_E_m

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

$\lambda$

lambda

Medida del ancho de la distribución de probabilidad

1/J^2

$\lambda_0$

lambda_0

Medida del ancho de la distribución de probabilidad total

1/J^2

$\Omega_E$

Omega_E

Numero de estados del sistema con la energía $E$

$P_E$

P_E

Probabilidad del sistema de tener una energía $E$

$P_0$

P_0

Probabilidad del sistema de tener una energía media $\bar{E}$

$T$

Temperatura del sistema

Calculations

Equation

Number

First, select the equation:

, then, select the variable:

Symbol

Equation

Solved

Translated

Calculations

Symbol

Equation

Solved

Translated

Variable

Given

Calculate

Target :

Equation

To be used

Equations

$ S \equiv k_B \ln \Omega $

S = k_B * ln( Omega )

(ID 3439)

$ S + S_h =max$

S + S_h =max

(ID 3440)

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

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

(ID 3442)

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

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

(ID 3443)

$P(E)=P(\bar{E})e^{-\lambda_0(E-\bar{E})^2/2}$

P(E)=P(\bar{E})e^{-\lambda_0(E-\bar{E})^2/2}

(ID 3444)

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

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

(ID 11549)

Examples

Forming a maximum

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)

ID:(437, 0)