Finding beta for the boltzman distribution.

[center o bass]

Hello! I'm trying to do a satisfactory derivation of the Boltzmann distribution. By using lagrange multipliers I've come as far as to prove that

$$P(i) = \frac{1}{Z} e^{-\beta E(i)}$$
where
$$Z = \sum_i e^{-\beta E(i)},$$

but how does one actually establish that
$$\beta = 1/kT?$$

 

[vanhees71]

Take a monatomic ideal gas and derive the mean energy,

$$U=-\frac{\partial \ln Z}{\partial \beta}$$

and compare with the definition of the temperature,

$$U=\frac{3}{2} N k T.$$

 

[Shunyata]

I think that Leonard Susskind does this derivation in one of his lectures on statistical mechanics that is available on youtube.

http://www.youtube.com/watch?v=H1Zbp6__uNw"

 

[center o bass]

Take a monatomic ideal gas and derive the mean energy

Ah, yes that is certainly a way to go. But how could that result possibly be general? Doesn't the distribution apply to any combination of systems who shares a total energy E and a number of particles N?

 

[PJC01]

Hello! It's great that you are working on deriving the Boltzmann distribution. The Boltzmann distribution is a fundamental concept in statistical mechanics and plays a crucial role in understanding the behavior of systems at the microscopic level.

To establish that \beta = 1/kT, we need to consider the thermodynamic definition of temperature. Temperature is defined as the measure of the average kinetic energy of the particles in a system. In statistical mechanics, this is related to the average energy of the system, which is given by the Hamiltonian H.

In the Boltzmann distribution, we have P(i) = \frac{1}{Z} e^{-\beta E(i)}, where P(i) is the probability of a particle being in a specific state i with energy E(i), and Z is the partition function.

To find the average energy of the system, we need to take the sum over all states i, weighted by their respective energies:

<H> = \sum_i P(i) E(i) = \frac{1}{Z} \sum_i e^{-\beta E(i)} E(i)

Using the definition of Z, we can rewrite this as:

<H> = \frac{1}{Z} \sum_i e^{-\beta E(i)} E(i) = \frac{1}{Z} \frac{\partial}{\partial \beta} \sum_i e^{-\beta E(i)} = -\frac{\partial}{\partial \beta} \ln Z

Now, using the definition of temperature as \frac{1}{kT} = \frac{\partial}{\partial \beta} \ln Z, we can rewrite this as:

<H> = -kT \frac{\partial}{\partial \beta} \ln Z

Comparing this with the definition of average energy in thermodynamics, <H> = kT, we can conclude that \beta = 1/kT.

I hope this helps in establishing the relationship between \beta and temperature in the Boltzmann distribution. Keep up the good work!