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valerio
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Not really.

You want to calculate the probability to find all the molecules in the left half of the volume.

Since the probability density is

$$\rho(\{q,p\}) = c_N \frac{e^{-\beta H(\{q,p\})}}{\Omega(U,V,N)}\ ,$$

where, $c_N$ is a constant and, to ensure the normalization, $$\Omega(U,V,N ) = c_N\int e^{-\beta H(\{q,p\})} dq^{3N} dp^{3N}$$

what you need to consider is

$$\text{Pr} =\int_{\{q,p\}_{V/2}} \rho (\{q,p\}) dq^{3N} dp^{3N} = c_N \frac{ \int_{\{q,p\}_{V/2}} e^{-\beta H(\{q,p\})} dq^{3N} dp^{3N}}{\Omega(U,V,N)}$$

where $\{q,p\}_{V/2}$ is the set of coordinates such that all the molecules are in the left half of the volume.

You can immediately see that

$$c_N \int_{\{q,p\}_{V/2}} e^{-\beta H(\{q,p\})} dq^{3N} dp^{3N} = \Omega(N,V/2,U)$$

So that you finally obtain the desired result.

Not really.

You want to calculate the probability to find all the molecules in the left half of the volume.

Since the probability density is

$$\rho(\{q,p\}) = c_N \frac{e^{-\beta H(\{q,p\})}}{\Omega(U,V,N)}\ ,$$

where, to ensure the normalization, $$\Omega(U,V,N ) = c_N\int e^{-\beta H(\{q,p\})} dq^{3N} dp^{3N}$$

what you need to consider is

$$\text{Pr} =\int_{\{q,p\}_{V/2}} \rho (\{q,p\}) dq^{3N} dp^{3N} = c_N \frac{ \int_{\{q,p\}_{V/2}} e^{-\beta H(\{q,p\})} dq^{3N} dp^{3N}}{\Omega(U,V,N)}$$

where $\{q,p\}_{V/2}$ is the set of coordinates such that all the molecules are in the left half of the volume.

You can immediately see that

$$c_N \int_{\{q,p\}_{V/2}} e^{-\beta H(\{q,p\})} dq^{3N} dp^{3N} = \Omega(N,V/2,U)$$

So that you finally obtain the desired result.

Not really.

You want to calculate the probability to find all the molecules in the left half of the volume.

Since the probability density is

$$\rho(\{q,p\}) = c_N \frac{e^{-\beta H(\{q,p\})}}{\Omega(U,V,N)}\ ,$$

where, $c_N$ is a constant and, to ensure the normalization, $$\Omega(U,V,N ) = c_N\int e^{-\beta H(\{q,p\})} dq^{3N} dp^{3N}$$

what you need to consider is

$$\text{Pr} =\int_{\{q,p\}_{V/2}} \rho (\{q,p\}) dq^{3N} dp^{3N} = c_N \frac{ \int_{\{q,p\}_{V/2}} e^{-\beta H(\{q,p\})} dq^{3N} dp^{3N}}{\Omega(U,V,N)}$$

where $\{q,p\}_{V/2}$ is the set of coordinates such that all the molecules are in the left half of the volume.

You can immediately see that

$$c_N \int_{\{q,p\}_{V/2}} e^{-\beta H(\{q,p\})} dq^{3N} dp^{3N} = \Omega(N,V/2,U)$$

So that you finally obtain the desired result.

Source Link
valerio
  • 16.5k
  • 1
  • 49
  • 87

Not really.

You want to calculate the probability to find all the molecules in the left half of the volume.

Since the probability density is

$$\rho(\{q,p\}) = c_N \frac{e^{-\beta H(\{q,p\})}}{\Omega(U,V,N)}\ ,$$

where, to ensure the normalization, $$\Omega(U,V,N ) = c_N\int e^{-\beta H(\{q,p\})} dq^{3N} dp^{3N}$$

what you need to consider is

$$\text{Pr} =\int_{\{q,p\}_{V/2}} \rho (\{q,p\}) dq^{3N} dp^{3N} = c_N \frac{ \int_{\{q,p\}_{V/2}} e^{-\beta H(\{q,p\})} dq^{3N} dp^{3N}}{\Omega(U,V,N)}$$

where $\{q,p\}_{V/2}$ is the set of coordinates such that all the molecules are in the left half of the volume.

You can immediately see that

$$c_N \int_{\{q,p\}_{V/2}} e^{-\beta H(\{q,p\})} dq^{3N} dp^{3N} = \Omega(N,V/2,U)$$

So that you finally obtain the desired result.