PHYS6562 F3 Daily Quaetion

Undistinguished particles

For N distinguishable particles, if at certain instant they occupy N states in the phase space, there are \(N!\) different choices of such configuration. The total volume for this system in phase space is the product of the volume for each particle in its own sub phase space.

\[\Omega_D = N! \space \omega_1 \omega_2 \omega_3 ... \omega_N\]

If the particles are now undistinguished, there is only one choice of such configuration without permutation.

\[\Omega_U = \omega_1 \omega_2 \omega_3 ... \omega_N\]

Therefore, \(\Omega_D = N!\space \Omega_U \).

For the mixing problem of black and white particles of the same pressure on both sides, if you cannot tell the difference between black and white ones, you cannot extract work from the mixing. Actully, you won’t know if they’ve mixed already or not. However, suppose the black and white particles are of different pressure, and you still cannot tell them apart. If there were a membrane which allows particles of lower pressure to penetrate through, this membrane then can tell the difference between the two sides without telling the particles apart. Then work can be extracted.

Now if a door can distinguish the black from the white particles and let through white ones, after very long time, the white particles will distribute themselves equally in the two sides, i.e. there will be \(\frac{1}{4} N\space white + 0 \space black\) in the left and \(\frac{1}{4} N \space white + \frac{1}{2} N \space black\) in the right. And we can extract work from it due to the pressure difference between the two sides.