# Microstates kT and entropy

• Glenn G
Glenn G
Hi community,
I'm trying to get my head round all these concepts.

So entropy is given by S= k ln W. where W is the number of microstates of a system. Know let's imagine there is a box containing a number of gas atoms let's say the gas atoms have a current position and velocity and say you can take a snapshot of this and record this data then this would correspond to a particular microstate. Let's say you could play and then repause a very small time later such that the position of the particles changes a tiny bit but their velocities have not changes as there have been no collisions and therefore transfer of momentum and kinetic energy, is this still the same microstate?? It's clearly a different microstate if the particles have collided.

I'm also intrigued by the term kT, has units of J per K, I read that in some senses this can act as a unit of measurement for energy levels in a system. Does it have anything to do with the spacing of energy levels i.e. The discrete nature of available energy levels or not?

Have a great weekend,
Glenn Gooch.

Stephen Tashi

Homework Helper
Gold Member
I can give a simple input here, but it is difficult to elaborate much. The microstates are counted as the volume of space (in momentum space and coordinate space) and the number of states in space ## \Delta^3 \vec{p} ## and ## \Delta^3 \vec{x} ## is ## \Delta n=(\Delta^3 \vec{p} \Delta^3 \vec{x})/h^3 ## where ## h ## is Planck's constant. This is for a single particle. For a number ## N ## of indistinguishable particles, you take the ## N ##th power of this divided by ## N! ## to count the states. If a particle is at a different location, it of course counts as a different state. ## \\ ## One additional comment is in statistical physics the partition function ## Z ## is often computed. If ## \zeta ## is the partition function for a single particle, ## Z ## for ## N ## particles is ## Z=\frac{\zeta^N}{N!} ##. The partition function for a single particle ## \zeta = \Sigma \, e^{\frac{-E_s}{kT}} ## summed over all states of the single particle. For the case of momentum and position states in a container (in the gaseous state), the sum over all states can be performed as an integral in momentum and coordinate space.

Glenn G
Glenn G
Thanks Charles that's very interesting.

Does this mean that seeing as the granularity in the change in position is the Planck's length (is this correct?) that even if only one particle moves by one Planck length then that's a different microstate? and that if this displacement can be in any random directions then wouldn't this mean that we are introducing an infinite number of possible microstates just for considering this one change OR seeing as there can't be an infinite number of microstates that the possible directional changes are also quantised (is there any such thing as the Planck angle?).

Homework Helper
Gold Member
Planck's constant has the dimensionality of ## (\Delta \vec{p})( \Delta \vec{x}) ##. It appears in Heisenberg's uncertainty principle as ## (\Delta p)( \Delta x) \geq \frac{\hbar}{2} ## where ## \hbar=\frac{h}{2 \pi} ##. If the direction changes, this is a change in ## \vec{p} ##. ## \\ ## Additional note: Angular momentum (mvr) is also quantized in units of ## \hbar ## or ## \hbar/2 ##. ## \\ ## One more item: If you are interested in the partition functions of for the gas of ## N ## particles, the book Fundamentals of Statistical and Thermal Physics by F. Reif has a good section about it. The single particle partition function is ## \zeta=\frac{V}{h^3}(2 \pi m kT)^{\frac{3}{2}} ##. F. Reif computes it with a rather straightforward integration in x and p space.

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