# B Microstates kT and entropy

1. Jan 14, 2017

### 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.

2. Jan 14, 2017

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.

3. Jan 14, 2017

### 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?).

4. Jan 14, 2017

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.

Last edited: Jan 14, 2017
5. Jan 14, 2017

### Stephen Tashi

The situation in quantum mechanics is clearer than the situation in classical mechanics, but it may be useful to consider the situation in classical statistical mechanics - useful because it is (to me) murky.

The "number of microstates" obviously depends on how you define the microstates. As far as I can tell, there is no standard definition for microstates that would specify each microstate as a unique subset of the (detailed) state space of a system. For example, suppose the detailed state of a system is given by a vector $(x_1,x_2,...x_n)$ where the $x_i$ take values in a continuous range of real numbers and specify the position and velocity of each particle in the system. I know of no standardized definition of "microstate" that would be so specific as to say things like "microstate number 37 consists of the set of all $(x_1,x_2,...x_n)$ such that the total kinetic energy of the particles is 28 J and $(x_1,x_2,...x_n$) satisifes $( 0.3m < x_1 \le 0.3002 m, 1.7 m < x_2 \le 1.7002 m ..., .0001 m/s < x_n < .0002 m/s$."

If you are thinking of the system being in a specific state $(x_1,x_2,....x_n)$ at a given time then you are (conceptually) contradicting a basic assumption of statistical mechanics that says the system has a non-trivial probability distribution of being in different microstates. If the system is conceptualized as being in a particular microstate at a given time, its probability of being there is 1 and its probability of being in a different microstate is 0. You might be able to force the concept of probability into the situation by saying that you sample the state of the system "at a random time" in some time interval.

My impression of the concept of a microstate in classical physics is that it is analogous to the concept of "dx" in elementary calculus. We don't conceive of "dx" as have a particular numerical value. We conceive of it as something that could be given a particular value when we do an approximation.

If my view is correct, then it would be interesting to discuss this further because the final goal of conceptualizing a finite number of microstates seems to be to derrive formulas by taking a limit as the number of microstates becomes infinite and the volume of each microstate becomes smaller.