Heat bath and canonical ensembles

[WWCY]

Hi all, I have encountered the idea of a heat bath but am slightly perplexed as to what it is.

There was a textbook example that looked to find the number density expression for gas molecules as a function of position (image below). It then said that the probability $P(z)$ of finding the particle at height $z$ was given by
$$P(z) \propto e^{- mgz / k_B T}$$

a) Does this not mean that the particle is drawing gravitational potential energy from the heat bath? What sort of "object" would this heat bath be?

b) I might be being slightly pedantic here but isn't the probability of an event occurring at any point $z$ equal to $0$ for continuous distributions? If so, is the "proper" way of obtaining the expression to consider the probability in the interval $[z, z+dz]$?

Many thanks in advance!

 

[Andrew Mason]

Hi all, I have encountered the idea of a heat bath but am slightly perplexed as to what it is.

There was a textbook example that looked to find the number density expression for gas molecules as a function of position (image below). It then said that the probability $P(z)$ of finding the particle at height $z$ was given by
$$P(z) \propto e^{- mgz / k_B T}$$

a) Does this not mean that the particle is drawing gravitational potential energy from the heat bath? What sort of "object" would this heat bath be?

b) I might be being slightly pedantic here but isn't the probability of an event occurring at any point $z$ equal to $0$ for continuous distributions? If so, is the "proper" way of obtaining the expression to consider the probability in the interval $[z, z+dz]$?

Many thanks in advance!View attachment 237713

I am not sure what you are referring to as a heat bath. The idea is to assume a column of the atmosphere all at the same temperature (so the speed distribution of the molecules follows a Maxwell-Boltzmann speed distribution). The molecules with the highest kinetic energy will be the ones that reach the highest elevation. So the author is just substituting mgz for mv^2/2 in the expression for the speed distribution.

AM

 

[WWCY]

Hi Andrew, thanks for the response

I am not sure what you are referring to as a heat bath. The idea is to assume a column of the atmosphere all at the same temperature (so the speed distribution of the molecules follows a Maxwell-Boltzmann speed distribution). The molecules with the highest kinetic energy will be the ones that reach the highest elevation. So the author is just substituting mgz for mv^2/2 in the expression for the speed distribution.

AM

In the book I'm reading, it states that the speed distribution was derived from the assumption that each particle constitutes a system, and is in thermal contact to a "heat bath" at constant T, in the form of the gas molecules around it. Collisions with such particles would transfer energy from the "heat bath" to the system/particle.

In the case of the above example, each gas particle is able to exchange gravitational potential energy. But what would be the equivalent of the "heat bath" with which this energy is exchanged, in such a scenario?

 

[Lord Jestocost]

In the case of the above example, each gas particle is able to exchange gravitational potential energy. But what would be the equivalent of the "heat bath" with which this energy is exchanged, in such a scenario?

The heat bath are - so to speak - all other molecules in the column of gas. A nice visualization can be found in the Feynman Lectures, chapter 40-1: http://www.feynmanlectures.caltech.edu/I_40.html

 

[WWCY]

Thanks for your response

The heat bath are - so to speak - all other molecules in the column of gas. A nice visualization can be found in the Feynman Lectures, chapter 40-1: http://www.feynmanlectures.caltech.edu/I_40.html

In this case, how does a gas particle "exchange" gravitational potential energy with other gas particles?

 

[Andrew Mason]

Thanks for your response
In this case, how does a gas particle "exchange" gravitational potential energy with other gas particles?

You don't have to have the molecules exchanging potential energies. The distribution of molecular speeds is derived from kinetic theory. That is all you need. The gas molecules can exchange kinetic energies through elastic collisions, but it doesn't matter because the molecules are treated as being indistinguishable.

AM

 

[Lord Jestocost]

In this case, how does a gas particle "exchange" gravitational potential energy with other gas particles?

Gravitational potential energy arises from the gravitational interaction between the gas molecules and the Earth. It is – so to speak – no attribute of the gas molecules themselves. It changes when a molecule is moving upwards or downwards in the gravitational field. Due to energy conservation, this leads to corresponding change in the kinetic energy of the considered molecule which can be transferred to other molecules through elastic collisions.

 

[WWCY]

I think I'm beginning to get it, many thanks!

 

[Philip Koeck]

Hi all, I have encountered the idea of a heat bath but am slightly perplexed as to what it is.

There was a textbook example that looked to find the number density expression for gas molecules as a function of position (image below). It then said that the probability $P(z)$ of finding the particle at height $z$ was given by
$$P(z) \propto e^{- mgz / k_B T}$$

a) Does this not mean that the particle is drawing gravitational potential energy from the heat bath? What sort of "object" would this heat bath be?

b) I might be being slightly pedantic here but isn't the probability of an event occurring at any point $z$ equal to $0$ for continuous distributions? If so, is the "proper" way of obtaining the expression to consider the probability in the interval $[z, z+dz]$?

Many thanks in advance!View attachment 237713

I wouldn't say you are pedantic at all, but the textbook you quote is definitely sloppy.
The probability that a molecule is exactly at height z is zero.
One should always state an interval.
So n(z) dz is the number density of molecules at heights between z and z + dz.
This turns out to be very important for example when you convert the Boltzmann distribution from f(E) dE to f(v) dv for example. Some textbooks get that wrong!

I would go for the "pedestrian derivation" starting after equation 4.21. That makes more sense.
However, the connection to canonical ensembles is interesting.