# Equilibrium chemical potential if the population is not conserved?

The chemical potential, $$\mu$$, is equal to the free energy $$F$$, per photon.

$$F = \mu = u-Ts$$

The energy per photon is $$u=h\nu$$ and $$s$$ is the entropy per photon. Say we have a process where an incoming photon ($$\nu_{1}$$) is absorbed by a dye and then emitted at ($$\nu_{2}$$). Then at equilibrium,

$$h\nu_{1} - s_{1} = h\nu_{2} - s_{2}$$

What happens at equilibrium if the dye immediately emits two (or more?) photons, so the total number of photons is no longer conserved? Can I still write that the chemical potentials of all the photons are equal as in the equation above?

An attempt at a solution:

The change in Gibbs free energy at equilibrium is $$\Delta G=0$$. Since this is a homogeneous system, the Gibbs free energy is given by

$$\Delta G = \sum_{i} \mu_{x}x_{i} - \sum_{k} \mu_{y}y_{k}$$

Where $$x_{i}$$ is the number of particles of product and $$y$$ is the number of particles reactant. If one photon just becomes one photon then $$i=k$$, and so

$$\sum_{i} \mu_{y}y_{i} = \sum_{k} \mu_{x}x_{k} \implies \mu_{x}=\mu_{y}$$

and so the chemical potentials match. Now, if one photon becomes two, so $$2k=i$$, then

$$\sum_{k} \mu_{y}y_{k} = \sum_{2k} \mu_{x}x_{k} \implies 2 \mu_{x}=\mu_{y}$$. This implies that

$$h\nu_{y} - s_{1} = 2 \times (h\nu_{2} - s_{2})$$

at equilibrium.

• I think this is more appropriate for physics stack exchange. That said, I would say that the equation should hold for individual photons, irrespective of origin (but tbh that is a guess). Aug 26 '21 at 16:34
• I thought someone would see an equivalence to a chemical reaction here (something splitting/generating two things). Not many things in physics which suddenly become two....
– Tomi
Aug 26 '21 at 16:40
• Good point, your intuition seems pretty sound. Sounds like the blackbody radiation problem. Aug 26 '21 at 18:53

This question is either confused or simply confusing. The concept of equilibrium implies reactant and products and energy interacting so that everything remains constant and cannot be applied to a one photon scenario unless that is intended to be allegorical. For a photon to interact with anything there must be an interchange of energy and momentum in such a manner that total energy and momentum are both conserved. This means that the energy and momentum of the photon are changed so the idea that chemical potential is constant must be incorrect. Also can entropy be considered with a single particle? If so the exact energy and momentum must enter the calculation and the resultant would probably be applied to the maximum of a temperature stabilized system. An example if an inner electron is ejected from an atom by an X or gamma ray. The resulting cascade of photons as outer electrons fell into successive empty orbitals will give a spectrum and the energy of the emitted light will add to the original energy if recoil etc is added in but the entropy of the emitted photons will be greater than that of the one Xray. This can be applied to any energy flux steady state system where the driving force is the actual entropy change of energy entering and leaving the system since mass, U, H and G are all constant. At equilibrium Delta G = 0. The scenario above seems to describe a steady state not an equilibrium.

• +1 for "allegorical" Aug 30 '21 at 13:19
• So, I do understand your point about equilibrium. For a true equilibrium to exist, photons must be able to interact. However, as photons do not have any first-order interactions, they cannot be in true equilibrium. However, they are still well described by the thermodynamic properties such as chemical potential and temperature (e.g. bose-einstein distribution, and black body radiation). – lets call it quasi equilibrium.
– Tomi
Sep 3 '21 at 14:48
• However, the condition of quasi equilibrium in a photon beam must be equivalent to a statement that all photons have, in addition to temperature, an identical chemical potential. Since the chemical potential is equal to the free energy 𝐹 per photon, 𝐹=𝜇=𝑢−𝑇𝑠, where for a photon 𝑢=ℎ𝜈 and 𝑠 is the entropy per photon. Your point about defining entropy per photon - we can easily determine the entropy of radiation (en.wikipedia.org/wiki/Photon_gas), and so the entropy per photon is simply the number differential of the number of photons $s = (\delta S/ \delta N)$.
– Tomi
Sep 3 '21 at 14:49
• Your comment about the allegorical nature I do not understand. It doesn't matter the process going on. A volume of photons in equilibrium can be changed into another volume of photons in equilibrium, however, this brings about irreversibility, which must introduce entropy generation (as required by the second law of thermodynamics). Process doesn't matter. Maybe this point in your answer does solve the question....the transformed beam cannot be in equilibrium, and so we cannot equate chemical potentials....although photons can be in (quasi) equilibrium which each other...
– Tomi
Sep 3 '21 at 14:49

At equilibrium, the number of each species is constant. So if a molecule falls appart into two fragments, these fragments would recombine at the same rate.

Thermodynamics is for looking at bulk properties, not individual particles. Pressure, temperature, entropy are not defined for single particles using equilibrium thermodynamics.

Chemical reactions involving photons are most often non-equilibrium processes, so while it is nice to use analogies, they might not deliver rigorous results in this case.

This reference might be a starting point for studying how to think about the entropy of photons: https://arxiv.org/ftp/arxiv/papers/1306/1306.1993.pdf

• The entropy can be determined per photon relatively easily, and we can quite easily think of the chemical potential of an excited charge state or even a photon, so I'm not sure it's true to say that these are not defined for individual particles. My question is if a reaction goes from A->2B, if the chemical potentials are not the same, what are they? I thought using photons would be easier for analysis.... Agreed regarding the point about non-equilibrium processes. But let us pretend they are!....
– Tomi
Aug 27 '21 at 13:26