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Why is the chemical potential negligible at equilibrium?

  1. Oct 8, 2014 #1
    I am currently reading cosmology and I read the statement "... chemical
    potential which is nearly always negligible when processes are in equilibrium."

    So the current understanding I got of chemical potential is through the law:
    $$dE = - pdV + TdS + \mu dN.$$

    Hence for processes which occur at constant volume and without heat transfer
    $$\frac{\partial E}{\partial N} = \mu$$
    indicating that $$\mu$$ is the work done on the system in transferring one particle to it.

    Now, why is this energy negligible at equilibrium?
  2. jcsd
  3. Oct 8, 2014 #2
    The statement is not quite correct. What they are trying to say depends on the context. But, if they are talking about equilibrium between phases, they meant to say that the chemical potential of a species is the same in all physical phases if the system is equilibrium. So the difference in chemical potential of a species between phase A and phase B is zero.

  4. Oct 9, 2014 #3
    Thanks. Would you happen to know when -- and the reasons why -- one might be able to neglect the chemical potential? For example: I have read that for ultra relativistic fluids one can neglect it (but I have not found any reasons for this).
  5. Oct 9, 2014 #4
    Sorry. Ultra relativistic fluid are out of my area of expertise. But, whenever something is negligible, it has to be negligible compared to something else. This is particularly relevant to chemical potential, since chemical potential is not an absolute quantity, but is reckoned relative to some reference state of the material. Maybe they are talking about the effects of changes in chemical potential (or variations in chemical potential) being negligible.

  6. Oct 11, 2014 #5


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    In general that's not enirely true. One relativistic fluid we investigate today is the matter created in ultrarelativistic heavy-ion collisions. There two atomic nuclei are smashed together at high energies (e.g., two gold nuclei at the Relativistic Heavy Ion Collider at the Brookhaven National Lab or two lead nuclei at the Large Hadron Collider at CERN, Geneva), and a lot of particles and antiparticles are created. At the highest energies a socalled Quark-Gluon Plasma is created at huge temperatures of about 500 MeV (usually we set the Boltzmann constant to 1 and measure temperatures in units of energy), which rapidly expands and cools down. What's measured are the hadrons which are formed from the quarks and gluons.

    This fireball can to a quite high precision be described as the collective fluid-like motion, using relativistic (ideal or to better precision viscous) hydrodynamics. In relativistic equilibrium thermodynamics the chemical potential has a slightly different meaning than in the non-relavivistic case. You must use it as a Lagrange parameter to keep the average of some conserved quantity rather than particle number. In this case this is the net-baryon number (being +1 for baryons and -1 for anti-baryons). The higher the collision enegy the more particles are newly created in the collision, and in each process you must produce as many baryons as anti-baryons. The small amount of baryons already present in the colliding nuclei then becomes negligible compared to these many baryon-anti-baryon pairs, and thus the baryo-chemical potential is close to 0 and the higher the initial temperature becomes.

    To the contrary, at not so high collision energies, the baryo-chemical potential is larger to keep the surplus of baryons compared to anti-baryons correct in the thermodynamic description. This is a regime being investigated at RHIC in the beam-energy-scan program right now and is the motivation to build a new accelerator here in Germany at the Helmholtz Heavy-Ion Research Center (GSI) in Darmstadt.
  7. Oct 12, 2014 #6


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    Interesting, Hendrik. I remember that in protons there are also much more sea quark-antiquark pairs than the three constituent quarks. So is there a mu=0 approximation possible also for elementary particles?
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