# Constant shift in the Gibbs free energy

As far as I know, measurable thermodynamic properties (pressure, temperature, volume, etc.) all depend on derivatives of the thermodynamic potentials. For instance,

$$1/T= (\partial S / \partial E)_{V,N}$$

and similar relations are available for all thermodynamic potentials and measurable quantities. The value of the thermodynamic potential itself should be irrelevant, because it depends on the zero point of the potential energy which, again, is an arbitrary choice.

I am, however, confused by the Gibbs free energy $$G$$. On one hand, the chemical potential is given by

$$\mu = (\partial G / \partial N)_{p,T}$$

Om the other hand, it can be proven that $$\mu$$ is the per-particle Gibbs free energy:

$$\mu = G/N$$

Now, if I add a constant $$G_0$$ to the Gibbs free energy, that clearly vanishes when taking the derivative. However, the constant matters in the chemical potential which now becomes $$\mu = G/N + G_0/N$$ so there will be an $$N$$-dependent term added. How do I reconcile the fact that the derivatives of the thermodynamic potentials matter with the above dependence?

• Nov 23, 2023 at 1:23

When you add a constant, the formula: $$\mu = \frac{G}{N}$$ is not valid anymore. Indeed, it uses the fact that $$G$$ is extensive: $$G(\lambda N,T,P) = \lambda G(N,T,P)$$ When you add your constant, you lose extensivity.
By large deviation arguments, you'd expect that extensivity holds in the thermodynamic limit, so it is not physically relevant to add this extra constant. You can reconcile the approaches if you assume that it is not a constant but: $$G_0 = Ng_0$$ with $$g_0$$ a constant.