Why do pressure and chemical potential depend on temperature, instead of having symmetric definitions?

Question

I'm following an introduction to statistical mechanics and have seen the following definitions for fundamental temperature, pressure and chemical potential respectively:

$$\frac{1}{\tau} := \left( \frac{\delta \sigma}{\delta U} \right)_{V,N}$$ $$p := \tau \left( \frac{\delta \sigma}{\delta V} \right)_{U,N}$$ $$\mu := -\tau \left( \frac{\delta \sigma}{\delta N} \right)_{U,V}$$

where $\sigma$ is entropy $U$ energy, $V$ volume and $N$ the number of particles.

First, I suppose the inverse in the temperature definition is there because we expect lower temperatures to draw energy. But I can't explain the factor $\tau$ in the definitions of pressure and chemical potential, nor the minus sign in the latter. This is the introduction I read:

Consider two systems A and B with entropies $\sigma_A$ and $\sigma_B$ and volumes $V_A$ and $V_B$. They are put into contact so that they can exchange volume, but not energy or particles. The maximal entropy will then be obtained at:

$$ 0 = \frac{d\sigma_{AB}}{dV_A} = \frac{d\sigma_A}{dV_A} - \frac{d\sigma_B}{dV_B} \Leftrightarrow \frac{d\sigma_A}{dV_A} = \frac{d\sigma_B}{dV_B} $$

We call this equilibrium quantity $p/\tau$. But why not just $p$?

Answer 1

This is thermodynamics in entropic formulation. The simple definitions (and more intuitive) of $\tau$,$p$ and $\mu$ are given in the energetic formulation. The entropic formulation is derived from the energetic one.

Think about $p/\tau$ as one function not as two separated entities.

In the energetic formulation U (internal energy of a system in equilibrium) is a function of $\sigma$,$V$ and $N$ and we define: $$\tau:=\frac{\partial U}{\partial \sigma}$$ $$p:=-\frac{\partial U}{\partial V}$$ $$\mu:=\frac{\partial U}{\partial N}$$

Equivalent to: $$dU=\tau d\sigma-pdV+\mu dN$$

With unjustified algebraic manipulations we obtain:

$$d\sigma=\frac{1}{\tau}dU+\frac{p}{\tau}dV-\frac{\mu}{\tau} dN$$

hence $$\frac{\partial \sigma}{\partial U}=\frac{1}{\tau}$$ $$\frac{\partial \sigma}{\partial V}=\frac{p}{\tau}$$ $$\frac{\partial \sigma}{\partial N}=-\frac{\mu}{\tau}$$