# Is $d U = 0$ at thermodynamic equilibrium?

At thermodynamic equilibrium, there is thermal, mechanical, and diffusive equilibrium. Does this imply:

$$d\mu = dT = dV = 0$$

$$dU = TdS - PdV \implies dU = TdS$$

Here, I know entropy is maximum, so perhaps $$dS=0$$ and hence $$dU = 0$$? I also don't think I can write $$d\mu = 0$$ as this may be sort of abusive treating of the chemical potential.

• If you are just sitting at equillibrium it would be surprising for anything to be changing, so $dX = 0$, for all $X$. None zero differentials impy change, which implies some sort of process is taking place. – By Symmetry May 2 at 12:04

So yes, at thermodynamic equilibrium $$\Delta U=0$$.