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Proof: Introduce an additional heat reservoir at an arbitrary temperature T₀, as well as N cycles with the following property: the j-th such cycle operates between the T₀ reservoir and the T_j reservoir, transferring energy dQ_j to the latter. From the above definition of temperature, the energy extracted from the T₀ reservoir by the j-th cycle is

d Q 0 , j = T 0 d Q j T j {\displaystyle dQ_{0,j}=T_{0}{\frac {dQ_{j}}{T_{j}}}\,\!}

Now consider one cycle of the heat engine, accompanied by one cycle of each of the smaller cycles. At the end of this process, each of the N reservoirs have zero net energy loss (since the energy extracted by the engine is replaced by the smaller cycles), and the heat engine has done an amount of work equal to the energy extracted from the T₀ reservoir,

W = ∑ j = 1 N d Q 0 , j = T 0 ∑ j = 1 N d Q j T j {\displaystyle W=\sum _{j=1}^{N}dQ_{0,j}=T_{0}\sum _{j=1}^{N}{\frac {dQ_{j}}{T_{j}}}\,\!}

If this quantity is positive, this process would be a perpetual motion machine of the second kind, which is impossible. Thus,

∑ i = 1 N d Q i T i ≤ 0 {\displaystyle \sum _{i=1}^{N}{\frac {dQ_{i}}{T_{i}}}\leq 0\,\!}

Now repeat the above argument for the reverse cycle. The result is

∑ i = 1 N d Q i T i = 0 {\displaystyle \sum _{i=1}^{N}{\frac {dQ_{i}}{T_{i}}}=0\,\!} (reversible cycles)

In mathematics, it is often desireable to express a functional relationship f ( x ) {\displaystyle f(x)\,} as a different function, whose argument is the derivative of f , rather than x . If we let y=df/dx  be the argument of this new function, then this new function is written f ⋆ ( y ) {\displaystyle f^{\star }(y)\,} and is called the Legendre transform of the original function.

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References

δQh is heat, δWx is irreversible work, so TdS=δQh+δWx. If both are zero, then dS=0. For the 3 possible walls, "T ins" means thermally insulated, and "No IW" means no irreversible work. "External" specifies the region external to the system.