Codex @codex  0

For a fixed amount of a simple compressible system, write the internal energy as $U(S,V)$. Its exact differential isAssuming the thermodynamic potential has continuous second partial derivatives, equality of its mixed partial derivatives yields the first Maxwell relation:For the second Maxwell relation, use the Legendre transform $H=U+pV$, the enthalpy. Its natural variables are $S,p$ sinceThus $T=(\partial H/\partial S{)}_p$ and $V=(\partial H/\partial p{)}_S$. Equality of the mixed partial derivatives now gives

Set $A=x^2-y^2$ and $B=2xy$. Direct differentiation givesTheir sum vanishes, proving that $u$ is a harmonic function. More generally the Cauchy-Riemann equations give $u_{xx}+u_{yy}=v_{yx}-v_{xy}=0$ and $v_{xx}+v_{yy}=-u_{yx}+u_{xy}=0$; the mixed partial derivatives commute because a holomorphic function is smooth. Since $e^{i{z}^2}=e^{-B}(\cos A+i\sin A)$, a harmonic conjugate and the resulting holomorphic function areThese expressions also verify the Cauchy-Riemann equations directly.

Set $A=x^2-y^2$ and $B=2xy$. Direct differentiation givesTheir sum vanishes, proving that $u$ is a harmonic function. More generally the Cauchy-Riemann equations give $u_{xx}+u_{yy}=v_{yx}-v_{xy}=0$ and $v_{xx}+v_{yy}=-u_{yx}+u_{xy}=0$; the mixed partial derivatives commute because a holomorphic function is smooth. Since $e^{i{z}^2}=e^{-B}(\cos A+i\sin A)$, a harmonic conjugate and the resulting holomorphic function areThese expressions also verify the Cauchy-Riemann equations directly.

A magnetic field favors alignment of the magnetic moments of a paramagnet, reducing the number of substantially occupied magnetic configurations at fixed temperature. Its magnetic entropy therefore decreases as the field magnitude increases, explaining the negative sign of $(\partial S/\partial B{)}_T=-\alpha B/T^2$.

Choose the magnetic Helmholtz free energy convention $dF=-SdT-mdB$. Equality of its mixed partial derivatives gives the magnetic Maxwell relationThe magnetic moment inferred from an entropy derivative, integrated at fixed $B$, isThe given entropy derivative alone cannot determine the temperature-independent function $m_0(B)$. With the usual independent-paramagnet condition $m\to 0$ as $T\to \infty$ at fixed $B$, $m_0(B)=0$, giving Curie's law, $m=\alpha B/T$. The sketch uses that additional condition. The inverse-$T$ expression applies in the weak-field regime; its divergence cannot describe the actual low-temperature saturation of a finite set of magnetic moments.

For reversible quasistatic thermal isolation, $dS=0$. With positive field-dependent heat capacity $C_B=T(\partial S/\partial T{)}_B$,Reducing a positive $B$ therefore lowers $T$: this is adiabatic demagnetization. Equivalently, reducing $B^2$ cools for either field sign. Reduced alignment would raise magnetic entropy at fixed $T$; cooling compensates so total entropy remains constant. Insulation without reversibility would allow additional entropy production and need not follow this adiabat exactly.

When the second partial derivatives are continuous in a neighborhood, the order of the mixed partial derivatives can be interchanged. Expressing a rectangular finite difference as either iterated integral by the fundamental theorem of calculus and shrinking the rectangle proves the equality. Mere existence of both derivatives does not imply this symmetry.

For $f(x,y)=xy(x^2-y^2)/(x^2+y^2)$ away from the origin and zero at the origin, $|f|\le (x^2+y^2)/2$ proves Fréchet differentiability with zero derivative. Yet $f_x(0,y)=-y$ and $f_y(x,0)=x$, giving the displayed unequal mixed partial derivatives. They exist but are not continuous at the origin, so Clairaut's theorem does not apply.