Verified Kerr-Newman family first law and Smarr identity

Each point of $\mathcal{P}$ either has $a^2+q^2>0$ or lies on the Schwarzschild slice $m>0$, $a=q=0$. In the first case, $m>0$ and $m^2>a^2+q^2>0$, so [F2] applies with $M=m$. On that entire slice, [F3] supplies the exact Schwarzschild horizon data. Thus the geometric functions used in the proposition agree with the selected-model horizon data everywhere on $\mathcal{P}$.

Put $s=\sqrt{m^2-a^2-q^2}>0$, so $r_{+}=m+s$ and $r_{-}=m-s$. Then $\mathrm{\Delta }(r):=r^2-2mr+a^2+q^2=(r-r_{+})(r-r_{-})$, and $\mathrm{\Delta }(r_{+})=0$. Differentiating this identity on the open parameter domain gives $(r_{+}-m)d{r}_{+}=r_{+}dm-ada-qdq\mathrm{.}$ Also $dA=8\pi (r_{+}d{r}_{+}+ada)$ and $dj=adm+mda$.

Multiply the proposed first-law right side by $D$ and substitute [F1]: $D(\frac{{\kappa }_{\mathrm{K}\mathrm{N}}}{8\pi }dA+{\mathrm{\Omega }}_{\mathrm{K}\mathrm{N}}dj+{\mathrm{\Phi }}_{\mathrm{K}\mathrm{N}}dq)=(r_{+}-m)(r_{+}d{r}_{+}+ada)+a(adm+mda)+q{r}_{+}dq\mathrm{.}$ Using step 1.2 to replace $r_{+}(r_{+}-m)d{r}_{+}$ by $r_{+}^{2}dm-a{r}_{+}da-q{r}_{+}dq$, the $da$ and $dq$ coefficients cancel, leaving $(r_{+}^2+a^2)dm=Ddm$. Since $D>0$, division yields the first identity.

For the Smarr identity, [F1] gives $D(\frac{{\kappa }_{\mathrm{K}\mathrm{N}}A}{4\pi }+2{\mathrm{\Omega }}_{\mathrm{K}\mathrm{N}}j+{\mathrm{\Phi }}_{\mathrm{K}\mathrm{N}}q)=(r_{+}-m)D+2m{a}^2+q^2{r}_{+}\mathrm{.}$ The horizon equation $\mathrm{\Delta }(r_{+})=0$ gives $q^2=2m{r}_{+}-r_{+}^2-a^2$. Substitution and expansion reduce the right side to $(r_{+}-m)(r_{+}^2+a^2)+2m{a}^2+r_{+}(2m{r}_{+}-r_{+}^2-a^2)=m(r_{+}^2+a^2)=mD\mathrm{.}$ Division by $D>0$ proves the Smarr identity.

Fix $(a_0,q_0)$ and let $I_{a_0,q_0}=(\sqrt{a_0^2+q_0^2},\mathrm{\infty })$ with the parametrization $m\mapsto (m,a_0,q_0)$. This is a nonempty connected open interval in the strict family, where ${\kappa }_{\mathrm{K}\mathrm{N}}>0$ and $D>0$. Here $da=dq=0$ but $dj=a_{0}dm$, so step 2.1 gives $dm=({\kappa }_{\mathrm{K}\mathrm{N}}\mathrm{/}(8\pi ))dA+(a_0^2\mathrm{/}D)dm$. Since $D=r_{+}^2+a_0^2$, rearrangement yields $({\kappa }_{\mathrm{K}\mathrm{N}}\mathrm{/}(8\pi ))dA=(r_{+}^2\mathrm{/}D)dm$. For the displayed functions $T_b,S_{b,s_{\ast }}$ one always has $T_{b}d{S}_{b,s_{\ast }}=({\kappa }_{\mathrm{K}\mathrm{N}}\mathrm{/}(8\pi ))dA$ by direct differentiation. Their product on this fixed-$(a,q)$ subfamily is therefore the asserted weighted form; it equals $dm$ when $a_0=0$. Since ${\kappa }_{\mathrm{K}\mathrm{N}}>0$ on this nonempty interval, unequal positive $b$ give different $T_b$ there, and unequal offsets give different $S_{b,s_{\ast }}$ without changing its differential.

Now fix $(j_0,q_0)$ and set $y_{\pm }:=(q_0^2\pm \sqrt{q_0^4+4j_0^2})\mathrm{/}2$, so $y_{-}\le 0\le y_{+}=m_0^2$. Since $m>m_0\ge 0$, $m>0$ and $a=j_0\mathrm{/}m$ is smooth. The strict family inequality is equivalent to $m^4-q_0^2{m}^2-j_0^2>0$. With $y=m^2>0$ this polynomial is $(y-y_{+})(y-y_{-})$; its second factor is positive, so the inequality holds exactly when $m>m_0$. Thus ${\iota }_{j_0,q_0}$ parametrizes the whole fixed-$(j,q)$ subfamily over the nonempty connected interval $J_{j_0,q_0}$. Its first coordinate is $m$, so it is injective with nonzero differential and the first-coordinate projection is its smooth inverse onto its image. All restricted functions are smooth and ${\kappa }_{\mathrm{K}\mathrm{N}}>0$. Along this parametrization $j=m(j_0\mathrm{/}m)=j_0$ and $q=q_0$, hence $dj=dq=0$; step 2.1 yields $dm=({\kappa }_{\mathrm{K}\mathrm{N}}\mathrm{/}(8\pi ))dA$. Apply [F4] on this interval with $E(m)=m$, $C=1\mathrm{/}(8\pi )$ and the restricted $A,{\kappa }_{\mathrm{K}\mathrm{N}}$. It gives the asserted work-free product and its distinct scales and offsets. Together with step 3.1, this proves both the weighted fixed-$(a,q)$ restriction and the work-free fixed-$(j,q)$ calibration without dropping a work term or assigning a physical meaning.