Kerr geometry and hidden symmetry — Examples

Setting $a=0$ gives $r_{+}=2M$, $\mathrm{\Sigma }=r^2$, $\mathrm{\Delta }=r(r-2M)$, and $g_{t\varphi }=0$. The remaining components are $g_{tt}=-(1-2M\mathrm{/}r)$, $g_{rr}=(1-2M\mathrm{/}r{)}^{-1}$, $g_{\theta \theta }=r^2$, and $g_{\varphi \varphi }=r^2{\sin }^2\theta$. These are exactly [F2] with $s=2m_{\ast }=2M$ and the same domain $r>2M$.

Since $\mathrm{\Phi }$ is Killing by [F3] and [F8], set $K_{ab}^{\mathrm{\Phi }}:={\mathrm{\Phi }}_a{\mathrm{\Phi }}_b$. The product rule gives ${\mathrm{\nabla }}_{(a}{K}_{bc)}^{\mathrm{\Phi }}=\frac{1}{3}[{\mathrm{\Phi }}_b({\mathrm{\nabla }}_a{\mathrm{\Phi }}_c+{\mathrm{\nabla }}_c{\mathrm{\Phi }}_a)+{\mathrm{\Phi }}_c({\mathrm{\nabla }}_a{\mathrm{\Phi }}_b+{\mathrm{\nabla }}_b{\mathrm{\Phi }}_a)+{\mathrm{\Phi }}_a({\mathrm{\nabla }}_b{\mathrm{\Phi }}_c+{\mathrm{\nabla }}_c{\mathrm{\Phi }}_b)]=0$. Thus $K^{\mathrm{\Phi }}$ is a reducible Killing tensor by [F4].

By [F3], ${\mathrm{\Omega }}_H=a\mathrm{/}(r_{+}^2+a^2)$, so ${\mathrm{\Omega }}_H=0$ at $a=0$. The vacuum statement follows by applying [F6] to this same metric and restricting its domain to $r>2M$.

Along an affine geodesic, $K_{ab}^{\mathrm{\Phi }}{v}^a{v}^b=({\mathrm{\Phi }}_a{v}^a{)}^2$ is constant by [F5]. This is the axial quadratic integral; it is reducible and is not identified with an irreducible Carter tensor.

Setting $a=0$ in [F7] gives $Q=p_{\theta }^2+L_z^2{\cos }^2\theta \mathrm{/}{\sin }^2\theta =p_{\theta }^2+L_z^2{\cot }^2\theta$. Since $1\mathrm{/}{\sin }^2\theta -{\cot }^2\theta =1$, this equals $L_{\mathrm{s}\mathrm{p}\mathrm{h}}^2-L_z^2$ under the definitions in the Example.

Because $T$ is spacelike by [F1], decompose it relative to $n$ as $T=\alpha n+w$, where $\alpha :=-g(T,n)$ and $w:=T+g(T,n)n\in n^{\perp }$. By [F2], if $\beta :=\sqrt{g(w,w)}$ then $\beta >\mathrm{\mid }\alpha \mathrm{\mid }$; set $e:=w\mathrm{/}\beta$, so $g(e,e)=1$, $g(n,e)=0$, and $T=\alpha n+\beta e$.

Set $\lambda :=-(n^{\mathrm{♭}}+e^{\mathrm{♭}})$ at $p$, where $n^{\mathrm{♭}}=g(n,\cdot )$ and $e^{\mathrm{♭}}=g(e,\cdot )$. Then $\lambda (n)=1$, $\lambda (e)=-1$, and $\lambda$ vanishes on the orthogonal complement of $\mathrm{span}\{n,e\}$. In the chosen chart $x:U\to x(U)$, define $\phi :U\to \mathbb{R}$ by $\phi (q):={\sum }_{a=0}^3{\lambda }_a(x^a(q)-x^a(p))$. This smooth local scalar field has $d{\phi }_p=\lambda$. Thus $g^{-1}(\lambda ,\lambda )=-1+1=0$ and $\lambda (T)=\alpha -\beta$.

By [F3], at $p$ the scalar stress tensor is $Q_{ab}=K({\lambda }_a{\lambda }_b-\frac{1}{2}{g}_{ab}{g}^{-1}(\lambda ,\lambda ))$. Using [F4] and $g^{-1}(\lambda ,\lambda )=0$, ${\rho }_T(p;n)=Q_{ab}{n}^a{T}^b=K\lambda (n)\lambda (T)=K(\alpha -\beta )<0$. Hence the stationary Killing-multiplier density is not nonnegative on all smooth scalar first-derivative data at $p$.

For any field that additionally satisfies ${\mathrm{\square }}_g\phi =0$, [F5] makes $Q$ divergence-free and [F6] makes the associated Killing current divergence-free. This is conditional; the example does not assert existence of an on-shell solution with the selected first derivative. Step 3.1 proves that this multiplier density is not nonnegative on all smooth scalar first-derivative data at $p$.