Flat metric components in a spherical coordinate chart

$\mathrm{\Phi }$ is smooth on the open set $W$ as a tuple of polynomials and trigonometric functions, and its differential has the matrix with rows obtained by differentiating the four coordinate functions; a direct computation of the $4\times 4$ determinant gives $\det D\mathrm{\Phi }(u)=r^2\sin \vartheta$, which is nonzero on $W$ because $r>0$ and $0<\vartheta <\pi$.

$\mathrm{\Phi }$ is injective on $W$: if $\mathrm{\Phi }(u)=\mathrm{\Phi }(u^{\mathrm{\prime }})$, then $(u^1{)}^2=(\mathrm{\Phi }(u{)}^1{)}^2+(\mathrm{\Phi }(u{)}^2{)}^2+(\mathrm{\Phi }(u{)}^3{)}^2=(u^{\mathrm{\prime }1}{)}^2$, so $u^1=u^{\mathrm{\prime }1}>0$; then $u^1\cos u^2=\mathrm{\Phi }(u{)}^3=u^1\cos u^{\mathrm{\prime }2}$ gives $\cos u^2=\cos u^{\mathrm{\prime }2}$ with $u^2,u^{\mathrm{\prime }2}\in (0,\pi )$, hence $u^2=u^{\mathrm{\prime }2}$; then $\sin u^2>0$ and the pair of first two spatial components gives $(\cos u^3,\sin u^3)=(\cos u^{\mathrm{\prime }3},\sin u^{\mathrm{\prime }3})$ with $u^3,u^{\mathrm{\prime }3}\in (0,2\pi )$, hence $u^3=u^{\mathrm{\prime }3}$; finally $u^0=u^{\mathrm{\prime }0}$ from the first component.

Let $z\in \mathrm{\Phi }(W)$ and take a single $u\in W$ with $\mathrm{\Phi }(u)=z$. Step 1.1 and [L2] supply open neighbourhoods $u\in O\subseteq W$ and $z\in V=\mathrm{\Phi }(O)$ on which $\mathrm{\Phi }{\mathrm{\mid }}_O$ has a smooth inverse. In particular $z\in V\subseteq \mathrm{\Phi }(W)$; since every image point has such an open neighbourhood, $\mathrm{\Phi }(W)$ is open. By step 1.2, the inverse on this one $V$ equals the unique set-theoretic inverse $y={\mathrm{\Phi }}^{-1}$ restricted to $V$. Consequently $y$ is smooth near each image point, hence smooth on $\mathrm{\Phi }(W)$, without selecting neighbourhoods simultaneously. It and its smooth inverse $\mathrm{\Phi }$ form a homeomorphism with smooth transitions to the identity chart, so [L3] gives the asserted smooth chart.

Apply the transformation law [L1] with the chart $y$ and $x(u):=\mathrm{\Phi }(u)$, so that $\frac{\mathrm{\partial }{x}^c}{\mathrm{\partial }{y}^a}(u)=D\mathrm{\Phi }(u{)}^c{}_a$. The first column of $D\mathrm{\Phi }$ is $(1,0,0,0)$, so $A_{a0}^{\mathrm{\prime }}=A_{0a}^{\mathrm{\prime }}=-1$ for $a=0$ and $0$ for $a\mathrm{\ne }0$. The remaining columns are the partial derivatives of $(r\sin \vartheta \cos \phi ,r\sin \vartheta \sin \phi ,r\cos \vartheta )$ with respect to $r,\vartheta ,\phi$, whose squared Euclidean norms are $1$, $r^2$ and $r^2{\sin }^2\vartheta$, and whose pairwise Euclidean inner products are $0$; since $A_{cd}=\mathrm{diag}(-1,1,1,1)$ and the spatial indices carry the positive signs, the transformation law gives $A_{11}^{\mathrm{\prime }}=1$, $A_{22}^{\mathrm{\prime }}=r^2$, $A_{33}^{\mathrm{\prime }}=r^2{\sin }^2\vartheta$ and $A_{ab}^{\mathrm{\prime }}=0$ for the remaining off-diagonal pairs.

The component functions of step 3.1 are smooth on $W$, so by [L1] the transformed family is a smooth covariant $2$-tensor field on $\mathrm{\Phi }(W)$; its component matrix has determinant $-r^4{\sin }^2\vartheta \mathrm{\ne }0$ on $W$ and diagonal entries of both signs, with exactly one negative entry, so the form has inertia $(3,1,0)$ and the chart expression defines a Lorentzian metric on this chart domain. This shows that a chart change without a change of the field produces non-constant components and that the Lorentzian character is preserved by the transformation law of [L1].