A rotating spatial basis is not an inertial frame: nonzero coordinate acceleration of a uniform motion

The basis field is $g$-orthonormal for every $s$: by the expansion and [L2], $g({\tilde{e}}_1,{\tilde{e}}_1)={\cos }^2(\omega s)+{\sin }^2(\omega s)=1$, $g({\tilde{e}}_2,{\tilde{e}}_2)={\sin }^2(\omega s)+{\cos }^2(\omega s)=1$, $g({\tilde{e}}_3,{\tilde{e}}_3)=1$ and $g({\tilde{e}}_1,{\tilde{e}}_2)=-\cos (\omega s)\sin (\omega s)+\sin (\omega s)\cos (\omega s)=0$, with the other mixed pairings vanishing; the orientation is unchanged because the determinant of the two-dimensional block is ${\cos }^2+{\sin }^2=1$ and the third axis is unchanged.

The worldline $\gamma$ with $F$-coordinate curve $x(s)=(1,0,0)$ is a worldline of the model: it is of class $C^{\mathrm{\infty }}$ and satisfies the parameter condition, and its coordinate velocity and acceleration in $F$ are $w=0$ and $a=0$; its spatial velocity vector in $F$ is $\underset{‾}{v}={\gamma }^{\mathrm{\prime }}(s)-u=0$.

Hence $R(s)\in SO(3)$ for every $s$: the columns of $R(s)$ are the coordinates of $({\tilde{e}}_1(s),{\tilde{e}}_2(s),{\tilde{e}}_3(s))$ and form a positively oriented orthonormal basis, so $R(s{)}^{\mathsf{T}}R(s)=I_3$ and $\det R(s)=1$ by [L3]; moreover $\sin$ and $\cos$ are continuous because differentiable. Inductively, if both are $C^k$, their derivatives $\cos$ and $-\sin$ are $C^k$ by [L1], so both are $C^{k+1}$. Thus they are $C^{\mathrm{\infty }}$; composing with $s\mapsto \omega s$ by [L7] gives $C^{\mathrm{\infty }}$ entries of $R$ and $R^{\mathsf{T}}$.

The rotating-basis coordinate curve is $y(s)=R(s{)}^{\mathsf{T}}x(s)=(\cos (\omega s),-\sin (\omega s),0)$, because $R(s{)}^{\mathsf{T}}$ has first column $(\cos (\omega s),-\sin (\omega s),0)$; differentiating with the chain rule [L7] and step 2.1 gives $y^{\mathrm{\prime }}(s)=\omega (-\sin (\omega s),-\cos (\omega s),0)$ and $y^{\mathrm{\prime }\mathrm{\prime }}(s)=-{\omega }^2(\cos (\omega s),-\sin (\omega s),0)=-{\omega }^{2}y(s)$, so $y^{\mathrm{\prime }\mathrm{\prime }}\equiv ̸0$ because $y$ is never the zero vector (its norm is $1$ by [L2]).

The identity of [L4] applies with $c=0$, $x^{\mathrm{\prime }\mathrm{\prime }}=0$ and, from $\mathrm{\Omega }=R^{\mathsf{T}}{R}^{\mathrm{\prime }}$ computed from the displayed $R$, with $\mathit{\omega }$ the constant vector $(0,0,\omega )$; by steps 2.1 and 3.1 the two sides read $R^{\mathsf{T}}{x}^{\mathrm{\prime }\mathrm{\prime }}=0$ and $y^{\mathrm{\prime }\mathrm{\prime }}+2\mathit{\omega }\times y^{\mathrm{\prime }}+{\mathit{\omega }}^{\mathrm{\prime }}\times y+\mathit{\omega }\times (\mathit{\omega }\times y)=0$, since $2\mathit{\omega }\times y^{\mathrm{\prime }}=2{\omega }^{2}y$, ${\mathit{\omega }}^{\mathrm{\prime }}\times y=0$ and $\mathit{\omega }\times (\mathit{\omega }\times y)=-{\omega }^{2}y$, so the displayed identity is satisfied on this example. Every term has acceleration dimension $\mathrm{\ell }{\tau }^{-2}$; the scalar $\omega$ retains the inverse-time dimension declared in the Example.

The result $y^{\mathrm{\prime }\mathrm{\prime }}\equiv ̸0$ of step 3.1, together with $x^{\mathrm{\prime }\mathrm{\prime }}\equiv 0$ in the inertial frame $F$, shows that the rotating basis is not an inertial frame of the model: were it inertial, its coordinate acceleration of the same worldline would have to vanish identically by clause 3 of [L5], contradicting step 3.1; equivalently, an inertial frame of the model has a constant spatial basis by [L5], and the basis field ${\tilde{e}}_j(s)$ is not constant. The example therefore delimits the frame assumptions: a time-dependent change of basis preserves simultaneity and the time coordinate but does not preserve the uniform-motion property, and the two extra terms displayed in step 4.1 are kinematic coordinate effects in this basis, not a force law.