Framework, quantities and geometry — Examples

The data of the Example satisfy [L1]: $\tau$ is linear and surjective (its value is the first coordinate), $K=\{0\}\times {\mathbb{R}}^3$ has dimension three, the standard inner product on $K$ is symmetric and positive definite, the standard orientation of ${\mathbb{R}}^3$ orients $K$, and $A={\mathbb{R}}^4$ is an affine space over itself.

The constructed frame $F^{\mathrm{\prime }}$ is inertial in the sense of [L2]: $\tau (u^{\mathrm{\prime }})=1$ because $u^{\mathrm{\prime }}=(1,-\frac{1}{2},0,0)$, and the basis $(e_1^{\mathrm{\prime }},e_2^{\mathrm{\prime }},e_3^{\mathrm{\prime }})=(e_1,e_2,e_3)$ is $g$-orthonormal and positively oriented; hence $F^{\mathrm{\prime }}$ is an adapted chart with adapted coordinates $(t^{\mathrm{\prime }},x^{\mathrm{\prime }})$.

The primed coordinates of the event $p=tu+{\sum }_j{x}^j{e}_j$ are obtained by expanding $p=o^{\mathrm{\prime }}+t^{\mathrm{\prime }}{u}^{\mathrm{\prime }}+{\sum }_j{x}^{\mathrm{\prime }j}{e}_j^{\mathrm{\prime }}$: comparing components gives $t^{\mathrm{\prime }}=t+2$ and $x^{\mathrm{\prime }}=(x^1+\frac{1}{2}t,x^2+3,x^3)=Rx+vt+a$, so the two frames differ by the transformation of [L3] with the displayed parameters $(b,a,v,R)$.

The prescribed worldline $\gamma$ is of class $C^{\mathrm{\infty }}$ and its $F$-time coordinate is $\tau (\gamma (s)-o)=s$, so the parameter condition holds; its coordinate curve is $x(s)$ with coordinate velocity $w=(\frac{1}{4},0,0)$ and coordinate acceleration $\alpha (s):=x^{\mathrm{\prime }\mathrm{\prime }}(s)=0$.

The primed description of the motion is $\eta (t^{\mathrm{\prime }})=x(t^{\mathrm{\prime }}-2)+v(t^{\mathrm{\prime }}-2)+a=(\frac{3}{4}{t}^{\mathrm{\prime }}-\frac{1}{2},5,0)$; hence ${\eta }^{\mathrm{\prime }}=(\frac{3}{4},0,0)=Rw+v$ and ${\eta }^{\mathrm{\prime }\mathrm{\prime }}(t^{\mathrm{\prime }})=(0,0,0)=R\alpha (t^{\mathrm{\prime }}-b)$, so the velocity and acceleration laws of [L3] hold on this motion, and the vanishing of the coordinate acceleration is frame-independent for it.

As a cross-check, at $t=2$ the $F$-coordinates are $(\frac{3}{2},2,0)$ and the $F^{\mathrm{\prime }}$-coordinates of the same event are $\eta (4)=(\frac{5}{2},5,0)$, which equals $x(2)+2v+a$ by the transformation of step 1.3.

Finally, the spatial velocity vectors confirm the same laws: in $F$ one has $\underset{‾}{v}={\gamma }^{\mathrm{\prime }}(s)-u=\frac{1}{4}{e}_1$, and in $F^{\mathrm{\prime }}$ one has ${\underset{‾}{v}}^{\mathrm{\prime }}={\gamma }^{\mathrm{\prime }}(s)-u^{\mathrm{\prime }}=\frac{1}{4}{e}_1+\frac{1}{2}{e}_1=\frac{3}{4}{e}_1$, whose primed-basis coordinates $(\frac{3}{4},0,0)$ equal the coordinate velocity ${\eta }^{\mathrm{\prime }}$ of step 2.1; the example thus verifies the transformation law and the velocity and acceleration laws on explicit data. It does not identify the displayed uniform motion with the motion of an unforced particle in any dynamical sense, which requires the force postulates of a later page.

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.

In the frame $F$ the worldline has coordinate velocity $w=x^{\mathrm{\prime }}=(\frac{1}{4},0,0)$: differentiating $x(s)=(1+\frac{1}{4}s,2,0)$ gives $w=(\frac{1}{4},0,0)$, independent of $s$.

In the frame $F^{\mathrm{\prime }}$ the primed coordinate curve of the same worldline is $\eta (t^{\mathrm{\prime }})=(\frac{3}{4}{t}^{\mathrm{\prime }}-\frac{1}{2},5,0)$, whose derivative is ${\eta }^{\mathrm{\prime }}(t^{\mathrm{\prime }})=(\frac{3}{4},0,0)$; this agrees with the velocity law ${\eta }^{\mathrm{\prime }}=Rw+v=(\frac{1}{4},0,0)+(\frac{1}{2},0,0)=(\frac{3}{4},0,0)$ of [L1].

The two coordinate velocities ${\eta }^{\mathrm{\prime }}=(\frac{3}{4},0,0)$ and $w=(\frac{1}{4},0,0)$ differ: their first components $\frac{3}{4}\mathrm{\ne }\frac{1}{4}$. The refuted statement therefore fails for this worldline and this pair of inertial frames; the correct law is $w^{\mathrm{\prime }}=Rw+v$ of clause 2 of Galilean transformation of worldlines: velocity and acceleration laws, frame-independence of zero coordinate acceleration, and the group of frame changes, which is what step 2.1 verified, and the coordinate acceleration $a=0$ is preserved, in accordance with clause 3 of the same theorem.

$\mathrm{\Delta }$ is closed, hence $Q_0={\mathbb{R}}^6\setminus \mathrm{\Delta }$ is open: for $x\mathrm{\notin }\mathrm{\Delta }$ the number $\delta :=\mathrm{\parallel }{x}_1-x_2{\mathrm{\parallel }}_2$ is positive, and every $y\in {\mathbb{R}}^6$ with $\mathrm{\parallel }y-x{\mathrm{\parallel }}_2<\delta \mathrm{/}4$ satisfies $\mathrm{\parallel }{y}_1-y_2{\mathrm{\parallel }}_2\ge \mathrm{\parallel }{x}_1-x_2{\mathrm{\parallel }}_2-\mathrm{\parallel }{y}_1-x_1{\mathrm{\parallel }}_2-\mathrm{\parallel }{y}_2-x_2{\mathrm{\parallel }}_2>\delta -\delta \mathrm{/}4-\delta \mathrm{/}4=\delta \mathrm{/}2>0$, so $y\mathrm{\notin }\mathrm{\Delta }$; the ball of radius $\delta \mathrm{/}4$ about $x$ therefore lies in $Q_0$, and $Q_0$ is open as required by [L1].

The differential of $G$: for $x\in {\mathbb{R}}^6$ and $u=(u_1,u_2)\in {\mathbb{R}}^6$ one has $G(x+su)-G(x)=2s⟨x_1-x_2,u_1-u_2{⟩}_2+s^2\mathrm{\parallel }{u}_1-u_2{\mathrm{\parallel }}_2^2$ by bilinearity of the inner product [L4], so $DG(x)(u)=2⟨x_1-x_2,u_1-u_2{⟩}_2$; this linear functional is nonzero exactly when $x_1\mathrm{\ne }{x}_2$, since it vanishes for all $u$ only if $x_1=x_2$, while if $x_1\mathrm{\ne }{x}_2$ then $u=(x_1-x_2,0)$ gives $DG(x)(u)=2\mathrm{\parallel }{x}_1-x_2{\mathrm{\parallel }}_2^2>0$; hence $DG(x)$ has rank $1$ exactly on the complement of $\mathrm{\Delta }$.

On the zero set the rank condition holds: if $G(x)=0$ then $\mathrm{\parallel }{x}_1-x_2{\mathrm{\parallel }}_2^2={\mathrm{\ell }}^2>0$, so $x_1\mathrm{\ne }{x}_2$ and $DG(x)$ is surjective by step 1.2; and $G^{-1}(0)\mathrm{\ne }\mathrm{\varnothing }$, for instance $x_1=0$ and $x_2=\mathrm{\ell }{e}_1$ give $G(x)={\mathrm{\ell }}^2-{\mathrm{\ell }}^2=0$; thus $0$ is a regular value of $G$ with nonempty level set.

By the regular level set theorem [L2], applied to the smooth map $G:{\mathbb{R}}^6\to \mathbb{R}$ with the regular value $0$, the set $Q=G^{-1}(0)$ is an embedded submanifold of ${\mathbb{R}}^6$ of dimension $6-1=5$; by [L1] the rigid-rod branch therefore has $5=3\cdot 2-1$ configuration degrees of freedom, in agreement with the identification of the same constraint in the cited treatment. With the masses $m_1,m_2$, the pair consisting of $Q$ and the inclusion is a particle configuration whose mass matrix is $M=\mathrm{diag}(m_1{I}_3,m_2{I}_3)$ and whose mass metric is that of [L3]; the constraint function $G$ has dimension ${\mathrm{\ell }}^2$, and rescaling it by any nonzero constant leaves both its zero set and the rank condition on that set unchanged, as recorded in [L1].

The time coordinate and worldline parameter have dimension $[t]=\tau$ and the spatial coordinates $[x_i]=\mathrm{\ell }$ by the conventions; the dimension of a bilinear product of two quantities is the product of their dimensions, since each component of the product is a sum of products of components to which the rule $[XY]=[X][Y]$ applies.

The coordinate velocity has $[w]=[d{x}_{\gamma }\mathrm{/}dt]=\mathrm{\ell }{\tau }^{-1}$ and the coordinate acceleration $[a]=[dw\mathrm{/}dt]=\mathrm{\ell }{\tau }^{-2}$, by the derivative rule and step 1.1.

The masses have $[m_i]=\mu$, so the mass operator has dimension $\mu$ by [L3]; the momentum $p=mw$ has $[p]=\mu \mathrm{\ell }{\tau }^{-1}$ and the product $ma$ has $[ma]=\mu \mathrm{\ell }{\tau }^{-2}$, both being products of the displayed dimensions.

The frame parameters have the dimensions of [L1]; each law of [L1] is dimensionally consistent: $[t^{\mathrm{\prime }}]=[t+b]=\tau$ because $[t]=[b]=\tau$; in $x^{\mathrm{\prime }}=Rx+vt+a_0$ the three summands have dimensions $\mathrm{\ell }$, $\mathrm{\ell }{\tau }^{-1}\cdot \tau =\mathrm{\ell }$ and $\mathrm{\ell }$; $[w^{\mathrm{\prime }}]=[Rw+v]=\mathrm{\ell }{\tau }^{-1}$ because $[Rw]=\mathrm{\ell }{\tau }^{-1}$ and $[v]=\mathrm{\ell }{\tau }^{-1}$; and $[a^{\mathrm{\prime }}]=[Ra]=\mathrm{\ell }{\tau }^{-2}$.

For the locally specified rotating-basis coordinates $y$ and the angular speed $\omega >0$ measured in radians per second, $[\omega ]={\tau }^{-1}$ under the radian convention, and the two cross-product expressions are consistent with the acceleration: $[\mathit{\omega }\times y]={\tau }^{-1}\mathrm{\ell }=\mathrm{\ell }{\tau }^{-1}=[y^{\mathrm{\prime }}]$ and $[\mathit{\omega }\times (\mathit{\omega }\times y)]={\tau }^{-2}\mathrm{\ell }=\mathrm{\ell }{\tau }^{-2}=[y^{\mathrm{\prime }\mathrm{\prime }}]$, where each component of a cross product is a sum of products of two components, so its dimension is the product of the dimensions.

Finally the locally defined two-particle constraint function has dimension ${\mathrm{\ell }}^2$; the level equation $G(x)=0$ relates a quantity of dimension ${\mathrm{\ell }}^2$ to the number $0$, which is admissible because $0$ is the neutral element of addition in every dimension, and multiplying $G$ by a nonzero constant of any fixed dimension produces the same zero set with the same ranks of $DG$ on it, as recorded with the definition of the configuration stage; all quantities and laws of this page are therefore dimensionally consistent under the stated conventions.

The standard chart is adapted and its coordinates are the standard ones: $o=0$, $u=(1,0,0,0)$ has $\tau (u)=1$, the standard basis is orthonormal and positively oriented, and for $p=(x^0,x^1,x^2,x^3)$ the expansion $p=o+x^{0}u+{\sum }_j{x}^j{e}_j$ exhibits the adapted coordinates as $t=x^0$, $x=(x^1,x^2,x^3)$ by [L1]. The assigned curve $\gamma (t)=tu$ is $C^{\mathrm{\infty }}$ on $\mathbb{R}$, has $F$-coordinates $(t,0,0,0)$ and satisfies $\tau (\gamma (s)-o)=s$. Thus its assignment to label $1$ with the specified fixed positive mass satisfies [L2]; worldline status uses that assignment, not the parameter condition alone.

The curve $\kappa (s)=2su$ has $\tau (\kappa (s)-o)=2s$, which differs from $s$ for every $s\mathrm{\ne }0$, for instance at $s=1$ the value is $2$; therefore $\kappa$ fails the parameter condition and is not a worldline of the model, although it is of class $C^{\mathrm{\infty }}$ and its image is the same time axis as that of $\gamma$; it is the reparametrization $\kappa (s)=\gamma (2s)$. The parameter condition is thus a genuine restriction, and the image of a curve does not determine whether it is a worldline.

Under a change of frame the parameter condition is consistent with the reparametrization: by [L3] the same worldline $\gamma$ satisfies $t^{\mathrm{\prime }}(\gamma (s))=s+b$, so the curve ${\gamma }_b(t^{\mathrm{\prime }}):=\gamma (t^{\mathrm{\prime }}-b)$, defined on the shifted interval, satisfies $t^{\mathrm{\prime }}({\gamma }_b(t^{\mathrm{\prime }}))=t^{\mathrm{\prime }}$ and is the description of the motion in $F^{\mathrm{\prime }}$ in the sense of [L2]; the time shift $b$ is absorbed into the reparametrization. For the explicit worldline of step 1.1 this reads ${\gamma }_b(t^{\mathrm{\prime }})=(t^{\mathrm{\prime }}-b)u$.

For the affine reparametrizations in one fixed frame, $\tau ({\kappa }_{\lambda ,\mu }(s)-o)=t(\gamma (\lambda s+\mu ))=\lambda s+\mu$ by [L1] and step 1.1, and the equation $\lambda s+\mu =s$ holds for all real $s$ exactly when $\lambda =1$ and $\mu =0$. Hence no nontrivial affine reparametrization preserves the parameter condition in a fixed frame: the parameter of a worldline is fixed to be the frame time, and a reparametrized curve with $\lambda \mathrm{\ne }1$ becomes a worldline only after the inverse reparametrization. The example confirms that both the parameter and the frame enter the definition of a worldline, as recorded in [L2].