Worldlines And Clocks — Examples

Worked computation. In an inertial chart of the model with ${\eta }_0=\mathrm{diag}(-1,1,1,1)$ let $c>0$ and $(a_1,a_2,a_3)\in {\mathbb{R}}^3$. For $t\in \mathbb{R}$ consider the coordinate worldline $x(t)=(ct,a_1,a_2,a_3)$ of a clock at rest at the spatial position $(a_1,a_2,a_3)$, parametrized by the coordinate time $t$. Then $x^{\mathrm{\prime }}(t)=(c,0,0,0)$, so $-\eta (x^{\mathrm{\prime }},x^{\mathrm{\prime }})=c^2$ and the proper-time function is $\tau (t)-\tau (t_0)=t-t_0$; the four-velocity is $U=x^{\mathrm{\prime }}\mathrm{/}{\tau }^{\mathrm{\prime }}=(c,0,0,0)$, with $\eta (U,U)=-c^2$, and the four-acceleration vanishes, $A=0$. Relative to the chart's observer $e_0$ the dimensionless four-velocity is $u=U\mathrm{/}c=(1,0,0,0)=e_0$, so $\gamma =-\eta (e_0,u)=1$ and the relative velocity is $w=ch(u)\mathrm{/}\gamma =0$: the clock is at rest in the chart.

• The computation is the model's notion of rest. "At rest in the chart" here means exactly that the relative velocity $w$ of clause (d) vanishes; it is a statement about the ideal observer $e_0$ of the chart and the worldline, not about any apparatus.

Worked computation. In an inertial chart with ${\eta }_0=\mathrm{diag}(-1,1,1,1)$ let $c>0$ and, for $t\in \mathbb{R}$, consider the coordinate worldline $x(t)=(ct,(3c\mathrm{/}5)t,0,0)$, parametrized by coordinate time, which describes a clock moving along the first spatial axis with speed $3c\mathrm{/}5$. Then $x^{\mathrm{\prime }}(t)=(c,3c\mathrm{/}5,0,0)$ and $\eta (x^{\mathrm{\prime }},x^{\mathrm{\prime }})=-c^2+(9\mathrm{/}25)c^2=-(16\mathrm{/}25)c^2$, so $\sqrt{-\eta (x^{\mathrm{\prime }},x^{\mathrm{\prime }})}=4c\mathrm{/}5$ and $\tau (t)-\tau (t_0)=(4\mathrm{/}5)(t-t_0)$: the moving clock runs slow by the factor $4\mathrm{/}5=1\mathrm{/}\gamma$ with $\gamma =5\mathrm{/}4$. The four-velocity is $U=x^{\mathrm{\prime }}\mathrm{/}{\tau }^{\mathrm{\prime }}=(5\mathrm{/}4)(c,3c\mathrm{/}5,0,0)=(5c\mathrm{/}4,3c\mathrm{/}4,0,0)$, which satisfies $\eta (U,U)=-25c^2\mathrm{/}16+9c^2\mathrm{/}16=-c^2$, and the four-acceleration vanishes because $U$ is constant: $A=0$. The dimensionless four-velocity is $u=U\mathrm{/}c=(5\mathrm{/}4,3\mathrm{/}4,0,0)$; the split of clause (d) with $n=e_0$ gives $\gamma =5\mathrm{/}4$ and relative velocity $w=(0,3c\mathrm{/}5,0,0)\in e_0^{\perp }$, whose spatial coordinates are exactly the coordinate velocity.

• All numbers are exact rationals. The computation uses only the diagonal form of $\eta$ and the rational values $3\mathrm{/}5$, $4\mathrm{/}5$, $5\mathrm{/}4$; no numerical approximation and no measurement is involved.

• The factor $1\mathrm{/}\gamma$ is a model statement. That the moving ideal clock runs slow by $4\mathrm{/}5$ is a statement about the model's proper time, not a reported measurement of any physical clock.

Worked computation. In an inertial chart with ${\eta }_0=\mathrm{diag}(-1,1,1,1)$ let $d:=(5,3,0,0)$, a future timelike displacement with $-\eta (d,d)=25-9=16$ and $\sqrt{-\eta (d,d)}=4$.

(a) The worldline $x(s):=sd$ on the parameter interval $[1,8]$ has $x^{\mathrm{\prime }}(s)=d$, integrand $\sqrt{-\eta (d,d)}=4$, and elapsed proper time $\tau (8)-\tau (1)=c^{-1}{\int }_1^{8}4ds=28\mathrm{/}c$.

(b) The reparametrization $\phi (u):=u^3$ is a strictly increasing $C^1$ surjection of $[1,2]$ onto $[1,8]$ whose derivative $3u^2$ is strictly positive throughout $[1,2]$; the reparametrized curve $y(u):=x(u^3)=u^{3}d$ has $y^{\mathrm{\prime }}(u)=3u^{2}d$, hence $\sqrt{-\eta (y^{\mathrm{\prime }},y^{\mathrm{\prime }})}=3u^2\sqrt{-\eta (d,d)}=12u^2$, and elapsed proper time $c^{-1}{\int }_1^{2}12u^{2}du=(4\mathrm{/}c)(8-1)=28\mathrm{/}c$. The two elapsed times agree, as Length-integral invariance under monotone C^1 maps and proper-time invariance under regular increasing maps requires, and the proper-time function of $x$ is $\tau (s)=(4\mathrm{/}c)(s-s_0)$ for the chosen base point $s_0\in [1,8]$, whose inverse is $\rho (\tau )=s_0+c\tau \mathrm{/}4$ on the corresponding proper-time interval.

(c) Boundary check: instead take the same straight curve $x(s)=sd$ on $[-1,1]$ and $\phi (u)=u^3$ on $[-1,1]$. This is a strictly increasing $C^1$ surjection, but $y^{\mathrm{\prime }}(0)=0$, so $y$ is not a worldline of Curves, regular causal curves and future timelike worldlines in affine Minkowski space. The broad length-integral identity nevertheless holds: ${\int }_{-1}^{1}12u^{2}du=8={\int }_{-1}^{1}4ds$. No proper-time function of $y$ is invoked outside its defining worldline domain.

• Two independent computations of the same number. The integrals $c^{-1}{\int }_1^{8}4ds$ and $c^{-1}{\int }_1^{2}12u^{2}du$ are computed separately, and their agreement is the concrete instance of Length-integral invariance under monotone C^1 maps and proper-time invariance under regular increasing maps; the cubic's derivative, rather than an inverse function, produces the second computation.

• The boundary case is deliberate. On $[-1,1]$ the derivative $3u^2$ vanishes at $u=0$, so the reparametrized curve has a zero tangent there and is not a worldline; the broad integral identity survives, and no proper-time function of $y$ is claimed.

Worked computation. In an inertial chart with ${\eta }_0=\mathrm{diag}(-1,1,1,1)$ let $c>0$, $R>0$ and let $\omega$ be real with $R\mathrm{\mid }\omega \mathrm{\mid }<c$. Here $\sin ,\cos :\mathbb{R}\to \mathbb{R}$ are the everywhere-convergent real power-series functions of Sine and cosine defined by their real power series; their derivatives are $\cos$ and $-\sin$ (The derivatives of sine and cosine are cosine and minus sine), and ${\sin }^2+{\cos }^2=1$ (Parity and the Pythagorean identity for sine and cosine). For $t\in \mathbb{R}$, consider the coordinate worldline $x(t)=(ct,R\cos (\omega t),R\sin (\omega t),0)$, parametrized by coordinate time: uniform circular motion in the first two spatial coordinates. Then $w(t)=(-R\omega \sin (\omega t),R\omega \cos (\omega t),0)$ has $\mathrm{\mid }w\mathrm{\mid }=R\mathrm{\mid }\omega \mathrm{\mid }$, so $-\eta (x^{\mathrm{\prime }},x^{\mathrm{\prime }})=c^2-R^2{\omega }^2$ and the proper-time factor is ${\tau }^{\mathrm{\prime }}=c^{-1}\sqrt{c^2-R^2{\omega }^2}=1\mathrm{/}\gamma$ with $\gamma :=(1-R^2{\omega }^2\mathrm{/}{c}^2{)}^{-1\mathrm{/}2}$; the four-velocity is $U=\gamma (c,w)$. Since $\mathrm{\mid }w\mathrm{\mid }$ is constant, $\gamma$ is constant; the coordinate acceleration is $a(t)=(-R{\omega }^2\cos (\omega t),-R{\omega }^2\sin (\omega t),0)$, with $\mathrm{\mid }a\mathrm{\mid }=R{\omega }^2$ and $w\cdot a=0$. By clause (e) of Norms of four-velocity and four-acceleration, and the observer rest-space decomposition the four-acceleration is $A=(0,{\gamma }^{2}a)$, so $A^0=0$, $\eta (U,A)=0$ and $\eta (A,A)={\gamma }^4{R}^2{\omega }^4\ge 0$. Its magnitude is ${\gamma }^{2}R{\omega }^2$; for $R>0$ it is spacelike exactly when $\omega \mathrm{\ne }0$, while $\omega =0$ is the static limit with $\gamma =1$, $U=(c,0,0,0)$ and $A=0$. Numerical specialization: if $R\omega =3c\mathrm{/}5$, then $\gamma =5\mathrm{/}4$ and the magnitude of the four-acceleration is ${\gamma }^{2}R{\omega }^2=(25\mathrm{/}16)R{\omega }^2=(15c\omega )\mathrm{/}16$.

• The zero-frequency case is retained deliberately. At $\omega =0$ the clock is static and its four-acceleration vanishes; the acceleration square $\eta (A,A)={\gamma }^4{R}^2{\omega }^4$ is nonnegative always and positive exactly for nonzero $\omega$ when $R>0$.

• Centripetal acceleration is orthogonal to the velocity. The identity $w\cdot a=0$ is what forces $A^0=0$ and keeps the four-acceleration purely spatial; it reflects the constant speed of the motion and is not an additional dynamical assumption.

Worked computation. In an inertial chart with ${\eta }_0=\mathrm{diag}(-1,1,1,1)$ let the observer be $n=e_0=(1,0,0,0)$ and let $u=(5\mathrm{/}4,3\mathrm{/}4,0,0)$, a future unit timelike vector: $\eta (u,u)=-25\mathrm{/}16+9\mathrm{/}16=-1$. Then $\gamma =-\eta (n,u)=5\mathrm{/}4\ge 1$, $h(u)=u+\eta (n,u)n=u-(5\mathrm{/}4)(1,0,0,0)=(0,3\mathrm{/}4,0,0)$ lies in $n^{\perp }=\mathrm{span}(e_1,e_2,e_3)$, and the relative velocity is $w=ch(u)\mathrm{/}\gamma =(0,3c\mathrm{/}5,0,0)$. One checks $u=\gamma (n+w\mathrm{/}c)=(5\mathrm{/}4)(1,3\mathrm{/}5,0,0)$ and $\mathrm{\mid }w{\mathrm{\mid }}^2=9c^2\mathrm{/}25=c^2(1-{\gamma }^{-2})=c^2(1-16\mathrm{/}25)$. Also $h(h(u))=h(u)$ and $\eta (n,h(u))=0$. This four-velocity is the one of the clock with coordinate velocity $(3c\mathrm{/}5,0,0)$ in Worked example: proper time and four-velocity of a straight worldline at beta = 3/5.