Uniform Galilean limit of the boost charts on bounded regions, and its failure on the whole chart domain

The exact time difference. For $v\mathrm{\ne }0$ and $C>\mathrm{\mid }v\mathrm{\mid }$, subtraction in $\mathbb{R}$ using [L1] gives $t^{\mathrm{\prime }}-t={\gamma }_C(t-\frac{\delta (v,x)}{C^2})-t=({\gamma }_C-1)t-{\gamma }_C\frac{\delta (v,x)}{C^2},$ for every $(t,x)\in \mathbb{R}\times S$.

The exact spatial difference. In the same case, subtraction in the vector space $S$ using [L1] gives $x^{\mathrm{\prime }}-(x-tv)=x+\frac{{\gamma }_C-1}{\mathrm{\mid }v{\mathrm{\mid }}^2}\delta (v,x)v-{\gamma }_{C}tv-x+tv=({\gamma }_C-1)(\frac{\delta (v,x)}{\mathrm{\mid }v{\mathrm{\mid }}^2}v-tv)=({\gamma }_C-1)(x_{\parallel }-tv),$ with $x_{\parallel }=\delta (v,x)v\mathrm{/}\mathrm{\mid }v{\mathrm{\mid }}^2$; if $v=0$ then ${\mathrm{\Phi }}_{0,C}=\mathrm{i}\mathrm{d}$ and $t^{\mathrm{\prime }}-t=0$, $x^{\mathrm{\prime }}-(x-tv)=x-x=0$. This proves (a).

The Taylor estimate for ${\gamma }_C$. Fix $z_0\in (0,1)$, let $0\le z\le z_0$ and put $\gamma =(1-z{)}^{-1\mathrm{/}2}$; by [L2] $\gamma =1+\frac{z}{2}+R(z)$ with $\mathrm{\mid }R(z)\mathrm{\mid }\le K_0{z}^2$, so $\gamma -1=\frac{z}{2}+R(z)\le \frac{z}{2}+K_0{z}^2\le (\frac{1}{2}+K_0{z}_0)z=A_{0}z,$ because $K_0\ge 0$ and $z\le z_0$.

The reciprocal bound for ${\gamma }_C$. With the same data, $0<1-z_0\le 1-z\le 1$ gives $(1-z_0{)}^{1\mathrm{/}2}\le (1-z{)}^{1\mathrm{/}2}$ by [L5]; if $1-z_0<1-z$ the strict reciprocal rule of [L5] gives $(1-z{)}^{-1\mathrm{/}2}<(1-z_0{)}^{-1\mathrm{/}2}$, and if $1-z_0=1-z$ the two quantities are equal, so in all cases $\gamma =(1-z{)}^{-1\mathrm{/}2}\le (1-z_0{)}^{-1\mathrm{/}2}={\mathrm{\Gamma }}_0$.

The domain inclusion. For $v\mathrm{\ne }0$ and $C\ge \mathrm{\mid }v\mathrm{\mid }\mathrm{/}\sqrt{z_0}$ one has $z=\mathrm{\mid }v{\mathrm{\mid }}^2\mathrm{/}{C}^2\le \mathrm{\mid }v{\mathrm{\mid }}^2\mathrm{/}(\mathrm{\mid }v{\mathrm{\mid }}^2\mathrm{/}{z}_0)=z_0$; hence steps 1.3 and 1.4 apply with this $z$.

The time bound. Let $v\mathrm{\ne }0$, $C>\mathrm{\mid }v\mathrm{\mid }$ and $(t,x)\in D_{T,R}$; by [L5], $(1-z{)}^{-1\mathrm{/}2}>1$ for $0<z<1$ and ${\gamma }_C=1$ for $z=0$, so ${\gamma }_C-1\ge 0$ and ${\gamma }_C\ge 1$. Taking absolute values in step 1.1 and using [L4], $\mathrm{\mid }{t}^{\mathrm{\prime }}-t\mathrm{\mid }\le ({\gamma }_C-1)\mathrm{\mid }t\mathrm{\mid }+{\gamma }_C\mathrm{\mid }\delta (v,x)\mathrm{\mid }\mathrm{/}{C}^2$, and [L3] plus $\mathrm{\mid }t\mathrm{\mid }\le T$, $\mathrm{\mid }x\mathrm{\mid }\le R$ give $\mathrm{\mid }{t}^{\mathrm{\prime }}-t\mathrm{\mid }\le ({\gamma }_C-1)T+{\gamma }_C\mathrm{\mid }v\mathrm{\mid }R\mathrm{/}{C}^2$.

The spatial bound. In the same case, step 1.2 and the homogeneity and triangle inequality of [L4] give $\mathrm{\mid }{x}^{\mathrm{\prime }}-(x-tv)\mathrm{\mid }=({\gamma }_C-1)\mathrm{\mid }{x}_{\parallel }-tv\mathrm{\mid }\le ({\gamma }_C-1)(\mathrm{\mid }{x}_{\parallel }\mathrm{\mid }+\mathrm{\mid }t\mathrm{\mid }\mathrm{\mid }v\mathrm{\mid })$; by [L3] $\mathrm{\mid }{x}_{\parallel }\mathrm{\mid }=\mathrm{\mid }\delta (v,x)\mathrm{\mid }\mathrm{/}\mathrm{\mid }v\mathrm{\mid }\le \mathrm{\mid }x\mathrm{\mid }$, so on $D_{T,R}$ we get $\mathrm{\mid }{x}^{\mathrm{\prime }}-(x-tv)\mathrm{\mid }\le ({\gamma }_C-1)(\mathrm{\mid }x\mathrm{\mid }+\mathrm{\mid }v\mathrm{\mid }T)\le ({\gamma }_C-1)(R+\mathrm{\mid }v\mathrm{\mid }T)$. This proves (b).

The case $v=0$. Here $G_0(t,x)=(t,x)$ and ${\mathrm{\Phi }}_{0,C}=\mathrm{i}\mathrm{d}$ for every $C>0$, so $\mathrm{\mid }{t}^{\mathrm{\prime }}-t\mathrm{\mid }=0$ and $\mathrm{\mid }{x}^{\mathrm{\prime }}-(x-0\cdot v)\mathrm{\mid }=\mathrm{\mid }{x}^{\mathrm{\prime }}-x\mathrm{\mid }=0$ for every $(t,x)$; hence every $\epsilon >0$ is met already with any $C_{\epsilon }>0$.

The explicit $O(C^{-2})$ bounds. Let $v\mathrm{\ne }0$, $C\ge \mathrm{\mid }v\mathrm{\mid }\mathrm{/}\sqrt{z_0}$ and $(t,x)\in D_{T,R}$. By step 2.1 and steps 1.3, 1.5, $\mathrm{\mid }{t}^{\mathrm{\prime }}-t\mathrm{\mid }\le ({\gamma }_C-1)T+{\gamma }_C\frac{\mathrm{\mid }v\mathrm{\mid }R}{C^2}\le A_{0}zT+{\mathrm{\Gamma }}_0\frac{\mathrm{\mid }v\mathrm{\mid }R}{C^2}=\frac{A_0\mathrm{\mid }v{\mathrm{\mid }}^{2}T+{\mathrm{\Gamma }}_0\mathrm{\mid }v\mathrm{\mid }R}{C^2},$ and by step 2.2 and steps 1.3, 1.5, $\mathrm{\mid }{x}^{\mathrm{\prime }}-(x-tv)\mathrm{\mid }\le ({\gamma }_C-1)(R+\mathrm{\mid }v\mathrm{\mid }T)\le A_{0}z(R+\mathrm{\mid }v\mathrm{\mid }T)=\frac{A_0\mathrm{\mid }v{\mathrm{\mid }}^2(R+\mathrm{\mid }v\mathrm{\mid }T)}{C^2},$ since $z=\mathrm{\mid }v{\mathrm{\mid }}^2\mathrm{/}{C}^2$.

The $\epsilon$-threshold for $v\mathrm{\ne }0$. Fix $\epsilon >0$, put $H=T+R+\mathrm{\mid }v\mathrm{\mid }T>0$, and choose $z_1$ with $0<z_1\le \min \{z_0,(2A_0{)}^{-1},\epsilon \mathrm{/}(2A_{0}H)\}$; the minimum is positive because $z_0,A_0,\epsilon ,H$ are positive. Let $C\ge \max \{\mathrm{\mid }v\mathrm{\mid }\mathrm{/}\sqrt{z_1},\sqrt{3\mathrm{\mid }v\mathrm{\mid }R\mathrm{/}\epsilon }\}$. Then $z\le z_1\le z_0$, so by step 1.3, ${\gamma }_C-1\le A_0{z}_1\le \frac{1}{2}$, whence ${\gamma }_C\le \frac{3}{2}$, and $A_0{z}_{1}T\le \frac{\epsilon }{2H}T\le \frac{\epsilon }{2}$ while $A_0{z}_1(R+\mathrm{\mid }v\mathrm{\mid }T)\le \frac{\epsilon }{2H}(R+\mathrm{\mid }v\mathrm{\mid }T)\le \frac{\epsilon }{2}$. Thus step 2.1 gives $\mathrm{\mid }{t}^{\mathrm{\prime }}-t\mathrm{\mid }\le A_0{z}_{1}T+\frac{3}{2}\frac{\mathrm{\mid }v\mathrm{\mid }R}{C^2}\le \frac{\epsilon }{2}+\frac{\epsilon }{2}=\epsilon$ and step 2.2 gives $\mathrm{\mid }{x}^{\mathrm{\prime }}-(x-tv)\mathrm{\mid }\le A_0{z}_1(R+\mathrm{\mid }v\mathrm{\mid }T)\le \frac{\epsilon }{2}\le \epsilon$ on $D_{T,R}$. Since $z_1\le z_0<1$ and $v\mathrm{\ne }0$, the number $C_{\epsilon }=\max \{\mathrm{\mid }v\mathrm{\mid }\mathrm{/}\sqrt{z_1},\sqrt{3\mathrm{\mid }v\mathrm{\mid }R\mathrm{/}\epsilon }\}$ satisfies $C_{\epsilon }\ge \mathrm{\mid }v\mathrm{\mid }\mathrm{/}\sqrt{z_1}>\mathrm{\mid }v\mathrm{\mid }$.

Uniform and pointwise convergence. By steps 3.2 and 2.3, for every fixed $v,T,R$ and every $\epsilon >0$ there is $C_{\epsilon }>\mathrm{\mid }v\mathrm{\mid }$ with $\max \{\mathrm{\mid }{t}^{\mathrm{\prime }}-t\mathrm{\mid },\mathrm{\mid }{x}^{\mathrm{\prime }}-(x-tv)\mathrm{\mid }\}\le \epsilon$ for all $(t,x)\in D_{T,R}$ and all $C\ge C_{\epsilon }$; taking the supremum over $D_{T,R}$ gives ${\sup }_{D_{T,R}}\max \{\mathrm{\mid }{t}^{\mathrm{\prime }}-t\mathrm{\mid },\mathrm{\mid }{x}^{\mathrm{\prime }}-(x-tv)\mathrm{\mid }\}\le \epsilon$, so the supremum tends to $0$ as $C\to \mathrm{\infty }$. Since ${\mathrm{\Phi }}_{v,C}(t,x)-(t,x-tv)$ has both components bounded by that maximum, ${\mathrm{\Phi }}_{v,C}(t,x)\to G_v(t,x)$ for each fixed $(t,x)$, which lies in $D_{T,R}$ for $T,R$ large; this proves (c).

Failure on the whole chart domain. Let $v\mathrm{\ne }0$ and $C>\mathrm{\mid }v\mathrm{\mid }$, so $z>0$, ${\gamma }_C>1$ and $x_{\parallel }$ is defined. For $\lambda >0$ take $(t,x)=(0,\lambda v\mathrm{/}\mathrm{\mid }v\mathrm{\mid })$; then $x_{\parallel }=\lambda v\mathrm{/}\mathrm{\mid }v\mathrm{\mid }$ and step 1.2 gives $x^{\mathrm{\prime }}-(x-0\cdot v)=({\gamma }_C-1)\lambda v\mathrm{/}\mathrm{\mid }v\mathrm{\mid }$, of norm $({\gamma }_C-1)\lambda$, which is unbounded as $\lambda \to \mathrm{\infty }$; hence ${\sup }_{\mathbb{R}\times S}\mathrm{\mid }{x}^{\mathrm{\prime }}-(x-tv)\mathrm{\mid }=+\mathrm{\infty }$. For $\mu >0$ take $(t,x)=(0,-\mu v)$; then $\delta (v,x)=-\mu \mathrm{\mid }v{\mathrm{\mid }}^2$ and step 1.1 gives $t^{\mathrm{\prime }}-t={\gamma }_C\mu \mathrm{\mid }v{\mathrm{\mid }}^2\mathrm{/}{C}^2$, which is unbounded as $\mu \to \mathrm{\infty }$; hence ${\sup }_{\mathbb{R}\times S}\mathrm{\mid }{t}^{\mathrm{\prime }}-t\mathrm{\mid }=+\mathrm{\infty }$. This proves (d), so the boundedness of $T,R$ is essential and no uniform limit holds on the whole chart domain.