Newtonian Bridge — Examples

The numeric data. With $z=1\mathrm{/}25$ and $z_0=z$, [L5] and [L6] give ${\gamma }_w=(\frac{24}{25}{)}^{-1\mathrm{/}2}=(\frac{25}{24}{)}^{1\mathrm{/}2}=\frac{5}{\sqrt{24}}=\frac{5}{2\sqrt{6}}=\frac{5\sqrt{6}}{12},$ and $K(z_0)=\frac{3}{8}(25\mathrm{/}24{)}^{5\mathrm{/}2}$; the statement records the ten-place rounded values of these exact expressions.

Clause (a). By [L2] with $z=1\mathrm{/}25$, the exact remainder and the exact bound are ${\gamma }_w-1-\frac{z}{2}=R(z)=\frac{5\sqrt{6}}{12}-\frac{51}{50},K(z_0)z^2=\frac{3}{8}(\frac{25}{24}{)}^{5\mathrm{/}2}\frac{1}{625},$ the remainder is positive because ${\gamma }_w^2=25\mathrm{/}24>(51\mathrm{/}50{)}^2$ (cross multiplication gives $62500>62424$) and both quantities are positive by [L5], so $({\gamma }_w-51\mathrm{/}50)({\gamma }_w+51\mathrm{/}50)>0$ with positive second factor, hence ${\gamma }_w>51\mathrm{/}50$; its absolute value is at most the bound by [L2], and the statement records the ten-place rounded values of the two exact expressions and of the fractional overestimate $K(z_0)z^2\mathrm{/}R(z)-1$.

Clause (c). By [L3] and [L4], $\mathrm{\mid }{q}_n(P)\mathrm{\mid }=m{\gamma }_w\mathrm{\mid }w\mathrm{\mid }=m\cdot \frac{5\sqrt{6}}{12}\cdot \frac{c}{5}=\frac{\sqrt{6}}{12}mc$ and $\mathrm{\mid }{N}_p(m,w)\mathrm{\mid }=m\mathrm{\mid }w\mathrm{\mid }=mc\mathrm{/}5$, so the exact deviation is $mc(\sqrt{6}\mathrm{/}12-1\mathrm{/}5)$; the bound of [L1] clause (c) is $(\frac{1}{2}+K(z_0)z)z\frac{mc}{5}=(\frac{1}{2}+\frac{K(z_0)}{25})\frac{1}{25}\frac{mc}{5},$ and it dominates the deviation; the statement records the ten-place rounded values of both exact expressions.

Clause (d). By [L1], $w=q_n(P)\mathrm{/}(m{\gamma }_w)$ exactly and $\mathrm{\mid }w-q_n(P)\mathrm{/}m\mathrm{\mid }=({\gamma }_w-1)\mathrm{\mid }w\mathrm{\mid }$, so the relative deviation is ${\gamma }_w-1=5\sqrt{6}\mathrm{/}12-1$, which by [L1] clause (d) is at most $(\frac{1}{2}+K(z_0)z)z=(\frac{1}{2}+K(z_0)\mathrm{/}25)\mathrm{/}25$; the statement records the ten-place rounded values of these two exact expressions.

Clause (b). By [L1] and [L3], $E_n(P)-m{c}^2=m{c}^2({\gamma }_w-1)=m{c}^2(5\sqrt{6}\mathrm{/}12-1)$ and $N_K(m,w)=\frac{1}{2}m(c\mathrm{/}5{)}^2=m{c}^2\mathrm{/}50$, so the exact deviation is $m{c}^2({\gamma }_w-1-\frac{1}{50})=m{c}^2(5\sqrt{6}\mathrm{/}12-51\mathrm{/}50)$, which is the remainder of step 2.1 and obeys the bound $K(z_0)zm\mathrm{\mid }w{\mathrm{\mid }}^2=m{c}^{2}K(z_0)z^2$ of [L1] clause (a); the statement records the ten-place rounded values of the deviation and of this bound.

Conclusion of the worked instance. The state satisfies the hypotheses of Specified low-speed momentum-energy correspondence between the relativistic measured quantities and the Newtonian comparison quantities with $z=1\mathrm{/}25<1$ and $z_0=z$; clauses (a)-(d) have been evaluated at the exact radical values ${\gamma }_w=5\sqrt{6}\mathrm{/}12$ and $K(z_0)=\frac{3}{8}(25\mathrm{/}24{)}^{5\mathrm{/}2}$, every displayed inequality of the statement is the corresponding exact inequality of The uniform second-order Taylor remainder of $z\mapsto (1-z{)}^{-1\mathrm{/}2}$ on $[0,z_0]$ or of the correspondence theorem at this data, and the decimals are the ten-place roundings of the displayed exact expressions. The numbers are a numerical illustration of the conditional correspondence for this one state and carry no empirical content or precision claim.

The exact deviations at the point. With $v=e_1$, $\mathrm{\mid }v\mathrm{\mid }=1$, $\delta (v,x)=2$, $x_{\parallel }=2e_1$ and $x-tv=e_1$, [L1] and [L3] give ${\gamma }_C=C\mathrm{/}\sqrt{C^2-1}$, $t^{\mathrm{\prime }}={\gamma }_C(1-\frac{2}{C^2})=(C^2-2)\mathrm{/}(C\sqrt{C^2-1})$ and $x^{\mathrm{\prime }}=2e_1+2({\gamma }_C-1)e_1-{\gamma }_C{e}_1={\gamma }_C{e}_1$, so the exact deviations are $d_t(C)=1-(C^2-2)\mathrm{/}(C\sqrt{C^2-1})$ and $d_x(C)={\gamma }_C-1$. Indeed, if $1<C^2\le 2$ then $t^{\mathrm{\prime }}\le 0<1$; if $C^2>2$, both numerator and denominator of $t^{\mathrm{\prime }}$ are positive and $(C^2-2{)}^2<C^2(C^2-1)$ since $4<3C^2$, so $t^{\mathrm{\prime }}<1$ again. Thus $d_t(C)>0$ for every $C>1$, including $C=10,30,100$, and the absolute value may be read as written.

The $\epsilon$-threshold data. Take $z_0=1\mathrm{/}4$. By [L3] and [L4], $K_0=\frac{3}{8}(3\mathrm{/}4{)}^{-5\mathrm{/}2}=\frac{3}{8}\cdot \frac{32}{9\sqrt{3}}=\frac{4\sqrt{3}}{9}$ and $A_0=\frac{1}{2}+K_0{z}_0=\frac{1}{2}+\frac{\sqrt{3}}{9}$; since $\sqrt{3}<9\mathrm{/}4$ by [L4], $A_0<\frac{1}{2}+\frac{1}{4}=\frac{3}{4}$. With $z_1=1\mathrm{/}900$ one has $z_1<z_0$, $A_0{z}_1<\frac{3}{4}\cdot \frac{1}{900}=\frac{1}{1200}<\frac{1}{2}$ and $A_{0}H{z}_1=4A_0{z}_1<\frac{1}{300}<\frac{1}{200}=\epsilon \mathrm{/}2$ for $\epsilon =1\mathrm{/}100$, so $z_1$ satisfies the hypothesis of [L2],(c).

The whole-region bound values. Instance of [L2],(b) with $T=1$, $R=2$, $\mathrm{\mid }v\mathrm{\mid }=1$: the bounds read $B_t(C)=({\gamma }_C-1)+2{\gamma }_C\mathrm{/}{C}^2$ and $B_x(C)=3({\gamma }_C-1)$, and $d_t(C)\le B_t(C)$, $d_x(C)\le B_x(C)$ hold at every $C>1$ by the lemma; the statement records the ten-place rounded values of the exact expressions $d_t(C),d_x(C),B_t(C),B_x(C)$ at $C=10$ and $C=100$.

The sufficient threshold. By [L2],(c) applied with $z_1=1\mathrm{/}900$, $\mathrm{\mid }v\mathrm{\mid }=1$, $R=2$ and $\epsilon =1\mathrm{/}100$, $C_{\epsilon }=\max \{1\mathrm{/}\sqrt{z_1},\sqrt{3\mathrm{\mid }v\mathrm{\mid }R\mathrm{/}\epsilon }\}=\max \{30,\sqrt{600}\}=30,$ because $\sqrt{600}<30$ by [L4]; since $30>1=\mathrm{\mid }v\mathrm{\mid }$, every $C\ge 30$ meets both error bounds of [L2],(b) on the whole region $D_{1,2}$.

The evaluations and ratios at the named scales. For $C=10$ one has ${\gamma }_{10}=10\mathrm{/}\sqrt{99}=10\mathrm{/}(3\sqrt{11})$, $d_t(10)=1-98\mathrm{/}(10\sqrt{99})$, $d_x(10)={\gamma }_{10}-1$; for $C=100$ one has ${\gamma }_{100}=100\mathrm{/}\sqrt{9999}$ and the corresponding expressions; for $C=30$ one has ${\gamma }_{30}=30\mathrm{/}\sqrt{899}$, $t^{\mathrm{\prime }}=898\mathrm{/}(30\sqrt{899})=449\mathrm{/}(15\sqrt{899})$ and the bound value $B_t(30)=451\mathrm{/}(15\sqrt{899})-1$. The two ratios of exact deviations are $\frac{d_t(10)}{d_t(100)}=\frac{(10\sqrt{99}-98)\cdot 10\sqrt{9999}}{\sqrt{99}(100\sqrt{9999}-9998)},\frac{d_x(10)}{d_x(100)}=\frac{{\gamma }_{10}-1}{{\gamma }_{100}-1},$ evaluated from these exact radicals before rounding; the statement records the ten-place rounded values of all these quantities, and each bound comparison at the named scales is the instance of [L2],(b) already recorded in step 2.1.

Conclusion of the worked instance. The exact differences, the uniform bounds at $C=10,30,100$, the ten-place rounded evaluations, the two ratios of exact deviations and the $\epsilon =1\mathrm{/}100$ threshold $C_{\epsilon }=30$ have been checked above from the exact radicals, and the constants $K_0=4\sqrt{3}\mathrm{/}9$, $A_0=1\mathrm{/}2+\sqrt{3}\mathrm{/}9<3\mathrm{/}4$ and the inequalities $z_1<z_0$, $A_0{z}_1<1\mathrm{/}1200$, $A_{0}H{z}_1<1\mathrm{/}300<\epsilon \mathrm{/}2$ hold with exact values; the auxiliary family ${\mathrm{\Phi }}_{v,C}$ is mathematical and no physical claim is made.

The fixed data and the family. Fix once and for all the future unit $n$ and the unit vector $e\in n^{\perp }$; such data exist because timelike vectors exist by the inertia $(3,1,0)$ of the form and any nonzero timelike vector can be normalized by dividing by the positive rational power $(-\eta (v,v){)}^{1\mathrm{/}2}$, and because $n^{\perp }$ is a three-dimensional positive definite space by [L2]. For every integer $N\ge 2$ set $w_N:=c\sqrt{1-N^{-2}}e$, $U_N:=cN(n+w_N\mathrm{/}c)$ and $P_N:=m{U}_N$, using the same $n$ and $e$ for every $N$.

The normalization of $U_N$. Since $e\in n^{\perp }$ is unit, $\eta (e,e)=\mathrm{\mid }e{\mathrm{\mid }}^2=1$ and $\eta (n,w_N)=c\sqrt{1-N^{-2}}\eta (n,e)=0$, so $\eta (w_N,w_N)=c^2(1-N^{-2})=\mathrm{\mid }{w}_N{\mathrm{\mid }}^2<c^2$. Hence $\eta (U_N,U_N)=c^2{N}^2(\eta (n,n)+2\eta (n,w_N\mathrm{/}c)+\eta (w_N\mathrm{/}c,w_N\mathrm{/}c))=c^2{N}^2(-1+(1-N^{-2}\left)\right)=-c^2,$ and the continuous segment $a(s)=n+s{w}_N\mathrm{/}c$, $0\le s\le 1$, has $\eta (a(s),a(s))=-1+s^2(1-N^{-2})<0$ throughout. The segment is continuous in any transported Euclidean norm because $\mathrm{\mid }a(s)-a(t){\mathrm{\mid }}_E=\mathrm{\mid }s-t\mathrm{\mid }\mathrm{\mid }{w}_N\mathrm{/}c{\mathrm{\mid }}_E$. By [L6], its endpoint lies in the same component as $n\in C^{+}$ (Time orientation, future and past timelike/null/causal vectors, and the closed cones). Thus $n+w_N\mathrm{/}c\in C^{+}$ and $U_N=cN(n+w_N\mathrm{/}c)$ is future timelike by positive scaling.

The relative velocity and Lorentz factor of $U_N$. The vector $u_N:=U_N\mathrm{/}c=N(n+w_N\mathrm{/}c)$ satisfies $\eta (u_N,u_N)=-1$ by step 2.1, so $u_N$ is future unit and ${\gamma }_N:=-\eta (n,u_N)=-N\eta (n,n)-N\eta (n,w_N)\mathrm{/}c=N\ge 1$. By [L2], the relative velocity of $U_N$ is $w=ch(u_N)\mathrm{/}{\gamma }_N=cNh(n+w_N\mathrm{/}c)\mathrm{/}N=ch(w_N\mathrm{/}c)=w_N$ and its Lorentz factor is $N$; consistently $\mathrm{\mid }{w}_N{\mathrm{\mid }}^2=c^2(1-N^{-2})=c^2(1-{\gamma }_N^{-2})$.

The kinetic energy and the comparison value. By [L3] applied to $P_N=m{U}_N$ with Lorentz factor ${\gamma }_N=N$ from step 3.1, $T_n(P_N)=m{c}^2(N-1),\frac{1}{2}m\mathrm{\mid }{w}_N{\mathrm{\mid }}^2=\frac{1}{2}m{c}^2(1-N^{-2}),$ so $T_n(P_N)-\frac{1}{2}m\mathrm{\mid }{w}_N{\mathrm{\mid }}^2=m{c}^2(N-1-\frac{1}{2}+\frac{1}{2N^2})=m{c}^2(N-\frac{3}{2}+\frac{1}{2N^2})\mathrm{.}$ For $N\ge 2$ this difference is positive, hence equal to its absolute value: $N-\frac{3}{2}+\frac{1}{2N^2}>N-\frac{3}{2}\ge \frac{1}{2}>0$ for every $N\ge 2$.

The relative deviation. The ratio of the difference of step 4.1 to the comparison value $\frac{1}{2}m\mathrm{\mid }{w}_N{\mathrm{\mid }}^2=\frac{1}{2}m{c}^2(1-N^{-2})$ equals $\frac{N-\frac{3}{2}+\frac{1}{2N^2}}{\frac{1}{2}(1-N^{-2})}=\frac{2N^3-3N^2+1}{N^2-1}=\frac{(N-1{)}^2(2N+1)}{(N-1)(N+1)}=\frac{(N-1)(2N+1)}{N+1}\ge N-1,$ using $(N-1{)}^2(2N+1)=2N^3-3N^2+1$ and $(2N+1)\ge (N+1)$ for $N\ge 0$.

Both sequences are unbounded. Let $B\ge 0$ be any real. By [L5] choose an integer $N>\max \{2,B\mathrm{/}(m{c}^2)+\frac{3}{2}\}$. Then by step 4.1, $\mid T_n(P_N)-\frac{1}{2}m\mathrm{\mid }{w}_N{\mathrm{\mid }}^2\mid =m{c}^2(N-\frac{3}{2}+\frac{1}{2N^2})>m{c}^2(N-\frac{3}{2})>B,$ so the bound $B$ fails at the admissible state $(n,P_N)$; and by step 5.1 the relative deviation at that $N$ is at least $N-1>B\mathrm{/}(m{c}^2)+\frac{1}{2}$, so it is unbounded as well. Since $B\ge 0$ was arbitrary and the construction uses one fixed future unit $n$, the refuted claim fails for every pair $(c,m>0)$.

Conclusion. The displayed family of states $P_N=m{U}_N$ in the stipulated model has $T_n(P_N)-\frac{1}{2}m\mathrm{\mid }{w}_N{\mathrm{\mid }}^2\to \mathrm{\infty }$ with the relative deviation also tending to infinity, so no real constant $B$ can bound $\mathrm{\mid }{T}_n(P)-\frac{1}{2}m\mathrm{\mid }w{\mathrm{\mid }}^2\mathrm{\mid }$ uniformly over all admissible states; a valid low-speed error bound necessarily depends on the admissible speed domain, as the constants $K(z_0)$ of Specified low-speed momentum-energy correspondence between the relativistic measured quantities and the Newtonian comparison quantities do, which is confirmed independently by the bounded $z\le z_0$ hypothesis of The uniform second-order Taylor remainder of $z\mapsto (1-z{)}^{-1\mathrm{/}2}$ on $[0,z_0]$. No experimental, measurement or other empirical claim is made.