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Worked uniform Galilean chart bound on a bounded region

Example

Take v=e1 (so ∣v∣=1), T=1, R=2 and the point (t,x)=(1,2e1) in Galilean limiting chart maps and bounded chart regions, and let Φv,C(1,2e1)=(t′,x′). For C>1, write gC=C/C2−1, so t′=(C2−2)/(CC2−1) and x′=gCe1, while Gv(1,2e1)=(1,e1). The exact point deviations and whole-region bounds are dt(C):=1−C2−2CC2−1≤Bt(C):=gC−1+2gCC2,dx(C):=gC−1≤Bx(C):=3(gC−1). Their rounded values are:

CgCt′dt(C)Bt(C)dx(C)Bx(C)
101.00503781530.98493705900.01506294100.02513857160.00503781530.0151134458
1001.00005000380.99984999370.00015000630.00025001380.00005000380.0001500113

Here g10=10/99=10/(311). The computed deviations shrink by factors dt(10)/dt(100)≈100.4154227043 and dx(10)/dx(100)≈100.7487483984, or about 100.4 and 100.7 to one decimal place, consistent with the O(C−2) rate of Uniform Galilean limit of the boost charts on bounded regions, and its failure on the whole chart domain,(c).

Finally, for tolerance ε=0.01, take z0=1/4. Then K0=43/9 and A0=1/2+3/9<3/4. Choose exactly z1=1/900. With H=T+R+∣v∣T=4, one has z1<z0, A0z1<1/1200<1/2 and A0Hz1<1/300<ε/2=1/200, so every threshold inequality is satisfied. The sufficient threshold is Cε=max⁡{1/z1,3∣v∣R/ε}=max⁡{30,600}=30>1=∣v∣. Thus C=30 is sufficient uniformly on D1,2. At the specified point γ30=30/899, ∣t′−t∣=1−449/(15899)≈0.0016674389 and ∣x′−(x−tv)∣=30/899−1≈0.0005560189, both below 0.01. The whole-region bound values at C=30 are approximately 0.0027794768 and 0.0016680568, also below 0.01. Computed values are rounded to ten places after the decimal point, except the explicitly stated one-place shrink factors; terminating rational data are exact and all inequalities compare exact values. No physical interpretation of the auxiliary Φv,C family is asserted.

Facts & Assumptions

Given: v=e1, T=1, R=2, the point (t,x)=(1,2e1), the maps Φv,C and Gv and the region D1,2 of Galilean limiting chart maps and bounded chart regions, and the bounds and threshold of Uniform Galilean limit of the boost charts on bounded regions, and its failure on the whole chart domain.

[L1]

In the frame data, ∣v∣=∣e1∣=1 and δ(v,x)=δ(e1,2e1)=2, and Φv,C(t,x)=(γC(t−δ(v,x)/C2), x+γC−1∣v∣2δ(v,x)v−γCtv) for v≠0 while Gv(t,x)=(t,x−tv) (Galilean limiting chart maps and bounded chart regions).

[L2]

With these data the lemma gives: the exact differences t′−t=(γC−1)t−γCδ(v,x)/C2 and x′−(x−tv)=(γC−1)(x∥−tv); the bounds ∣t′−t∣≤(γC−1)T+γC∣v∣R/C2 and ∣x′−(x−tv)∣≤(γC−1)(R+∣v∣T); and the sufficient threshold Cε=max⁡{∣v∣/z1,3∣v∣R/ε} for any admissible z1 (Uniform Galilean limit of the boost charts on bounded regions, and its failure on the whole chart domain).

[L3]

Rational powers: γC=(1−1/C2)−1/2=(C2/(C2−1))1/2=C/C2−1 for C>1, and (3/4)−5/2=(4/3)5/2=(4/3)2(4/3)1/2 (Rational powers ar of a positive base, Laws of rational exponents).

[L4]

Square roots:   is the nonnegative root of Square roots exist: a unique a≥0 with (a)2=a; the positives are {x2:x≠0}, and for 0≤a<b one has a<b; in particular 3<9/4 because 3<81/16, and 600<30 because 600<900. Also (4/3)5/2=(16/9)(2/3)=32/(93) (Square roots exist: a unique a≥0 with (a)2=a; the positives are {x2:x≠0}, Monotonicity of r↦ar and of a↦ar).

[L5]

Order arithmetic in R: inequalities may be multiplied by positive numbers and added, and 0<1/300<1/200 (Ordered field): (b+d)−(a+c)=(b−a)+(d−c) and λb−λa=λ(b−a) are nonnegative when a≤b, c≤d, λ>0, by positive-cone closure, with equality handled separately.

Verification

technique · direct
1.1givenL1L3L5algebra

The exact deviations at the point. With v=e1, ∣v∣=1, δ(v,x)=2, x∥=2e1 and x−tv=e1, [L1] and [L3] give γC=C/C2−1, t′=γC(1−2C2)=(C2−2)/(CC2−1) and x′=2e1+2(γC−1)e1−γCe1=γCe1, so the exact deviations are dt(C)=1−(C2−2)/(CC2−1) and dx(C)=γC−1. Indeed, if 1<C2≤2 then t′≤0<1; if C2>2, both numerator and denominator of t′ are positive and (C2−2)2<C2(C2−1) since 4<3C2, so t′<1 again. Thus dt(C)>0 for every C>1, including C=10,30,100, and the absolute value may be read as written.

1.2givenL3L4L5algebra

The ε-threshold data. Take z0=1/4. By [L3] and [L4], K0=38(3/4)−5/2=38⋅3293=439 and A0=12+K0z0=12+39; since 3<9/4 by [L4], A0<12+14=34. With z1=1/900 one has z1<z0, A0z1<34⋅1900=11200<12 and A0Hz1=4A0z1<1300<1200=ε/2 for ε=1/100, so z1 satisfies the hypothesis of [L2],(c).

2.1step 1.1L2algebra

The whole-region bound values. Instance of [L2],(b) with T=1, R=2, ∣v∣=1: the bounds read Bt(C)=(γC−1)+2γC/C2 and Bx(C)=3(γC−1), and dt(C)≤Bt(C), dx(C)≤Bx(C) hold at every C>1 by the lemma; the statement records the ten-place rounded values of the exact expressions dt(C),dx(C),Bt(C),Bx(C) at C=10 and C=100.

2.2step 1.2L2L4algebra

The sufficient threshold. By [L2],(c) applied with z1=1/900, ∣v∣=1, R=2 and ε=1/100, Cε=max⁡{1/z1,3∣v∣R/ε}=max⁡{30,600}=30, because 600<30 by [L4]; since 30>1=∣v∣, every C≥30 meets both error bounds of [L2],(b) on the whole region D1,2.

3.1step 1.1step 2.1L3algebra

The evaluations and ratios at the named scales. For C=10 one has γ10=10/99=10/(311), dt(10)=1−98/(1099), dx(10)=γ10−1; for C=100 one has γ100=100/9999 and the corresponding expressions; for C=30 one has γ30=30/899, t′=898/(30899)=449/(15899) and the bound value Bt(30)=451/(15899)−1. The two ratios of exact deviations are dt(10)dt(100)=(1099−98)⋅10999999 (1009999−9998),dx(10)dx(100)=γ10−1γ100−1, 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.

4.1step 2.2step 3.1L2given∎

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 ε=1/100 threshold Cε=30 have been checked above from the exact radicals, and the constants K0=43/9, A0=1/2+3/9<3/4 and the inequalities z1<z0, A0z1<1/1200, A0Hz1<1/300<ε/2 hold with exact values; the auxiliary family Φv,C is mathematical and no physical claim is made.

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