Problem Statement
極低光速世界における運動観測者の雷撃観測と時空遅延効果 問題文
光速 c c c S S S v s = 351.0 m/sv_s = 351.0 \text{ m/s} v s = 351.0 m/s c = 501.0 m/sc = 501.0 \text{ m/s} c = 501.0 m/s γ = 1 / 1 − ( u / c ) 2 \gamma = 1/\sqrt{1-(u/c)^2} γ = 1/ 1 − ( u / c ) 2
大気（系 S S S u = 400.8 m/su = 400.8 \text{ m/s} u = 400.8 m/s S ′ S' S ′ x ′ x' x ′ x x x
列車が x x x x x x S S S x = L x = L x = L L > 0 L > 0 L > 0 t = 0 t = 0 t = 0
列車の先頭にいる観測者は、自身の持つ時計（系 S ′ S' S ′ t ′ = t 1 ′ t' = t'_1 t ′ = t 1 ′ t ′ = t 2 ′ t' = t'_2 t ′ = t 2 ′ Δ t ′ = t 2 ′ − t 1 ′ \Delta t' = t'_2 - t'_1 Δ t ′ = t 2 ′ − t 1 ′
大気の静止系 S S S L L L Δ t ′ \Delta t' Δ t ′ L Δ t ′ \frac{L}{\Delta t'} Δ t ′ L
制約
光速（系 S S S c = 501.0 m/sc = 501.0 \text{ m/s} c = 501.0 m/s
音速（系 S S S v s = 351.0 m/sv_s = 351.0 \text{ m/s} v s = 351.0 m/s
列車の速度（系 S S S u = 400.8 m/su = 400.8 \text{ m/s} u = 400.8 m/s
時刻 t = 0 t = 0 t = 0 x = 0 x = 0 x = 0 t ′ = 0 t' = 0 t ′ = 0 S S S S ′ S' S ′
列車および軌道は完全に直線であり、落雷位置は軌道の延長線上にあるものとみなせる。空気抵抗や電磁気的な影響による列車の減速、および雷の進展に伴う位置の変化はすべて無視できるものとする。
入力形式
求める比 L Δ t ′ \frac{L}{\Delta t'} Δ t ′ L m/s \text{m/s} m/s p , q p, q p , q p q \frac{p}{q} q p p p p q q q p + q p + q p + q
解説
1. 観測者のローレンツ因子 γ \gamma γ
大気静止系 S S S u = 400.8 m/su = 400.8 \text{ m/s} u = 400.8 m/s c = 501.0 m/sc = 501.0 \text{ m/s} c = 501.0 m/s β \beta β β = u c = 400.8 501.0 = 0.80 = 4 5 \beta = \frac{u}{c} = \frac{400.8}{501.0} = 0.80 = \frac{4}{5} β = c u = 501.0 400.8 = 0.80 = 5 4 γ \gamma γ γ = 1 1 − β 2 = 1 1 − 0.64 = 1 0.60 = 5 3 \gamma = \frac{1}{\sqrt{1 - \beta^2}} = \frac{1}{\sqrt{1 - 0.64}} = \frac{1}{0.60} = \frac{5}{3} γ = 1 − β 2 1 = 1 − 0.64 1 = 0.60 1 = 3 5
2. 大気静止系 S S S
系 S S S t = 0 t=0 t = 0 x = L x=L x = L x ( t ) = u t x(t) = ut x ( t ) = u t
光の到達（イベント1） :
光は大気中を速度 − c -c − c x = L x=L x = L x light( t ) = L − c t x_{\text{light}}(t) = L - ct x light ( t ) = L − c t t 1 t_1 t 1 u t 1 = L − c t 1 ut_1 = L - ct_1 u t 1 = L − c t 1 t 1 = L c + u t_1 = \frac{L}{c + u} t 1 = c + u L x 1 = u t 1 = u L c + u x_1 = ut_1 = \frac{uL}{c + u} x 1 = u t 1 = c + u uL
音の到達（イベント2） :
音は媒質である大気（系 S S S − v s -v_s − v s x sound( t ) = L − v s t x_{\text{sound}}(t) = L - v_s t x sound ( t ) = L − v s t t 2 t_2 t 2 u t 2 = L − v s t 2 ut_2 = L - v_s t_2 u t 2 = L − v s t 2 t 2 = L v s + u t_2 = \frac{L}{v_s + u} t 2 = v s + u L x 2 = u t 2 = u L v s + u x_2 = ut_2 = \frac{uL}{v_s + u} x 2 = u t 2 = v s + u uL
3. 列車静止系 S ′ S' S ′
列車内の観測者が「自身の時計」で測る時刻 t ′ t' t ′ t ′ = γ ( t − u x c 2 ) t' = \gamma \left( t - \frac{ux}{c^2} \right) t ′ = γ ( t − c 2 ux ) S S S
光の観測時刻 t 1 ′ t'_1 t 1 ′ :
t 1 ′ = γ ( t 1 − u x 1 c 2 ) = γ t 1 ( 1 − u 2 c 2 ) = γ t 1 ( 1 − β 2 ) = t 1 γ t'_1 = \gamma \left( t_1 - \frac{u x_1}{c^2} \right) = \gamma t_1 \left( 1 - \frac{u^2}{c^2} \right) = \gamma t_1 (1 - \beta^2) = \frac{t_1}{\gamma} t 1 ′ = γ ( t 1 − c 2 u x 1 ) = γ t 1 ( 1 − c 2 u 2 ) = γ t 1 ( 1 − β 2 ) = γ t 1 t 1 ′ = 1 γ ⋅ L c + u t'_1 = \frac{1}{\gamma} \cdot \frac{L}{c + u} t 1 ′ = γ 1 ⋅ c + u L
音の観測時刻 t 2 ′ t'_2 t 2 ′ :
t 2 ′ = γ ( t 2 − u x 2 c 2 ) = γ t 2 ( 1 − u 2 c 2 ) = γ t 2 ( 1 − β 2 ) = t 2 γ t'_2 = \gamma \left( t_2 - \frac{u x_2}{c^2} \right) = \gamma t_2 \left( 1 - \frac{u^2}{c^2} \right) = \gamma t_2 (1 - \beta^2) = \frac{t_2}{\gamma} t 2 ′ = γ ( t 2 − c 2 u x 2 ) = γ t 2 ( 1 − c 2 u 2 ) = γ t 2 ( 1 − β 2 ) = γ t 2 t 2 ′ = 1 γ ⋅ L v s + u t'_2 = \frac{1}{\gamma} \cdot \frac{L}{v_s + u} t 2 ′ = γ 1 ⋅ v s + u L
運動する観測者から見ても、光と音の観測時刻がそれぞれ大気静止系における時間に 1 γ \frac{1}{\gamma} γ 1 x = u t x = ut x = u t
4. 列車内で観測される時間差 Δ t ′ \Delta t' Δ t ′
列車内で観測される光と音の時間差 Δ t ′ \Delta t' Δ t ′ Δ t ′ = t 2 ′ − t 1 ′ = L γ ( 1 v s + u − 1 c + u ) = L γ ( c + u ) − ( v s + u ) ( v s + u ) ( c + u ) = L γ c − v s ( v s + u ) ( c + u ) \Delta t' = t'_2 - t'_1 = \frac{L}{\gamma} \left( \frac{1}{v_s + u} - \frac{1}{c + u} \right) = \frac{L}{\gamma} \frac{(c + u) - (v_s + u)}{(v_s + u)(c + u)} = \frac{L}{\gamma} \frac{c - v_s}{(v_s + u)(c + u)} Δ t ′ = t 2 ′ − t 1 ′ = γ L ( v s + u 1 − c + u 1 ) = γ L ( v s + u ) ( c + u ) ( c + u ) − ( v s + u ) = γ L ( v s + u ) ( c + u ) c − v s
求めたい比 L Δ t ′ \frac{L}{\Delta t'} Δ t ′ L L Δ t ′ = γ ( v s + u ) ( c + u ) c − v s \frac{L}{\Delta t'} = \gamma \frac{(v_s + u)(c + u)}{c - v_s} Δ t ′ L = γ c − v s ( v s + u ) ( c + u )
5. 数値の代入と既約分数化
与えられた具体的な物理定数を代入します。
c − v s = 501.0 − 351.0 = 150.0 c - v_s = 501.0 - 351.0 = 150.0 c − v s = 501.0 − 351.0 = 150.0 v s + u = 351.0 + 400.8 = 751.8 v_s + u = 351.0 + 400.8 = 751.8 v s + u = 351.0 + 400.8 = 751.8 c + u = 501.0 + 400.8 = 901.8 c + u = 501.0 + 400.8 = 901.8 c + u = 501.0 + 400.8 = 901.8 γ = 5 3 \gamma = \frac{5}{3} γ = 3 5
これらを代入して計算を進めます。
L Δ t ′ = 5 3 ⋅ 751.8 × 901.8 150.0 = 1 3 ⋅ 751.8 × 901.8 30.0 = 751.8 × 901.8 90.0 \frac{L}{\Delta t'} = \frac{5}{3} \cdot \frac{751.8 \times 901.8}{150.0} = \frac{1}{3} \cdot \frac{751.8 \times 901.8}{30.0} = \frac{751.8 \times 901.8}{90.0} Δ t ′ L = 3 5 ⋅ 150.0 751.8 × 901.8 = 3 1 ⋅ 30.0 751.8 × 901.8 = 90.0 751.8 × 901.8 10 10 10 L Δ t ′ = 7518 × 9018 9000 \frac{L}{\Delta t'} = \frac{7518 \times 9018}{9000} Δ t ′ L = 9000 7518 × 9018 9 9 9 6 6 6
9018 ÷ 9000 = 1002 1000 9018 \div 9000 = \frac{1002}{1000} 9018 ÷ 9000 = 1000 1002 L Δ t ′ = 7518 × 1002 1000 \frac{L}{\Delta t'} = \frac{7518 \times 1002}{1000} Δ t ′ L = 1000 7518 × 1002
さらに 2 2 2
7518 ÷ 2 = 3759 7518 \div 2 = 3759 7518 ÷ 2 = 3759 1000 ÷ 2 = 500 1000 \div 2 = 500 1000 ÷ 2 = 500 1002 ÷ 2 = 501 1002 \div 2 = 501 1002 ÷ 2 = 501 500 ÷ 2 = 250 500 \div 2 = 250 500 ÷ 2 = 250
これにより、式は以下のようになります。
L Δ t ′ = 3759 × 501 250 = 1883259 250 \frac{L}{\Delta t'} = \frac{3759 \times 501}{250} = \frac{1883259}{250} Δ t ′ L = 250 3759 × 501 = 250 1883259
分母の因数は 250 = 2 × 5 3 250 = 2 \times 5^3 250 = 2 × 5 3 2 2 2 5 5 5 1883259 1883259 1883259 9 9 9 2 2 2 5 5 5 p q \frac{p}{q} q p
p = 1883259 p = 1883259 p = 1883259 q = 250 q = 250 q = 250
自然数化
求めた既約分数の分子 p p p q q q p + q = 1883259 + 250 = 1883509 p + q = 1883259 + 250 = 1883509 p + q = 1883259 + 250 = 1883509
解答：1883509