Exercises — Reference frames — Galilean transformations

(b) ${\vec{a}}^{\prime }=\vec{a}$. Humne kya kiya: derivation chain ki aakhri line yaad ki. Kyun: velocity constant $\vec{V}$ se differ karti hai; ek constant ko differentiate karne se zero aata hai, isliye shift acceleration level pe disappear ho jaati hai. Options (a) aur (c) velocity shift ($\vec{V}$) aur position shift ($\vec{V}t$) ko mix up karte hain — woh ek aur do derivatives neeche rehte hain.

(i) $S^{\prime }$ mein passenger rest pe hai: $u^{\prime }=0$. (ii) Ground velocity $u=u^{\prime }+V=0+15=15\text{ m/s east}$. Kyun: "train pe rest pe hona" ka matlab hai primed frame mein zero velocity; ground observer phir bhi unhe train ki speed se aage jaate dekhta hai. Yahi "kis ke relative" ka poora matlab hai.

$u=u^{\prime }+V=1.5+1.2=2.7\text{ m/s in }+x.$ Kyun plain addition kaam karti hai: dono velocities ek hi direction ($+x$) mein point karti hain, isliye vector equation same signs ki scalar addition mein collapse ho jaati hai.

$+x$ ko positive lo. Toh $u^{\prime }=-1.5$, $V=+1.2$: $u=u^{\prime }+V=-1.5+1.2=-0.3\text{ m/s}.$ Negative sign ka matlab hai person abhi bhi $-x$ mein drift karta hai (belt unke saath nahi reh sakta), $0.3\text{ m/s}$ se. Kyun sign-tracking matter karta hai: opposite directions opposite signs ban jaate hain; tab algebra khud hi net direction bata deta hai.

Humein $u=0$ (ground) diya gaya hai aur $u^{\prime }$ chahiye. $u=u^{\prime }+V$ rearrange karo: $u^{\prime }=u-V=0-20=-20\text{ m/s},\text{ i.e. train ke relative }20\text{ m/s west.}$ Kyun: ground ke liye frozen dikhne ke liye, ball ko train pe exactly train ki speed se peeche move karna padega — dono shifts cancel ho jaate hain.

$+x$ = east, $+y$ = north set karo. Toh ${\vec{u}}^{\prime }=(0,8)$ aur $\vec{V}=(6,0)$. $\vec{u}={\vec{u}}^{\prime }+\vec{V}=(0+6,8+0)=(6,8)\text{ m/s}.$ Speed ($6$ aur $8$ legs wale right triangle pe Pythagoras — figure dekho): $|\vec{u}|=\sqrt{6^2+8^2}=\sqrt{100}=10\text{ m/s}.$ Direction. Angle $\theta$ north-of-east satisfy karta hai $\tan \theta =\frac{\text{north leg}}{\text{east leg}}=\frac{8}{6}$, isliye $\theta ={\tan }^{-1}\left(\frac{8}{6}\right)\approx 53.13^{\circ }\text{ north of east}.$ Kyun $\tan$ aur ${\tan }^{-1}$? Right triangle pe ground-velocity arrow hypotenuse hai; $\tan \theta$ = opposite/adjacent encode karta hai ki woh arrow north ki taraf kitna "steep" jhukta hai. ${\tan }^{-1}$ reverse sawaal ka jawaab deta hai "kis angle ki yeh steepness hai?" Dono components positive hain, isliye $\vec{u}$ quadrant I mein hai aur naive ${\tan }^{-1}$ ko koi correction ki zaroorat nahi.

${\vec{u}}^{\prime }=(-8,6)$, $\vec{V}=(6,0)$: $\vec{u}=(-8+6,6+0)=(-2,6)\text{ m/s}.$ Speed: $|\vec{u}|=\sqrt{(-2{)}^2+6^2}=\sqrt{40}=2\sqrt{10}\approx 6.32\text{ m/s}.$ Direction. $x$-component negative hai, $y$-component positive hai → arrow quadrant II (up-left) mein point karta hai. Naive calculator value ${\tan }^{-1}(6/-2)={\tan }^{-1}(-3)\approx -71.57^{\circ }$ quadrant IV mein land karta hai — galat, kyunki $\tan$ har $180^{\circ }$ pe repeat hota hai aur quadrant II ko quadrant IV se alag nahi bata sakta. $180^{\circ }$ add karke fix karo: $\theta =-71.57^{\circ }+180^{\circ }=108.43^{\circ }\text{ from }+x.$ Fix kyun: $\tan$ do opposite directions ko ek number pe collapse karta hai; sign pair $(-,+)$ humein quadrant II pin karta hai, isliye hum principal value ko us quadrant mein shift karte hain.

$+x$ = east. $u_A=+32$, $u_B=-18$. B ka frame $S^{\prime }$ ka role play karta hai jahan $\vec{V}=u_B$: $u_{A/B}=u_A-u_B=32-(-18)=50\text{ m/s (eastward, toward B).}$ Kyun difference nahi, sum aaya: woh opposite directions mein move karte hain, isliye negative subtract karna add karta hai — closing speed dono road-speeds ka sum hai.

${\vec{u}}_{A/B}={\vec{u}}_A-{\vec{u}}_B=(20-0,0-15)=(20,-15)\text{ m/s}.$ Speed: $\sqrt{20^2+15^2}=\sqrt{625}=25\text{ m/s}$. Direction: $x>0$, $y<0$ → quadrant IV (down-right); yahan raw ${\tan }^{-1}$ valid hai: $\theta ={\tan }^{-1}\left(\frac{-15}{20}\right)\approx -36.87^{\circ }\text{ from }+x.$ Kyun quadrant IV ko fix ki zaroorat nahi: ${\tan }^{-1}$ naturally quadrant-I aur quadrant-IV angles return karta hai, aur sign pair $(+,-)$ quadrant IV confirm karta hai.

Mouse se bahar ki taraf kaam karte hue $u=u^{\prime }+V$ do baar apply karo: $u_{\text{trolley/ground}}=V_2+V_1=3+4=7\text{ m/s east},$ $u_{\text{mouse/ground}}=u^{\prime \prime }+u_{\text{trolley/ground}}=2+7=9\text{ m/s east}.$ Kyun yeh cleanly chain hota hai: saari motions collinear (east) hain, isliye har Galilean shift ek simple add hai; frame velocities stack ho jaate hain.

$\vec{V}=\vec{0}$ ke saath: ${\vec{r}}^{\prime }=\vec{r}-\vec{0}\cdot t=\vec{r},{\vec{u}}^{\prime }=\vec{u},{\vec{a}}^{\prime }=\vec{a}.$ Har measurement agree karti hai. Physically: koi relative motion na hone wale do observers ek hi inertial frame share karte hain — transformation identity hai. Yeh sanity-check limiting case hai: koi relative motion nahi, translate karne ko kuch nahi.

${\vec{u}}^{\prime }=\vec{u}-\vec{V}(t)$ se start karo aur differentiate karo. Ab $\vec{V}$ constant nahi hai, isliye uski derivative survive karti hai: ${\vec{a}}^{\prime }=\vec{a}-\frac{d\vec{V}}{dt}=\vec{a}-\vec{A}.$ Acceleration ab invariant nahi hai. Rearrange karte hue, $m{\vec{a}}^{\prime }=m\vec{a}-m\vec{A}={\vec{F}}_{\text{real}}-m\vec{A}$; extra $-m\vec{A}$ ek aisa force lagta hai jiska koi source nahi — ek pseudo-force. Yeh kyun matter karta hai: constant $\vec{V}$ magic ingredient tha (dekho Inertial and non-inertial frames). Ise drop karo aur Newton's second law mein fictitious terms aa jaate hain.

Correction denominator compute karo: $\frac{u^{\prime }V}{c^2}=\frac{300\times 300}{(3\times 10^8{)}^2}=\frac{9\times 10^4}{9\times 10^{16}}=10^{-12}.$ Relativistic answer hai $u=\frac{600}{1+10^{-12}}\approx 600\times (1-10^{-12})$, $600\text{ m/s}$ se lagbhag $6\times 10^{-10}\text{ m/s}$ ka farq — $10^{12}$ mein ek part. Kyun Galileo yahan safe hai: correction $u^{\prime }V/c^2$ ke scale par hota hai, jo everyday speeds pe absurdly tiny hai. Sirf tab jab speeds $c$ ke paas pahunchte hain, $t^{\prime }=t$ assumption fail hoti hai aur tumhe Lorentz transformation switch karna padta hai.

Relative shift acceleration ke liye kyun disappear ho jaati hai lekin velocity ke liye nahi? ::: Kyunki $\vec{V}$ constant hai: velocity $\vec{V}$ se differ karti hai, lekin $d\vec{V}/dt=0$, isliye acceleration kisi bhi cheez se differ nahi karta. Har derivative shift $\vec{V}t\to \vec{V}\to 0$ se $t$ ki ek power hatata hai.