Visual walkthrough — Reference frames — Galilean transformations

1.1.22 · D2 · Physics › Measurement, Vectors & Kinematics › Reference frames — Galilean transformations

Hum puri tarah sirf do ideas use karte hain: (1) ek arrow jo ek jagah se doosri jagah point karta hai (ek "vector"), aur (2) do arrows jodne ka head-to-tail rule. Dono Vectors — addition and components mein hain — agar arrows naye lagte hain toh wahan ek nazar daalo.

---

Step 1 — Do observers, do rulers

KYA. Hum do observers draw karte hain. Pehle ko $S$ kaho (socho: koi zameen par khada hai). Doosre ko $S^{\prime }$ kaho — ise "S-prime" padho, woh chhota tick mark sirf itna matlab hai "doosra wala" (socho: koi train mein baitha hai). Dono ke paas ek ruler $x$-direction mein aur ek $y$-direction mein rakha hai, saath mein ek clock bhi.

KYUN. Koi measurement meaningless hai jab tak hum yeh na batayein kisne measure kiya. Position aur time sirf numbers hain kisi ke ruler aur clock ke relative. Toh kisi bhi formula se pehle, humein do rulers pakke karne padenge. Reference frame ka yahi toh poora point hai — reference frame.

PICTURE. Ground observer ka origin $O$ hai (magenta). Train observer ka origin $O^{\prime }$ hai (violet). Jis instant $t=0$ ho, hum deliberately unhe slide karte hain taaki dono origins ek upar ek baith jayein — yeh sirf ek constant bachata hai aur kuch bhi kharcha nahi hota.

---

Step 2 — Train ka origin door drift karta hai

KYA. Train ek constant velocity se chalti hai, ek arrow jise hum $\vec{V}$ kehte hain (orange). "Constant" matlab arrow ki length ya direction kabhi nahi badlti. Jab $t$ time tick ho jaata hai, ground ke hisaab se train ka origin $O^{\prime }$ kahan slide ho gaya?

KYUN. Aage ki sab bookkeeping do origins ke beech ke gap ke baare mein hai. Yahi gap do observers mein farq hai, isliye pehle ise compute karte hain. Aur iska rule kinematics ka sabse basic rule hai: distance = velocity × time.

PICTURE. $O$ se shuru karke, origin $O^{\prime }$ arrow $\vec{V}t$ ki tip tak slide ho gaya hai.

$\underset{\text{origins ke beech gap}}{\underbrace{\vec{V}t}}=\underset{\begin{array}{c}\text{train ki velocity}\\ \text{(fixed arrow)}\end{array}}{\underbrace{\vec{V}}}\times \underset{\begin{array}{c}\text{clock par}\\ \text{time}\end{array}}{\underbrace{t}}$

Term by term padhein: $\vec{V}$ drift ki direction aur speed set karta hai; $t$ se multiply karne par woh arrow utna lamba hota jaata hai jitna zyada time beeche. $t=0$ par arrow ki length zero hai — origins milte hain, bilkul jaisa Step 1 mein arrange kiya tha.

---

Step 3 — Ek event, do nazariye se dekha

KYA. Ab kuch hota hai — ek firecracker pops karta hai. Yeh ek event hai: ek definite jagah par ek definite time par. Ground uski position $\vec{r}$ kehta hai; train usi pop ki position $\vec{r}{}^{\prime }$ kehta hai. Hum in do arrows ka relationship chahte hain.

KYUN. Yahi derivation ka dil hai, aur isme koi physics nahi lagti — yeh pure geometry hai. Event tak $O$ se pahunchne ke liye, ya toh seedha wahan jao ($\vec{r}$), ya ek detour lo: pehle train ke origin tak chalo ($\vec{V}t$), phir wahan se event tak chalo ($\vec{r}{}^{\prime }$). Dono raaste ek hi jagah khatam hote hain, isliye arrows match hone chahiye.

PICTURE. Teen arrows ka triangle dekho. Head-to-tail rule kehta hai seedha arrow do-leg path ke barabar hai:

$\underset{\begin{array}{c}\text{event jaise ground}\\ \text{se }O\text{ ne dekha}\end{array}}{\underbrace{\vec{r}}}=\underset{\begin{array}{c}O\to O^{\prime }\\ \text{(Step 2 se)}\end{array}}{\underbrace{\vec{V}t}}+\underset{\begin{array}{c}{O}^{\prime }\to \text{event}\\ \text{(train ka jawab)}\end{array}}{\underbrace{\vec{r}{}^{\prime }}}$

---

Step 4 — Train ke coordinates ke liye rearrange karo

KYA. Ab hum triangle equation ko $\vec{r}{}^{\prime }$ ke liye solve karte hain — primed observer ka jawab un quantities mein jo ground measure kar sakta hai.

KYUN. Practice mein hum usually ground numbers ($\vec{r}$, $t$, $\vec{V}$) jaante hain aur train ke numbers chahte hain. $\vec{V}t$ ko equals sign ke doosri taraf le jaane se exactly yahi hota hai.

PICTURE. $\vec{V}t$ subtract karne ka matlab hai "orange arrow ke saath waapis chalo" — wahi triangle, ulta padha.

$\boxed{\underset{\text{train ki position}}{\underbrace{\vec{r}{}^{\prime }}}=\underset{\text{ground ki position}}{\underbrace{\vec{r}}}-\underset{\begin{array}{c}\text{origin drift ko}\\ \text{undo karo}\end{array}}{\underbrace{\vec{V}t}},\underset{\begin{array}{c}\text{clocks agree karte hain}\\ \text{(Galilean bet)}\end{array}}{\underbrace{t^{\prime }=t}}}$

Components mein $\vec{V}=V\hat{x}$ ke saath (train $x$ ke saath chal rahi hai):

$x^{\prime }=x-Vt,y^{\prime }=y,z^{\prime }=z,t^{\prime }=t.$

Sirf woh $x$ jo $\vec{V}$ ke saath line up karta hai shift hota hai; sideways directions untouched rehti hain — seedha triangle se padho.

---

Step 5 — Pehla derivative: velocities kaise transform hoti hain

KYA. Velocity yeh hai ki position arrow har second kitna change hota hai. Hum ise $\frac{d}{dt}$ se likhte hain, jo sirf "per unit time rate of change" ka shorthand hai. Ise poore position rule par apply karo.

KYUN derivative yahan use karein? Kyunki velocity defined hai position ke time-rate-of-change ke roop mein. Hum style ke liye koi fancy tool nahi chun rahe — "velocities kaise compare hoti hain?" ka sawaal literally yeh sawaal hai "position relation har second kaise badlta hai?", aur $\frac{d}{dt}$ woh operation hai jo exactly yeh answer karta hai. Koi aur tool per-second rate measure nahi karta.

PICTURE. $\vec{r}{}^{\prime }=\vec{r}-\vec{V}t$ ka $\frac{d}{dt}$ lo. Kyunki $t^{\prime }=t$, dono observers ki clocks same rate par chalti hain, isliye "per second" differentiate karna dono ke liye same matlab rakhta hai. $\vec{r}$ ki rate of change ground velocity $\vec{u}$ hai; $\vec{r}{}^{\prime }$ ki train velocity $\vec{u}{}^{\prime }$ hai; aur $\vec{V}t$ ki rate of change sirf $\vec{V}$ hai (kyunki $\vec{V}$ fixed arrow hai):

$\underset{\begin{array}{c}\text{event ki velocity}\\ \text{train ke hisaab se}\end{array}}{\underbrace{\vec{u}{}^{\prime }}}=\underset{\begin{array}{c}\text{event ki velocity}\\ \text{ground ke hisaab se}\end{array}}{\underbrace{\vec{u}}}-\underset{=\vec{V}}{\underbrace{\frac{d(\vec{V}t)}{dt}}}⟹\boxed{\vec{u}=\vec{u}{}^{\prime }+\vec{V}}$

Box padhein: ground velocity train ki apni reading plus train ki velocity hai. Velocities exactly ek $\vec{V}$ se differ karti hain — $\vec{V}t$ gap se $t$ ki ek power "knock off" ho gayi.

---

Step 6 — Doosra derivative: acceleration shared hai

KYA. Acceleration yeh hai ki velocity arrow har second kitna badlta hai — $\frac{d}{dt}$ ek aur baar apply karo, ab velocity rule par.

KYUN. Step 5 jaisi hi logic, ek level upar: acceleration defined hai velocity ke rate of change ke roop mein, isliye phir se differentiate karte hain. Crucial fact yeh hai ki $\vec{V}$ constant hai, aur kisi cheez ka rate of change jo kabhi nahi badlti woh zero hota hai.

PICTURE. $\vec{u}=\vec{u}{}^{\prime }+\vec{V}$ differentiate karo:

$\underset{\text{ground ki accel.}}{\underbrace{\vec{a}}}=\underset{\text{train ki accel.}}{\underbrace{\vec{a}{}^{\prime }}}+\underset{=0\text{(constant!)}}{\underbrace{\frac{d\vec{V}}{dt}}}⟹\boxed{\vec{a}{}^{\prime }=\vec{a}}$

Extra shift ab bilkul gayab ho gaya. Dono observers identical acceleration measure karte hain. Kyunki mass dono ke liye same hai, Newton ka $\vec{F}=m\vec{a}$ dono frames mein same padha jaata hai — yahi Galilean relativity hai.

---

Step 7 — Degenerate cases (kuch bhi uncovered nahi)

KYA. Aao har knob ko extreme par push karein aur check karein ki pictures phir bhi hold karti hain.

KYUN. Jo rule tum trust karo woh apne edge cases survive karna chahiye. Agar koi special input triangle tod deta, toh hum use sach mein nahi samjhe hote.

PICTURE. Usi triangle ke char collapses:

• $t=0$ — gap $\vec{V}t$ ki length zero hai. Triangle ek point mein flat ho jaata hai: $\vec{r}{}^{\prime }=\vec{r}$. Do observers shuru ke instant mein agree karte hain, hamare Step-1 setup ke mutabiq.
• $\vec{V}=\vec{0}$ — train kabhi nahi chalti. Toh $\vec{V}t=\vec{0}$ har time par: $\vec{r}{}^{\prime }=\vec{r}$, $\vec{u}{}^{\prime }=\vec{u}$. Same frame, same answers.
• Event at $O^{\prime }$ (train ke origin par firecracker) — toh $\vec{r}{}^{\prime }=\vec{0}$ aur $\vec{r}=\vec{V}t$: event sirf drifting origin ko track karta hai, Step 2 recover karta hai.
• $\vec{V}$ not constant (accelerating train) — ab $\frac{d\vec{V}}{dt}=\vec{A}\ne 0$, toh Step 6 deta hai $\vec{a}{}^{\prime }=\vec{a}-\vec{A}$. Frame non-inertial hai aur pseudo-forces appear hoti hain. Inertial and non-inertial frames dekho — constant $\vec{V}$ woh magic ingredient tha.

---

Ek-picture summary

Poori derivation ek triangle hai jo do baar differentiate kiya gaya. Position full drift arrow $\vec{V}t$ se differ karti hai; velocity constant arrow $\vec{V}$ se differ karti hai; acceleration kisi se nahi.

Recall Feynman retelling — poora walkthrough simple shabdon mein

Socho tum aur ek dost. Tumhara dost ek train mein chadh jaata hai jo steady speed se chalti hai. Seeti bajne par ($t=0$) tum kaandhe se kaandha milakar khade ho, isliye tum dono agree karte ho ki sab kuch kahan hai. Jaise train door jaati hai, tumhare beech ka gap exactly "train-speed times seconds" se badhta hai — yahi orange arrow hai. Ab kahin ek firecracker pops karta hai. Yeh pata karne ke liye ki tum kahan sochte ho woh hai, ya toh seedha wahan chalo, ya pehle apne dost ke paas jao aur phir firecracker tak — same trip hai, toh woh arrows add up hone chahiye. Woh adding-up hi position rule hai. "Firecracker kitni tez chal raha hai?" poocho — yeh sirf yeh hai ki uska arrow har second kitna badlta hai. Kyunki gap steadily badhta hai, tumhare dono jawaab exactly train ki speed se differ karte hain: velocities ek $\vec{V}$ se differ karti hain. "Yeh kitni tez speed up karta hai?" poocho — jo differences steady the woh ab kuch contribute nahi karte, kyunki ek steady train-speed khud speed up nahi karta. Toh tum dono same acceleration paate ho. Woh shared number ki wajah se train par aur zameen par physics ke same laws kaam karte hain — aur yeh sab ek triangle se nikla, ek baar draw kiya aur teen baar padha.

---

Recall checkpoints

Recall

Position rule ko arrows (vectors) kyun chahiye, sirf number subtraction kyun nahi? ::: Kyunki event train ki motion ke sideways ho sakta hai; arrows sideways parts ko equal rakhte hain aur sirf along-$\vec{V}$ part ko automatically shift karte hain. Chain $\vec{V}t\to \vec{V}\to 0$ mein, ek step right move karne ke liye kaunsa operation chahiye? ::: Time-derivative $d/dt$ — har ek $t$ ki ek power peel off karta hai. Kaunsa single assumption transformation ko Lorentz ki jagah Galilean banata hai? ::: Absolute time, $t^{\prime }=t$ (sab clocks identically tick karte hain). Agar train $\vec{A}$ se accelerate kare, toh Step 6 kya banta hai? ::: $\vec{a}{}^{\prime }=\vec{a}-\vec{A}$ — frame non-inertial hai aur pseudo-forces appear hoti hain. $t=0$ par do observers position par kyun agree karte hain? ::: Drift arrow $\vec{V}t$ ki length zero hoti hai, isliye $\vec{r}{}^{\prime }=\vec{r}$.

---

Connections

• Vectors — addition and components — head-to-tail triangle Step 3 ka engine hai.
• Relative velocity — Step 5 rearranged: ${\vec{u}}_{A/B}={\vec{u}}_A-{\vec{u}}_B$.
• Inertial and non-inertial frames — Step 7 ka accelerating-train case.
• Newton's laws of motion — invariant hai kyunki Step 6 ki wajah se.
• Special relativity — Lorentz transformation — $t^{\prime }=t$ ko replace karta hai jab $v\sim c$ ho.