Dimensional analysis — checking equations, deriving relations: Visual walkthrough

1.1.3 · D2 · Physics › Measurement, Vectors & Kinematics › Dimensional analysis — checking equations, deriving relation

Step 1 — Pendulum se miliye aur har part ka naam rakho

WHAT. Ek pendulum ek choti bhaari ball (the bob) hoti hai jo ek string se latki hoti hai, use sideways kheencha jaata hai aur chodh diya jaata hai taaki yeh jhoolti rahe. Hum un teen cheezein ko naam dete hain jinse yeh bani hai:

$ℓ$ (letter "ell") — string ki length, metres mein measure ki gayi. Iska dimension $[L]$ hai.
$m$ — bob ki mass (usme kitna stuff hai), kilograms mein. Iska dimension $[M]$ hai.
$g$ — gravity ki strength: girne wali cheezein kitni tezi se speed up karti hain. Iska dimension $[L T^{- 2}]$ hai (har second, ek length per second gain hoti hai).

Jo cheez hum chahte hain woh hai $T$ — period, ek full swing aage aur waapis aane ka time. Iska dimension $[T]$ (ek time) hai.

Yeh teen ingredients WHY? Kyunki yeh woh single knobs hain jo koi bhi apparatus par ghumaata. Tum string length badal sakte ho, ball swap kar sakte ho, ya experiment Moon par kar sakte ho (alag $g$ ). Baaki sab kuch (colour, string ki thickness) plausibly timing se koi lena-dena nahi rakhta, isliye hum bet lagate hain ki period sirf $ℓ$ , $m$ , $g$ se bani hai.

PICTURE. Swing aur uske labelled parts.

Figure — Dimensional analysis — checking equations, deriving relations

Step 2 — Unknown powers ke saath ek guess banao

WHAT. Hum guess karte hain ki period in ingredients ka ek product hai, har ek kisi unknown power tak raised hai:

$T = k \cdot [L]^{a} ℓ^{a} \cdot [M]^{b} m^{b} \cdot [L T^{- 2}]^{c} g^{c}$

Symbols ko left se right padhte hue:

$T$ — woh period jise hum dhoondh rahe hain.
$k$ — ek plain number jiska koi dimension nahi (jaise $2$ ya $π$ ). Isme koi units nahi hain, isliye homogeneity ise kabhi dekh nahi sakti. Hum shuru se hi maante hain ki yeh method $k$ nahi dhoondh sakta.
$a, b, c$ — unknown powers jinhe hum solve karna chahte hain. Agar $a = \frac{1}{2}$ ho toh matlab $ℓ$ hai; agar $b = 0$ ho toh matlab mass bilkul bahar nikal jaati hai.

Product of powers kyun, sum nahi? Kyunki " $ℓ + g$ " jaisa sum hume ek length aur ek acceleration add karne par majboor karega — alag dimensions, homogeneity se forbidden. Ek product kabhi kinds ko mix nahi karta; yeh unhe sirf multiply karta hai. Isliye ek single-term guess ke liye dimensionally safe shape sirf powers ka product hai. Yahi wajah hai ki derivation trick single-term laws par kaam karti hai aur sums par fail hoti hai (parent ki limitations dekho).

PICTURE. Guess ko ek machine ke roop mein draw kiya gaya: teen dials $a, b, c$ jinhe hum tab tak ghumaate hain jab tak output kind $[T]$ se match na kare.

Step 3 — Guess ko pure dimensions mein badlo

WHAT. Ab hum har quantity ko uski kind — uske dimension — se replace karte hain, kyunki homogeneity numbers ke baare mein nahi, kinds ke baare mein statement hai. $[T] = T$ , $[ℓ] = L$ , $[m] = M$ , $[g] = L T^{- 2}$ substitute karte hue:

$T^{1} = L^{a} M^{b} (L T^{- 2})^{c}$

$k$ kyun drop karo? Kyunki $k$ dimensionless hai — iska "kind" $M^{0} L^{0} T^{0}$ hai, ek dimension equation mein $1$ ka ek invisible factor. Yahan yeh kuch contribute nahi karta (aur yahi wajah hai ki yeh hamesha hidden rehta hai).

Ab hum bracket expand karte hain. Power $c$ , $(L T^{- 2})^{c}$ ke andar dono letters ko hit karti hai:

$(L T^{- 2})^{c} = L^{c} T^{- 2 c}$

toh poori right side ban jaati hai

$T^{1} = M^{b} L^{a + c} T^{- 2 c}$

Right side par term by term:

$M^{b}$ — saari mass, ek $M$ par ikatthi.
$L^{a + c}$ — length do sources se aati hai: string ( $a$ ) aur gravity ke andar chhupi hui length ( $c$ ). Woh add hoti hain.
$T^{- 2 c}$ — right side par sirf time gravity ke "per second squared" se aata hai, isliye iska power $- 2 c$ hai.

PICTURE. Teen alag bins — ek $M$ bin, ek $L$ bin, ek $T$ bin — jinmein powers daali ja rahi hain.

Step 4 — Har base dimension ko alag-alag match karo

WHAT. Left side $M^{0} L^{0} T^{1}$ hai (period pure time hai: koi mass nahi, koi length nahi, ek time). Right side $M^{b} L^{a + c} T^{- 2 c}$ hai. Teen same letters ke do products ke equal hone ke liye, har letter ki powers ek-se-ek match honi chahiye:

M : L : T : b = 0 a + c = 0 - 2 c = 1

Har line ko ek sentence ki tarah padhte hue:

$b = 0$ — right par mass power ( $b$ ) left par mass power ( $0$ ) ke barabar honi chahiye.
$a + c = 0$ — right par total length power left ki length power ( $0$ ) ke barabar honi chahiye.
$- 2 c = 1$ — right par time power left ki ( $1$ ) ke barabar hona chahiye.

Yeh poora engine kyun hai. Har base dimension ke liye ek equation apne aap nikal aayi. Humne teen unknowns $a, b, c$ se shuruat ki aur homogeneity ne hume exactly teen equations diye — ek perfectly determined system. (Agar humne chaar ingredients guess kiye hote toh bhi sirf teen equations milte: underdetermined, aur trick ruk jaati. Yeh parent mein limitation #2 hai.)

PICTURE. Left–right "balance scale", har base dimension ke liye ek pan, dikhata hai ki kaunsi powers kaunse ke barabar honi chahiye.

Step 5 — Teen chote equations solve karo

WHAT. Unhe easy order mein solve karo:

$b = 0$ — turant ho gaya.
$- 2 c = 1 \Rightarrow c = - \frac{1}{2}$ — dono sides ko $- 2$ se divide karo.
$a + c = 0 \Rightarrow a = - c = + \frac{1}{2}$ — $c$ ko cross karo.

Toh $a = \frac{1}{2}$ , $b = 0$ , $c = - \frac{1}{2}$ .

$b = 0$ headline kyun hai. Right par $m$ ke alaawa kuch bhi koi mass nahi le jaata tha, aur left bhi koi mass nahi le jaati thi — isliye $M$ -axis balance karne ka ek hi tarika hai ki zero mass use karo. Mathematics force karta hai ki pendulum period bob ke weight ko ignore kare. Usi string par bowling ball ya marble girano: same swing time.

$a$ aur $c$ opposite kyun nikalte hain. Length axis ko zero tak cancel karna tha, aur time axis ne $c = - \frac{1}{2}$ fix kiya; length conservation ( $a + c = 0$ ) ne phir ise mirror karke $a = + \frac{1}{2}$ bana diya. Lambi string → bada $ℓ$ → dheemi swing; zyada gravity → denominator mein bada $g$ → tezi se swing.

PICTURE. Step 2 ke teen dials ab apne solved values par locked hain, ek swing sketch ke saath "longer = slower, stronger gravity = faster."

Step 6 — Law assemble karo

WHAT. Solved powers ko guess $T = k ℓ^{a} m^{b} g^{c}$ mein waapis dalo:

$T = k ℓ^{1/2} m^{0} g^{- 1/2}$

Ab har factor simplify karo:

$ℓ^{1/2} = ℓ$ — numerator mein length ka square-root.
$m^{0} = 1$ — mass gayab ho jaati hai, jaisa promise kiya tha.
$g^{- 1/2} = 1/ g$ — denominator mein gravity ka square-root.

$T = k \frac{ℓ}{g}$

$k$ kyun nahi milta. Boxed law form mein sahi hai, lekin $k$ ek bare number hai jo dimensions ko invisible hai. Full physics (ya ek experiment) reveal karta hai $k = 2 π$ , jo textbook ka $T = 2 π ℓ / g$ deta hai. Dimensional analysis ne hume zero physics se 90% tak pahuncha diya.

PICTURE. Final formula jisme har piece annotated hai aur swing par uska effect dikha hai.

Step 7 — Degenerate aur limiting cases (inhe kabhi skip mat karo)

WHAT. Formula ko uske extremes par test karo taaki sure ho sake ki yeh kabhi break na ho:

$ℓ \to 0$ (koi string nahi): $T = k 0/ g = 0$ . Bina length ke pendulum ki koi swing nahi — instant. ✅ Sensible.
$g \to 0$ (deep space, weightless): $ℓ / g \to \infty$ , isliye $T \to \infty$ . Bina gravity ke koi restoring pull nahi hai, isliye yeh kabhi waapis nahi aata — infinite period. ✅ Bilkul sahi, aur denominator mein $g$ hi yeh blow-up produce karta hai.
$g$ bada (ek bhaari planet): bada denominator ⇒ chota $T$ ⇒ tezi se swing. ✅
$m$ kuch bhi ho: $m^{0} = 1$ regardless, isliye feather-bob aur lead-bob ek barabar string par tie karte hain. ✅ Yeh mass-independence hai, phir se confirm hua.

Yeh kyun dikhao. Ek formula apne edge cases survive karna chahiye warna woh trustworthy nahi hai. Har limit yahan physical common sense se match karti hai aur waapis us exponent ( $a = + \frac{1}{2}$ vs $c = - \frac{1}{2}$ ) ki taraf point karti hai jo responsible hai — length upar, gravity neeche.

PICTURE. Do mini-panels: $ℓ \to 0$ collapse (period zero ho jaati hai) aur $g \to 0$ runaway (period infinity tak shoot karti hai).

Ek-picture summary

Ek board par poora safar: guess → dimensions → three-axis matching → solve → assemble → check limits.

Recall Feynman: walkthrough plain words mein

Main jaanna chahta tha ki ek swing ko aage-peechhe jaane mein kitna time lagta hai. Maine swing dekhi aur sirf teen dials dikhे jinhe main ghuma sakta tha: string ki length, ball ka weight, aur gravity kitni zor se kheenchti hai. Toh maine guess kiya ki answer woh teen cheezein hain jo ek saath multiply hoti hain, har ek kisi mystery power tak — ek product, kabhi sum nahi, kyunki tumhe ek length ko ek gravity se add karne ki ijazat nahi hai. Phir maine saare numbers hataa diye aur sirf kinds rakhe: length-kind, weight-kind, time-kind. Rule "tum sirf matching kinds equate kar sakte ho" teen choti demands mein split ho gayi — ek weight ke liye, ek length ke liye, ek time ke liye. Weight demand ne kaha "zero weight use karo," isliye ball ki mass simply matter nahi karti — bhaari ya hafla, same swing. Time demand ne gravity power ko minus-a-half pin kar diya, aur length ko usse mirror karke plus-a-half banana pada. Pieces ko waapis snap karne par mila: swing time length ke square-root ki tarah badhta hai aur gravity ke square-root ki tarah ghatata hai. Ek cheez jo yeh trick mujhe nahi bata sakti woh hai aage ek plain multiplying number — woh two-pi nikla, jo tum ek experiment se seekhte ho, bookkeeping se nahi.

Connections

Dimensional analysis — checking equations, deriving relations — parent; yeh page uska central result poori tarah draw out kiya gaya hai.
Buckingham Pi theorem — "base dimensions count karo taaki apne equations count kar sako" ke peeche ki general theory.
Equations of motion (kinematics) — jahan same homogeneity check galat formulas pakadta hai.
Units and the SI system — woh units jo $ℓ$ , $g$ , $T$ par actual numbers rakhte hain.
Vectors — components and addition — ek aur jagah jahan "sirf like kinds add karo" quietly algebra govern karta hai.