Covariant and contravariant components: Question bank

4.10.8 · D5 · Maths › Advanced Topics (Elite Level) › Covariant aur contravariant components

Ye page Covariant and contravariant components ke traps ko target karta hai. Prerequisites jo click karne layak hain: Dual (reciprocal) basis, Tensors and index notation, Change of basis and transformation laws, Inner product spaces.

Ground rules (traps se pehle ye padho)

Neeche sab kuch 2 dimensions mein hai jab tak prompt aur na kahe: ek flat plane par bana hua ek skew basis ${e_{1}, e_{2}}$ . 2D mein rakhne se hum har projection dekh sakte hain.

Figure — Covariant and contravariant components

Definition Dual (reciprocal) basis — aur

kyun exist karta hai ${e_{1}, e_{2}}$ ke saath ek doosra pair ${e^{1}, e^{2}}$ define karo is rule se $e^{i} \cdot e_{j} = δ_{j}^{i} .$ Literally padho: $e^{1}$ , $e_{2}$ ke perpendicular hai (unka dot $0$ hai) aur is tarah scaled hai ki $e^{1} \cdot e_{1} = 1$ ; similarly $e^{2} ⊥ e_{1}$ . Hum ise hamesha kyun bana sakte hain? Kyunki basis non-parallel hai, " $e_{2}$ ke perpendicular" ek definite direction pick karta hai aur scaling condition length fix karti hai — ek unique solution exist karta hai. Dual basis exactly woh tool hai jo shadow numbers $v_{i}$ ko honest expansion coefficients mein badalta hai, kyunki $v = v_{i} e^{i} because v \cdot e_{j} = (v_{i} e^{i}) \cdot e_{j} = v_{i} δ_{j}^{i} = v_{j},$ jo definition $v_{j} = v \cdot e_{j}$ se match karta hai. Toh $v = v^{i} e_{i} = v_{i} e^{i}$ ek coincidence nahi hai — dual basis design hi aise kiya gaya tha ki doosra expansion kaam kare. Figure mein ${e^{i}}$ (amber) ${e_{i}}$ (white) se rotated baithe hue dikhaye gaye hain.

True or false — justify karo

Neeche har prompt ek claim hai. Reveal T/F aur kyun kehta hai — "kyun" hi point hai.

Ek vector

v

ke alag covariant aur contravariant components hain, isliye ye aslmein do alag vectors hain.

False.

v

ek geometric arrow hai jo coordinates ko ignore karta hai;

v^{i}

aur

v_{i}

usi arrow ki do readings hain (

v = v^{i} e_{i} = v_{i} e^{i}

), jaise ek hi point ko polar aur Cartesian dono addresses dena.

Agar basis orthonormal hai, toh covariant aur contravariant components identical hain.

True. Tab

g_{ij} = e_{i} \cdot e_{j} = δ_{ij}

, isliye lowering

v_{i} = g_{ij} v^{j}

kuch nahi karta aur

v_{i} = v^{i}

. Isliye school mein ye distinction kabhi mention nahi hoti.

Metric

g_{ij}

hamesha identity matrix hoti hai.

False. Ye identity sirf orthonormal basis ke liye hai. Skew ya stretched basis ke liye iske off-diagonal entries

e_{i} \cdot e_{j}

(

i \neq = j

) nonzero hote hain aur diagonal entries axis lengths encode karti hain.

g_{ij}

symmetric hai kyunki dot product commutative hai.

True.

g_{ij} = e_{i} \cdot e_{j} = e_{j} \cdot e_{i} = g_{j i}

, isliye ek real inner-product metric hamesha symmetric hoti hai. Hamare 2D setting mein iska matlab hai ki iske 4 entries mein se sirf 3 independent hain (

g_{12} = g_{21}

Covariant components

v_{i}

aslmein dual basis mein

v

ke contravariant components hain.

True. Kyunki

v = v_{i} e^{i}

, numbers

v_{i}

literally

{e^{i}}

ke along parallel-expansion coefficients hain. "Covariant" aur "contravariant" do alag bases ke against dekha gaya ek hi idea hai.

Index raise karke phir lower karne par original components wapis milte hain.

True.

g_{ij} (g^{j k} v_{k}) = (g_{ij} g^{j k}) v_{k} = δ_{i}^{k} v_{k} = v_{i}

kyunki

g^{ij}

defined hai

g_{ij}

ke inverse ke roop mein aur

δ_{i}^{k}

index filter ki tarah kaam karta hai. Dono operations exact inverses hain.

g^{ij}

ek genuinely naya object hai jise

g_{ij}

se alag compute karna hoga.

Half-false.

g^{ij}

poori tarah

g_{ij}

se determined hota hai — ye iska matrix inverse hai (

g^{ik} g_{k j} = δ_{j}^{i}

). Ye independent data nahi hai; ye same geometry doosri taraf se padhi gayi hai (aur ye sirf isliye exist karta hai kyunki basis non-degenerate hai).

Error dhundho

Har line ek plausible lekin galat move batati hai. Reveal flaw ka naam aur fix batata hai.

v

ke expansion coefficients paane ke liye bas

v \cdot e_{i}

compute karo."

Skew basis mein galat.

v \cdot e_{i} = v_{i}

covariant (perpendicular/shadow) numbers deta hai, expansion coefficients

v^{i}

nahi.

v^{i}

paane ke liye

v^{i} e_{i} = v

solve karo ya raise karo:

v^{i} = g^{ij} v_{j}

"Squared length

v^{i} v^{i}

hai (dono components squared sum karo)."

Galat. Do up-indices sum karna secretly assume karta hai

g_{ij} = δ_{ij}

. Invariant hai

∣ v ∣^{2} = v^{i} v_{i} = g_{ij} v^{i} v^{j}

— ek up ek down ke saath paired.

"Covariant components basis ke inversely transform hote hain, kyunki 'co' ka matlab opposite hai."

Galat. Covariant basis ke saath transform hota hai (matrix

A

\tilde{v}_{i} = A^{j}_{i} v_{j}

); contravariant

A^{- 1}

ke saath transform hota hai, basis ke against. Prefix change of basis ke under behaviour describe karta hai, aur ye intuitive guess ka ulta hai.

"Kyunki indices sirf labels hain, main

v^{i} w^{i}

likh ke sum kar sakta hoon."

Galat. Sirf up–down contraction

v^{i} w_{i}

coordinate-free hai (ek true scalar).

v^{i} w^{i}

ka value basis change karne par badal jaata hai, isliye ye meaningful invariant nahi hai — hamesha staircase rakhna, ek up + ek down.

"Ek scalar ka gradient ek contravariant vector hai, kyunki ye 'kisi taraf point' karta hai."

Galat.

\partial_{i} f = \partial f / \partial x^{i}

ek lower index carry karta hai — ye naturally ek covector hai. Ise ek contravariant displacement-type vector banana ho toh raise karo:

(\nabla f)^{i} = g^{ij} \partial_{j} f

(dekho General relativity — raising and lowering indices).

"Polar coordinates mein covariant aur contravariant gradient same hain, jaise Cartesian mein."

Galat. Curvilinear coordinates mein metric identity nahi hai (

g_{θ θ} = r^{2}

), isliye index raise karna entries ko rescale karta hai. Cartesian mein ye coincidence special hai, general nahi.

g_{ij} v^{i} v^{j}

aur

g^{ij} v_{i} v_{j}

alag lengths dete hain."

Galat. Dono

∣ v ∣^{2}

ke barabar hain. Kyunki

v_{i} = g_{ij} v^{j}

, substitution se dikhta hai

g^{ij} v_{i} v_{j} = g^{ij} g_{ik} v^{k} g_{j l} v^{l} = δ_{k}^{j} g_{j l} v^{k} v^{l} = g_{ij} v^{i} v^{j}

. Same scalar (ye

δ

j \to k

filter karta hai).

Why questions

Ek hi vector ke liye hume do tarah ke components ki zaroorat kyun hai?

Kyunki non-orthonormal basis mein "

v

mein kitna

e_{1}

hai" ke do honest answers hote hain — slide-along-axes (parallel) vs cast-a-shadow (perpendicular) — aur dono alag hote hain. Orthonormal bases ye chhupa dete hain kyunki dono agree karte hain.

Dot product

e_{i} \cdot e_{j}

ko "metric" naam dene ki sahi cheez kyun hai?

Kyunki jab tum index lower karte ho toh substitution

v_{i} = (v^{j} e_{j}) \cdot e_{i} = v^{j} (e_{j} \cdot e_{i})

e_{i} \cdot e_{j}

unavoidably appear hota hai. Ye exactly woh data hai (lengths aur angles) jo up-indices ko down-indices mein convert karne ke liye chahiye.

Ek invariant scalar mein ek up aur ek down index kyun pair hona chahiye?

Basis change ke under up-index

A^{- 1}

se transform hota hai aur down-index

A

se; unka product cancel ho jaata hai (

A^{- 1} A = I

, yani

δ

), ek aisa number bachta hai jis par sab observers agree karte hain. Do same-height indices cancel nahi hote.

"Contra" aur "co" naam components se kyun attached hain, vector se kyun nahi?

Vector fixed hai; sirf uske numbers basis change par respond karte hain. "Contra" un numbers ko describe karta hai jo

v

ko fixed rakhne ke liye basis ke opposite move karte hain; "co" un numbers ko describe karta hai jo basis ke saath move karte hain.

Dual basis vector

e^{1}

e_{2}

ke perpendicular kyun point karta hai?

Kyunki

e^{1} \cdot e_{2} = δ_{2}^{1} = 0

perpendicularity force karta hai, aur

e^{1} \cdot e_{1} = 1

iska length fix karta hai. Ye exactly woh hai jo

v_{i} = v \cdot e_{i}

{e^{i}}

ke along coefficients banata hai.

Physically covariant/contravariant machinery kyun matter karti hai?

Curved spacetime aur curvilinear frames mein metric genuinely non-trivial hoti hai, isliye raising/lowering hi ek sahi tarika hai coordinate-independent quantities jaise lengths, energies, aur physical laws banane ka.

Edge cases

Jab basis orthonormal hoti jaati hai toh covariant/contravariant distinction ka kya hota hai?

Ye smoothly collapse hoti hai: off-diagonal

g_{ij} \to 0

aur diagonal

\to 1

, isliye

v_{i} \to v^{i}

aur

g^{ij} \to g_{ij} \to δ_{ij}

. Dono readings merge hokar ek "school" component ban jaate hain.

Agar dono basis vectors almost same direction mein point karein (almost parallel)?

Basis almost degenerate ho jaati hai:

det g \to 0

, isliye

g^{ij}

blow up karta hai aur contravariant components bahut bade aur numerically unstable ho jaate hain. Geometrically, arrow mein thoda sa change karne ke liye near-parallel axes ke along bahut bade coefficients chahiye hote hain.

Agar do basis vectors actually parallel hon (linearly dependent)?

Tab ye basis hi nahi hai:

g

singular hai,

g^{- 1}

exist nahi karta, aur tum indices raise nahi kar sakte. Kuch vectors ka koi expansion nahi hoga aur dooson ke infinitely many expansions honge — poora framework ek genuine basis maangta hai.

Zero vector ke liye, uske covariant aur contravariant components kya hain?

Kisi bhi basis mein dono mein sab zero:

0 = 0 \cdot g^{ij} = g_{ij} \cdot 0

. Zero vector woh ek case hai jahan "kaun sa projection?" kabhi matter nahi karta.

Agar basis orthogonal ho lekin normalized nahi (axes perpendicular lekin alag lengths), toh kya

v_{i} = v^{i}

Nahi. Tab

g_{ij}

diagonal hai lekin identity nahi (

g_{ii} = ∣ e_{i} ∣^{2} \neq = 1

), isliye

v_{i} = g_{ii} v^{i}

har component ko rescale karta hai. Orthogonality off-diagonal mixing ko khatam karti hai lekin length rescaling ko nahi.

Ek dimension mein, covariant aur contravariant mein koi fark hai?

Haan, agar single basis vector unit length nahi hai:

g_{11} = ∣ e_{1} ∣^{2}

, isliye

v_{1} = g_{11} v^{1}

. Distinction 1D mein bhi survive karti hai jab bhi axis stretched ho.

Recall Final gut-check

Ek saanson mein kaho: contravariant = parallel projection = index up = basis ke against transform hota hai ( $A^{- 1}$ ); covariant = perpendicular projection = index down = basis ke saath transform hota hai ( $A$ ); metric $g_{ij}$ tumhe staircase se neeche le jaata hai aur $g^{ij}$ upar le jaata hai.