Covariant and contravariant components: Worked examples

4.10.8 · D3 · Maths › Advanced Topics (Elite Level) › Covariant and contravariant components

Yeh page parent topic ka drill hall hai. Hum do rules lete hain — parallel projection (contravariant, index upar) aur perpendicular projection (covariant, index neeche) — aur inhe har tarah ki basis geometry pe run karte hain jo ek problem mein aa sakti hai. Agar yeh symbols pehle kabhi nahi dekhe, pehle parent padho; yahan sirf compute karte hain.

The scenario matrix

Inhi components ke baare mein har problem asliyat mein neeche ke kisi ek cell jaisi hoti hai. Hum har cell ko ek labelled example se hit karenge.

Cell	Basis geometry	Kya stressed hai	Example
A	Orthonormal (square)	Distinction collapse ho jaata hai, $v_{i} = v^{i}$	Ex 1
B	Orthogonal lekin unequal lengths (rectangular)	$g_{ij}$ diagonal hai par $\neq = δ_{ij}$ ; sirf scaling	Ex 2
C	Skew, axes ke beech acute angle	$g_{12} > 0$ ; full split	Ex 3
D	Skew, axes ke beech obtuse angle	$g_{12} < 0$ ; components mein sign flip	Ex 4
E	Dual (reciprocal) basis explicitly banaya	Dekho $v_{i}$ "kahaan rehta hai"	Ex 5
F	Degenerate / limiting (axes → parallel)	$g$ singular ho jaata hai, $g^{- 1}$ blow up karta hai	Ex 6
G	Curvilinear (polar) — position-dependent basis	gradient covariant hota hai	Ex 7
H	Word problem + exam twist (invariant length, wrong-pairing trap)	kyun up-with-down zaroori hai	Ex 8

Prerequisites jo aap khule rakhna chahein: Dual (reciprocal) basis, Inner product spaces, Change of basis and transformation laws, Curvilinear coordinates (polar, spherical).

Cell A — orthonormal basis (sanity anchor)

Worked example Example 1 — distinction gayab ho jaati hai

Basis $e_{1} = (1, 0)$ , $e_{2} = (0, 1)$ (ordinary square paper). Vector $v = (3, 4)$ . $v^{i}$ aur $v_{i}$ nikalo.

Forecast: abhi andaza lagao — kya dono number sets alag honge? Apna guess likho.

Step 1 — metric. $g_{11} = e_{1} \cdot e_{1} = 1$ , $g_{12} = e_{1} \cdot e_{2} = 0$ , $g_{22} = 1$ . Toh $g = (1001)$ . Yeh step kyun? Metric sab kuch decide karta hai; components se pehle geometry compute karo.

Step 2 — contravariant. Solve karo $v^{1} (1, 0) + v^{2} (0, 1) = (3, 4) \Rightarrow v^{1} = 3, v^{2} = 4$ . Yeh step kyun? Contravariant = parallelogram (walk-along) coefficients.

Step 3 — covariant. $v_{1} = v \cdot e_{1} = 3$ , $v_{2} = v \cdot e_{2} = 4$ . Yeh step kyun? Covariant = perpendicular projection, definition $v_{i} = v \cdot e_{i}$ .

Verify: $v_{i} = g_{ij} v^{j} = δ_{ij} v^{j} = v^{i}$ . Sach mein $(3, 4) = (3, 4)$ . Sanity anchor holds: orthonormal $\Rightarrow$ equal.

Cell B — orthogonal lekin unequal lengths

Worked example Example 2 — rectangular paper (pure scaling)

Basis $e_{1} = (2, 0)$ , $e_{2} = (0, 3)$ — perpendicular, lekin lamba. Vector $v = (4, 3)$ .

Forecast: axes abhi bhi perpendicular hain, toh kya $v_{i} = v^{i}$ hoga? (Dhyan raho — "perpendicular" matlab "orthonormal" nahi hota.)

Step 1 — metric. $g_{11} = 4$ , $g_{12} = 0$ , $g_{22} = 9$ . Toh $g = (4009)$ , diagonal hai par identity nahi. Yeh step kyun? Off-diagonals zero hain (perpendicular) phir bhi diagonals $\neq = 1$ (lambe axes) — yahi exactly "scaling only" cell hai.

Step 2 — contravariant. $v^{1} (2) = 4 \Rightarrow v^{1} = 2$ ; $v^{2} (3) = 3 \Rightarrow v^{2} = 1$ . Yeh step kyun? $e_{1}$ ke $2$ steps aur $e_{2}$ ke $1$ step chalke $(4, 3)$ tak pahunchte hain.

Step 3 — covariant. $v_{1} = v \cdot e_{1} = (4) (2) + (3) (0) = 8$ ; $v_{2} = (4) (0) + (3) (3) = 9$ . Yeh step kyun? Dot karne se axis ki length pick up hoti hai — wahin $g_{ii}$ ka factor chhupa hota hai.

Verify: $v_{1} = g_{11} v^{1} = 4 \cdot 2 = 8$ ✓, $v_{2} = g_{22} v^{2} = 9 \cdot 1 = 9$ ✓. Toh perpendicular axes ke saath bhi dono component sets alag hain ( $v^{i} \neq = v_{i}$ ) — yeh myth ki "perpendicular kaafi hai" bust ho gayi.

Cell C — skew basis, acute angle

Figure — Covariant and contravariant components

Worked example Example 3 — workhorse skew case

Basis $e_{1} = (1, 0)$ , $e_{2} = (1, 1)$ (inke beech angle $45°$ , acute). Vector $v = (2, 1)$ . (Yeh parent ke Example 1 jaisa hai — ise puri tarah karo taaki yeh tumhara apna bane.)

Forecast: kyunki $g_{12} = e_{1} \cdot e_{2} = 1 > 0$ , dono component sets alag honge aur covariant $v_{2}$ , $v^{2}$ se zyada hoga. Andaza lagao kitna zyada.

Step 1 — metric. $g_{11} = 1$ , $g_{12} = 1$ , $g_{22} = 2$ : $g = (1112)$ , $det g = 1$ , toh $g^{- 1} = (2 - 1 - 1 1)$ . Yeh step kyun? $g_{12} = 1$ off-diagonal ek acute skew ki fingerprint hai — geometry pehle.

Step 2 — contravariant (parallel). Solve karo $v^{1} + v^{2} = 2, v^{2} = 1 \Rightarrow v^{1} = 1, v^{2} = 1$ . Figure dekho: red dashed lines parallel to axes slide karke tip tak pahunchti hain. Yeh step kyun? Parallelogram rule = walk-along coefficients.

Step 3 — covariant (perpendicular). $v_{1} = v \cdot e_{1} = 2$ ; $v_{2} = v \cdot e_{2} = 2 + 1 = 3$ . Figure mein yeh green perpendicular drops hain. Yeh step kyun? Definition $v_{i} = v \cdot e_{i}$ ; lambe, slanted $e_{2}$ pe drop isliye $v_{2} = 3 > v^{2} = 1$ hai.

Verify: $v_{i} = g_{ij} v^{j}$ : $v_{1} = 1 \cdot 1 + 1 \cdot 1 = 2$ ✓, $v_{2} = 1 \cdot 1 + 2 \cdot 1 = 3$ ✓. Reverse: $v^{i} = g^{ij} v_{j}$ : $v^{1} = 2 \cdot 2 - 1 \cdot 3 = 1$ ✓, $v^{2} = - 2 + 3 = 1$ ✓.

Cell D — skew basis, obtuse angle (sign flip)

Worked example Example 4 — obtuse axes se

$g_{12} < 0$ hota hai Basis $e_{1} = (1, 0)$ , $e_{2} = (- 1, 1)$ (angle $135°$ , obtuse). Vector $v = (1, 2)$ .

Forecast: axes ab ek dusre se door jhuk rahe hain, toh $g_{12} < 0$ . $v_{1}$ aur $v^{1}$ ke beech difference ka sign predict karo.

Step 1 — metric. $g_{11} = 1$ , $g_{12} = e_{1} \cdot e_{2} = (1) (- 1) + (0) (1) = - 1$ , $g_{22} = (- 1)^{2} + 1^{2} = 2$ . Toh $g = (1 - 1 - 1 2)$ , $det g = 1$ , $g^{- 1} = (2111)$ . Yeh step kyun? Negative $g_{12}$ obtuse skew ki fingerprint hai — yeh genuinely naya cell hai.

Step 2 — contravariant. Solve karo $v^{1} - v^{2} = 1, v^{2} = 2 \Rightarrow v^{1} = 3, v^{2} = 2$ . Check karo: $3 (1, 0) + 2 (- 1, 1) = (1, 2)$ ✓. Yeh step kyun? $e_{1}$ ke saath $3$ chalo, phir backward-leaning $e_{2}$ ke saath $2$ — figure mein tip exactly $v$ pe land karti hai.

Step 3 — covariant. $v_{1} = v \cdot e_{1} = (1) (1) + (2) (0) = 1$ ; $v_{2} = v \cdot e_{2} = (1) (- 1) + (2) (1) = 1$ . Yeh step kyun? Backward-pointing axis pe perpendicular drop number ko shrink kar sakti hai — yahan $v_{1} = 1 < v^{1} = 3$ , Cell C ke difference sign ka ulta.

Verify: $v_{i} = g_{ij} v^{j}$ : $v_{1} = 1 \cdot 3 + (- 1) \cdot 2 = 1$ ✓, $v_{2} = - 1 \cdot 3 + 2 \cdot 2 = 1$ ✓. Toh obtuse geometry ( $g_{12} < 0$ ) relationship ko flip kar deti hai: covariant ab contravariant se undershoot karta hai.

Cell E — dual (reciprocal) basis explicitly banana

Worked example Example 5 — covariant numbers "kahaan rehte hain"

Ex 3 ka acute basis reuse karo: $e_{1} = (1, 0)$ , $e_{2} = (1, 1)$ . Dual basis $e^{1}, e^{2}$ banao aur confirm karo ki $v = v_{i} e^{i}$ $v_{1} = 2, v_{2} = 3$ ke saath.

Forecast: $e^{1}$ ko $e_{2}$ ke perpendicular hona chahiye. Kyunki $e_{2} = (1, 1)$ up-right point karta hai, $e^{1}$ ki direction guess karo.

Step 1 — dual defining rule. $e^{i} \cdot e_{j} = δ_{j}^{i}$ . 2D mein dual yeh hai: $e^{1} ⊥ e_{2}$ , $e^{1} \cdot e_{1} = 1$ . Yeh step kyun? Wahi perpendicularity covariant components ke exist karne ki poori wajah hai — yeh contravariant components hain is basis mein.

Step 2 — $e^{1} = (a, b)$ ke liye solve karo. $e_{2} = (1, 1)$ ke perp: $a + b = 0$ . Aur $e^{1} \cdot e_{1} = a = 1$ . Toh $a = 1, b = - 1$ : $e^{1} = (1, - 1)$ . Yeh step kyun? Do scalar conditions 2 unknowns ko exactly fix karte hain.

Step 3 — $e^{2} = (c, d)$ ke liye solve karo. $e_{1} = (1, 0)$ ke perp: $c = 0$ . Aur $e^{2} \cdot e_{2} = c + d = 1 \Rightarrow d = 1$ . Toh $e^{2} = (0, 1)$ . Yeh step kyun? Wahi recipe, doosra axis.

Step 4 — $v$ reassemble karo. $v_{1} e^{1} + v_{2} e^{2} = 2 (1, - 1) + 3 (0, 1) = (2, - 2 + 3) = (2, 1) = v$ ✓. Yeh step kyun? Yahi punchline hai: covariant numbers $(2, 3)$ dual grid mein walk-along coefficients hain, wahi arrow reproduce karte hain.

Verify: $e^{1} \cdot e_{1} = 1$ , $e^{1} \cdot e_{2} = (1) (1) + (- 1) (1) = 0$ , $e^{2} \cdot e_{1} = 0$ , $e^{2} \cdot e_{2} = 1$ — sab $δ_{j}^{i}$ ✓. Aur $2 (1, - 1) + 3 (0, 1) = (2, 1)$ ✓.

Cell F — degenerate / limiting (axes collapse together)

Worked example Example 6 — kya toot ta hai jab axes parallel ho jaate hain

Family $e_{1} = (1, 0)$ , $e_{2} = (1, ε)$ with $ε \to 0^{+}$ (doosra axis pehle ki taraf neeche jhukta hai). Metric watch karo.

Forecast: jab $ε \to 0$ toh dono axes merge ho jaate hain. Kya aapko lagta hai $g$ invertible rahega? $g^{- 1}$ ka kya hoga?

Step 1 — $ε$ ke function ke roop mein metric. $g_{11} = 1$ , $g_{12} = 1$ , $g_{22} = 1 + ε^{2}$ . Toh $det g = (1) (1 + ε^{2}) - 1^{2} = ε^{2}$ . Yeh step kyun? Determinant axes se bana parallelogram ka "volume" hai; exactly tab vanish hona chahiye jab yeh align ho jaayein.

Step 2 — inverse blow up karta hai. $g^{- 1} = \frac{1}{ε ^{2}} (1 + ε^{2} - 1 - 1 1)$ . Jab $ε \to 0$ toh har entry $\to \infty$ . Yeh step kyun? Indices raise karne ke liye $g^{- 1}$ chahiye; agar basis degenerate ho, toh raising impossible ho jaati hai — mathematically $det g \to 0$ se encode hota hai.

Step 3 — geometric meaning. $v = (0, 1)$ ke saath: contravariant ko chahiye $v^{1} (1, 0) + v^{2} (1, ε) = (0, 1)$ , jo deta hai $v^{2} = 1/ ε \to \infty$ , $v^{1} = - 1/ ε \to - \infty$ . Yeh step kyun? Near-parallel axes se door ek tip tak pahunchne ke liye enormous, opposite distances chalna padta hai — components explode kar dete hain chahe $v$ unit length ka ho.

Verify: $det g = ε^{2}$ ; $ε = 0.1$ par, $det g = 0.01$ , aur $v^{2} = 1/0.1 = 10$ , $v^{1} = - 10$ . Sanity: $- 10 (1, 0) + 10 (1, 0.1) = (0, 1)$ ✓. Rule learned: ek basis usable hai tabhi jab $det g \neq = 0$ ; degeneracy = metric singular ho jaata hai.

Cell G — curvilinear (polar): gradient covariant hota hai

Worked example Example 7 — polar coordinates, position-dependent metric

Polar coordinates $x^{1} = r$ , $x^{2} = θ$ with line element $d s^{2} = d r^{2} + r^{2} d θ^{2}$ , toh metric hai $g_{ij} = (10 0 r^{2})$ . Lo $f (r, θ) = r^{2}$ . Covariant gradient $\partial_{i} f$ aur contravariant (true vector) gradient $(\nabla f)^{i}$ nikalo.

Forecast: guess karo ki raised gradient ke $θ$ -component mein $r^{2}$ ka factor aayega ya $1/ r^{2}$ ka.

Step 1 — metric read off karo. $g_{11} = 1$ , $g_{22} = r^{2}$ (radius $r$ par angles arc length $r d θ$ span karte hain), $g_{12} = 0$ . Inverse: $g^{11} = 1$ , $g^{22} = 1/ r^{2}$ . Yeh step kyun? Yahan $g$ position par depend karta hai — Cells A–F ke muqable yeh essential naya feature hai.

Step 2 — covariant gradient. $\partial_{r} f = 2 r$ , $\partial_{θ} f = 0$ . Toh $\partial_{i} f = (2 r, 0)$ . Yeh step kyun? $\partial_{i} f = \partial f / \partial x^{i}$ naturally lower index carry karta hai — ek contravariant coordinate ke respect mein differentiation ek covector produce karta hai (dekho Tensors and index notation).

Step 3 — raise karke true gradient vector pao. $(\nabla f)^{i} = g^{ij} \partial_{j} f$ : $(\nabla f)^{1} = g^{11} \partial_{r} f = 2 r$ ; $(\nabla f)^{2} = g^{22} \partial_{θ} f = (1/ r^{2}) (0) = 0$ . Yeh step kyun? $g^{ij}$ se raise karne ke baad hi gradient physically steepest ascent ki taraf point karta hai (General relativity — raising and lowering indices).

Verify: $f = r^{2}$ ke liye, physically $∣\nabla f ∣ = ∣ df / d r ∣ = 2 r$ (radial). Compute karo $∣\nabla f ∣^{2} = g_{ij} (\nabla f)^{i} (\nabla f)^{j} = 1 \cdot (2 r)^{2} + r^{2} \cdot 0 = 4 r^{2}$ , toh magnitude $= 2 r$ ✓. Cartesian mein $f = x^{2} + y^{2}$ , $\nabla f = (2 x, 2 y)$ , $∣\nabla f ∣ = 2 x^{2} + y^{2} = 2 r$ ✓ — same invariant.

Cell H — word problem + exam twist (kyun up pairs with down)

Worked example Example 8 — "wrong pairing" trap

Word problem. Ek surveyor skew grid $e_{1} = (1, 0)$ , $e_{2} = (1, 1)$ lay out karta hai. Ek pipeline $v = (2, 1)$ ke roop mein chalti hai. Usse pipeline ki true length-squared (ek inner-product invariant) chahiye. Ek jaldi mein student $\sum_{i} (v^{i})^{2}$ compute karta hai. Dikhao yeh galat kyun hai aur sahi number do.

Forecast: naive $\sum (v^{i})^{2} = 1^{2} + 1^{2} = 2$ . Lekin $(2, 1)$ ki honest length $5$ hai. Kaun sahi hai?

Step 1 — invariant rule. Ek coordinate-free scalar mein up ko down ke saath contract karna hota hai: $∣ v ∣^{2} = v^{i} v_{i} = g_{ij} v^{i} v^{j}$ . Yeh step kyun? Sirf up-with-down ek change of basis mein survive karta hai ( $A$ aur $A^{- 1}$ cancel ho jaate hain). Do-ups-summed yeh galat assumption karta hai ki $g = δ$ .

Step 2 — correct contraction. Ex 3 se: $v^{i} = (1, 1)$ , $v_{i} = (2, 3)$ . Toh $v^{i} v_{i} = (1) (2) + (1) (3) = 5$ . Yeh step kyun? Yahi pairing rule maangta hai.

Step 3 — trap ka number. $\sum_{i} (v^{i})^{2} = 1 + 1 = 2$ ; $\sum_{i} (v_{i})^{2} = 4 + 9 = 13$ . Dono $5$ ke barabar nahi — dono basis-dependent junk hain. Yeh step kyun? Concretely dikhane ke liye ki dono same-level sums fail karte hain.

Verify: Background-Cartesian truth: $v = (2, 1)$ , $∣ v ∣^{2} = 2^{2} + 1^{2} = 5$ ✓. $v^{i} v_{i} = 5$ ke barabar ✓. Pipeline $5 \approx 2.236$ units ki hai. Exam moral: hamesha ek index up + ek down — staircase sum karo.

Recall Main kis cell mein hoon? (fast triage)

Diagonal $g$ with all $1$ s? ::: Cell A — components coincide karte hain. Diagonal $g$ , entries $\neq = 1$ ? ::: Cell B — perpendicular but scaled; $v_{i} \neq = v^{i}$ . $g_{12} > 0$ ? ::: Cell C — acute skew; covariant overshoot karta hai. $g_{12} < 0$ ? ::: Cell D — obtuse skew; covariant undershoot karta hai. $det g \to 0$ ? ::: Cell F — degenerate; $g^{- 1}$ blow up karta hai, raising possible nahi. $g$ position par depend karta hai? ::: Cell G — curvilinear; gradient ko raise karna padega.