Covariant and contravariant components: Exercises — Hinglish deep dive

4.10.8 · D4 · Maths › Advanced Topics (Elite Level) › Covariant aur contravariant components

Poore page mein "background" ordinary Cartesian $x y$ -space hai, aur skew basis ka matlab hai aisi basis vectors jo na sab unit length ki hain aur na sab $9 0^{\circ}$ par hain. Geometric problems ke liye hum ek hi running skew basis reuse karte hain taaki ek hi picture par intuition build ho:

$e_{1} = (2, 0), e_{2} = (1, 2) .$

Figure — Covariant and contravariant components

Kisi vector ka parallel (parallelogram) reading contravariant components $v^{i}$ (index upar) deta hai; perpendicular (drop-a-shadow) reading covariant components $v_{i}$ (index neeche) deta hai. Har problem mein yeh picture apne dimaag mein rakho.

Level 1 — Recognition

Recall Solution L1.1

(a) Contravariant $v^{1}$ — yeh literally ek expansion coefficient hai (parallel projection). (b) Covariant $v_{2} = v \cdot e_{2}$ — covariant component ki definition hai (perpendicular projection). (c) Covariant $v_{1}$ — "perpendicular shadow" orthogonal-projection rule hai. (d) Contravariant — "axes par chalna" parallelogram rule hai.

Recall Solution L1.2

$g_{11} = e_{1} \cdot e_{1} = 2 \cdot 2 + 0 \cdot 0 = 4$ . $g_{12} = e_{1} \cdot e_{2} = 2 \cdot 1 + 0 \cdot 2 = 2$ (aur $g_{21} = g_{12} = 2$ symmetry se). $g_{22} = e_{2} \cdot e_{2} = 1 \cdot 1 + 2 \cdot 2 = 5$ . $g = (4225) .$ Yeh pehle kyun? Metric tensor basis ki har length aur angle store karta hai — baad ke har formula isi se draw karta hai.

Level 2 — Application

Recall Solution L2.1

(a) Contravariant — parallelogram system solve karo $v^{1} e_{1} + v^{2} e_{2} = v$ : ${2 v^{1} + 1 v^{2} = 4 0 v^{1} + 2 v^{2} = 2 \Rightarrow v^{2} = 1, 2 v^{1} + 1 = 4 \Rightarrow v^{1} = \frac{3}{2} .$ Toh $v^{1} = \frac{3}{2}, v^{2} = 1$ . System kyun solve karo? Kyunki "har slanted axis par kitna jaana" exactly woh coefficients hain jo $v$ rebuild karte hain. (b) Covariant — har basis vector se dot karo (definition): $v_{1} = v \cdot e_{1} = 4 \cdot 2 + 2 \cdot 0 = 8, v_{2} = v \cdot e_{2} = 4 \cdot 1 + 2 \cdot 2 = 8.$ Toh $v_{1} = 8, v_{2} = 8$ . Dhyaan do $v^{i} \neq = v_{i}$ — basis skew hai, isliye dono readings genuinely differ karti hain.

Recall Solution L2.2

$v_{1} = g_{11} v^{1} + g_{12} v^{2} = 4 \cdot \frac{3}{2} + 2 \cdot 1 = 6 + 2 = 8 ✓$ $v_{2} = g_{21} v^{1} + g_{22} v^{2} = 2 \cdot \frac{3}{2} + 5 \cdot 1 = 3 + 5 = 8 ✓$ Dono L2.1(b) se match karte hain. Yeh metric apna kaam kar raha hai: $g_{ij}$ mein stored geometry ko mix karke upper index ko "lower" karta hai.

Level 3 — Analysis

Recall Solution L3.1

(a) $det g = 4 \cdot 5 - 2 \cdot 2 = 16$ . $2 \times 2$ matrix ke liye, $(a c b d)^{- 1} = \frac{1}{d e t} (d - c - b a)$ : $g^{- 1} = \frac{1}{16} (5 - 2 - 2 4) = (\frac{5}{16} - \frac{1}{8} - \frac{1}{8} \frac{1}{4}) .$ (b) Raising: $v^{1} = g^{11} v_{1} + g^{12} v_{2} = \frac{5}{16} \cdot 8 - \frac{1}{8} \cdot 8 = \frac{40}{16} - 1 = \frac{5}{2} - 1 = \frac{3}{2} ✓$ $v^{2} = g^{21} v_{1} + g^{22} v_{2} = - \frac{1}{8} \cdot 8 + \frac{1}{4} \cdot 8 = - 1 + 2 = 1 ✓$ Raising aur lowering perfect inverses hain — jaise hone chahiye, kyunki $g^{ik} g_{k j} = δ_{j}^{i}$ .

Recall Solution L3.2

$g^{- 1} g = (\frac{5}{16} - \frac{1}{8} - \frac{1}{8} \frac{1}{4}) (4225) = (\frac{20}{16} - \frac{2}{8} - \frac{4}{8} + \frac{2}{4} \frac{10}{16} - \frac{5}{8} - \frac{2}{8} + \frac{5}{4}) = (1001) .$ $δ_{j}^{i}$ Kronecker delta hai: $1$ jab $i = j$ , $0$ otherwise — woh machine jo kehti hai "raise karke lower karne se kuch nahi badalta." Dekho Tensors and index notation.

Level 4 — Synthesis

Recall Solution L4.1

(a) Cartesian truth: $∣ v ∣^{2} = 4^{2} + 2^{2} = 16 + 4 = 20$ . (b) Contraction (up with down): $v^{1} = \frac{3}{2}, v^{2} = 1, v_{1} = 8, v_{2} = 8$ use karke, $v^{i} v_{i} = v^{1} v_{1} + v^{2} v_{2} = \frac{3}{2} \cdot 8 + 1 \cdot 8 = 12 + 8 = 20 ✓$ Scalar basis-independent hai kyunki humne ek upper ek lower index pair kiya. (c) Both up (illegal): $v^{1} v^{1} + v^{2} v^{2} = (\frac{3}{2})^{2} + 1^{2} = \frac{9}{4} + 1 = \frac{13}{4} = 3.25 \neq = 20$ . Yeh meaningless hai — yeh secretly $g = I$ assume karta hai. Dekho Inner product spaces kyun metric components ke beech hona zaroori hai.

Recall Solution L4.2

Dual basis hai $e^{i} = g^{ij} e_{j}$ . Har ek ko Cartesian vector ke roop mein compute karo: $e^{1} = g^{11} e_{1} + g^{12} e_{2} = \frac{5}{16} (2, 0) - \frac{1}{8} (1, 2) = (\frac{10}{16} - \frac{1}{8}, - \frac{2}{8}) = (\frac{1}{2}, - \frac{1}{4}) .$ $e^{2} = g^{21} e_{1} + g^{22} e_{2} = - \frac{1}{8} (2, 0) + \frac{1}{4} (1, 2) = (- \frac{1}{4} + \frac{1}{4}, \frac{1}{2}) = (0, \frac{1}{2}) .$ Quick sanity: $e^{1} \cdot e_{1} = \frac{1}{2} \cdot 2 + (- \frac{1}{4}) \cdot 0 = 1 ✓$ ; $e^{1} \cdot e_{2} = \frac{1}{2} \cdot 1 - \frac{1}{4} \cdot 2 = 0 ✓$ . Ab covariant components $v_{1} = 8, v_{2} = 8$ se rebuild karo: $v_{1} e^{1} + v_{2} e^{2} = 8 (\frac{1}{2}, - \frac{1}{4}) + 8 (0, \frac{1}{2}) = (4, - 2) + (0, 4) = (4, 2) = v ✓$ Toh covariant components ARE dual basis ke saath respect mein contravariant components hain — woh promised symmetry.

Level 5 — Mastery

Recall Solution L5.1

(a) Contravariant — basis ke against. Agar basis vectors length mein double ho jaate hain, toh usi tip tak pahunchne ke liye tumhe aadha use karna padega: $\tilde{v}^{1} = \frac{1}{2} v^{1} = \frac{3}{4}, \tilde{v}^{2} = \frac{1}{2} v^{2} = \frac{1}{2} .$ Check: $\tilde{v}^{1} \tilde{e}_{1} + \tilde{v}^{2} \tilde{e}_{2} = \frac{3}{4} (4, 0) + \frac{1}{2} (2, 4) = (3, 0) + (1, 2) = (4, 2) = v ✓$ . Yeh $A^{- 1} = \frac{1}{2} I$ se scale hue — contravariant components basis change se ladte hain. (b) Covariant — basis ke saath. $\tilde{v}_{i} = v \cdot \tilde{e}_{i} = v \cdot (2 e_{i}) = 2 v_{i}$ : $\tilde{v}_{1} = 2 \cdot 8 = 16, \tilde{v}_{2} = 2 \cdot 8 = 16.$ Yeh $A = 2 I$ se scale hue — covariant components basis change ke saath cooperate karte hain. Dekho Change of basis and transformation laws. (c) Length invariance: $\tilde{v}^{i} \tilde{v}_{i} = \frac{3}{4} \cdot 16 + \frac{1}{2} \cdot 16 = 12 + 8 = 20$ — L4.1 se identical. $A$ aur $A^{- 1}$ factors exactly cancel ho jaate hain. Yahi cancellation poore up–down staircase ke hone ka reason hai.

Recall Solution L5.2

(a) Covariant (natural) gradient — differentiate karo, index neeche rehta hai: $\partial_{1} f = \frac{\partial ( r ^{2} )}{\partial r} = 2 r, \partial_{2} f = \frac{\partial ( r ^{2} )}{\partial θ} = 0.$ Toh $\partial_{i} f = (2 r, 0)$ . Neeche kyun? Gradient "per unit coordinate step change" measure karta hai — yeh ek displacement khata hai, isliye yeh ek covector hai. Dekho Curvilinear coordinates (polar, spherical). (b) Inverse metric se raise karo $g^{ij} = (10 0 1/ r^{2})$ : $(\nabla f)^{1} = g^{11} \partial_{1} f = 1 \cdot 2 r = 2 r, (\nabla f)^{2} = g^{22} \partial_{2} f = \frac{1}{r ^{2}} \cdot 0 = 0.$ Toh $(\nabla f)^{i} = (2 r, 0)$ . (c) $θ$ -component $0$ hai kyunki $f = r^{2}$ angle ke saath vary nahi karta. Agar karta, toh $1/ r^{2}$ ka factor essential hota — yeh "per radian" ko "per unit arc length" mein convert karta hai. Yeh exactly wahi raising/lowering machinery hai jo concrete bani; Cartesian coordinates mein $g^{ij} = δ^{ij}$ yeh sab chupa deta hai.

Recall Jaane se pehle one-line self-tests

Skew basis mein, $v \cdot e_{i}$ kaun se components deta hai? ::: Covariant $v_{i}$ (neeche).
$v_{i}$ se $v$ rebuild karne ke liye kaun sa basis? ::: Dual basis $e^{i}$ : $v = v_{i} e^{i}$ .
Kaun sa contraction invariant hai, $v^{i} v_{i}$ ya $v^{i} v^{i}$ ? ::: $v^{i} v_{i}$ (ek upar, ek neeche).
Basis lengths double karo — $v^{i}$ ka kya hoga? ::: Yeh aadhe ho jaate hain ( $A^{- 1}$ ); covariant $v_{i}$ double hote hain ( $A$ ).