Tensor analysis — scalars, vectors, rank-2 tensors: Visual walkthrough

4.10.7 · D2 · Maths › Advanced Topics (Elite Level) › Tensor analysis — scalars, vectors, rank-2 tensors

Neeche sab kuch ek seedhe sawaal par tika hai: agar main apne axes naye sire se banaun, toh ek physical cheez ke liye jo numbers maine likhe hain unka kya hoga? Hum iska jawab pehle ek dot ke liye denge, phir ek displacement ke liye, phir ek slope ke liye, phir ek do-slot wali machine ke liye.

Step 1 — Stage: ek hi kamre ke upar do rulers

KYA. Hamare paas ek flat kamra hai (ek plane). Main apne axes bichhaata hun aur ek point ke coordinates ko $x^{1}, x^{2}$ kehta hun. Mera dost tilted, alag scale wale axes bichhata hai aur usi point ko $\tilde{x}^{1}, \tilde{x}^{2}$ kehta hai (tilde ~ ka matlab sirf "dost ka version" hai).

KYUN. Ek tensor ko define kiya jaata hai is baat se ki numbers in dono descriptions ke beech kaise badalte hain. Toh kisi bhi formula se pehle, hume do grids ko ek hi physical kamre ke upar baithaa ke dekhna hoga.

PICTURE. Dekho . Ek kala dot. Do coordinate meshes. Dot kabhi nahi hilta — sirf labels badalti hain jo hum usse lagate hain. Woh aandolan-se-mukt dot tensors ka pura naitik paath hai: cheez asli hai; components sirf hisaab-kitaab hain.

Step 2 — Woh ek gadget jo humein chahiye: Jacobian, khichain ke roop mein drawn

KYA. Mere grid mein ek chota sa qadam lo, $d x^{k}$ (ek tiny displacement, upper index kyunki yeh ek displacement hai). Dost ke numbers mein wahi qadam kitna bada hai, $d \tilde{x}^{i}$ ? Jawab hai Jacobian:

J^{i}_{k} = kitni tezi se dost ka \tilde{x}^{i} badhta hai jab main apna x^{k} thoda sa hilata hun \frac{\partial x ~ ^{i}}{\partial x ^{k}}

KYUN yahi tool, koi aur nahi? Hume grids ke beech tiny changes convert karni hain. Tiny changes ke liye exact converter derivative hai — yahi literally ek derivative hota hai: local conversion rate. Hum partial derivative $\partial \tilde{x}^{i} / \partial x^{k}$ use karte hain kyunki kai coordinates hain aur hum ek ek ki rate chahte hain. Yeh hai Jacobian.

PICTURE. dikhata hai ek chota sa blue square mere grid mein jo dost ke grid mein ek tilted parallelogram ban jaata hai. Jacobian bilkul wahi local "mera chota square kaise stretch aur shear hua" matrix hai.

Step 3 — Contravariant vectors: displacement gold standard set karta hai

KYA. Ek displacement $d x^{k}$ sabse bharosemand vector hai jo hamare paas hai — yeh do asli points ke beech ek asli arrow hai. Chain rule apply karo dekhne ke liye ki uske components kaise badalte hain:

dost ke numbers mein qadam d \tilde{x}^{i} = J^{i}_{k} = converter \frac{\partial x ~ ^{i}}{\partial x ^{k}} mere numbers mein qadam (sum over k) d x^{k}

Jiska bhi component yeh behaviour copy kare, use hum naam dete hain contravariant, upper index:

\tilde{V}^{i} = \frac{\partial x ~ ^{i}}{\partial x ^{k}} V^{k}

KYUN. Hum rule guess nahi karte — hum use churaate hain kisi aise cheez se jis par hum pehle se believe karte hain (ek displacement). Yeh seedha tarika hai: "vector" ko define karo "woh jo us ek arrow ki tarah transform kare jis par hum shak nahi kar sakte." Repeated index $k$ (ek baar upar $V^{k}$ mein, ek baar neeche $J$ mein) ka matlab hai sum over $k$ — Einstein summation convention.

PICTURE. : wahi narangi arrow. Ek mote dost-grid par (bade boxes) arrow kam boxes span karta hai → uske components chhote ho jaate hain. Numbers shrink karte hain jab ruler bada hota hai: woh ruler-ke-ulta behaviour hi contra-variant shabd ka matlab hai.

Step 4 — Covariant vectors: ek slope dusre taraf transform hota hai

KYA. Ab ek scalar $ϕ$ lo (ek point par ek number, jaise temperature) aur uski slopes $\partial ϕ / \partial x^{k}$ (gradient components). Chain rule phir se — lekin dhyan do kaun sa variable kahan baitha hai:

dost ki slope \frac{\partial ϕ}{\partial x ~ ^{i}} = (J^{- 1})^{k}_{i} = inverse converter \frac{\partial x ^{k}}{\partial x ~ ^{i}} meri slope (sum over k) \frac{\partial ϕ}{\partial x ^{k}}

Jiska bhi component yeh copy kare use hum naam dete hain covariant, lower index:

\tilde{W}_{i} = \frac{\partial x ^{k}}{\partial x ~ ^{i}} W_{k}

KYUN yahan inverse Jacobian? Kyunki derivative coordinate ke saath respect to hai: $\tilde{x}^{i}$ ab denominator mein hai, toh chain rule $\partial x^{k} / \partial \tilde{x}^{i}$ — ulta converter — nikaalti hai. Ek displacement mein naya coordinate upar tha; gradient mein woh neeche hai. Wahi chain rule, ulta slot.

PICTURE. : temperature contour lines. Ek mote grid par temperature har box mein zyada chadhti hai → gradient ke components bade ho jaate hain. Numbers ruler ke saath badhte hain: co-variant, "basis ke saath vary karta hai."

Step 5 — Rank 2: do slots jodo, ek ek Jacobian har ek ke liye

KYA. Ek rank-2 tensor do slots wali machine hai. Uska transformation sirf Step 3 (ya 4) hai jo ek ek baar har slot par apply hota hai — ek converter factor har ek index ke liye:

\tilde{T}^{ij} = slot 1: k \to i convert karta hai \frac{\partial x ~ ^{i}}{\partial x ^{k}} slot 2: l \to j convert karta hai \frac{\partial x ~ ^{j}}{\partial x ^{l}} mere components (sum over k, l) T^{k l}

KYUN. Koi nayi idea nahi hai — ek do-slot object do ek-slot rules ko staple karke banaya gaya hai. Har index "exactly ek Jacobian factor kheenchta hai" (master rule). Upper indices $J$ kheenchte hain; lower indices $J^{- 1}$ kheenchte hain; mixed tensors unhe mix karte hain.

PICTURE. : do arrows ek box mein feed hote hain, har ek alag alag rotate hota hai, toh box ki poori grid of numbers reshuffle ho jaati hai. Pure rotation ke liye Jacobian ek rotation matrix $R$ hai, uska inverse $R^{⊤}$ hai (rotations orthonormal hote hain, toh inverse = transpose), aur do-factor rule compress hokar neat matrix sandwich ban jaata hai:

\tilde{T} = R T R^{⊤} .

Step 6 — Kyun contraction survive karta hai: converters cancel ho jaate hain

KYA. "Contract" ka matlab hai ek upper index ko ek lower ke barabar set karo aur sum karo, jaise $A^{i}_{i}$ (yeh trace hai). Coordinates ke change ke neeche do dragged converters milte hain:

J \frac{\partial x ~ ^{i}}{\partial x ^{k}} J^{- 1} \frac{\partial x ^{l}}{\partial x ~ ^{i}} = chain rule unhe jodti hai \frac{\partial x ^{l}}{\partial x ^{k}} = δ_{k}^{l}

KYUN. Shared index $i$ par sum karne se $J$ aur $J^{- 1}$ multiply hote hain — aur chain rule ki wajah se woh $δ_{k}^{l}$ (1 agar $l = k$ , nahi toh 0) mein collapse ho jaate hain. Saari coordinate-dependence gaayab ho jaati hai. Toh ek fully contracted tensor ek sachcha scalar hai — har frame mein same number padhta hai.

PICTURE. : ek forward-stretch arrow jo turant apne inverse se undo ho jaata hai, tumhe bilkul wahi pahuncha deta hai jahan se chale (identity $δ$ ). Woh "wahan-aur-wapas = kuch nahi" $J J^{- 1} = δ$ ki picture hai.

Step 7 — Woh edge cases jinpar kabhi mat phislo

KYA & KYUN. Ek derivation tabhi bharosemand hai jab woh apni extremes mein survive kare. Char cases yaad rakho:

PICTURE. har case ko uske grid ke saath pair karta hai.

Identity change ( $\tilde{x} = x$ ). Tab $J = δ$ , aur har rule $\tilde{T} = T$ padhta hai: koi relabelling nahi, kuch nahi hilta. Sanity floor.
Kronecker delta $δ_{j}^{i}$ (isotropic). Ise mixed rule se pass karo: $\tilde{δ}_{j}^{i} = \frac{\partial x ~ ^{i}}{\partial x ^{k}} \frac{\partial x ^{l}}{\partial x ~ ^{j}} δ_{l}^{k} = \frac{\partial x ~ ^{i}}{\partial x ^{k}} \frac{\partial x ^{k}}{\partial x ~ ^{j}} = δ_{j}^{i}$ . Har frame mein same — chain rule ise collapse karti hai (yeh Step 6 ka cancellation phir se use ho raha hai).
Ek plain data table (tensor NAHI hai). Times-table ya spreadsheet ke saath koi $\tilde{x}^{i} (x)$ attached nahi hai, toh koi converter drag karne ke liye nahi → koi transformation law nahi → tensor nahi. Sirf numbers kabhi kaafi nahi hote.
Ordinary derivative $\partial_{k} V^{i}$ (curved grids mein fail karta hai). $\partial_{k}$ position-dependent Jacobian ko bhi hit karta hai, ek extra non-tensor term nikaalke. Fix hai covariant derivative $\nabla_{k} V^{i}$ , jo us junk ko cancel karne ke liye Christoffel symbols add karta hai.

Ek-picture summary

Ek physical object do grids ke upar. Ek displacement forward converter $J$ sikhata hai (contravariant, index upar). Ek gradient inverse converter $J^{- 1}$ sikhata hai (covariant, index neeche). Ek rank-2 tensor ek ek converter har slot ke liye staple karta hai → $\tilde{T} = R T R^{⊤}$ . Ek upar aur ek neeche pair karo aur woh cancel ho jaate hain → ek invariant scalar. Woh akela diagram poori theory hai.

Recall Feynman retelling — ek 12-saal ke bachche ko batao

Socho zameen par ek dot hai. Tum square tiles kheeenchte ho; tumhara dost usi zameen par tirchi tiles kheenchta hai. Dot ko koi fark nahi padta — lekin tum dono jo tile-numbers likhte ho woh alag alag hain. Tumhare numbers ko dost ke numbers mein translate karne ka rule ek choti "stretch machine" hai jise Jacobian kehte hain.

Ek arrow (tiles ke paas ek qadam) chhote numbers leta hai jab tiles bade hote hain — yeh ek contravariant vector hai, yeh tiles se ladhta hai. Ek slope (temperature har tile mein kitni tezi se chadhti hai) bade numbers leta hai jab tiles bade hote hain — yeh ek covariant vector hai, yeh tiles ke saath chalta hai. Ek stretch machine use karta hai, doosra stretch machine ko ulta chalata hai.

Ek rank-2 tensor ek do-bahu wali machine hai, toh tum stretch machine ko har ek bahu ke liye ek baar chalate ho — ek rotation ke liye yeh neat sandwich $R T R^{⊤}$ ban jaata hai. Aur agar kabhi tum ek arrow-bahu ko ek slope-bahu se pair karo aur unhe add karo, toh stretch aur un-stretch perfectly cancel ho jaate hain aur tumhe ek plain number milta hai jis par sab agree karte hain — jaise trace, jo $6$ hi raha chahe hum axes kaise bhi ghoomayen. Woh agreement hi poora point hai: ek tensor woh object hai jiska asli content tiles ke change ke baad bhi zinda rehta hai.

Recall Quick self-check

Ek covariant index inverse Jacobian kyun kheenchta hai? ::: Kyunki woh naye coordinate ke saath respect to differentiate karne se aata hai, toh naya coordinate denominator mein baithta hai aur chain rule $\partial x^{k} / \partial \tilde{x}^{i}$ nikaalti hai. Trace invariant kyun hota hai? ::: Ek upper ko ek lower index ke saath contract karna $J$ ko $J^{- 1}$ se multiply karta hai, jo $δ_{k}^{l}$ mein collapse ho jaate hain, saari coordinate dependence mitaate hue. Kya ek multiplication table ek tensor hai? ::: Nahi — uske saath koi coordinate-change rule attached nahi hai; tensors hain numbers plus ek transformation law.