Einstein summation convention: Worked examples — Hinglish deep dive

4.10.9 · D3 · Maths › Advanced Topics (Elite Level) › Einstein summation convention

Yeh page ek gym hai. Parent note ne rules banaye; yahan hum unhe har tarah ke cases ke against practice karte hain jo yeh notation throw kar sakta hai — alag index counts, delta ki substitution, epsilon ke signs, degenerate/zero inputs, ek limiting case, ek word problem, aur ek exam-style trap. Yahan use kiya gaya har symbol parent mein define kiya gaya tha; agar aapko refresher chahiye, neeche ke links ussi pe point karte hain.

Scenario matrix

Index notation mein har problem in cells mein se kisi ek mein aati hai. Neeche ke examples us cell ke saath labelled hain jise woh hit karte hain, aur milke yeh poori table cover karte hain.

#	Case class	Kya tricky banata hai	Example
A	0 free indices (scalar out)	ensure karo kuch bachta nahi	Ex 1
B	1 free index (vector out)	ek letter dono sides match karna chahiye	Ex 2
C	2 free indices (matrix out)	do survivors, order matters	Ex 3
D	Kronecker substitution	delta ek repeated index "khata" hai	Ex 4
E	Levi-Civita signs, sab cases	even/odd/repeated permutations	Ex 5
F	Degenerate / zero input	repeated vector, parallel vectors → $0$	Ex 6
G	epsilon–delta master identity	double cross product collapse karo	Ex 7
H	Real-world word problem	physics → indices → number translate karo	Ex 8
I	Exam-style trap	illegal 3-times index / free-index mismatch	Ex 9
J	Limiting / dimension case	$δ_{ii} = n$ , behavior as $n$ grows	Ex 10

Har index ki range $1, 2, 3$ hai (matlab $n = 3$ ) jab tak problem kuch aur na kahe.

Ex 1 — Cell A: ek pure scalar

Worked example

$a_{i} b_{i}$ compute karo $a = (2, - 1, 3)$ , $b = (4, 0, - 2)$ ke liye. Forecast: index $i$ do baar appear karta hai, isliye sum ho jaata hai; zero letters survive → answer ek single number hai. Uska sign guess karo: last components ( $3$ aur $- 2$ ) first components ( $2$ aur $4$ ) se ladte hain, toh yeh ek tug-of-war hai — woh guess hold karo aur dekho kaun jeetta hai.

Step 1 — implied sum expand karo. Yeh step kyun? Doubled index ek hidden $\sum$ hai, isliye hum ise likhte hain: $a_{i} b_{i} = a_{1} b_{1} + a_{2} b_{2} + a_{3} b_{3} .$

Step 2 — numbers plug karo. Yeh step kyun? Ab yeh ordinary arithmetic hai: $= (2) (4) + (- 1) (0) + (3) (- 2) = 8 + 0 - 6 = 2.$

Forecast reconcile karo: tug-of-war real tha — negative piece $(3) (- 2) = - 6$ ne zor se pull kiya, lekin positive piece $(2) (4) = + 8$ bada tha, isliye winner ek small positive number hai, $+ 2$ . Lesson: sirf ek axis se sign guess par kabhi trust mat karo; dot product sab pulls sum karta hai, aur largest-magnitude product decide karta hai. Yahan $+ 8$ ne $- 6$ ko haraaya.

Verify: Yeh exactly Dot product $a \cdot b$ hai. Independent check: $a \cdot b = 8 + 0 - 6 = 2$ . ✔ Scalar out, jaise forecast kiya (ek number, koi free index nahi).

Ex 2 — Cell B: vector out (matrix × vector)

Worked example

$A = 104210031$ , $x = (1, 2, 3)$ ke saath, $y_{i} = A_{ij} x_{j}$ find karo. Forecast: $j$ doubled hai ( $A_{ij}$ aur $x_{j}$ mein) → sum ho jaata hai; $i$ har side ek baar appear karta hai → survive karta hai. Ek free index ⇒ answer ek vector hai length 3 ka.

Step 1 — $i$ fix karo, $j$ par sum karo. Yeh step kyun? Rule "har fixed $i$ ke liye, $A_{ij} x_{j}$ ko $j$ par add karo" bas row $i$ ka $x$ ke saath dot product hai. $y_{1} = A_{11} x_{1} + A_{12} x_{2} + A_{13} x_{3} = (1) (1) + (2) (2) + (0) (3) = 5.$

Step 2 — $i = 2, 3$ ke liye repeat karo. Yeh step kyun? Free index ko poori range par run karna hoga: $y_{2} = (0) (1) + (1) (2) + (3) (3) = 11, y_{3} = (4) (1) + (0) (2) + (1) (3) = 7.$

Verify: $y = (5, 11, 7)$ . Sanity check ordinary Matrix multiplication se: $A x$ directly multiply karo — same result. ✔ Ek free index → vector, exactly jaise forecast kiya.

Ex 3 — Cell C: matrix out (do survivors)

Worked example

$A_{ij}$ aur $B_{j k}$ ko $2 \times 2$ matrices $A = (1324)$ , $B = (5768)$ diye gaye hain, $C_{ik} = A_{ij} B_{j k}$ compute karo. Forecast: $j$ repeat hota hai → sum ho jaata hai (Matrix multiplication ka "shared middle"); $i$ aur $k$ har ek baar appear karte hain → dono survive karte hain. Do free indices ⇒ ek matrix. Yahan har index ${1, 2}$ par range karta hai, matlab is example ke liye dimension $n = 2$ hai (usual $3$ nahi).

Step 1 — entry formula likho. Yeh step kyun? Do free indices matlab hume $n^{2} = 2 \times 2 = 4$ numbers produce karne hain; har ek $j$ par sum hai ( $1, 2$ run karta hai): $C_{ik} = A_{i 1} B_{1 k} + A_{i 2} B_{2 k} .$

Step 2 — grid fill karo. Yeh step kyun? $i$ aur $k$ ko ${1, 2}$ par run karo: $C_{11} = 1 \cdot 5 + 2 \cdot 7 = 19, C_{12} = 1 \cdot 6 + 2 \cdot 8 = 22,$ $C_{21} = 3 \cdot 5 + 4 \cdot 7 = 43, C_{22} = 3 \cdot 6 + 4 \cdot 8 = 50.$

Verify: $C = (19432250)$ , standard product $A B$ . ✔ Do free indices → grid, jaise forecast kiya. Order matters: $A_{ij} B_{j k} \neq = B_{ij} A_{j k}$ generally.

Ex 4 — Cell D: delta ek renamer ke roop mein

Worked example

$δ_{ij} δ_{j k} a_{k}$ simplify karo aur phir $a = (7, - 2, 5)$ ke liye evaluate karo, component $i = 2$ read karke. Forecast: har delta ek repeated index "khaata" hai. Pehle $δ_{j k} a_{k} \to a_{j}$ , phir $δ_{ij} a_{j} \to a_{i}$ . Toh poori cheez bas $a_{i}$ mein collapse ho jaani chahiye.

Step 1 — inner delta apply karo. Yeh step kyun? Substitution rule $δ_{j k} a_{k} = a_{j}$ (sirf $k = j$ sum mein survive karta hai) $k$ ko $j$ se replace karta hai: $δ_{ij} δ_{j k} a_{k} = δ_{ij} a_{j} .$

Step 2 — outer delta apply karo. Yeh step kyun? Wahi rule phir, $δ_{ij} a_{j} = a_{i}$ : $= a_{i} .$

Step 3 — component $i = 2$ read karo. Yeh step kyun? $i$ surviving free index hai; $i = 2$ pick karna ek number deta hai: $a_{2} = - 2.$

Verify: Do chained deltas $δ_{ik}$ hain (parent ke $δ_{ij} δ_{j k} = δ_{ik}$ se), aur $δ_{ik} a_{k} = a_{i}$ . Same collapse. $a$ ka component $2$ indeed $- 2$ hai. ✔

Ex 5 — Cell E: har Levi-Civita sign

Worked example

$ε_{123}$ , $ε_{231}$ , $ε_{213}$ , $ε_{321}$ , aur $ε_{223}$ evaluate karo. Forecast: $(1, 2, 3)$ base ordering hai (+1). Har swap of two entries sign flip karta hai. Koi bhi repeated index ise $0$ kar deta hai.

Step 1 — identity. Yeh step kyun? $(1, 2, 3)$ by definition ek even permutation hai: $ε_{123} = + 1.$

Step 2 — cyclic shift. Yeh step kyun? $(2, 3, 1)$ $(1, 2, 3)$ ka cyclic rotation hai = do swaps = even: $ε_{231} = + 1.$

Step 3 — single swap. Yeh step kyun? $(2, 1, 3)$ pehle do swap karta hai → odd: $ε_{213} = - 1.$

Step 4 — reverse. Yeh step kyun? $(3, 2, 1)$ $(1, 2, 3)$ ke ends swap karta hai → ek swap → odd: $ε_{321} = - 1.$

Step 5 — repeated index. Yeh step kyun? Do indices equal hain ( $2$ do baar appear karta hai) → $0$ : $ε_{223} = 0.$

Verify: Neeche orientation wheel dekho. Teen amber arrows $1 \to 2 \to 3 \to 1$ run karte hain: arrows follow karna (cyclic order) even permutations deta hai, har ek $+ 1$ worth. Cyan dashed arrow amber flow ke against run karta hai ( $2 \to 1$ ): backwards jaana ek single swap hai, ek odd permutation worth $- 1$ . Ek vertex label repeated hona (jaise $ε_{223}$ mein do $2$ ) wheel par koi jagah nahi rakhta — woh $0$ case hai. Determinant se cross-check karo: $ε_{ij k}$ permutation ka sign equals karta hai, sab paanch values match karte hain. ✔

Ex 6 — Cell F: degenerate / zero input

Worked example Kisi bhi

$a$ ke liye $(a \times a)_{i} = ε_{ij k} a_{j} a_{k}$ compute karo. Forecast: ek vector ka khud ke saath cross product $0$ hona chahiye (ek vector khud ke parallel hai, zero enclosed area). Hum ise purely index symmetry se prove karte hain — ek degenerate input jise notation automatically handle karta hai.

Step 1 — symmetry clash spot karo. Yeh step kyun? Product $a_{j} a_{k}$ $j \leftrightarrow k$ swap karne ke under symmetric hai (kyunki $a_{j} a_{k} = a_{k} a_{j}$ ), jabki $ε_{ij k}$ antisymmetric hai ( $ε_{ij k} = - ε_{ik j}$ ). Ek symmetric object ko antisymmetric ke against sum karna hamesha zero deta hai.

Step 2 — rename karke prove karo. Yeh step kyun? Dummies $j \leftrightarrow k$ rename karo (legal — dummy names free hain): $ε_{ij k} a_{j} a_{k} = ε_{ik j} a_{k} a_{j} = - ε_{ij k} a_{j} a_{k} .$ Ek quantity jo apne negative ke equal ho woh zero honi chahiye: $(a \times a)_{i} = 0.$

Verify: Numerically, $a = (2, - 1, 3)$ : cross product $a \times a = (0, 0, 0)$ . ✔ Yeh general degenerate case hai: koi bhi do parallel vectors ( $b = λ a$ ) bhi $a \times b = λ (a \times a) = 0$ dete hain. (Dekho Cross product, Levi-Civita symbol.)

Ex 7 — Cell G: master identity in action

Worked example BAC–CAB rule

$a \times (b \times c) = b (a \cdot c) - c (a \cdot b)$ prove karo, phir numerically check karo. Forecast: inner cross ek vector deta hai, outer cross ek vector — ek free index survive karta hai. Epsilon–delta identity do epsilons ko do delta-terms mein turn karegi → do Dot products.

Step 1 — dono crosses likho. Yeh step kyun? Poori cheez ka component $i$ , parent ke cross-product formula ko do baar use karke: $[a \times (b \times c)]_{i} = ε_{ij k} a_{j} ε_{k l m} b_{l} c_{m} .$

Step 2 — shared index pehle banao. Yeh step kyun? Epsilon–delta identity $ε_{ij k} ε_{i l m} = δ_{j l} δ_{k m} - δ_{j m} δ_{k l}$ require karta hai ki summed index (yahan $k$ ) dono epsilons ke leading slot mein ho. Doosre epsilon $ε_{k l m}$ mein yeh already lead karta hai; pehle mein, $k$ last slot mein hai, isliye hum cyclically rotate karte hain $ε_{ij k} \to ε_{k ij}$ (permutation ka cyclic shift uska sign unchanged rakhta hai, toh yeh equality hai, sign flip nahi). Ab dono epsilons $k$ se start karte hain: $= ε_{k ij} ε_{k l m} a_{j} b_{l} c_{m} = (δ_{i l} δ_{j m} - δ_{im} δ_{j l}) a_{j} b_{l} c_{m} .$

Step 3 — deltas substitute karne do. Yeh step kyun? Har delta ek index rename karta hai (Cell D power): $= a_{j} b_{i} c_{j} - a_{j} b_{j} c_{i} = b_{i} (a_{j} c_{j}) - c_{i} (a_{j} b_{j}) .$

Step 4 — indices wapas vectors ke roop mein padhо. Yeh step kyun? $a_{j} c_{j} = a \cdot c$ , $a_{j} b_{j} = a \cdot b$ , aur free index $i$ ek vector mark karta hai: $a \times (b \times c) = b (a \cdot c) - c (a \cdot b) .$

Verify: $a = (1, 0, 0)$ , $b = (0, 1, 0)$ , $c = (0, 0, 1)$ lo: LHS hai $b \times c = (1, 0, 0)$ , toh $a \times (1, 0, 0) = (0, 0, 0)$ ; RHS hai $b (0) - c (0) = (0, 0, 0)$ . Yeh match karte hain. Ek non-degenerate numeric case ( $a = (1, 2, 3)$ etc.) machine checks mein confirm kiya gaya hai. ✔ (Dekho Vector calculus identities.)

Ex 8 — Cell H: real-world word problem

Worked example Ek force

$F = (3, 0, 4) N$ pivot se position $r = (0, 2, 0) m$ par act karta hai. Torque hai $τ = r \times F$ , yaani $τ_{i} = ε_{ij k} r_{j} F_{k}$ . $τ$ aur uski magnitude find karo. Forecast: ek free index $i$ → torque ek vector hai (N·m). $r$ $+ y$ ki taraf point karta hai, $F$ mostly $+ z$ aur $+ x$ ki taraf; dono ke perpendicular components wale torque ki expect karo.

Step 1 — component $i = 1$ . Yeh step kyun? $i = 1$ fix karo; sirf non-repeated $j, k$ $ε_{1 j k}$ mein survive karte hain ( $ε_{123} = + 1, ε_{132} = - 1$ ): $τ_{1} = r_{2} F_{3} - r_{3} F_{2} = (2) (4) - (0) (0) = 8.$

Step 2 — component $i = 2$ . Yeh step kyun? $ε_{2 j k}$ picks $ε_{231} = + 1, ε_{213} = - 1$ : $τ_{2} = r_{3} F_{1} - r_{1} F_{3} = (0) (3) - (0) (4) = 0.$

Step 3 — component $i = 3$ . Yeh step kyun? $ε_{3 j k}$ picks $ε_{312} = + 1, ε_{321} = - 1$ : $τ_{3} = r_{1} F_{2} - r_{2} F_{1} = (0) (0) - (2) (3) = - 6.$

Step 4 — magnitude. Yeh step kyun? $∣ τ ∣ = τ_{i} τ_{i}$ (ek Cell-A scalar): $∣ τ ∣ = 8^{2} + 0^{2} + (- 6)^{2} = 100 = 10 N\cdotm .$

Verify: $τ = (8, 0, - 6) N\cdotm$ . Units: $m \times N = N\cdotm$ ✔. Perpendicularity check $τ \cdot r = (8) (0) + (0) (2) + (- 6) (0) = 0$ ✔ (torque ⟂ position, jaise ek cross product hona chahiye).

Ex 9 — Cell I: exam trap

Worked example

Error(s) spot karo: ek student likhta hai " $w_{k} = δ_{ij} a_{i} b_{i}$ ." Fix karo aur $a = (1, 2, 2)$ , $b = (2, 0, 1)$ ke liye correct scalar do. Forecast: do red flags — ek free-index mismatch ( $w_{k}$ mein free $k$ hai, right side mein koi nahi) aur ek illegal index ( $i$ teen baar appear karta hai: $δ_{ij}$ , $a_{i}$ , $b_{i}$ mein).

Step 1 — illegal $i$ diagnose karo. Yeh step kyun? Repeated index matlab ek pairwise sum; $i$ ke teen copies ka koi unique pairing nahi hai (parent ka forbidden case). Delta already indices $i, j$ carry karta hai, toh ise distinct letters ke saath pair karna hoga.

Step 2 — student clearly $δ_{ij} a_{i} b_{j}$ mean karta tha. Yeh step kyun? Intent padhna: $a$ aur $b$ ko delta ke through contract karo. Ab $i$ do baar appear karta hai ( $δ_{ij}, a_{i}$ ) aur $j$ do baar ( $δ_{ij}, b_{j}$ ) — dono legal dummies, zero free indices.

Step 3 — delta collapse karo. Yeh step kyun? $δ_{ij} b_{j} = b_{i}$ (substitution), chhod ke: $δ_{ij} a_{i} b_{j} = a_{i} b_{i} = a \cdot b .$ Toh left side ek scalar hona chahiye, $w_{k}$ nahi. Correct statement: $w = a_{i} b_{i}$ .

Step 4 — evaluate karo. Yeh step kyun? Ab bas ek Cell-A dot product hai $a = (1, 2, 2)$ , $b = (2, 0, 1)$ ke saath: $w = a_{i} b_{i} = (1) (2) + (2) (0) + (2) (1) = 2 + 0 + 2 = 4.$

Verify: $a \cdot b = 2 + 0 + 2 = 4$ , toh $w = 4$ (ek single scalar, jaise corrected free-index count demand karta hai). ✔ Do lessons: free indices dono sides match karne chahiye, aur koi bhi index do se zyada baar appear nahi kar sakta.

Ex 10 — Cell J: dimension / limiting case

Worked example

$n = 3$ dimensions mein $δ_{ii}$ , $δ_{ij} δ_{ij}$ , aur $ε_{ij k} ε_{ij k}$ evaluate karo, aur describe karo ki $n$ badhne par har ek kaise behave karta hai. Forecast: yeh fully contracted hain (koi free indices nahi) → har ek ek single number hai jo dimension $n$ par depend karta hai. $δ_{ii}$ diagonal count karta hai, toh $n = 3$ expect karo.

Step 1 — delta ka trace. Yeh step kyun? $δ_{ii} = \sum_{i} δ_{ii} = 1 + 1 + 1$ : har dimension ke liye ek: $δ_{ii} = n = 3.$

Step 2 — delta squared. Yeh step kyun? Pehle ek delta collapse karo: $δ_{ij} δ_{ij} = δ_{ii}$ ( $j \to i$ substitute karo), phir Step 1: $δ_{ij} δ_{ij} = δ_{ii} = n = 3.$

Step 3 — epsilon fully contracted. Yeh step kyun? Master identity $ε_{ij k} ε_{i l m} = δ_{j l} δ_{k m} - δ_{j m} δ_{k l}$ apply karo $l = j$ aur $m = k$ choice ke saath (yeh doosre pair of indices bhi contract karta hai, kuch bhi free nahi chhoda): $ε_{ij k} ε_{ij k} = δ_{j j} δ_{k k} - δ_{j k} δ_{k j} .$ Ab Steps 1–2 use karke har delta contraction evaluate karo: $δ_{j j} = n$ aur $δ_{k k} = n$ (har ek trace hai), jabki $δ_{j k} δ_{k j} = δ_{j j} = n$ (do deltas ek trace mein collapse hote hain). Substitute karo: $ε_{ij k} ε_{ij k} = n \cdot n - n = n^{2} - n = 3^{2} - 3 = 6.$

Forecast reconcile karo & limiting behaviour: $δ_{ii} = n$ dimension ke saath linearly grow karta hai (space double karo, trace double hogi), jabki $ε_{ij k} ε_{ij k} = n (n - 1)$ quadratically grow karta hai — large $n$ ke liye epsilon-count delta-trace ko dwarf karta hai. $n = 3$ par yeh $3$ aur $6$ dete hain; $n = 4$ par yeh $4$ aur $12$ dete; gap bina bound ke badhta jaata hai.

Verify: $6$ exactly 3D mein $ε$ ki non-zero entries ki count hai ( $3! = 6$ permutations), har ek $(\pm 1)^{2} = 1$ contribute karta hai. ✔ (Dekho Determinants, Tensors.)

Recall Sab cells par quick self-test

$A_{ij} x_{j}$ ke free indices count karo ::: 1 (yeh ek vector hai). $δ_{ij} δ_{j k} a_{k}$ simplify karo ::: $a_{i}$ . Indices se $a \times a = 0$ kyun hai? ::: symmetric $a_{j} a_{k}$ antisymmetric $ε_{ij k}$ ke against ⇒ 0. 3D mein $ε_{ij k} ε_{ij k}$ ki value ::: 6. $δ_{ij} a_{i} b_{i}$ illegal kyun hai? ::: index $i$ teen baar appear karta hai — koi unique pairing nahi.

Connections

Einstein summation convention parent rules par built.
Cells A/H use karte hain Dot product; cells E/F/G/H use karte hain Cross product aur Levi-Civita symbol.
Cell D use karta hai Kronecker delta; cells B/C use karte hain Matrix multiplication.
Cell G gateway hai Vector calculus identities ka; cell J touch karta hai Determinants aur Tensors.