Einstein summation convention: Exercises — Hinglish deep dive

4.10.9 · D4 · Maths › Advanced Topics (Elite Level) › Einstein summation convention

Throughout, indices $1, 2, 3$ tak run karte hain (teen dimensions) jab tak koi problem aur na kahe, aur do helper objects hain Kronecker delta $δ_{ij}$ aur Levi-Civita symbol $ε_{ij k}$ .

Recall Do facts jo tum baar baar reuse karoge

Substitution: $δ_{ij} a_{j} = a_{i}$ (delta repeated index ka naam badal deta hai). Epsilon–delta master identity: $ε_{ij k} ε_{i l m} = δ_{j l} δ_{k m} - δ_{j m} δ_{k l}$ (summed index shared first slot $i$ hai).

Level 1 — Recognition

Exercise 1.1

Har expression ke liye, free indices aur dummy (summed) indices list karo, aur bolo ki yeh kis tarah ka object hai (scalar, vector, ya matrix): (a) $a_{i} b_{i}$ (b) $A_{ij} x_{j}$ (c) $A_{ij} B_{j k}$ (d) $ε_{ij k} a_{j} b_{k}$ .

Recall Solution

Rule: ek index jo ek term mein exactly twice appear kare woh summed (dummy) hai; jo exactly once appear kare woh free hai. Free indices ki sankhya = rank = object ka type (0→scalar, 1→vector, 2→matrix).

(a) $i$ twice appear karta hai → dummy. Free: koi nahi. → scalar.
(b) $j$ twice appear karta hai ( $A_{ij}$ aur $x_{j}$ mein) → dummy. $i$ ek baar → free. → vector (ek free index).
(c) $j$ twice → dummy. $i$ ek baar, $k$ ek baar → free. Do free indices → matrix.
(d) $j$ twice, $k$ twice → dono dummy. $i$ ek baar → free. → vector.

Exercise 1.2

Inme se konsa summation convention mein legal hai, aur kyun? (a) $a_{i} b_{i} c_{i}$ (b) $a_{i} b_{i} + c_{j} d_{j}$ (c) $y_{k} = A_{ij} x_{j}$ (d) $δ_{ii}$ .

Recall Solution

(a) Illegal. $i$ teen baar appear karta hai. Repeated index ka matlab hai pairwise sum; teen copies se koi unique pairing nahi milti. Agar sach mein woh mean karte ho toh likho $\sum_{i} a_{i} b_{i} c_{i}$ .
(b) Legal. Har term mein alag se ek valid doubled index hai (pehle mein $i$ , doosre mein $j$ ). Dummy names apne term ke liye local hote hain, isliye koi clash nahi.
(c) Illegal. Left side mein free index $k$ hai; right side mein free index $i$ hai. Free indices dono sides pe identical hone chahiye. Fix: $y_{i} = A_{ij} x_{j}$ .
(d) Legal. $δ_{ii} = \sum_{i} δ_{ii} = n = 3$ . Ek scalar.

Level 2 — Application

Exercise 2.1

$a = (1, 2, 3)$ aur $b = (4, 5, 6)$ lo. $a_{i} b_{i}$ haath se compute karo aur bolo yeh kya represent karta hai.

Recall Solution

$a_{i} b_{i} = \sum_{i = 1}^{3} a_{i} b_{i} = (1) (4) + (2) (5) + (3) (6) = 4 + 10 + 18 = 32$ . Yeh Dot product $a \cdot b$ hai — koi free index nahi, isliye ek single number, jaisa dot product hona chahiye.

Exercise 2.2

$a = (1, 2, 3)$ , $b = (4, 5, 6)$ ke saath, $a \times b$ ka pehla component $(a \times b)_{i} = ε_{ij k} a_{j} b_{k}$ use karke compute karo. Phir teeno components compute karo.

Recall Solution

$i = 1$ set karo. Sirf nonzero $ε_{1 j k}$ hain $ε_{123} = + 1$ aur $ε_{132} = - 1$ : $(a \times b)_{1} = ε_{123} a_{2} b_{3} + ε_{132} a_{3} b_{2} = a_{2} b_{3} - a_{3} b_{2} = (2) (6) - (3) (5) = 12 - 15 = - 3 .$ Similarly $(a \times b)_{2} = a_{3} b_{1} - a_{1} b_{3} = (3) (4) - (1) (6) = 12 - 6 = 6$ , aur $(a \times b)_{3} = a_{1} b_{2} - a_{2} b_{1} = (1) (5) - (2) (4) = 5 - 8 = - 3$ . To $a \times b = (- 3, 6, - 3)$ . Neeche figure mein dekho ki har $ε$ ka sign kya pick out karta hai.

Exercise 2.3

$δ_{ij} a_{i} b_{j}$ ko simplify karke ek jaani-pehchaani expression mein badlo.

Recall Solution

Delta pe substitution use karo. $δ_{ij}$ ek repeated index ko "kha" leta hai: isse $b_{j}$ pe apply karne se $δ_{ij} b_{j} = b_{i}$ milta hai. To $δ_{ij} a_{i} b_{j} = a_{i} (δ_{ij} b_{j}) = a_{i} b_{i} .$ Yeh dot product $a \cdot b$ hai. (Equivalently, delta $i = j$ force karta hai, double sum ko ek single sum mein collapse karta hai.)

Level 3 — Analysis

Exercise 3.1

Prove karo ki $δ_{ij} δ_{j k} = δ_{ik}$ .

Recall Solution

Implied sum likho: $δ_{ij} δ_{j k} = \sum_{j} δ_{ij} δ_{j k}$ . Pehle factor mein sirf $j = i$ nonzero hai, jo $δ_{ik}$ contribute karta hai; baaki sab $j$ ke liye $0$ milta hai. Isliye $δ_{ij} δ_{j k} = δ_{ik}$ . (Interpretation: identity matrix ko khud se multiply karne par identity milti hai.)

Exercise 3.2

Dikhao ki $ε_{ij k} a_{j} a_{k} = 0$ kisi bhi vector $a$ ke liye. (Yeh statement hai $a \times a = 0$ .)

Recall Solution

Key yeh hai ki $ε_{ij k}$ $j, k$ mein antisymmetric hai (unhe swap karne se sign flip hota hai), jabki $a_{j} a_{k}$ symmetric hai (unhe swap karne se kuch nahi badalta). Dummy pair $j \leftrightarrow k$ rename karo (allowed — dummy names free hain): $ε_{ij k} a_{j} a_{k} = ε_{ik j} a_{k} a_{j} = - ε_{ij k} a_{j} a_{k},$ $ε_{ik j} = - ε_{ij k}$ aur $a_{k} a_{j} = a_{j} a_{k}$ use karke. Jo quantity apne aap ke negative ke barabar ho woh $0$ hona chahiye.

Exercise 3.3

$ε_{ij k} ε_{ij k}$ evaluate karo (teeno indices summed).

Recall Solution

Master identity use karo last two slots matched karke. $ε_{ij k} ε_{i l m} = δ_{j l} δ_{k m} - δ_{j m} δ_{k l}$ mein $l = j$ , $m = k$ set karo: $ε_{ij k} ε_{ij k} = δ_{j j} δ_{k k} - δ_{j k} δ_{k j} .$ Ab $δ_{j j} = 3$ , $δ_{k k} = 3$ , to pehla term $9$ hai. Doosra: $δ_{j k} δ_{k j} = δ_{j k} δ_{j k} = δ_{j j} = 3$ . Isliye $ε_{ij k} ε_{ij k} = 9 - 3 = 6 .$ (Sanity check: $ε$ ke exactly $6$ nonzero components hain, har ek $\pm 1$ , aur squaring se $+ 1$ milta hai har baar → $6$ .)

Level 4 — Synthesis

Exercise 4.1

BAC–CAB rule $a \times (b \times c) = b (a \cdot c) - c (a \cdot b)$ index notation use karke prove karo.

Recall Solution

$i$ -th component se shuru karo aur inner cross product ke liye ek fresh dummy introduce karo: $[a \times (b \times c)]_{i} = ε_{ij k} a_{j} (b \times c)_{k} = ε_{ij k} a_{j} ε_{k l m} b_{l} c_{m} .$ Doosre epsilon ko cycle karo taaki shared index pehle aaye: $ε_{k l m} = ε_{k ij}$ ? Nahi — hum pehle wale ko cycle karte hain. Kyunki $ε_{ij k} = ε_{k ij}$ (cyclic shift even hai), $ε_{ij k} = ε_{k ij}$ rewrite karo taaki dono $k$ pehle slot mein share karein: $= ε_{k ij} ε_{k l m} a_{j} b_{l} c_{m} .$ Master identity apply karo shared index $k$ ke saath: $ε_{k ij} ε_{k l m} = δ_{i l} δ_{j m} - δ_{im} δ_{j l}$ : $= (δ_{i l} δ_{j m} - δ_{im} δ_{j l}) a_{j} b_{l} c_{m} .$ Distribute karo aur substitution use karo. Pehla term: $δ_{i l} δ_{j m} a_{j} b_{l} c_{m} = b_{i} (a_{j} c_{j}) = b_{i} (a \cdot c)$ . Doosra term: $δ_{im} δ_{j l} a_{j} b_{l} c_{m} = c_{i} (a_{j} b_{j}) = c_{i} (a \cdot b)$ . Isliye $[a \times (b \times c)]_{i} = b_{i} (a \cdot c) - c_{i} (a \cdot b),$ jo exactly $b (a \cdot c) - c (a \cdot b)$ hai. $■$

Exercise 4.2

Scalar identity $(a \times b) \cdot (c \times d) = (a \cdot c) (b \cdot d) - (a \cdot d) (b \cdot c)$ prove karo.

Recall Solution

Dono cross products index form mein likho (unhe dotting index $i$ share karna hai): $(a \times b)_{i} (c \times d)_{i} = (ε_{ij k} a_{j} b_{k}) (ε_{i l m} c_{l} d_{m}) = ε_{ij k} ε_{i l m} a_{j} b_{k} c_{l} d_{m} .$ Dono epsilons already pehle slot $i$ share kar rahe hain, to master identity seedha apply karo: $= (δ_{j l} δ_{k m} - δ_{j m} δ_{k l}) a_{j} b_{k} c_{l} d_{m} .$ Pehla term: $δ_{j l} δ_{k m} a_{j} b_{k} c_{l} d_{m} = (a_{l} c_{l}) (b_{m} d_{m}) = (a \cdot c) (b \cdot d)$ . Doosra term: $δ_{j m} δ_{k l} a_{j} b_{k} c_{l} d_{m} = (a_{m} d_{m}) (b_{l} c_{l}) = (a \cdot d) (b \cdot c)$ . Isliye result hai $(a \cdot c) (b \cdot d) - (a \cdot d) (b \cdot c)$ . $■$ ( $c = a, d = b$ set karne se Lagrange's identity $∣ a \times b ∣^{2} = ∣ a ∣^{2} ∣ b ∣^{2} - (a \cdot b)^{2}$ recover hoti hai.)

Level 5 — Mastery

Exercise 5.1

Index notation aur master identity use karke, vector-calculus identity prove karo $\nabla \times (\nabla \times F) = \nabla (\nabla \cdot F) - \nabla^{2} F .$ Yahan $\nabla$ operator $\partial_{i} = \frac{\partial}{\partial x _{i}}$ ki tarah kaam karta hai, aur $(\nabla \times F)_{i} = ε_{ij k} \partial_{j} F_{k}$ . Dekho Vector calculus identities.

Recall Solution

$i$ -th component likho, inner curl ke liye ek fresh dummy set ke saath: $[\nabla \times (\nabla \times F)]_{i} = ε_{ij k} \partial_{j} (\nabla \times F)_{k} = ε_{ij k} \partial_{j} (ε_{k l m} \partial_{l} F_{m}) .$ Constants bahar nikalo ( $ε$ sirf numbers hain, aur derivatives unke saath commute karte hain): $= ε_{ij k} ε_{k l m} \partial_{j} \partial_{l} F_{m} .$ Pehle epsilon ko cycle karo taaki $k$ aage aaye: $ε_{ij k} = ε_{k ij}$ , to $= ε_{k ij} ε_{k l m} \partial_{j} \partial_{l} F_{m} = (δ_{i l} δ_{j m} - δ_{im} δ_{j l}) \partial_{j} \partial_{l} F_{m} .$ Pehla term: $δ_{i l} δ_{j m} \partial_{j} \partial_{l} F_{m} = \partial_{m} \partial_{i} F_{m} = \partial_{i} (\partial_{m} F_{m}) = \partial_{i} (\nabla \cdot F) = [\nabla (\nabla \cdot F)]_{i} .$ Doosra term: $δ_{im} δ_{j l} \partial_{j} \partial_{l} F_{m} = \partial_{j} \partial_{j} F_{i} = (\partial_{j} \partial_{j}) F_{i} = \nabla^{2} F_{i} .$ Combine karne par: $[\nabla \times (\nabla \times F)]_{i} = [\nabla (\nabla \cdot F)]_{i} - \nabla^{2} F_{i}$ , yaani $\nabla \times (\nabla \times F) = \nabla (\nabla \cdot F) - \nabla^{2} F . ■$

Exercise 5.2

Dikhao ki $ε_{ij k} \partial_{j} \partial_{k} f = 0$ kisi bhi twice-differentiable scalar field $f$ ke liye (yaani $\nabla \times \nabla f = 0$ ).

Recall Solution

Wahi symmetry argument jo Exercise 3.2 mein tha, ab derivatives ke saath. Symbol $ε_{ij k}$ $j \leftrightarrow k$ ke under antisymmetric hai; double derivative $\partial_{j} \partial_{k} f$ symmetric hai (mixed partials commute: $\partial_{j} \partial_{k} f = \partial_{k} \partial_{j} f$ ). Dummies $j \leftrightarrow k$ rename karo: $ε_{ij k} \partial_{j} \partial_{k} f = ε_{ik j} \partial_{k} \partial_{j} f = - ε_{ij k} \partial_{j} \partial_{k} f .$ Apne negative ke barabar ⇒ $0$ . To kisi bhi gradient ka curl zero hota hai.

Exercise 5.3

Ek $3 \times 3$ matrix $A$ ka determinant index form mein express karo aur $A = diag (2, 3, 4)$ ke liye verify karo. Dekho Determinants.

Recall Solution

Determinant hai $det A = ε_{ij k} A_{1 i} A_{2 j} A_{3 k}$ ( $i, j, k$ par sum; $ε$ signed permutation orientation deta hai, exactly jo Determinants ko chahiye). Sirf $(1, 2, 3)$ ke permutations survive karte hain, aur diagonal matrix ke liye sirf nonzero entries $A_{11}, A_{22}, A_{33}$ hain, jo $i = 1, j = 2, k = 3$ force karte hain: $det A = ε_{123} A_{11} A_{22} A_{33} = (+ 1) (2) (3) (4) = 24 .$ Diagonal ke product se match karta hai. ✔

Recall Self-test checklist

Ladder run kiya? ::: L1 recognise · L2 compute · L3 manipulate deltas/epsilons · L4 prove named vector identities · L5 prove calculus identities aur determinants — sab sirf index notation se.

Connections

Parent: Einstein summation convention.
Built on Dot product, Cross product, Kronecker delta, Levi-Civita symbol.
Extends to Matrix multiplication, Determinants, Tensors, aur Vector calculus identities.