Tensor analysis — scalars, vectors, rank-2 tensors: Exercises

4.10.7 · D4 · Maths › Advanced Topics (Elite Level) › Tensor analysis — scalars, vectors, rank-2 tensors

Shuru karne se pehle, ek line ka notation refresher jo is page par use hota hai, taaki koi bhi symbol bina explanation ke na aaye.

Level 1 — Recognition

Exercise 1.1

Tumhe chaar $2 \times 2$ grids of numbers diye gaye hain. Inme se kaun sa ek rank-2 tensor ho sakta hai aur kaun sa bilkul nahi ho sakta, sirf unki definition ke basis par?

(a) Ek solid mein kisi point par stress.
(b) Ek restaurant menu ka price matrix (rows = dishes, columns = day of week).
(c) Kronecker delta $δ_{j}^{i}$ .
(d) Arc length $d s^{2} = g_{ij} d x^{i} d x^{j}$ se bana metric $g_{ij}$ .

Recall Solution

Test shape nahi hai, behaviour hai. Ek grid tensor tabhi hai jab uske entries mein coordinates change hone par ek transformation law ho.

(a) Tensor. Stress define hota hai ki forces surfaces ke across kaise act karte hain; axes rotate karo aur components $\tilde{T} = R T R^{⊤}$ follow karte hain. ✔
(b) Tensor nahi. "Dishes" aur "days" coordinate axes nahi hain; koi rule nahi hai jo prices ko bataye ki jab kuch rotate karo toh kaise change karein. Yeh sirf ek table hai. ✘
(c) Tensor (isotropic). $\tilde{δ}_{j}^{i} = \frac{\partial x ~ ^{i}}{\partial x ^{k}} \frac{\partial x ^{k}}{\partial x ~ ^{j}} = δ_{j}^{i}$ — har frame mein same. ✔
(d) Tensor. $d s^{2}$ ek invariant scalar hai aur $d x^{i}$ contravariant hai, isliye $g_{ij}$ ko covariant rank 2 hona hi padega (quotient theorem). ✔

Exercise 1.2

Har object ko rank aur index type se classify karo: (i) temperature $ϕ$ , (ii) velocity $V^{i}$ , (iii) ek gradient $\partial_{i} ϕ$ , (iv) inertia tensor $I^{ij}$ .

Recall Solution

(i) $ϕ$ — rank 0, ek scalar. $n^{0} = 1$ component.
(ii) $V^{i}$ — rank 1, contravariant (upper index, $d x^{i}$ ki tarah transform hota hai).
(iii) $\partial_{i} ϕ$ — rank 1, covariant (lower index, inverse Jacobian ke saath gradient ki tarah transform hota hai).
(iv) $I^{ij}$ — rank 2, contravariant (do upper indices, $n^{2}$ components).

Level 2 — Application

Exercise 2.1

Flat 2D Cartesian coordinates mein ek contravariant vector ke components $V^{k} = (2, 3)$ hain. Axes ko $θ = 9 0^{\circ}$ se rotate karo using

R = (cos θ sin θ - sin θ cos θ) .

Naye frame mein components $\tilde{V}^{i}$ nikalo. Figure dekho.

Figure — Tensor analysis — scalars, vectors, rank-2 tensors

Recall Solution

KYA: hum contravariant rule apply karte hain $\tilde{V}^{i} = \frac{\partial x ~ ^{i}}{\partial x ^{k}} V^{k} = R^{i}_{k} V^{k}$ , yani matrix $R$ times column $V$ . Kyun $R$ aur $R^{⊤}$ nahi: naye coordinates $\tilde{x}$ old se $R$ ke through milte hain, aur ek contravariant (upper) index forward Jacobian $\partial \tilde{x} / \partial x$ leta hai, jo exactly $R$ hai. $θ = 9 0^{\circ}$ par: $cos 9 0^{\circ} = 0, sin 9 0^{\circ} = 1$ , isliye $R = (01 - 1 0)$ .

\tilde{V} = (01 - 1 0) (23) = (- 3 2) .

KAISA DIKHTA HAI: arrow khud kabhi nahi hila (figure mein amber colour); humne sirf axes ko quarter-turn ghuma diya, isliye uski readings $(2, 3)$ se $(- 3, 2)$ ho gayi. Length $2^{2} + 3^{2} = 13$ unchanged rahi — rotation length preserve karta hai.

Exercise 2.2

Usi Cartesian frame mein ek covariant vector ke components $W_{k} = (2, 3)$ hain. Usi $θ = 9 0^{\circ}$ se rotate karo. $\tilde{W}_{i}$ nikalo aur confirm karo ki yeh contravariant answer se sirf isliye match karta hai kyunki rotation orthonormal hai.

Recall Solution

KYA: covariant rule $\tilde{W}_{i} = \frac{\partial x ^{k}}{\partial x ~ ^{i}} W_{k} = (R^{- 1})^{k}_{i} W_{k}$ , jo matrix form mein $\tilde{W} = (R^{- 1})^{⊤} W = R W$ hai orthonormal $R$ ke liye, kyunki $R^{- 1} = R^{⊤}$ isliye $(R^{- 1})^{⊤} = R$ .

\tilde{W} = (01 - 1 0) (23) = (- 3 2) .

Kyun match kiya: ek rotation ke liye inverse Jacobian transpose ke barabar hota hai, isliye upper aur lower vectors identically move karte hain. Ek non-orthonormal (jaise polar ya skewed) frame mein woh alag hote — dekho Metric tensor and Riemannian geometry.

Level 3 — Analysis

Exercise 3.1

Symmetric stress tensor $T = (3113)$ (MPa) lo. Explicitly dikhao ki uska trace $T^{i}_{i}$ , $4 5^{\circ}$ rotation $R = \frac{1}{2} (11 - 1 1)$ ke under invariant hai, $\tilde{T} = R T R^{⊤}$ compute karke aur uska trace lekar.

Recall Solution

KYA: dono slots rotate karo. $R^{⊤} = \frac{1}{2} (1 - 1 11)$ . Pehle $T R^{⊤} = \frac{1}{2} (3113) (1 - 1 11) = \frac{1}{2} (2 - 2 44)$ . Phir

\tilde{T} = R (T R^{⊤}) = \frac{1}{2} (11 - 1 1) (2 - 2 44) = \frac{1}{2} (4008) = (2004) .

Trace pehle: $3 + 3 = 6$ . Trace baad mein: $2 + 4 = 6$ . ✔ Invariant. KYU: trace ek contraction $T^{i}_{i}$ hai; contraction ek forward aur inverse Jacobian ko $δ_{k}^{l}$ mein jod deta hai, isliye transformation factors cancel ho jaate hain aur result ek genuine scalar ban jaata hai. KAISA DIKHTA HAI: off-diagonal $1$ (shear) gayab ho gaya — hum principal axes par pahunch gaye (figure dekho). Rotation ne sirf re-label kiya; physics (trace $6$ , eigenvalues $2, 4$ ) untouched hai.

Exercise 3.2

$T = (5425)$ ko uske symmetric part $S_{ij}$ aur antisymmetric part $A_{ij}$ mein split karo, aur verify karo ki $T = S + A$ .

Recall Solution

Use karo $S_{ij} = \frac{1}{2} (T_{ij} + T_{j i})$ aur $A_{ij} = \frac{1}{2} (T_{ij} - T_{j i})$ .

S = \frac{1}{2} [(5425) + (5245)] = (5335), A = \frac{1}{2} [(5425) - (5245)] = (01 - 1 0) .

Check: $S + A = (5425) = T$ . ✔ Kyun matter karta hai: symmetry coordinate-independent hai — ek symmetric tensor har frame mein symmetric rehta hai, isliye yeh split kisi bhi rotation mein survive karta hai.

Level 4 — Synthesis

Exercise 4.1

Quotient theorem ko ek simple case mein prove karo: maano $A^{ij}$ ek unknown grid hai, lekin har covariant vector $V_{j}$ ke liye object $B^{i} \equiv A^{ij} V_{j}$ ek genuine contravariant vector hai. Dikhao ki $A^{ij}$ ko contravariant rank-2 tensor ki tarah transform hona hi hoga.

Recall Solution

KYA hum jaante hain: $B^{i}$ contravariant hai, isliye $\tilde{B}^{i} = \frac{\partial x ~ ^{i}}{\partial x ^{k}} B^{k}$ , aur $V_{j}$ covariant hai, isliye $\tilde{V}_{j} = \frac{\partial x ^{m}}{\partial x ~ ^{j}} V_{m}$ . Dono frames mein same equation likho:

\tilde{A}^{ij} \tilde{V}_{j} = \tilde{B}^{i} = \frac{\partial x ~ ^{i}}{\partial x ^{k}} B^{k} = \frac{\partial x ~ ^{i}}{\partial x ^{k}} A^{k m} V_{m} .

$V_{m}$ ko covariant rule ke inverse se rewrite karo. Kyunki $\tilde{V}_{j} = \frac{\partial x ^{m}}{\partial x ~ ^{j}} V_{m}$ , left par free index $j$ already $\tilde{V}_{j}$ carry karta hai; left par $\tilde{V}_{j} = \frac{\partial x ^{m}}{\partial x ~ ^{j}} V_{m}$ substitute karo:

\tilde{A}^{ij} \frac{\partial x ^{m}}{\partial x ~ ^{j}} V_{m} = \frac{\partial x ~ ^{i}}{\partial x ^{k}} A^{k m} V_{m} .

Kyun agla step kaam karta hai: yeh har $V_{m}$ ke liye true hai, isliye hum dono sides se $V_{m}$ strip kar sakte hain (coefficients equate karo):

\tilde{A}^{ij} \frac{\partial x ^{m}}{\partial x ~ ^{j}} = \frac{\partial x ~ ^{i}}{\partial x ^{k}} A^{k m} .

$\tilde{A}^{ij}$ ke liye solve karo dono sides ko $\frac{\partial x ~ ^{n}}{\partial x ^{m}}$ se multiply karke aur sum karke; kyunki $\frac{\partial x ^{m}}{\partial x ~ ^{j}} \frac{\partial x ~ ^{n}}{\partial x ^{m}} = δ_{j}^{n}$ (chain rule), left collapse ho jaata hai $\tilde{A}^{in}$ mein:

\tilde{A}^{in} = \frac{\partial x ~ ^{i}}{\partial x ^{k}} \frac{\partial x ~ ^{n}}{\partial x ^{m}} A^{k m} .

Yeh exactly contravariant rank-2 law hai. Toh $A^{ij}$ pehle se ek tensor tha — humne yeh uske entries check kiye bina prove kar diya. Yeh woh litmus test hai jo the parent note mein reference kiya gaya hai.

Exercise 4.2

Polar coordinates $(r, θ)$ mein metric $g_{ij}$ nikalo, Cartesian $x = r cos θ, y = r sin θ$ se start karke, using $g_{ij} = \frac{\partial x ^{a}}{\partial u ^{i}} \frac{\partial x ^{a}}{\partial u ^{j}}$ (Cartesian $a$ par sum, jahan $u^{1} = r, u^{2} = θ$ ).

Recall Solution

Kyun yeh formula: flat space mein $d s^{2} = d x^{2} + d y^{2}$ hai; $d x, d y$ ko $d r, d θ$ mein express karke aur collect karne par $g_{ij}$ automatically mil jaata hai. Jacobian columns tangent vectors hain. Partial derivatives:

\frac{\partial x}{\partial r} = cos θ, \frac{\partial y}{\partial r} = sin θ, \frac{\partial x}{\partial θ} = - r sin θ, \frac{\partial y}{\partial θ} = r cos θ .

Ab columns dot karo:

g_{r r} = cos^{2} θ + sin^{2} θ = 1, g_{r θ} = cos θ (- r sin θ) + sin θ (r cos θ) = 0,

g_{θ θ} = (- r sin θ)^{2} + (r cos θ)^{2} = r^{2} .

g_{ij} = (10 0 r^{2}) \Rightarrow d s^{2} = d r^{2} + r^{2} d θ^{2} .

KAISA DIKHTA HAI: radius $r$ par $d θ$ move karne se $r d θ$ length ka arc sweep hota hai, isliye $θ$ -slot "cost" karta hai $r^{2}$ — yahi woh $r^{2}$ hai $g_{θ θ}$ mein. Dekho Covariant derivative and Christoffel symbols ki coordinates ki is curvature ke baad kya force hota hai.

Level 5 — Mastery

Exercise 5.1

Dikhao ki ordinary derivative $\partial_{k} V^{i}$ tensor nahi hai, use transform karke aur ek extra offending term exhibit karke. Batao ki kaun sa object ise fix karta hai.

Recall Solution

KYA: pehle $V^{i}$ transform karo, phir differentiate karo. Hamare paas hai $\tilde{V}^{i} = \frac{\partial x ~ ^{i}}{\partial x ^{a}} V^{a}$ . Ab $\tilde{x}^{k}$ ke respect se differentiate karo, chain rule use karke $\frac{\partial}{\partial x ~ ^{k}} = \frac{\partial x ^{b}}{\partial x ~ ^{k}} \frac{\partial}{\partial x ^{b}}$ (do $x$ -dependent factors par product rule):

\frac{\partial V ~ ^{i}}{\partial x ~ ^{k}} = \frac{\partial x ^{b}}{\partial x ~ ^{k}} \frac{\partial x ~ ^{i}}{\partial x ^{a}} \frac{\partial V ^{a}}{\partial x ^{b}} + extra, non-tensorial \frac{\partial x ^{b}}{\partial x ~ ^{k}} \frac{\partial ^{2} x ~ ^{i}}{\partial x ^{b} \partial x ^{a}} V^{a} .

Kyun toot jaata hai: ek true mixed rank-2 tensor mein sirf pehla term hota ( $\partial_{b} V^{a}$ par do Jacobians). Doosra term coordinate map ka ek second derivative carry karta hai. Curved ya curvilinear coordinates mein woh term nonzero hota hai, isliye $\partial_{k} V^{i}$ tensor ki tarah transform nahi karta. The fix: covariant derivative $\nabla_{k} V^{i} = \partial_{k} V^{i} + Γ^{i}_{k m} V^{m}$ mein Christoffel symbol $Γ$ add hota hai, jiski apni transformation ek equal-and-opposite extra term supply karti hai, offender ko cancel karke tensor character restore karti hai.

Exercise 5.2

Einstein field equations $G_{μν} = \frac{8 π G}{c ^{4}} T_{μν}$ do rank-2 tensors ko equate karte hain. Sirf jo upar prove kiya ussi se explain karo, kyun dono sides ko same type ke tensors hone chahiye equation ko physically meaningful hone ke liye, aur agar nahi hote toh kya galat hota.

Recall Solution

Argument (teen earned facts):

Ex 2.1–2.2 se, tensor components frame-to-frame Jacobian factors ke saath change hote hain; sirf woh combination jo transformation-consistent hai, coordinate-independent hoti hai.
Physics ka ek law depend nahi karna chahiye ki kisi insaan ne kaunse coordinates likhe — har frame mein same conclusion hona chahiye.
Agar $G_{μν}$ ek tarah transform karta aur $T_{μν}$ doosre tarah, toh ek frame mein true equation ek coordinate change ke baad false ho jaati (dono sides alag-alag Jacobian factors pick up karti aur match nahi karti). Isliye dono sides covariant rank-2 tensors hain: $G_{μν}$ (spacetime ki curvature) aur $T_{μν}$ (stress–energy). Unki equality frame-independent hai — dekho General Relativity — Einstein field equations aur source side ke liye, Stress and strain tensors (continuum mechanics). Yahi slogan "yeh tensor hai agar yeh tensor ki tarah transform ho" ka ultimate payoff hai.

Recall Self-test summary

Ek-line invariant jo har level ne test kiya ::: L1 shape ≠ tensor; L2 ek index → ek Jacobian; L3 trace/eigenvalues similarity invariants hain; L4 quotient theorem ko "for all V" chahiye; L5 ordinary derivative fail karta hai, covariant derivative fix karta hai. $T^{i}_{i}$ frame-independent kyun hai? ::: Contraction ek forward aur inverse Jacobian ko $δ_{k}^{l}$ mein pair karta hai, transformation factors cancel ho jaate hain — result ek scalar hai. Kaun sa extra object $\partial_{k} V^{i}$ ko rescue karta hai? ::: Christoffel symbol $Γ^{i}_{k m}$ covariant derivative $\nabla_{k} V^{i}$ ke andar.