Tensor analysis — scalars, vectors, rank-2 tensors: Worked examples

4.10.7 · D3 · Maths › Advanced Topics (Elite Level) › Tensor analysis — scalars, vectors, rank-2 tensors

Shuru karne se pehle, un do objects ki ek yaad-dahaani jo hum baar baar use karte hain, seedhi baat mein:

Definition Index notation, seedhe alfaaz mein (formulas se pehle yeh padho)

Ek rank-2 tensor ki entries mein do labels hote hain. Hum $T_{ij}$ likhte hain (ya $T^{ij}$ , ya mixed $T^{i}_{j}$ ) jahan $i, j$ mein se har ek 2D mein $1, 2$ par chalta hai. Socho $i$ = row, $j$ = column, toh $T_{11}$ top-left hai, $T_{12}$ top-right, aur aise hi.

Ek lower index (jaise $T_{ij}$ ) ko covariant kehte hain; ek upper index (jaise $T^{ij}$ ) ko contravariant. Is page ki Cartesian, rotation-only duniya mein dono ke numbers same hain — yeh distinction sirf curved/non-orthonormal coordinates mein kaattta hai (dekho Ex 6). Hum dono notations sirf isliye rakhte hain kyunki parent note unhe use karta hai.
Einstein summation convention ka matlab: jab bhi koi same index ek baar upar aur ek baar neeche kisi product mein aaye, uske upar sum karo. Toh $A^{ij} V_{j}$ silently matlab hai $\sum_{j} A^{ij} V_{j}$ . Koi $Σ$ symbol nahi likhte; repeated index hi add karne ka instruction hai.
Example: $tr T = T^{1}_{1} + T^{2}_{2}$ bas "dono diagonal entries ko add karo" hai.

Scenario matrix

Ek rank-2 tensor problem mein jo bhi pehli baar seekhne waale ko aata hai, woh in cells mein sort ho jaata hai. Har column ek alag tarah ki mushkil hai; neeche ke examples un cell(s) ke saath label hain jo woh cover karte hain.

Cell	Case class	Kya tricky hai	Covered by
A	Symmetric tensor, "nice" rotation ( $4 5^{\circ}$ )	diagonalising, principal axes	Ex 1
B	Symmetric tensor, "ugly" angle ( $3 0^{\circ}$ )	messy $cos, sin$ , invariants phir bhi hold karte hain	Ex 2
C	Antisymmetric tensor	off-diagonal only, rotation uske saath kya karta hai	Ex 3
D	Degenerate input: zero / isotropic tensor	kya $0$ ya $λ I$ ek tensor hai?	Ex 4
E	Sign cases: negative & mixed-sign entries	tension vs compression, trace/det ka sign	Ex 5
F	Limiting / blow-up coordinate (polar $r \to 0$ )	metric degenerate ho jaata hai, $det g \to 0$	Ex 6
G	Real-world word problem	physics → tensor mein translate karo	Ex 7
H	Exam twist: quotient theorem / "kya yeh tensor hai?"	brute force ke bina prove karo	Ex 8

Invariants jo hum almost har example mein sanity-check karte hain, woh teen numbers hain jo rotation kabhi nahi badal sakta:

Example 1 — Cell A: symmetric tensor, clean $4 5^{\circ}$

Worked example

$T = (3113)$ MPa ko $θ = 4 5^{\circ}$ se rotate karo Pehle forecast: dono $1$ 's shear hain. Guess: ek smart rotation ke baad, kya shear khatam ho sakta hai? Agar haan, toh diagonal par kya aayega?

Step 1 — $4 5^{\circ}$ ke liye $R$ likho.

=\frac{1}{\sqrt2}\begin{pmatrix}1&-1\\1&1\end{pmatrix}.$$ *Yeh step kyun?* $\cos45^\circ=\sin45^\circ=\tfrac{1}{\sqrt2}$; rotation template mein plug karo. **Step 2 — Rank-2 rule $\tilde T=RTR^{\top}$ apply karo.** $$RT=\frac{1}{\sqrt2}\begin{pmatrix}1&-1\\1&1\end{pmatrix}\begin{pmatrix}3&1\\1&3\end{pmatrix} =\frac{1}{\sqrt2}\begin{pmatrix}2&-2\\4&4\end{pmatrix}.$$ *Yeh step kyun?* $T$ ki har index ek rotation factor "khaati" hai; matrix multiplication exactly wahi summation hai shared index par. **Step 3 — $R^{\top}$ se multiply karo.** $$\tilde T=\frac{1}{\sqrt2}\begin{pmatrix}2&-2\\4&4\end{pmatrix}\cdot\frac{1}{\sqrt2}\begin{pmatrix}1&1\\-1&1\end{pmatrix} =\begin{pmatrix}2&0\\0&4\end{pmatrix}.$$ *Yeh step kyun?* Doosra $R^{\top}$ *dusri* slot ko rotate karta hai. Off-diagonals khatam ho gaye → yeh ==principal axes== hain. (Eigenvalues $2,4$ ab diagonal par hain; kaunsa top-left par aata hai yeh rotation direction par depend karta hai — yahan $45^\circ$ counter-clockwise pehle $2$ rakhta hai.) ![[deepdives/dd-maths-4.10.07-d3-s01.png]] **Figure s01.** Purple ellipse $T$ ki "stress ellipse" hai; uske dono axes Step 3 mein mile principal directions ki taraf point karte hain. Magenta arrow wala axis eigenvalue $4$ ke saath hai, orange arrow wala eigenvalue $2$ ke saath. Isko aise padho: frame ko in arrows ke saath align karna hi woh kaam hai jo off-diagonal shear ko khatam karta hai. **Verify:** $\operatorname{tr}T=3+3=6$ aur $\operatorname{tr}\tilde T=2+4=6$ ✔. $\det T=3\cdot3-1\cdot1=8$, $\det\tilde T=2\cdot4=8$ ✔. $T$ ke eigenvalues: solve karo $(3-\lambda)^2-1=0\Rightarrow\lambda=4,2$ — exactly nayi diagonal. MPa units throughout ✔.

Example 2 — Cell B: symmetric tensor, ugly $3 0^{\circ}$

Worked example Wohi

$T = (3113)$ , ab $θ = 3 0^{\circ}$ Forecast: $3 0^{\circ}$ par hum principal axes par nahi hain (woh $4 5^{\circ}$ par the), toh off-diagonal nonzero hona chahiye lekin trace phir bhi $6$ rehna chahiye. Nayi shear ka sign guess karo.

Step 1 — $3 0^{\circ}$ ke exact numbers. Inhe exact rakho: $cos 3 0^{\circ} = \frac{3}{2}$ , $sin 3 0^{\circ} = \frac{1}{2}$ . $R = (\frac{3}{2} \frac{1}{2} - \frac{1}{2} \frac{3}{2}) .$ Yeh step kyun? Hum surds ( $3$ ) rakhte hain decimals ki jagah taaki invariants end mein clean integers banein — "ugly angle" ka matlab irrational entries hai, sloppy arithmetic nahi.

Step 2 — Rotation ke under symmetric $2 \times 2$ ke liye closed form use karo. $T = (a b b a)$ ke liye, $\tilde{T}_{11} = a + b sin 2 θ, \tilde{T}_{22} = a - b sin 2 θ, \tilde{T}_{12} = b cos 2 θ .$ Yeh formula kyun sahi hai (derivation): $\tilde{T} = R T R^{⊤}$ ko $R = (c s - s c)$ (likhte hain $c = cos θ, s = sin θ$ ) aur $T$ ki dono diagonal entries par $a$ ke saath multiply karo. Top-left entry ban jaati hai $\tilde{T}_{11} = a c^{2} + 2 b cs + a s^{2} = a (c^{2} + s^{2}) + b (2 cs) = a + b sin 2 θ$ , using $c^{2} + s^{2} = 1$ aur double-angle identity $2 cs = sin 2 θ$ . Similarly $\tilde{T}_{12} = b (c^{2} - s^{2}) = b cos 2 θ$ (double-angle $c^{2} - s^{2} = cos 2 θ$ ) aur $\tilde{T}_{22} = a - b sin 2 θ$ . Isliye sirf $2 θ$ aata hai — axes ko $θ$ se rotate karne par stress pattern $2 θ$ se rotate hota hai. $a = 3, b = 1, 2 θ = 6 0^{\circ}$ ke saath: $sin 6 0^{\circ} = \frac{3}{2}$ , $cos 6 0^{\circ} = \frac{1}{2}$ .

Step 3 — Exact values plug in karo.

\approx\begin{pmatrix}3.8660&0.5\\0.5&2.1340\end{pmatrix}\text{ MPa}.$$ *Yeh step kyun?* Exact form se trace clearly $6$ dikhta hai; decimals sirf sizes padhne ke liye hain. Off-diagonal $\tfrac12$ leftover shear hai kyunki $30^\circ$ principal angle nahi hai. **Verify (exact):** trace $=(3+\tfrac{\sqrt3}{2})+(3-\tfrac{\sqrt3}{2})=6$ ✔ (surds cancel ho gaye). $\det=(3+\tfrac{\sqrt3}{2})(3-\tfrac{\sqrt3}{2})-(\tfrac12)^2=9-\tfrac34-\tfrac14=8$ ✔. Off-diagonal $\tfrac12\neq0$ confirm karta hai hum principal axes par *nahi* hain — Ex 1 ke saath consistent.

Example 3 — Cell C: antisymmetric tensor

Worked example Antisymmetric

$A = (0 - 2 20)$ ko kisi bhi $θ$ se rotate karo (use karo $6 0^{\circ}$ ) Forecast: antisymmetric matlab $A_{ij} = - A_{j i}$ . Test karne layak ek guess: kya ek antisymmetric tensor rotation ke baad bhi antisymmetric rehta hai, aur kya number $2$ bacha rehta hai?

Step 1 — Split yaad karo. Koi bhi tensor $= \frac{1}{2} (T + T^{⊤}) + \frac{1}{2} (T - T^{⊤})$ : symmetric + antisymmetric. $A$ purely antisymmetric part hai. Yeh step kyun? Symmetry type coordinate-independent hai, toh rotation $A$ ko symmetric piece mein leak nahi kar sakta — yeh ek strong prediction hai test karne ke liye.

Step 2 — Pehle dikhao $\tilde{A}$ antisymmetric hai. $R$ orthogonal hone par, $(R A R^{⊤})^{⊤} = R A^{⊤} R^{⊤} = R (- A) R^{⊤} = - R A R^{⊤}$ , toh $\tilde{A} = - \tilde{A}^{⊤}$ . Yeh step kyun? Structure pehle prove karne se pata chalta hai $\tilde{A} = (0 - c c 0)$ ek single number $c$ ke liye; hume sirf woh ek $c$ dhundhna hai.

Step 3 — $c$ explicitly compute karo. $R = (cos θ sin θ - sin θ cos θ)$ likho aur $\tilde{A} = R A R^{⊤}$ multiply karo. Top-right entry hai

=\underbrace{(\cos\theta)(A_{12})(\cos\theta)}_{\text{from }R_{1\cdot},\,R^{\top}_{\cdot2}} +\underbrace{(-\sin\theta)(A_{21})(-\sin\theta)}_{} =A_{12}\cos^2\theta - A_{21}\sin^2\theta.$$ Kyunki $A_{21}=-A_{12}$, yeh ban jaata hai $A_{12}\cos^2\theta+A_{12}\sin^2\theta=A_{12}(\cos^2\theta+\sin^2\theta)$. Ab Pythagorean identity $\cos^2\theta+\sin^2\theta=1$ isko collapse kar deti hai $c=A_{12}=2$ par. $$\tilde A=\begin{pmatrix}0&2\\-2&0\end{pmatrix}.$$ *Yeh step kyun?* Entry ko term-by-term likhne se *exactly dikhai deta hai* $\cos^2+\sin^2$ kahan se aata hai — yeh antisymmetry $A_{21}=-A_{12}$ hai jo minus ko plus mein badal deti hai. Akela component ek rotational **invariant** hai (2D "curl"/spin). **Verify:** kisi bhi antisymmetric $2\times2$ ka trace $0$ hota hai: $0+0=0$ pehle aur baad ✔. $\det A=0\cdot0-2\cdot(-2)=4$; $\det\tilde A=4$ ✔. Antisymmetry preserved ✔.

Example 4 — Cell D: degenerate inputs (zero & isotropic)

Worked example Kya zero tensor

$0$ ek tensor hai? Kya $λ I$ frame-independent hai? Forecast: guess karo ki $(0000)$ aur $(5005)$ har frame mein apni form rakhte hain ya nahi.

Step 1 — Zero tensor. $\tilde{0} = R 0 R^{⊤} = 0$ har $R$ ke liye. Yeh step kyun? Kisi bhi cheez ko zero matrix se multiply karne par zero aata hai; transformation law trivially satisfy hoti hai, toh $0$ ek valid (rank-2) tensor hai.

Step 2 — Isotropic tensor $T = λ I$ . $\tilde{T} = R (λ I) R^{⊤} = λ R R^{⊤} = λ I .$ Yeh step kyun? $R R^{⊤} = I$ orthonormality se; scalar $λ$ pass through ho jaata hai. Toh ek pure "pressure" state (equal normal stress, koi shear nahi) har rotated axis par same dikhta hai.

Figure s02. Purple circle isotropic tensor

λ I

ki stress "ellipse" hai — ek perfect circle kyunki dono eigenvalues equal hain. Magenta arrows ka pair ek frame hai; orange pairs rotated frames hain. Notice karo ki har frame har direction mein same radius read karta hai: iska yehi matlab hai "har frame mein same components."

Verify: $λ = 5$ ke saath: trace $= 10$ dono frames mein ✔; $det = 25$ dono frames mein ✔; eigenvalues dono $5$ (repeated) ✔. Zero tensor: trace $0$ , det $0$ , invariant ✔.

Example 5 — Cell E: sign cases (compression, mixed sign)

Worked example

$T = (- 4 0 06)$ MPa — ek axis compression mein, ek tension mein. $4 5^{\circ}$ rotate karo. Forecast: ek axis par negative (compression), doosre par positive (tension). Guess karo: kya $4 5^{\circ}$ frame shear create karta hai, aur trace/determinant ka sign kya hai?

Step 1 — Signs physically padho. $- 4$ = andar dhakelna (compression), $+ 6$ = bahar kheenchna (tension). Trace $= - 4 + 6 = 2$ (net tension). $det = - 24 < 0$ : opposite signs $\Rightarrow$ ek saddle stress state. "Saddle" kyun? Jab dono eigenvalues ke opposite signs hon (equivalently $det T < 0$ ), toh material ek principal axis ke saath khiincha ja raha hai jabki doosre ke saath squeeze ho raha hai — geometrically stress "surface" ek direction mein rise karti hai aur perpendicular direction mein girr jaati hai, exactly Pringle/horse-saddle ki shape. Yeh mixed-sign state ka tensor signature hai.

Step 2 — $4 5^{\circ}$ rotate karo closed form use karke (ab $b = 0$ , lekin $a_{1} \neq = a_{2}$ ): diagonal $(p 0 0 q)$ ke liye, $\tilde{T} = \frac{1}{2} (p + q p - q p - q p + q)$ . $p = - 4, q = 6$ ke saath: $\tilde{T} = (1 - 5 - 5 1)$ MPa. Yeh step kyun? Principal axes se door rotate karna shear create karta hai — Ex 1 ka bilkul ulta. $4 5^{\circ}$ direction maximum shear $\frac{∣ q - p ∣}{2} = 5$ MPa feel karta hai (sign rotation direction par depend karta hai; magnitude $5$ physically matter karta hai).

Verify: trace $= 1 + 1 = 2$ ✔ (match karta hai). $det = 1 - 25 = - 24$ ✔ (sign preserved, abhi bhi saddle). $\tilde{T}$ ke eigenvalues: $1 \pm 5 = {6, - 4}$ — original ke same ✔.

Example 6 — Cell F: limiting / blow-up coordinate (polar $r \to 0$ )

Worked example Polar metric

$g_{ij} = (10 0 r^{2})$ jab $r \to 0$ Forecast: metric distance encode karta hai $d s^{2} = d r^{2} + r^{2} d θ^{2}$ . Guess karo ki origin $r = 0$ par kya gadbad hoti hai — kya yeh space mein real hole hai, ya sirf coordinates fail ho rahi hain?

Step 1 — Metric ka determinant. $det g = 1 \cdot r^{2} - 0 = r^{2}$ . Yeh step kyun? $det g$ area element $r d r d θ$ hai; agar $det g \to 0$ toh coordinate patch degenerate ho raha hai (ek poora $θ$ ka circle ek point par collapse ho jaata hai).

Step 2 — Limit $r \to 0$ lo. $det g \to 0$ , toh $g$ non-invertible ho jaata hai: inverse metric $g^{ij}$ jo upar define hua woh ab exist nahi karta, aur tum $θ$ index raise nahi kar sakte. Origin par angle $θ$ undefined hai. Yahan opening definitions se upper-vs-lower distinction finally kaatta hai: $V_{i} = g_{ij} V^{j}$ ko backwards chalane ke liye ek invertible $g$ chahiye. Yeh step kyun? Ek metric ko $g^{ij}$ define karne ke liye invertible hona chahiye; yahan failure ek coordinate singularity hai, curvature singularity nahi — flat plane apne centre par perfectly smooth hai.

Figure s03. Magenta curve

det g = r^{2}

hai jo radius

r

ke against plot hua hai. Yeh ek khush positive parabola hai har jagah siwaay violet dot

r = 0

ke, jahan woh zero touch karta hai. Woh single touch-point coordinate singularity hai: polar chart origin par "pinch shut" ho jaata hai —

θ

values ka ek poora circle ek point mein crowd ho jaata hai — even though flat plane underneath wahan perfectly smooth hai.

Verify: $r = 2$ par, $det g = 4 > 0$ (theek hai); $r = 0$ par, $det g = 0$ (degenerate) ✔. Cartesian $g = δ_{ij}$ se compare karo: $det = 1$ har jagah — plane ko khud koi problem nahi, confirm karta hai ki singularity polar chart mein rehti hai, metric tensor story se match karta hai.

Example 7 — Cell G: real-world word problem

Worked example Thin square plate ka moment-of-inertia tensor

Ek thin flat plate ke centre ke baare mein principal moments hain $I_{x} = 2, I_{y} = 8 kg\cdotm^{2}$ , koi product of inertia nahi. Engineer usse $3 0^{\circ}$ rotated mount karta hai. Nayi axes ke saath mehsoosi product of inertia $I_{x y}$ kya hai? Forecast: principal axes se door rotate karne par product of inertia on ho jaana chahiye. Uska rough size guess karo.

Step 1 — Rank-2 tensor ki tarah model karo. $I = (I_{x} 0 0 I_{y}) = (2008)$ . Yeh step kyun? Inertia $\tilde{I} = R I R^{⊤}$ obey karta hai — yeh ek rank-2 tensor hai (stress cases D/E jaisi same machinery), toh Ex 2 mein derive kiye closed forms verbatim apply hote hain.

Step 2 — Diagonal tensor ke liye off-diagonal formula. Ex 2 derivation se $b = 0$ ke saath original mein lekin distinct diagonal entries ke saath, general diagonal-tensor result hai $\tilde{I}_{x y} = \frac{1}{2} (I_{x} - I_{y}) sin 2 θ$ . $2 θ = 6 0^{\circ}$ , $sin 6 0^{\circ} = \frac{3}{2}$ ke saath: $\tilde{I}_{x y} = \frac{1}{2} (2 - 8) \cdot \frac{3}{2} = - \frac{3 3}{2} \approx - 2.598 kg\cdotm^{2}$ , matlab magnitude $2.598$ . Yeh step kyun? Yeh exactly physics ki pooshak mein Ex 5 wala shear-generation formula hai.

Step 3 — Nayi diagonal. $\tilde{I}_{xx} = \frac{I _{x} + I _{y}}{2} - \frac{I _{y} - I _{x}}{2} cos 2 θ = 5 - 3 (0.5) = 3.5$ , aur $\tilde{I}_{y y} = 5 + 1.5 = 6.5 kg\cdotm^{2}$ . Yeh step kyun? Same closed form Ex 2 se jahan average $\frac{I _{x} + I _{y}}{2}$ $a$ ki jagah play karta hai; state average $\pm$ ek rotated swing mein split ho jaati hai.

Verify: trace $= 3.5 + 6.5 = 10 = 2 + 8$ ✔ (invariant). $det = 3.5 \cdot 6.5 - 2.59 8^{2} = 22.75 - 6.75 = 16 = 2 \cdot 8$ ✔. Product of inertia magnitude $\approx 2.6 kg\cdotm^{2}$ forecast ke anusaar nonzero hai; units kg·m² throughout ✔.

Example 8 — Cell H: exam twist (quotient theorem)

Worked example Prove karo

$A^{ij}$ ek tensor hai sirf yeh deke ki $A^{ij} V_{j}$ ek vector hai har covector $V_{j}$ ke liye Forecast: tumhe nahi bataya gaya ki $A$ kaise transform karta hai. Guess karo ki kya output ka behavior $A$ par tensor law force karta hai ya nahi.

Step 1 — Jo pata hai woh naam do. Har $V_{j}$ ke liye, quantity $U^{i} = A^{ij} V_{j}$ (Einstein convention se $j$ par sum hua) ek contravariant vector ki tarah transform hota hai: $\tilde{U}^{i} = \frac{\partial x ~ ^{i}}{\partial x ^{k}} U^{k}$ . Yeh step kyun? Summation convention ka matlab hai $A^{ij} V_{j}$ , $j$ par sum hota hai; hypothesis is contraction ke baare mein ek statement hai.

Step 2 — Dono sides nayi frame mein likho. $\tilde{U}^{i} = \tilde{A}^{ij} \tilde{V}_{j}$ aur saath hi $\tilde{U}^{i} = \frac{\partial x ~ ^{i}}{\partial x ^{k}} A^{k l} V_{l}$ . Covector rule $V_{l} = \frac{\partial x ~ ^{m}}{\partial x ^{l}} \tilde{V}_{m}$ use karke, substitute karo aur equate karo. Yeh step kyun? Hum same physical vector $\tilde{U}^{i}$ ke dono expressions ko agree karne par force karte hain.

Step 3 — Kyunki $\tilde{V}_{j}$ arbitrary hai, coefficients match karte hain. $\tilde{A}^{ij} = \frac{\partial x ~ ^{i}}{\partial x ^{k}} \frac{\partial x ~ ^{j}}{\partial x ^{l}} A^{k l} .$ Yeh step kyun? "Har $V$ ke liye sach" arbitrary factor ko khatam karta hai aur precisely contravariant rank-2 transformation law chhoddta hai — toh $A^{ij}$ hai ek tensor. QED.

Verify (numeric sanity): lo $A = (3113)$ aur covector $V = (1, 0)$ , toh $U = A V = (3, 1)$ . Ab $9 0^{\circ}$ se $R = (01 - 1 0)$ se rotate karo. Output directly transform karne par: $\tilde{U} = R U = (- 1, 3)$ . Pieces transform karne par: $\tilde{A} = R A R^{⊤} = (3 - 1 - 1 3)$ aur $\tilde{V} = R V = (0, 1)$ , toh $\tilde{A} \tilde{V} = (- 1, 3)$ . Dono routes same $\tilde{U} = (- 1, 3)$ dete hain ✔ — exactly wahi jo quotient theorem promise karta hai: contraction se pehle ya baad transform karna agree karta hai, toh $A$ ek tensor hona chahiye.

Recall Self-test

$tr T$ kisi bhi rotation mein kyun bachta hai? ::: Kyunki $R T R^{⊤}$ ek similarity transform hai aur trace similarity ke under invariant hai (Jacobians $δ$ mein contract ho jaate hain). Ek diagonal tensor ke $4 5^{\circ}$ rotation par induced shear magnitude kya hai? ::: $\frac{1}{2} ∣ q - p ∣$ , maximum shear — off-axis rotations off-diagonals create karte hain. Kya $r \to 0$ par polar mein $det g \to 0$ ek real singularity hai? ::: Nahi — yeh ek coordinate singularity hai; flat plane smooth hai, sirf polar chart degenerate hoti hai.