Tensor analysis — scalars, vectors, rank-2 tensors: Question bank

4.10.7 · D5 · Maths › Advanced Topics (Elite Level) › Tensor analysis — scalars, vectors, rank-2 tensors

Shuru karne se pehle ek quick vocabulary refresher (taaki koi bhi symbol yahan anjaan na lage):

Ek tensor = numbers ka ek box plus ek rule jo batata hai ki jab tum apne coordinate axes re-draw karo to wo numbers kaise change hote hain.
Upper index $V^{i}$ = contravariant (transforms like a displacement arrow $d x^{i}$ ).
Lower index $W_{i}$ = covariant (transforms like a gradient $\partial ϕ / \partial x^{i}$ ).
Metric tensor $g_{ij}$ = wo covariant rank-2 object jo squared distance measure karta hai, $d s^{2} = g_{ij} d x^{i} d x^{j}$ . Ye indices raise/lower bhi karta hai ( $V_{i} = g_{ij} V^{j}$ ). Flat Cartesian axes mein $g_{ij} = δ_{ij}$ (the identity) hota hai, lekin curved ya non-orthonormal axes mein (e.g. polar, $g_{ij} = diag (1, r^{2})$ ) ye identity nahi hota — dekho Metric tensor and Riemannian geometry.
Contraction = ek upper aur ek lower index ko pair karke sum karna (dekho Einstein summation convention) — isse rank 2 se kam ho jaati hai.
Rotation matrix $R$ = coordinate-change map jab axes sirf rotate karte hain; iske liye "undo" ( $R^{- 1}$ ) equals "flip across the diagonal" ( $R^{⊤}$ ).

Neeche ke do pictures is poore page ki index bookkeeping fix karte hain — traps se pehle inhe ek nazar zaroor dekho.

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

True or false — justify

TF1. "Mere grocery prices ka

3 \times 3

table per shop ek rank-2 tensor hai kyunki uske do indices hain."

False — do indices kaafi nahi hain; use coordinate change ke under tensor law se transform karna zaroori hai, aur grocery prices ka aisa koi rule nahi hota. Ye sirf ek table hai.

TF2. "Ek scalar koi bhi single number hota hai."

False — scalar wo number hota hai jo kisi bhi coordinate change ke under ek point par same value rakhta hai (

\tilde{ϕ} = ϕ

). "5 seb" ek number hai lekin "velocity ka 1st component" scalar nahi hai kyunki wo axes ke saath change karta hai.

TF3. "Cartesian coordinates mein

V^{i}

aur

V_{i}

identical numbers rakhte hain, isliye upper/lower bekaar hai."

Ye generally False hai — ye sirf isliye match karte hain kyunki wahan metric

g_{ij} = δ_{ij}

hai. Polar ya kisi bhi non-orthonormal frame mein

V_{i} = g_{ij} V^{j}

V^{i}

se alag hota hai, isliye ye distinction real hai.

TF4. "Stress tensor ke eigenvalues tab change ho jaate hain jab main apne axes rotate karta hoon."

False —

\tilde{T} = R T R^{⊤}

ek similarity transform hai, aur similarity eigenvalues, trace aur determinant preserve karti hai. Ye hi frame-independent physics hai (dekho Linear algebra — change of basis & similarity transforms).

TF5. "Trace

T^{i}_{i}

frame ke saath individual entries ki tarah depend karta hai."

False — trace ek contraction hai, isliye Jacobian aur uska inverse cancel hokar

δ

dete hain, jo ek genuine scalar chhod jaata hai; e.g.

T = (3113)

ka trace

3 + 3 = 6

hai, aur kisi bhi rotation ke baad do diagonal entries ka sum

6

hi rehta hai.

TF6. "Kronecker delta

δ_{j}^{i}

ek tensor hai jiske components kuch frames mein alag hote hain."

False — ye isotropic hai:

\frac{\partial x ~ ^{i}}{\partial x ^{k}} \frac{\partial x ^{k}}{\partial x ~ ^{j}} = δ_{j}^{i}

, isliye ye literally har frame mein

δ

hi hota hai.

TF7. "Ek covariant tensor

A_{ij}

ko ek contravariant tensor

B^{ij}

mein add karne se ek tensor milta hai."

False — sirf same-type tensors ko hi add kar sakte ho; dono opposite Jacobian factors lete hain, isliye unka sum kisi consistent transformation rule ke saath nahi aata.

TF8. "Ek tensor symmetric hai (

T_{ij} = T_{j i}

) ya nahi, ye coordinate change karne par flip ho sakta hai."

False — covariant

T_{ij}

type ke liye symmetry coordinate-independent hai: same Jacobian factor dono slots par baitha hai, isliye

i \leftrightarrow j

swap karna transformation ke saath commute karta hai.

TF9. "Gradient

\partial ϕ / \partial x^{i}

ek contravariant vector hai."

False — ye covariant hai (lower index): chain rule deta hai

\partial / \partial \tilde{x}^{i} = (\partial x^{k} / \partial \tilde{x}^{i}) \partial / \partial x^{k}

, jo inverse Jacobian hai, aur ye covariant behaviour define karta hai.

TF10. "Ek vector ka ordinary partial derivative

\partial_{k} V^{i}

ek rank-2 tensor hai."

False — differentiate karne par position-dependent Jacobian ek extra non-tensorial term de deta hai. Tensor character wapas laane ke liye tumhe covariant derivative

\nabla_{k} V^{i}

chahiye.

TF11. "Levi-Civita symbol

ε_{ij k}

ek ordinary tensor ki tarah transform karta hai."

False — ye ek pseudotensor hai: orientation-reversing change (ek reflection,

det J < 0

) ke under ye ek extra minus sign uthata hai jo true tensor rule produce nahi karta. Isliye cross products mirror reflection ke under flip karte hain.

Spot the error

SE1. Claim: "

\tilde{T} = R T R^{- 1}

for rotating a rank-2 tensor."

Proper rotations ke liye numerically koi error nahi — kyunki wahan

R^{- 1} = R^{⊤}

hai, dono same matrix dete hain. Baat interpretive hai:

R^{⊤}

likhna explicit karta hai ki ek index inverse Jacobian leta hai, aur ye wo form hai jo non-orthogonal changes mein generalize hoti hai jahan

R^{- 1} \neq = R^{⊤}

SE2. Claim: "

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

is the contravariant rule."

Error: Jacobian ulta hai. Contravariant forward Jacobian

\partial \tilde{x}^{i} / \partial x^{k}

use karta hai; jo likha gaya hai wo actually covariant/inverse factor hai.

SE3. Claim: "Contraction

A^{ij}_{j}

keeps the rank at 3 because we still see three index letters."

Error: repeated upper–lower pair ko sum karna us pair ko hata deta hai; rank 2 se kam hoti hai, isliye

A^{ij}_{j}

ek rank-1 object hai (ek free index

i

SE4. Claim: "To prove

A^{ij}

is a tensor I check that

A^{ij} V_{j}

is a vector for one specific

V_{j}

Error: Quotient Theorem chahta hai ki ye har vector

V_{j}

ke liye hold kare. Ek single choice lucky coincidence ho sakti hai aur kuch prove nahi karti.

SE5. Claim: "

d s^{2} = g_{ij} d x^{i} d x^{j}

shows

d s^{2}

is a rank-2 object because

g_{ij}

has rank 2."

Error: dono indices do

d x

's ke against contracted hain, isliye

d s^{2}

ek rank-0 scalar hai (invariant length). Yahi invariance force karti hai ki

g_{ij}

covariant ho.

SE6. Claim: "Since

g_{ij} = δ_{ij}

everywhere, arc length is always

d x^{2} + d y^{2}

Error:

g_{ij} = δ_{ij}

sirf flat Cartesian mein hold karta hai. Polar mein

g_{ij} = diag (1, r^{2})

hai, jo deta hai

d s^{2} = d r^{2} + r^{2} d θ^{2}

(dekho Metric tensor and Riemannian geometry).

SE7. Claim: "A tensor equation true in one frame might be false in another, so tensors are unreliable."

Error ulta hai: kyunki har term same tarike se transform karti hai, jo tensor equation ek frame mein true hai wo sab frames mein true hai — tensors use karne ka yahi poora point hai physics laws ke liye.

SE8. Claim: "

det (g_{ij})

is a scalar, so

det g

transforms trivially."

Error:

det g

coordinate change ke under ek factor

(det J)^{- 2}

uthata hai, isliye

det g

weight

1

ki tensor density hai, plain scalar nahi — wahi extra

det J

exactly

det g d^{n} x

ko ek invariant volume element banata hai.

Why questions

WHY1. Contravariant aur covariant components ek doosre ke inverse Jacobians kyun use karte hain?

Taaki contraction

W_{i} V^{i}

mein dono Jacobian factors cancel hokar

δ

dein, ek true scalar chhod ke — wahi invariance jo hum demand karte hain.

WHY2. Stress tensor ko sirf

R T

ki jagah

R T R^{⊤}

se kyun rotate karte hain?

Ek rank-2 tensor ke do index slots hote hain; har slot ko alag se rotate karna padta hai, ek

R

se aur ek

R^{⊤}

se (uska inverse Jacobian). Sirf ek factor sirf ek slot transform karta.

WHY3. Quotient Theorem hume

g_{ij}

ko bina hand-check kiye tensor declare karne ki permission kyun deta hai?

Kyunki

g_{ij} d x^{i} d x^{j}

arbitrary vector

d x^{i}

ke liye invariant hai; aisa sirf tab possible hai jab

g_{ij}

matching covariant Jacobians uthata ho — theorem hume transformation rule muft mein de deta hai.

WHY4. Kisi bhi rank-2 tensor ko symmetric + antisymmetric parts mein hamesha split kyun kar sakte hain, aur ye useful kyun hai?

Identity

T_{ij} = \frac{1}{2} (T_{ij} + T_{j i}) + \frac{1}{2} (T_{ij} - T_{j i})

pure algebra hai, aur har part alag se ek tensor hai jiska symmetry har coordinate change ke baad survive karta hai — isliye split ka physical meaning hai (e.g. strain vs rotation in Stress and strain tensors (continuum mechanics)).

WHY5. Hume

4 5^{\circ}

rotation ke baad stress ke off-diagonals ka vanish hona kyun matter karta hai?

Zero off-diagonals matlab un axes mein koi shear nahi — humne principal axes dhundh liye, aur diagonal entries coordinate-independent eigenvalues hain, jo real load hai jo material feel karta hai.

WHY6. Curved space mein ordinary derivative kyun kaafi nahi hoti, aur Christoffel symbols ko andar aana kyun padta hai?

Basis vectors khud har point se point mein turn/stretch karte hain;

\partial_{k}

us motion ko dekh nahi sakta, isliye hum correction terms (Christoffel symbols) add karte hain changing basis ko track karne ke liye — covariant derivative. Ye General Relativity — Einstein field equations ke peeche ki machinery hai.

WHY7. Tensor transformation laws mein chain rule har jagah kyun aata hai?

Har Jacobian factor

\partial \tilde{x} / \partial x

hi purane aur naye coordinates ke beech chain-rule link hai; tensors un objects (

d x^{i}

\partial_{i} ϕ

) se bane hain jinki transformation chain rule already dictate karta hai (dekho Jacobian and the multivariable chain rule).

Edge cases

EC1. Rank-0 case: ek coordinate change ke under ek scalar ko kitne Jacobian factors multiply karte hain?

Zero — bina kisi index ke koi Jacobian drag karne wala nahi hota, isliye exactly

\tilde{ϕ} = ϕ

(pure invariance) milta hai.

EC2. Zero tensor: kya all-zeros array ek valid tensor hai, aur kya ye har frame mein zero rehta hai?

Haan — zeros par koi bhi Jacobian factors apply karne se zeros hi milte hain, isliye zero tensor genuinely har frame mein zero hai (isliye tensor equations "= 0" frame-safe hoti hain).

EC3. Degenerate change of coordinates: agar Jacobian singular ho (determinant zero) to kya toot jaata hai?

Map invertible nahi hota, isliye

\partial x^{k} / \partial \tilde{x}^{i}

exist nahi karta aur covariant/mixed rules collapse ho jaate hain — us point par aisi coordinates mein legally change nahi kar sakte.

EC4. Ek tensor jo symmetric aur antisymmetric dono ho (

T_{ij} = T_{j i}

aur

T_{ij} = - T_{j i}

): wo kya hoga?

Ye zero tensor hona zaroori hai, kyunki dono conditions force karti hain

T_{ij} = - T_{ij}

, isliye

T_{ij} = 0

har jagah.

EC5. Identity-like rotation (

R = I

, i.e. koi rotation nahi):

R T R^{⊤}

kya deta hai aur ye ek accha sanity check kyun hai?

T

unchanged return karta hai, confirm karta hai ki "kuch na karo" coordinate change components ko waise hi chhod deta hai — kisi bhi transformation rule par ek mandatory consistency check.

EC6. Highest-symmetry frame: ek symmetric rank-2 tensor ke liye kya sab off-diagonals zero wala frame hamesha mil sakta hai?

Haan — ek real symmetric tensor ek orthogonal rotation se diagonalisable hota hai (uske principal axes), isliye aisa frame hamesha exist karta hai; diagonal entries invariant eigenvalues hain.

EC7. One-dimensional space (

n = 1

): ek rank-

r

tensor ke kitne components hote hain, aur ye kya reveal karta hai?

Exactly

n^{r} = 1^{r} = 1

— har rank ek single number mein collapse ho jaata hai, jo dikhata hai ki upper/lower distinction tab hi kaam aata hai jab

n \geq 2

ho.

EC8. Orientation flip (ek reflection,

det J < 0

): ek true tensor aur ek pseudotensor/density mein yahan kya fark hai?

Ek true tensor plain Jacobian rule follow karta hai; ek pseudotensor (jaise

ε_{ij k}

) ek extra

sign (det J)

uthata hai, aur ek density

det J

ki ek power uthata hai — dono wo tell-tale corner cases hain jahan "transforms like a tensor" subtly violate hota hai.

Recall Aage badhne se pehle self-test

Upar har answer cover karo aur reason re-derive karo, verdict nahi. A grid of numbers is a tensor only when… ::: it obeys the tensor transformation law under coordinate change. Contraction changes rank by… ::: minus 2 (one upper + one lower index removed). Symmetric-and-antisymmetric forces the tensor to be… ::: identically zero. The Levi-Civita symbol is not a true tensor because… ::: it flips sign under orientation-reversing transforms (a pseudotensor).

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