Metric tensor — raising - lowering indices: Question bank

4.10.10 · D5 · Maths › Advanced Topics (Elite Level) › Metric tensor — raising - lowering indices

Shuru karne se pehle, ek shared vocabulary reminder taaki koi bhi symbol unearned na lage:

Spot the error

Har line mein ek mistake hai. Use naam do, phir fix reveal karo.

∥ V ∥^{2} = \sum_{μ} (V^{μ})^{2}

har coordinate system mein."

Galat. Invariant norm

∥ V ∥^{2} = g_{μν} V^{μ} V^{ν}

hai; naive sum-of-squares sirf tab kaam karta hai jab

g_{μν} = δ_{μν}

ho.

A^{μ} B^{μ}

A

aur

B

ka dot product hai."

Illegal contraction — dono indices up hain. Invariant

A^{μ} B_{μ} = g_{μν} A^{μ} B^{ν}

hai, ek up ko ek down ke saath sum karo.

"Index raise karne ke liye,

g_{μν}

se contract karo."

Tumhe inverse

g^{μν}

se contract karna chahiye. Down-indexed

g_{μν}

lower karta hai; ise raise karne ke liye use karna dimensionally aur algebraically galat hai jab tak

g

apna khud ka inverse na ho.

V_{θ} = g_{θ θ} V^{θ} = r^{2} V^{θ}

, isliye

V^{θ} = r^{2} V_{θ}

Raise step inverted hai. Kyunki

g^{θ θ} = 1/ r^{2}

, hum paate hain

V^{θ} = g^{θ θ} V_{θ} = V_{θ} / r^{2}

, na ki

r^{2} V_{θ}

T^{μν} = g_{μν} T^{μν}

ek saath dono indices lower kar deta hai."

Ek index sirf do baar appear ho sakta hai (ek up, ek down) agar wo summed ho; yahan

μ, ν

dono free aur dummy ke roop mein appear ho rahe hain — ek notation clash hai. Ek fresh dummy ke saath ek ek karke index lower karo:

T^{μ}_{ν} = g_{ν α} T^{μα}

"Kyunki

g^{μα} g_{α ν} = δ^{μ}_{ν}

, metric apne aap ka inverse hai."

Woh relation sirf yeh kehta hai ki yeh dono matrix inverses hain;

g^{μν} = g_{μν}

sirf tab hota hai jab

g^{2} = 1

, jaise Minkowski

diag (- 1, 1, 1, 1)

, lekin polar coordinates ke liye nahi.

g_{μν} = diag (- 1, 1, 1, 1)

mein,

U^{μ} = (2, 1, 0, 0)

ka norm

2^{2} + 1^{2} = 5

hai."

Time part mein minus sign hai:

U_{μ} U^{μ} = (- 1) (2)^{2} + (1) (1)^{2} = - 4 + 1 = - 3

. Signature bhulna classic galti hai.

"Metric hi vectors define karta hai."

Nahi — vectors metric ke bina bhi exist karte hain. Metric geometry add karta hai (lengths, angles, aur up/down translation). Vectors aur unke dual covectors pehle define hote hain; metric sirf unhe connect karta hai.

Why questions

Ek ya do sentences mein jawab do, phir reveal karo.

Humein do tarah ke components ki zaroorat kyun hai?

Kyunki ek general basis orthonormal nahi hoti: ek vector ko basis steps ginke measure karna (

V^{μ}

) aur basis directions par project karke (

V_{μ}

) alag numbers deta hai, aur dono useful hain.

Metric ko specifically basis vectors ke dot products ke roop mein kyun define kiya jata hai?

Kyunki squared length

g_{μν} V^{μ} V^{ν}

V \cdot V = (V^{μ} e_{μ}) \cdot (V^{ν} e_{ν})

expand karne se nikalta hai; dot products

e_{μ} \cdot e_{ν}

exactly wahi hain jo components se lengths paane ke liye chahiye.

Inverse metric raise kyun karta hai, metric khud kyun nahi?

Lowering

g_{μν}

se multiplication hai; raising ise undo karna chahiye, aur ek matrix multiply ko undo karne ke liye matrix inverse

g^{μν}

chahiye.

Summed (dummy) index hamesha ek up aur ek down kyun appear karta hai?

Ek valid contraction ek vector slot ko ek dual slot se pair karta hai (Dual space and covectors); yahi woh ek combination hai jo coordinate-independent scalar produce karta hai, Einstein summation convention se guarantee hota hai.

Polar coordinates mein raise karne par

θ

-component ko

1/ r^{2}

factor kyun milta hai?

Basis vector

e_{θ}

ki length

r

hai, isliye

g_{θ θ} = r^{2}

; iska inverse

g^{θ θ} = 1/ r^{2}

us stretched basis vector ko compensate karta hai jab

V_{θ} \to V^{θ}

convert hota hai.

High-school physics classes covariant vs contravariant kyun kabhi mention nahi karti?

Woh exclusively orthonormal Cartesian coordinates mein kaam karte hain jahan

g_{μν} = δ_{μν}

hota hai, isliye

V^{μ} = V_{μ}

aur distinction invisible hai.

Invariant dot product coordinate-independent kyun hai jabki

V^{μ}

aur

g_{μν}

dono coordinates ke saath change karte hain?

Changes cancel ho jaate hain:

g_{μν}

pair

V^{μ} V^{ν}

ke opposite transform karta hai, scalar

g_{μν} A^{μ} B^{ν}

ko fixed rakhta hai — yahi ek genuine scalar ki defining property hai.

Christoffel symbols mein

g_{μν}

ki values ki jagah derivatives kyun hoti hain?

Raising/lowering metric ki values use karta hai, lekin basis (aur hence components) ek point se doosre point tak kaise change karta hai, yeh describe karne ke liye metric kaise change hota hai — uski derivatives — chahiye.

Edge cases

Degenerate / limiting situations handle karo. Reasoning ke baad reveal karo.

Raising/lowering ka kya hoga jab

g_{μν} = δ_{μν}

(flat orthonormal) ho?

Operations identity ban jaate hain:

V_{μ} = V^{μ}

. Yeh wahi ek regime hai jahan distinction gayab ho jati hai, isliye yeh beginner ka default hai.

Agar metric singular hai (determinant zero) — kya tab bhi indices raise kar sakte ho?

Nahi. Zero determinant ka matlab hai

g^{μν}

exist nahi karta, isliye lowering ko undo karne ka koi tarika nahi; geometry degenerate hai aur lengths/angles wahan well defined nahi hain.

Minkowski space mein ek null (lightlike) vector ka norm kya hai?

Exactly zero:

g_{μν} V^{μ} V^{ν} = 0

jabki

V^{μ} \neq = 0

. Ek indefinite metric non-zero vectors ko zero length dene ki ijazat deta hai — Euclidean space mein yeh impossible hai.

Zero vector

V^{μ} = (0, 0, \dots)

lo. Uske lowered components kya hain?

Sab zero:

V_{μ} = g_{μν} \cdot 0 = 0

chahe metric kuch bhi ho. Zero vector har dialect mein same hota hai.

2D polar coordinates mein,

r \to 0

par

g^{θ θ} = 1/ r^{2}

ka kya hota hai?

Yeh blow up ho jata hai (

\to \infty

), origin par coordinate singularity signal karta hai —

θ

direction ek point par ill-defined hai, yeh geometry ki failure nahi balki coordinates ki hai.

Agar ek vector purely ek orthonormal axis ke saath point karta hai, toh kya lowering use change karta hai?

Sirf us axis ke diagonal entry se: jaise Minkowski mein purely-time vector

U^{μ} = (2, 0, 0, 0)

lower hokar

U_{μ} = (- 2, 0, 0, 0)

ban jata hai — ek sign flip, baaki untouched.

g_{μν} V^{μ} V^{ν} = 0

jabki

V \neq = 0

hona space ke baare mein kya batata hai?

Metric indefinite hai (jaise Minkowski spacetime); ek positive-definite inner product space mein norm sirf zero vector ke liye zero hota hai.

Recall Jaane se pehle quick self-audit

Agar tum yeh nahi bata sakte ki ek contraction mein ek up aur ek down index kyun chahiye, aur raising ke liye inverse metric kyun chahiye, toh parent note ki derivations dobara padho — upar ke har trap mein in dono rules mein se ek hi disguise mein hai.

Connections

Metric tensor — raising - lowering indices (parent)
Tensors — definition and transformation laws
Dual space and covectors
Line element and ds^2
Minkowski spacetime
Christoffel symbols
Inner product spaces
Einstein summation convention

Question bank — Metric tensor — raising - lowering indices

True or false — justify