Worked examples — Metric tensor — raising - lowering indices

4.10.10 · D3 · Maths › Advanced Topics (Elite Level) › Metric tensor — raising - lowering indices

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Scenario matrix

Har metric jiske saath tum kabhi bhi lower/raise karoge woh in case classes mein se ek mein aata hai. Raising/lowering ka saara kaam sirf $g_{\mu \nu }$ ki shape ke saath badalta hai, isliye hum shape ke hisaab se organize karte hain.

Cell	Case class	$g$ mein kya khas hai	Cover karta hai
A	Identity / Cartesian orthonormal	$g_{\mu \nu }={\delta }_{\mu \nu }$	Ex 1
B	Diagonal, sab $+$, unequal	pure rescaling, koi sign flip nahi	Ex 2
C	Diagonal mein ek $-$ (indefinite)	ek component sign flip karta hai	Ex 3
D	Coordinate-dependent diagonal (polar)	entries position $r$ par depend karti hain	Ex 4
E	Off-diagonal (slots ko couple karta hai)	cross terms $g_{12}\ne 0$	Ex 5
F	Degenerate / zero row (breakdown)	$\det g=0$, koi inverse nahi	Ex 6
G	Rank-2 tensor, ek slot ek baar	mixed index $T^{\mu }{}_{\nu }$	Ex 7
H	Real-world word problem	kisi physical setup se $g$ padhna	Ex 8
I	Exam twist: raise back / consistency	round-trip $g^{-1}g=1$	Ex 9

Hum har cell hit karenge. Har example batata hai ki woh kaun sa cell fill karta hai.

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Ex 1 — Cell A: woh case jahan kuch hota hi nahi

1. Rule likho. $V_{\mu }=g_{\mu \nu }{V}^{\nu }$. Ye step kyun? Ye definition hai; hum hamesha yahi se shuru karte hain — naam lete hain kaun si machine chalani hai.
2. $\mu =1$ karo: $V_1=g_{11}{V}^1+g_{12}{V}^2=(1)(3)+(0)(4)=3.$ Ye step kyun? Repeated index $\nu$ par sum karo; is row mein sirf $g_{11}$ non-zero hai.
3. $\mu =2$ karo: $V_2=g_{21}{V}^1+g_{22}{V}^2=(0)(3)+(1)(4)=4.$ Ye step kyun? Same, doosri row.
4. Result: $V_{\mu }=(3,4)=V^{\mu }$.

Verify: norm dono taraf se: $g_{\mu \nu }{V}^{\mu }{V}^{\nu }=3^2+4^2=25$, aur $V_{\mu }{V}^{\mu }=3\cdot 3+4\cdot 4=25$. Consistent, aur ordinary $\Vert V{\Vert }^2$ ke barabar. ✓ Yahi woh Mistake hai jiske baare mein parent warn karta hai: upper aur lower sirf yahan identical dikhte hain.

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Ex 2 — Cell B: pure rescaling (diagonal, sab positive, unequal)

1. Lower karo. $V_1=g_{11}{V}^1=4\cdot 5=20$; $V_2=g_{22}{V}^2=9\cdot 2=18.$ Ye step kyun? Diagonal metric ⇒ har lower component bas apna upper component hai jo diagonal entry se scale hua hai.
2. Inverse banao. Diagonal matrix ke liye, $g^{\mu \nu }=\mathrm{diag}(1/4,1/9)$. Ye step kyun? Raising ke liye $g^{\mu \nu }$ chahiye, aur diagonal $g$ ke liye inverse entry-by-entry reciprocals hote hain.
3. Wapas raise karo. $V^1=g^{11}{V}_1=\frac{1}{4}\cdot 20=5$; $V^2=g^{22}{V}_2=\frac{1}{9}\cdot 18=2.$ Ye step kyun? Round-trip original wapas lana chahiye — ek built-in sanity check.

Verify: raising ne lowering undo kar di: $(5,2)\to (20,18)\to (5,2)$. ✓ Aur $g^{11}{g}_{11}=\frac{1}{4}\cdot 4=1={\delta }^1{}_1$. ✓ Koi sign flip nahi kyunki saari diagonal entries positive hain (woh sirf Cell C mein hota hai).

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Ex 3 — Cell C: sign flip (indefinite metric)

1. Har slot lower karo. $p_0=g_{00}{p}^0=(-1)(5)=-5$; $p_1=(1)(3)=3$; $p_2=p_3=0.$ Ye step kyun? Minkowski spacetime metric diagonal hai, isliye har slot ek entry use karta hai — lekin time entry $-1$ hai, toh time component sign flip karta hai. Yahi flip ek indefinite metric ka poora physical content hai.
2. Contract karo. $p_{\mu }{p}^{\mu }=p_0{p}^0+p_1{p}^1=(-5)(5)+(3)(3)=-25+9=-16.$ Ye step kyun? Invariant length ek valid contraction hona chahiye: ek upar, ek neeche. Ye number coordinate-independent hai.

Verify: metric seedha use karke, $g_{\mu \nu }{p}^{\mu }{p}^{\nu }=-E^2+p_x^2=-25+9=-16$. ✓ Particle physics mein ye $-m^2$ ke barabar hai (jahan $c=1$), toh $m^2=16$, $m=4$. Sign negative hai kyunki vector timelike hai — bilkul sahi regime.

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Ex 4 — Cell D: coordinate-dependent metric (polar)

1. Inverse metric. $g^{\mu \nu }=\mathrm{diag}(1,1/r^2)$. $r=2$ par: $g^{\theta \theta }=1/4$. Ye step kyun? Raising ke liye $g^{\mu \nu }$ chahiye; yahan ye jagah par depend karta hai — Cells A–C ke mukable mein ye naya hai.
2. Radial raise karo. $V^r=g^{rr}{V}_r=1\cdot 6=6.$ Ye step kyun? $g^{rr}=1$, toh radial direction plain Cartesian jaisa behave karta hai.
3. Angular raise karo. $V^{\theta }=g^{\theta \theta }{V}_{\theta }=\frac{1}{4}\cdot 8=2.$ $1/r^2$ kyun? Figure mein red arrow dekho: basis vector ${\mathbf{e}}_{\theta }$ ki physical length $r=2$ hai, $1$ nahi. Ek covariant component pehle se us length ka factor le kar chalta hai, toh raising mein use wapas divide karna padta hai — isliye $1/r^2$.

Verify: norm representation se independent hona chahiye. $V_{\mu }{V}^{\mu }=V_r{V}^r+V_{\theta }{V}^{\theta }=6\cdot 6+8\cdot 2=36+16=52$. Cross-check $g^{\mu \nu }{V}_{\mu }{V}_{\nu }=1\cdot 6^2+\frac{1}{4}\cdot 8^2=36+16=52$ se. ✓

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Ex 5 — Cell E: off-diagonal metric (coupled slots)

1. Lower karo, DONO terms rakhte hue. $V_1=g_{11}{V}^1+g_{12}{V}^2=2\cdot 3+1\cdot 1=7.$ $V_2=g_{21}{V}^1+g_{22}{V}^2=1\cdot 3+2\cdot 1=5.$ Ye step kyun? Off-diagonal $g_{12}\ne 0$ components ko couple karta hai — cross term drop karna classic error hai. Har lower component saare upper components ko mix karta hai.
2. Metric invert karo. $\left(\begin{array}{cc}a& b\\ b& d\end{array}\right)$ ke liye, inverse $=\frac{1}{ad-b^2}\left(\begin{array}{cc}d& -b\\ -b& a\end{array}\right)$. Yahan $\det =2\cdot 2-1=3$, toh $g^{\mu \nu }=\frac{1}{3}\left(\begin{array}{cc}2& -1\\ -1& 2\end{array}\right)$. Ye step kyun? Raising ke liye true matrix inverse chahiye, entries ke reciprocals nahi (woh shortcut sirf tab kaam karta hai jab off-diagonals zero hon).
3. Wapas raise karo. $V^1=\frac{1}{3}(2\cdot 7-1\cdot 5)=\frac{1}{3}(9)=3$; $V^2=\frac{1}{3}(-1\cdot 7+2\cdot 5)=\frac{1}{3}(3)=1.$ Ye step kyun? Round-trip $(3,1)$ recover karna chahiye.

Verify: $(3,1)\to (7,5)\to (3,1)$. ✓ Aur $g^{\mu \alpha }{g}_{\alpha \nu }$: row-1 = $\frac{1}{3}(2\cdot 2-1\cdot 1,2\cdot 1-1\cdot 2)=\frac{1}{3}(3,0)=(1,0)$. ✓ Norm: $V_{\mu }{V}^{\mu }=7\cdot 3+5\cdot 1=26$; aur $g_{\mu \nu }{V}^{\mu }{V}^{\nu }=2\cdot 9+2\cdot 3\cdot 1+2\cdot 1=18+6+2=26$. ✓

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Ex 6 — Cell F: degenerate metric (breakdown case)

1. Lowering abhi bhi kaam karta hai. $V_1=1\cdot 3+2\cdot 1=5$; $V_2=2\cdot 3+4\cdot 1=10.$ Ye step kyun? Lowering sirf $g_{\mu \nu }$ se multiply karta hai; koi inverse nahi chahiye, toh ye hamesha defined hai.
2. Raise karne ki koshish karo. $\det g=1\cdot 4-2\cdot 2=0$ compute karo. Ye step kyun? Raising ke liye $g^{\mu \nu }=g^{-1}$ chahiye, aur zero determinant wali matrix ka koi inverse nahi hota. Raising operation undefined hai.
3. Interpret karo. Row 2, row 1 ka $2\times$ hai: doosri basis "direction" mein koi independent length information nahi hai. Inner product space par ek genuine metric non-degenerate ($\det g\ne 0$) hona chahiye — precisely taaki translation dono directions mein ho sake.

Verify: $\det \left(\begin{array}{cc}1& 2\\ 2& 4\end{array}\right)=0$, toh koi inverse nahi ⇒ raising impossible. Ye woh boundary/degenerate cell hai jise reader ko pehchanna chahiye: raise karne se pehle $\det g\ne 0$ check karo. ✓ (Notice karo ki $V_{\mu }=(5,10)$ khud $1:2$ mein proportional hai, degeneracy ko echo karta hua.)

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Ex 7 — Cell G: ek rank-2 tensor, ek baar mein ek slot

1. Ek slot ke liye rule. $T^{\mu }{}_{\nu }=g_{\nu \alpha }{T}^{\mu \alpha }$ — metric sirf doosre index ke saath pair karta hai. Ye step kyun? Ek baar mein ek index raise/lower karo; har $g$ exactly ek slot convert karta hai (parent §4).
2. Column $\nu =0$ ($g_{00}=-1$ use karta hai): $T^0{}_0=g_{00}{T}^{00}=-2$; $T^1{}_0=g_{00}{T}^{10}=-4.$ Ye step kyun? $0$-slot ko lower karna us poore column ko $-1$ se multiply karta hai.
3. Column $\nu =1$ ($g_{11}=1$ use karta hai): $T^0{}_1=T^{01}=3$; $T^1{}_1=T^{11}=5.$ Ye step kyun? $g_{11}=1$ $1$-column ko unchanged chhod deta hai.
4. Result: $T^{\mu }{}_{\nu }=\left(\begin{array}{cc}-2& 3\\ -4& 5\end{array}\right).$

Verify: trace $T^{\mu }{}_{\mu }=T^0{}_0+T^1{}_1=-2+5=3$ — ye contracted scalar invariant hai. Direct check: $g_{\mu \nu }{T}^{\mu \nu }=g_{00}{T}^{00}+g_{11}{T}^{11}=-2+5=3$. ✓

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Ex 8 — Cell H: real-world word problem

1. Metric padho. Perpendicular axes scales $2$ aur $3$ ke saath: $g_{\mu \nu }=\mathrm{diag}(2^2,3^2)=\mathrm{diag}(4,9)$. Ye step kyun? $g_{\mu \nu }={\mathbf{e}}_{\mu }\cdot {\mathbf{e}}_{\nu }$; har basis vector ki squared physical length diagonal par baith jaati hai.
2. True squared length. $\Vert V{\Vert }^2=g_{\mu \nu }{V}^{\mu }{V}^{\nu }=4\cdot 3^2+9\cdot 2^2=36+36=72.$ Ye step kyun? Invariant length metric use karta hai, na ki naive $\sum (V^{\mu }{)}^2$ — woh galat $9+4=13$ deta.
3. Length. $\Vert V\Vert =\sqrt{72}=6\sqrt{2}\approx 8.485$ metres.
4. Covariant components. $V_1=g_{11}{V}^1=4\cdot 3=12$; $V_2=g_{22}{V}^2=9\cdot 2=18.$

Verify: covariant form se length cross-check karo: $V_{\mu }{V}^{\mu }=12\cdot 3+18\cdot 2=36+36=72$. ✓ Naive (galat) answer $13$ hota — metric matter karta hai, exactly wahi Mistake point.

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Ex 9 — Cell I: exam twist, consistency round-trip

1. Key identity state karo. $g^{\mu \beta }{g}_{\beta \nu }={\delta }^{\mu }{}_{\nu }$ (inverse-metric ki defining relation). Ye step kyun? Yahi WOH ek fact hai jo raising ko lowering undo karne deta hai — metric aur uska inverse true matrix inverses hain.
2. Apply karo. $g^{\mu \beta }(g_{\beta \nu }{V}^{\nu })=(g^{\mu \beta }{g}_{\beta \nu })V^{\nu }={\delta }^{\mu }{}_{\nu }{V}^{\nu }=V^{\mu }.$ Ye step kyun? Matrix multiplication ki associativity humein pair ko $\delta$ mein collapse karne deti hai, aur Kronecker delta simply relabel karta hai.
3. Numeric spot-check $(a,b)=(3,1)$ ke saath: Ex 5 se humne pehle hi dekha $(3,1)\to (7,5)\to (3,1)$. ✓

Verify: Is specific $g$ ke liye, $g^{\mu \beta }{g}_{\beta \nu }=\frac{1}{3}\left(\begin{array}{cc}2& -1\\ -1& 2\end{array}\right)\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)=\delta$. ✓

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Recall Answers

1. Cell A ($g=\delta$, Cartesian orthonormal) — metric identity hai, toh machine kuch nahi badlti.
2. $p_{\mu }=(-5,3,0,0)$, $p_{\mu }{p}^{\mu }=-16$.
3. Off-diagonal $g_{12}\ne 0$ slots couple karta hai: $V_1=g_{11}{V}^1+g_{12}{V}^2$ dono mix karta hai.
4. $\det g\ne 0$ (non-degenerate) — warna koi inverse $g^{\mu \nu }$ exist nahi karta (Cell F).

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Connections

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

Case Map