Exercises — Metric tensor — raising - lowering indices

Metric diagonal hai. WHAT hum karte hain: ek diagonal matrix ko invert karne ke liye har diagonal entry ka reciprocal lete hain. WHY yeh kaam karta hai: ek diagonal matrix ke liye, $g^{\mu \nu }{g}_{\nu \rho }={\delta }^{\mu }{}_{\rho }$ alag-alag equations mein toot jaata hai $g^{ii}{g}_{ii}=1$ (no sum), toh $g^{ii}=1/g_{ii}$. $g^{\mu \nu }=\left(\begin{array}{cc}1& 0\\ 0& \frac{1}{4}\end{array}\right).$ Ek index raise karne ke liye tum up-indexed inverse metric $g^{\mu \nu }$ use karte ho.

• (i) $A^{\mu }{B}_{\mu }$ — valid. Repeated index $\mu$ ek baar up, ek baar down hai. Yeh invariant dot product hai.
• (ii) $A^{\mu }{B}^{\mu }$ — ek tensor expression ke roop mein valid nahi. Index $\mu$ dono baar up appear karta hai; koi metric pairing nahi kar raha, toh yeh basis-dependent bakwaas hai.
• (iii) $g_{\mu \nu }{A}^{\mu }{B}^{\nu }$ — valid. Metric missing "down" provide karta hai taaki $\mu$ aur $\nu$ dono up–down pair ho jayein. Actually (iii) equals (i), kyunki $g_{\mu \nu }{B}^{\nu }=B_{\mu }$.

$V_{\mu }=g_{\mu \nu }{V}^{\nu }$. Identity metric ke saath sirf diagonal survive karta hai: $V_1=g_{11}{V}^1=(1)(3)=3,V_2=g_{22}{V}^2=(1)(-5)=-5.$ Toh $V_{\mu }=(3,-5)=V^{\mu }$. Reveal: jab $g=1$, lowering kuch bhi nahi badalta — yahi reason hai ki school dono mein kabhi distinguish nahi karta.

Down-indexed metric se lower karo: $V_r=g_{rr}{V}^r=(1)(1)=1,V_{\theta }=g_{\theta \theta }{V}^{\theta }=(r^2)(3)=(4)(3)=12.$ Toh $V_{\mu }=(1,12)$. WHY factor 4: ${\mathbf{e}}_{\theta }$ ki length $r=2$ hai, $1$ nahi; metric entry $g_{\theta \theta }=r^2=4$ us stretch ko dot product mein build karta hai.

Diagonal ⇒ har component independently lower hoti hai: $p_0=g_{00}{p}^0=(-1)(5)=-5,p_1=(1)(3)=3,p_2=0,p_3=(1)(4)=4.$ Toh $p_{\mu }=(-5,3,0,4)$. Sirf time component ka sign flip hota hai — yeh indefinite (Lorentzian) metric ki fingerprint hai.

Cross terms alive hain, toh har row mein dono metric entries contribute karti hain: $V_1=g_{11}{V}^1+g_{12}{V}^2=(2)(1)+(1)(4)=6,$ $V_2=g_{21}{V}^1+g_{22}{V}^2=(1)(1)+(2)(4)=9.$ Toh $V_{\mu }=(6,9)$. WHY dono terms: $g_{12}\ne 0$ ka matlab hai ${\mathbf{e}}_1$ aur ${\mathbf{e}}_2$ perpendicular nahi hain, toh $\mathbf{V}$ ko ${\mathbf{e}}_1$ pe project karne par $V^2$-direction ka ek piece bhi aata hai.

2×2 ka Inverse $\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ hai $\frac{1}{ad-bc}\left(\begin{array}{cc}d& -b\\ -c& a\end{array}\right)$. Determinant $=2\cdot 2-1\cdot 1=3$: $g^{\mu \nu }=\frac{1}{3}\left(\begin{array}{cc}2& -1\\ -1& 2\end{array}\right).$ Ab $V_{\nu }=(6,9)$ ke saath $V^{\mu }=g^{\mu \nu }{V}_{\nu }$ se raise karo: $V^1=\frac{1}{3}(2\cdot 6+(-1)\cdot 9)=\frac{1}{3}(12-9)=1,$ $V^2=\frac{1}{3}((-1)\cdot 6+2\cdot 9)=\frac{1}{3}(-6+18)=4.$ Toh $V^{\mu }=(1,4)$ ✓ — lowering phir raising identity hai, jaise $g^{\mu \alpha }{g}_{\alpha \nu }={\delta }^{\mu }{}_{\nu }$ guarantee karta hai.

(i) $g_{\mu \nu }{V}^{\mu }{V}^{\nu }=2(1{)}^2+2g_{12}{V}^1{V}^2+2(4{)}^2$. Carefully: $=g_{11}(1)(1)+g_{12}(1)(4)+g_{21}(4)(1)+g_{22}(4)(4)=2+4+4+32=42.$ (ii) $V_{\mu }=(6,9)$ use karte hue: $V_{\mu }{V}^{\mu }=V_1{V}^1+V_2{V}^2=(6)(1)+(9)(4)=6+36=42$ ✓. Norm $42$ same object hai chahe kisi bhi tarah se dekho — wahi invariance puri wajah hai ki metric exist karta hai.

Pehla slot lower karne ka matlab hai $T$ ka pehla index $g$ ke saath contract hoga. Kyunki $g$ diagonal hai, $g_{\mu \alpha }{T}^{\alpha \nu }=g_{\mu \mu }{T}^{\mu \nu }$ (no sum), yaani row $\mu$ ko $g_{\mu \mu }$ se multiply karo:

• Row 1 ($\mu =1$): $\times g_{11}=1$ → $(0,2)$.
• Row 2 ($\mu =2$): $\times g_{22}=4$ → $(12,4)$. $T_{\mu }{}^{\nu }=\left(\begin{array}{cc}0& 2\\ 12& 4\end{array}\right).$

Doosra slot lower karne se column $\nu$ ko $g_{\nu \nu }$ se multiply hota hai:

• Column 1 ($\nu =1$): $\times g_{11}=1$ → $\left(\begin{array}{c}0\\ 12\end{array}\right)$.
• Column 2 ($\nu =2$): $\times g_{22}=4$ → $\left(\begin{array}{c}8\\ 16\end{array}\right)$. $T_{\mu \nu }=\left(\begin{array}{cc}0& 8\\ 12& 16\end{array}\right).$ WHY ek index at a time: har $g$ exactly ek free slot ko ek dummy ke saath pair karta hai; sequentially karne se guarantee hoti hai ki tum sirf wahi slot convert karo jo tum chahte ho.

Pehle $T^{\mu }{}_{\nu }=g_{\nu \beta }{T}^{\mu \beta }$ form karo ($T^{\mu \nu }$ ka doosra index lower karo): column $\nu$ ko $g_{\nu \nu }$ se multiply karo:

• Column 1 $\times 1$: $(0,3{)}^{\top }$. Column 2 $\times 4$: $(8,4{)}^{\top }$. $T^{\mu }{}_{\nu }=\left(\begin{array}{cc}0& 8\\ 3& 4\end{array}\right).$ Ab $\mu =\nu$ contract karo (diagonal sum karo): $T^{\mu }{}_{\mu }=0+4=4$. Yeh invariant trace hai.

Left se shuru karo aur regroup karo (contraction ki associativity): $g^{\mu \beta }\left(g_{\beta \nu }{V}^{\nu }\right)=\left(g^{\mu \beta }{g}_{\beta \nu }\right)V^{\nu }.$ WHAT/WHY: humne dono metrics ko bracket kiya kyunki unka product inverse metric ki defining relation hai: $g^{\mu \beta }{g}_{\beta \nu }={\delta }^{\mu }{}_{\nu }.$ Kronecker delta ${\delta }^{\mu }{}_{\nu }$ tab $1$ hai jab $\mu =\nu$, warna $0$; ise $V^{\nu }$ ke saath contract karne par index simply rename ho jaata hai: ${\delta }^{\mu }{}_{\nu }{V}^{\nu }=V^{\mu }.■$ Koi component values use nahi hue — yeh har metric ke liye hold karta hai, diagonal ho ya nahi, curved ho ya flat.

Lower karo: $p_{\mu }=g_{\mu \nu }{p}^{\nu }=(-E,E,0,0)$ (time component flip hoti hai). $p_{\mu }{p}^{\mu }=(-E)(E)+(E)(E)+0+0=-E^2+E^2=0.$ Meaning: invariant norm $0$ hai — momentum null (lightlike) hai. Yeh encode karta hai $m^2=-p_{\mu }{p}^{\mu }=0$: photon massless hai. Metric ka minus sign exactly wahi hai jo ek nonzero-energy object ko zero invariant length deta hai.

$A$ lower karo: $A_{\mu }=(-3,1,0,0)$. Phir $A_{\mu }{B}^{\mu }=(-3)(2)+(1)(2)+0+0=-6+2=-4.$ Metric se directly cross-check karo: $g_{\mu \nu }{A}^{\mu }{B}^{\nu }=-A^0{B}^0+A^1{B}^1=-(3)(2)+(1)(2)=-6+2=-4.✓$ Dono routes $-4$ dete hain; negative sign kehta hai vectors ka "time overlap" unke "space overlap" pe dominate karta hai.