Exercises — Christoffel symbols — intro

Minus ${\partial }_m{g}_{ij}$ par baithta hai — ==dummy index $m$== ke respect mein derivative, wahi $m$ jo inverse metric $g^{km}$ se tied hai.

Kyun matter karta hai: do plus terms $g$ ko free lower indices $i$ aur $j$ ke respect mein differentiate karte hain; minus term summed index ke respect mein differentiate karta hai. Inhe mix up karne se signs flip ho jaate hain aur galat "fictitious forces" produce hoti hain. Mnemonic: "half-inverse, two plus, one minus."

Upper index $k$ $2$ values leta hai. Lower pair $(i,j)$ symmetric hai, isliye $2\times 2=4$ ordered pairs ki jagah hum unordered pairs count karte hain: $(1,1),(1,2),(2,2)$ = $3$ choices.

Total independent components $=2\times 3=\boxed{6}$.

($n$ dimensions mein count hai $n\cdot \frac{n(n+1)}{2}$.)

True. Master formula mein har term ek derivative $\partial g$ hai. Agar har $g_{ij}$ constant hai toh ${\partial }_i{g}_{mj}={\partial }_j{g}_{mi}={\partial }_m{g}_{ij}=0$, isliye poora bracket $0$ hai aur ${\Gamma }_{ij}^k=0$ sabhi indices ke liye.

Yahi exactly Cartesian case hai $g_{ij}={\delta }_{ij}$ jo parent note ke Worked Example 2 mein hai.

Sirf $g_{\theta \theta }=r^2$ ek coordinate par depend karta hai; uska only nonzero derivative hai ${\partial }_r{g}_{\theta \theta }=2r$. Baaki sab ke derivatives zero hain.

${\Gamma }_{rr}^r$: $\frac{1}{2}{g}^{rr}({\partial }_r{g}_{rr}+{\partial }_r{g}_{rr}-{\partial }_r{g}_{rr})=\frac{1}{2}(1)(0)=0$.

${\Gamma }_{r\theta }^r$: $\frac{1}{2}{g}^{rr}({\partial }_r{g}_{r\theta }+{\partial }_{\theta }{g}_{rr}-{\partial }_r{g}_{r\theta })=0$.

${\Gamma }_{\theta \theta }^r$: $\frac{1}{2}{g}^{rr}({\partial }_{\theta }{g}_{r\theta }+{\partial }_{\theta }{g}_{r\theta }-{\partial }_r{g}_{\theta \theta })=\frac{1}{2}(1)(0+0-2r)=\boxed{-r}$.

${\Gamma }_{rr}^{\theta }$: $\frac{1}{2}{g}^{\theta \theta }({\partial }_r{g}_{\theta r}+{\partial }_r{g}_{\theta r}-{\partial }_{\theta }{g}_{rr})=0$.

${\Gamma }_{r\theta }^{\theta }$: $\frac{1}{2}{g}^{\theta \theta }({\partial }_r{g}_{\theta \theta }+{\partial }_{\theta }{g}_{\theta r}-{\partial }_{\theta }{g}_{r\theta })=\frac{1}{2}\cdot \frac{1}{r^2}\cdot 2r=\boxed{\frac{1}{r}}$.

${\Gamma }_{\theta \theta }^{\theta }$: $\frac{1}{2}{g}^{\theta \theta }({\partial }_{\theta }{g}_{\theta \theta }+{\partial }_{\theta }{g}_{\theta \theta }-{\partial }_{\theta }{g}_{\theta \theta })=\frac{1}{2}{g}^{\theta \theta }(0)=0$.

Nonzero: ${\Gamma }_{\theta \theta }^r=-r$ aur ${\Gamma }_{r\theta }^{\theta }={\Gamma }_{\theta r}^{\theta }=\frac{1}{r}$. Baaki charon zero hain.

Inverse metric: $g^{\theta \theta }=1/a^2$, $g^{\varphi \varphi }=1/(a^2{\sin }^2\theta )$. Sirf coordinate-dependent component hai $g_{\varphi \varphi }=a^2{\sin }^2\theta$, jiska ${\partial }_{\theta }{g}_{\varphi \varphi }=2a^2\sin \theta \cos \theta$ hai.

${\Gamma }_{\varphi \varphi }^{\theta }$: $k=\theta ,i=j=\varphi$. $\frac{1}{2}{g}^{\theta \theta }({\partial }_{\varphi }{g}_{\theta \varphi }+{\partial }_{\varphi }{g}_{\theta \varphi }-{\partial }_{\theta }{g}_{\varphi \varphi })=\frac{1}{2}\cdot \frac{1}{a^2}(0+0-2a^2\sin \theta \cos \theta )=\boxed{-\sin \theta \cos \theta }.$

${\Gamma }_{\theta \varphi }^{\varphi }$: $k=\varphi ,i=\theta ,j=\varphi$. $\frac{1}{2}{g}^{\varphi \varphi }({\partial }_{\theta }{g}_{\varphi \varphi }+{\partial }_{\varphi }{g}_{\varphi \theta }-{\partial }_{\varphi }{g}_{\theta \varphi })=\frac{1}{2}\cdot \frac{1}{a^2{\sin }^2\theta }\cdot 2a^2\sin \theta \cos \theta =\boxed{\cot \theta }.$

${\Gamma }_{\theta \theta }^r=-r$ ke saath: $\ddot{r}+(-r){\dot{\theta }}^2=0⟹\ddot{r}=r{\dot{\theta }}^2.$ Ruko — poora geodesic sum hai $\ddot{r}+{\Gamma }_{\theta \theta }^r{\dot{\theta }}^2=\ddot{r}-r{\dot{\theta }}^2=0$, isliye $\boxed{\ddot{r}-r{\dot{\theta }}^2=0}.$ Interpretation: yeh Newton ka "no force, straight line" hai polar coordinates mein likha hua. Term $-r{\dot{\theta }}^2$ centripetal acceleration hai — woh piece jo tum rotating description mein "fictitious force" bolte. Yeh purely isliye appear karta hai kyunki basis vectors ${\hat{e}}_r,{\hat{e}}_{\theta }$ rotate karte hain; koi real force nahi hai. Christoffel symbol wahi fictitious term hai.

Nonzero polar symbols: ${\Gamma }_{\theta \theta }^r=-r$, ${\Gamma }_{r\theta }^{\theta }={\Gamma }_{\theta r}^{\theta }=1/r$. Baaki sab zero.

• ${\partial }_r{\Gamma }_{\theta \theta }^r={\partial }_r(-r)=-1$.
• ${\partial }_{\theta }{\Gamma }_{r\theta }^r={\partial }_{\theta }(0)=0$.
• ${\Gamma }_{rm}^r{\Gamma }_{\theta \theta }^m$: $m$ par sum karo. ${\Gamma }_{rr}^r=0,{\Gamma }_{r\theta }^r=0$, isliye yeh poora term $0$ hai.
• ${\Gamma }_{\theta m}^r{\Gamma }_{r\theta }^m$: $m$ par sum karo. Sirf ${\Gamma }_{\theta \theta }^r=-r$ nonzero hai $m=\theta$ ke saath, ${\Gamma }_{r\theta }^{\theta }=1/r$ ke saath paired. Term $=(-r)(1/r)=-1$.

Assemble karo: $R_{\theta r\theta }^r=(-1)-0+0-(-1)=-1+1=\boxed{0}.$

Moral: plane flat hai. Nonzero $\Gamma$'s (polar coordinates se) curvature combination ke andar perfectly cancel ho gaye. $\Gamma \ne 0$ ka matlab kabhi curved nahi — sirf $\partial \Gamma +\Gamma \Gamma$, invariantly kiya hua, curvature detect karta hai.

Components constant hain, isliye har ${\partial }_j{V}^k=0$. Sirf $\Gamma$ terms survive karte hain.

${\nabla }_{\theta }{V}^{\theta }={\partial }_{\theta }{V}^{\theta }+{\Gamma }_{i\theta }^{\theta }{V}^i$. $i$ par sum karo: $=0+{\Gamma }_{r\theta }^{\theta }{V}^r+{\Gamma }_{\theta \theta }^{\theta }{V}^{\theta }=\frac{1}{r}\cdot 1+0\cdot 0=\boxed{\frac{1}{r}}.$

${\nabla }_{\theta }{V}^r={\partial }_{\theta }{V}^r+{\Gamma }_{i\theta }^r{V}^i$. $i$ par sum karo: $=0+{\Gamma }_{r\theta }^r{V}^r+{\Gamma }_{\theta \theta }^r{V}^{\theta }=0\cdot 1+(-r)\cdot 0=\boxed{0}.$

Interpretation: field $V^r=1$ outward unit radial field ${\hat{e}}_r$ hai. Jab tum $\theta$ direction mein swing karte ho, ${\hat{e}}_r$ physically ${\hat{e}}_{\theta }$ direction mein rotate karta hai. Covariant derivative yeh real turning dekhta hai even though number $V^r$ kabhi change nahi hota. Nonzero ${\nabla }_{\theta }{V}^{\theta }=1/r$ exactly woh rotation record karta hai. Constant components ≠ curvilinear grid par constant vector.

$x$-chart mein, $d{s}^2=d{x}^2$, isliye $\Gamma =0$ (constant metric).

Transform karo. $x=\sqrt{\tilde{x}}$ ke saath, $\frac{dx}{d\tilde{x}}=\frac{1}{2\sqrt{\tilde{x}}}$. Metric transform karta hai (dekho Tensor Transformation Laws): $\tilde{g}=g{\left(\frac{dx}{d\tilde{x}}\right)}^2=1\cdot \frac{1}{4\tilde{x}}=\frac{1}{4\tilde{x}}.$ Inverse: ${\tilde{g}}^{-1}=4\tilde{x}$. 1D mein master formula read karta hai $\tilde{\Gamma }=\frac{1}{2}{\tilde{g}}^{-1}{\partial }_{\tilde{x}}\tilde{g}$ (teen bracket terms collapse hokar ek ho jaate hain kyunki saare indices equal hain): $\tilde{\Gamma }=\frac{1}{2}(4\tilde{x})\cdot {\partial }_{\tilde{x}}\left(\frac{1}{4\tilde{x}}\right)=\frac{1}{2}(4\tilde{x})\cdot \left(-\frac{1}{4{\tilde{x}}^2}\right)=-\frac{1}{2\tilde{x}}\ne 0.$ Isliye $\boxed{\tilde{\Gamma }=-\frac{1}{2\tilde{x}}}$. Ek perfectly flat line ne sirf coordinate stretch se nonzero Christoffel symbol acquire kar liya. Yahi non-tensor behaviour hai: agar $\Gamma$ tensor hota, toh $\Gamma =0$ ek chart mein har chart mein $\tilde{\Gamma }=0$ force karta. Nahi karta.

(a) Metric: $g_{rr}=1,g_{\theta \theta }=f(r{)}^2,g_{r\theta }=0$; inverse $g^{rr}=1,g^{\theta \theta }=1/f^2$. Sirf ${\partial }_r{g}_{\theta \theta }=2f{f}^{\prime }$ nonzero hai.

${\Gamma }_{\theta \theta }^r=\frac{1}{2}{g}^{rr}(-{\partial }_r{g}_{\theta \theta })=\frac{1}{2}(1)(-2f{f}^{\prime })=\boxed{-f{f}^{\prime }}.$

${\Gamma }_{r\theta }^{\theta }=\frac{1}{2}{g}^{\theta \theta }\left({\partial }_r{g}_{\theta \theta }\right)=\frac{1}{2}\cdot \frac{1}{f^2}\cdot 2f{f}^{\prime }=\boxed{\frac{f^{\prime }}{f}}.$

Baaki sab vanish karte hain (check karo: har remaining bracket mein sirf constants ke ya $g_{rr}=1$ ke derivatives hain). Yeh polar results recover karta hai $f=r$ ke saath: $-f{f}^{\prime }=-r$, $f^{\prime }/f=1/r$. ✓

(b) $f=\sin r$, $f^{\prime }=\cos r$ ke saath: ${\Gamma }_{\theta \theta }^r=-\sin r\cos r,{\Gamma }_{r\theta }^{\theta }=\frac{\cos r}{\sin r}=\cot r.$ (Ye Problem 2.2 ke sphere result se match karte hain $a=1$ aur $\theta \to r$ ke saath: $-\sin \theta \cos \theta$ aur $\cot \theta$. ✓)

(c) Radial geodesic: $\ddot{r}+{\Gamma }_{\theta \theta }^r{\dot{\theta }}^2=0$, yaani $\boxed{\ddot{r}-f{f}^{\prime }{\dot{\theta }}^2=0}\text{(general)},\ddot{r}-\sin r\cos r{\dot{\theta }}^2=0\text{(sphere)}.$ Term $-f{f}^{\prime }{\dot{\theta }}^2$ generalised centripetal acceleration hai: motion "chahta hai" axis ki taraf ya usse door curve kare ek rate se jo set hota hai circle radius $f(r)$ kitni tezi se change karta hai iske hisaab se. Sphere par equator ke paas ($r=\pi /2$) hume milta hai $\cos r=0$, isliye term vanish karta hai — equator ke around great circles ko koi radial pull nahi chahiye, exactly as expected.

Kya tum memory se kar sakte ho: master formula likho? ::: ${\Gamma }_{ij}^k=\frac{1}{2}{g}^{km}({\partial }_i{g}_{mj}+{\partial }_j{g}_{mi}-{\partial }_m{g}_{ij})$ Do nonzero polar symbols ka naam batao? ::: ${\Gamma }_{\theta \theta }^r=-r,{\Gamma }_{r\theta }^{\theta }=1/r$ Explain karo ki flat polar space mein nonzero $\Gamma$ kyun hai? ::: Polar grid ke basis arrows rotate karte hain; $\Gamma$ yeh track karta hai, lekin Riemann combination phir bhi zero pe cancel ho jaata hai.