Worked examples — Christoffel symbols — intro

4.10.11 · D3 · Maths › Advanced Topics (Elite Level) › Christoffel symbols — intro

Yeh page ek drill hai. Parent note ne tumhe woh ek formula diya tha jo tumhe memorise karna hai. Yahan hum har tarah ka input usme daalte hain — flat aur curved, diagonal aur off-diagonal metrics, zero cases, degenerate cases, ek word problem, aur ek exam trap — taaki is page ke baad koi bhi scenario tumhe surprise na kar sake.

Sab kuch parent se liya hua ek boxed formula pe tikaa hai:

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Scenario matrix

Kuch bhi work karne se pehle, aao hum har woh alag cheez list karein jo galat ho sakti hai ya badal sakti hai. Neeche har worked example uss cell ke saath tagged hai jisme woh aata hai.

Cell	Kya test karta hai	Example
A. Flat / trivial	constant metric $\Rightarrow$ saare $\Gamma =0$	Ex 1
B. Diagonal, ek non-constant entry	classic polar radial symbol	Ex 2
C. Diagonal, derivative ka sign	jab ek $\Gamma$ positive vs negative aata hai	Ex 3
D. Off-diagonal metric	$g^{km}$ indices mix karta hai; sum mein do terms bachti hain	Ex 4
E. Har jagah coordinate-dependent (sphere)	multiple non-zero $\Gamma$, latitude sign flips	Ex 5
F. Degenerate / limiting input	$r\to 0$ par kya hota hai (coordinate breakdown)	Ex 6
G. Word problem (physics)	ek $\Gamma$ se "fictitious force" padhna	Ex 7
H. Exam twist (is-it-a-tensor trap)	conformal rescale, symmetry check	Ex 8

Goal yeh hai: Ex 8 ke baad tum kisi bhi naye problem ko in rows mein se ek mein place kar sako aur jaano ki woh kaun sa trick maangta hai.

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Ex 1 — Cell A: flat sanity check

Forecast: aage padhne se pehle answer guess karo. Ek constant ki derivative kya hoti hai?

1. Formula likhо $k=x,i=j=y$ ke saath. ${\Gamma }_{yy}^x=\frac{1}{2}{g}^{xm}({\partial }_y{g}_{my}+{\partial }_y{g}_{my}-{\partial }_m{g}_{yy}).$ Yeh step kyun? Hum bas requested indices ko machine mein substitute kar rahe hain; abhi kuch clever nahi hai.

2. Note karo ki har metric component ek constant number hai. Kyunki $g_{yy}=1$ har jagah hai, ${\partial }_m{g}_{yy}=0$. Kyunki $g_{my}$ constant hai, ${\partial }_y{g}_{my}=0$. Yeh step kyun? Formula mein hamesha sirf metric ki derivatives hoti hain. Ek constant metric ki saari derivatives zero hain — isliye poora bracket collapse ho jaata hai.

3. Conclude karo. ${\Gamma }_{yy}^x=\frac{1}{2}{g}^{xm}\cdot 0=\boxed{0}.$ Wahi argument yahan har Christoffel symbol ko zero kar deta hai.

Verify: units/logic — basis vectors $\hat{x},\hat{y},\hat{z}$ har point par ek hi direction mein point karte hain, isliye woh "turn" nahi kar sakte, isliye unke turn-rate cards ($\Gamma$) saare zero padhne chahiye. Consistent. ✔

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Ex 2 — Cell B: classic polar radial symbol

Forecast: formula mein kaunsi ek derivative non-zero ho sakti hai?

Figure dekho: ${\mathbf{e}}_r$ (burnt orange) arrows baahir ki taraf point karte hain aur ${\mathbf{e}}_{\theta }$ (teal) sideways point karte hain. Jab tum $\theta$ swing karte ho, notice karo ki ${\mathbf{e}}_{\theta }$ origin ki taraf andar swing karta hai — iske change ka ek component $-{\mathbf{e}}_r$ ke along hai. Woh inward lean precisely ${\Gamma }_{\theta \theta }^r$ hai.

1. $k=r,i=j=\theta$ substitute karo. ${\Gamma }_{\theta \theta }^r=\frac{1}{2}{g}^{rm}({\partial }_{\theta }{g}_{m\theta }+{\partial }_{\theta }{g}_{m\theta }-{\partial }_m{g}_{\theta \theta }).$ Yeh step kyun? Hum ek specific symbol target kar rahe hain — "output basis" $r$ hai (upper index $k=r$) aur "kaun sa basis vector, kis direction mein" ka pair $\theta ,\theta$ hai — isliye hum woh exact labels machine mein daalne se pehle place karte hain.

2. $m$ par sum khatam karo. Metric diagonal hai, isliye $g^{rm}$ sirf $m=r$ ke liye non-zero hai, jo $g^{rr}=1$ deta hai. Yeh step kyun? Diagonal inverse metric ka matlab hai "sirf matching index bachega" — yahi diagonal problems ko fast banata hai. ${\Gamma }_{\theta \theta }^r=\frac{1}{2}(1)({\partial }_{\theta }{g}_{r\theta }+{\partial }_{\theta }{g}_{r\theta }-{\partial }_r{g}_{\theta \theta }).$

3. Har derivative evaluate karo. $g_{r\theta }=0$ isliye iski $\theta$-derivative $0$ hai. Aur $g_{\theta \theta }=r^2$, isliye ${\partial }_r{g}_{\theta \theta }=2r$. ${\Gamma }_{\theta \theta }^r=\frac{1}{2}(0+0-2r)=\boxed{-r}.$ Yeh step kyun? Minus sign ${\partial }_m$ term par hai (woh jo $g^{km}$ se tied hai) — mnemonic yaad karo "do plus, ek minus."

Verify: geometric picture ne kaha ki ${\mathbf{e}}_{\theta }$ ki change $-{\mathbf{e}}_r$ ki taraf lean karti hai (ek negative radial component), jo minus sign se match karta hai. Physically yeh centripetal term $-r{\dot{\theta }}^2$ hai. ✔

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Ex 3 — Cell C: sign dekho — ek positive Christoffel symbol

Forecast: kya yeh positive hoga ya negative? Guess karo, phir check karo.

1. $k=\theta ,i=r,j=\theta$ substitute karo. ${\Gamma }_{r\theta }^{\theta }=\frac{1}{2}{g}^{\theta m}({\partial }_r{g}_{m\theta }+{\partial }_{\theta }{g}_{mr}-{\partial }_m{g}_{r\theta }).$ Yeh step kyun? Is baar "output basis" $\theta$ hai (upper index $k=\theta$) aur direction/vector pair $r,\theta$ hai; humein woh precise slots feed karni padengi — poora exercise yeh dekhna hai ki ek alag index choice kaise surviving derivative ko aur is tarah sign ko change karti hai.

2. Diagonal $\Rightarrow$ sirf $m=\theta$ bachta hai: $g^{\theta \theta }=1/r^2$. ${\Gamma }_{r\theta }^{\theta }=\frac{1}{2}\cdot \frac{1}{r^2}({\partial }_r{g}_{\theta \theta }+{\partial }_{\theta }{g}_{\theta r}-{\partial }_{\theta }{g}_{r\theta }).$ Yeh step kyun? Bilkul Ex 2 ki tarah: inverse metric diagonal hai, isliye $g^{\theta m}$ zero hai jab tak $m=\theta$ na ho. $m=r$ term drop karna legitimate sirf isliye hai kyunki $g^{\theta r}=0$ yahan — ek off-diagonal metric (Ex 4) ke liye hum aisa nahi kar sakte.

3. Evaluate karo. ${\partial }_r{g}_{\theta \theta }=2r$; baaki do mein $g_{\theta r}=0$ aata hai, isliye woh $0$ hain. ${\Gamma }_{r\theta }^{\theta }=\frac{1}{2}\cdot \frac{1}{r^2}\cdot 2r=\boxed{\frac{1}{r}}.$ Yeh step kyun? Yahan non-zero derivative ek plus slot mein baith gayi hai (yeh ${\partial }_i{g}_{mj}$ hai, ${\partial }_m{g}_{ij}$ nahi), isliye answer positive hai — Ex 2 ke opposite sign, haalaanki yeh wahi ${\partial }_r{r}^2=2r$ se aaya.

Verify: jaise $r$ badhta hai, $1/r$ kam hota hai — origin se door coordinate lines "kam curved" hote hain, isliye twist-rate girta hai. Sensible. Yeh Coriolis-jaisa term hai. ✔

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Ex 4 — Cell D: ek off-diagonal metric (sum actually sums)

Forecast: inverse metric mein ab off-diagonal entries hain, isliye $m$-sum mein do terms hain. Lekin pehle metric ko dhyan se dekho — kya koi cheez actually vary ho rahi hai?

1. Sum ko explicitly $m=1,2$ pe likhо. ${\Gamma }_{22}^1=\frac{1}{2}\left[g^{11}\right(2{\partial }_2{g}_{12}-{\partial }_1{g}_{22})+g^{12}(2{\partial }_2{g}_{22}-{\partial }_2{g}_{22}\left)\right].$ Yeh step kyun? Off-diagonal $g^{12}$ ke saath, hum $m=2$ term nahi chhod sakte jaise diagonal cases mein kiya tha.

2. Har entry ek constant hai, isliye har derivative $0$ hai. ${\Gamma }_{22}^1=\frac{1}{2}[g^{11}\cdot 0+g^{12}\cdot 0]=\boxed{0}.$ Yeh step kyun? Yeh cell jo trap sikhaata hai: off-diagonal metric apne aap Christoffel symbols create nahi karta. Sirf coordinate-dependence karta hai. Ek constant metric — chahe skewed ho — phir bhi flat hai aur saare $\Gamma =0$ hain.

Verify: Cartesian se ek constant-coefficient linear coordinate change basis vectors ki direction fix rakhta hai, isliye koi twisting nahi, isliye $\Gamma =0$. Ex 1 ki logic se match karta hai. ✔ (Setup mein $g^{12}$ term ka present hona zaroori tha yeh point banane ke liye ki woh sirf isliye vanish hota hai kyunki derivatives vanish hoti hain, metric nahi.)

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Ex 5 — Cell E: sphere — kai symbols, ek sign jo latitude ke saath flip karta hai

Forecast: ek sphere par, East jaana poles ke paas "sasta" hota hai. In do symbols mein se kaun sa $\theta \to 0$ par blow up karna chahiye?

Figure longitude ki lines ko pole ki taraf ek saath squeeze hote dikhata hai. Metric coefficient ${\sin }^2\theta$ (plum curve) measure karta hai ki $\varphi$ ki ek unit kitna real distance deti hai — yeh equator par sabse zyada hai ($\theta =\pi /2$) aur poles par zero.

1. ${\Gamma }_{\varphi \varphi }^{\theta }$: set karo $k=\theta ,i=j=\varphi$, diagonal $\Rightarrow g^{\theta \theta }=1$. ${\Gamma }_{\varphi \varphi }^{\theta }=\frac{1}{2}(1)(0+0-{\partial }_{\theta }{g}_{\varphi \varphi })=-\frac{1}{2}{\partial }_{\theta }({\sin }^2\theta )=-\frac{1}{2}\cdot 2\sin \theta \cos \theta =\boxed{-\sin \theta \cos \theta }.$ Yeh step kyun? Polar Ex 2 jaisi hi shape: sirf ek surviving derivative ${\partial }_{\theta }{g}_{\varphi \varphi }$ hai, jo minus slot mein hai.

2. Pure range mein uska sign padhо. $0<\theta <\frac{\pi }{2}$ ke liye (northern hemisphere) $\cos \theta >0$ isliye yeh negative hai; equator par $\theta =\frac{\pi }{2}$ par yeh $\boxed{0}$ hai; $\frac{\pi }{2}<\theta <\pi$ ke liye (southern) $\cos \theta <0$ isliye yeh positive flip karta hai. Yeh step kyun? Cell E demand karta hai ki hum coordinate ke saare cases cover karein — sign genuinely equator par change hota hai, aur wahan exactly vanish hota hai.

3. ${\Gamma }_{\theta \varphi }^{\varphi }$: set karo $k=\varphi ,i=\theta ,j=\varphi$, diagonal $\Rightarrow g^{\varphi \varphi }=1/{\sin }^2\theta$. ${\Gamma }_{\theta \varphi }^{\varphi }=\frac{1}{2}\cdot \frac{1}{{\sin }^2\theta }{\partial }_{\theta }({\sin }^2\theta )=\frac{1}{2}\cdot \frac{2\sin \theta \cos \theta }{{\sin }^2\theta }=\boxed{\cot \theta }.$ Yeh step kyun? Ab hum "East-tracking" symbol target kar rahe hain: upper index $k=\varphi$, direction/vector pair $\theta ,\varphi$. Ex 3 ki tarah diagonal inverse metric sirf $m=\varphi$ chhodta hai, aur ek surviving derivative ${\partial }_{\theta }{g}_{\varphi \varphi }$ ek plus slot mein hai, isliye yeh symbol positive hai jahan $\cos \theta >0$.

4. Limiting behaviour. Jaise $\theta \to 0^{+}$ (north pole ki taraf jaate hue), $\cot \theta \to +\infty$: yeh symbol blow up karta hai. Yeh step kyun? Wahi forecast answer hai — East-tracking symbol pole par diverge karta hai kyunki "East" wahan infinitely fast spin karta hai (ek coordinate singularity, real nahi).

Verify: equator par $\theta =\pi /2$: $\cos (\pi /2)=0$ isliye ${\Gamma }_{\varphi \varphi }^{\theta }=0$ aur $\cot (\pi /2)=0$ — dono symbols vanish ho jaate hain, us fact se match karta hai ki equator ke paas sphere flat Cartesian paper jaisa dikhta hai. ✔

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Ex 6 — Cell F: degenerate limit $r\to 0$

Forecast: $r=0$ ke paas $1/r$ physically kya quantity hai, aur kya space actually wahan singular hai?

1. Limit lo. $\underset{r\to 0^{+}}{\lim }{\Gamma }_{r\theta }^{\theta }=\underset{r\to 0^{+}}{\lim }\frac{1}{r}=+\infty .$ Yeh step kyun? Humein limiting inputs check karne chahiye — origin exactly woh jagah hai jahan polar coordinates degenerate hain ($\theta$ wahan undefined hai).

2. Diagnose karo: kya yeh geometry hai ya coordinates? Usi point par ${\Gamma }_{\theta \theta }^r=-r$ compute karo: ${\lim }_{r\to 0}(-r)=0$, perfectly finite. Yeh step kyun? Agar space singular hota, toh koi bhi coordinate choice finite geometric invariants nahi deta. Yahan plane manifestly flat aur smooth hai origin par — Cartesian coordinates kuch bhi special nahi dekhte.

3. Conclude karo. $1/r$ ka divergence ek coordinate artefact hai: $r=0$ par "$\theta$-increasing" direction undefined hai (saare angles wahan milte hain), isliye iska turn-rate card $\infty$ padhta hai. Yeh exactly Ex 5 mein sphere ke pole jaisa hai.

Verify: origin par Cartesian mein switch karo — wahan saare $\Gamma =0$ hain (Ex 1). Ek quantity jo ek chart mein $\infty$ hai aur doosre mein $0$, woh geometrically real nahi ho sakti; woh coordinate-borne hai. Yeh confirm karta hai ki $\Gamma$ ek tensor nahi hai (ek tensor jo ek basis mein zero ho, woh saare bases mein zero hota hai). ✔

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Ex 7 — Cell G: word problem — ek fictitious force padhna

Forecast: Ex 2 aur Ex 3 ke kaunse do Christoffel numbers dikhne chahiye, aur har ek kitni baar?

1. Radial ($k=r$) equation. Sirf non-zero ${\Gamma }^r$ hai ${\Gamma }_{\theta \theta }^r=-r$. $\ddot{r}+{\Gamma }_{\theta \theta }^r\dot{\theta }\dot{\theta }=\ddot{r}-r{\dot{\theta }}^2=0.$ Yeh step kyun? Geodesic equation mein $k=r$ ke saath plug karo; sirf $\theta \theta$ term Ex 2 se bachti hai.

2. Angular ($k=\theta$) equation. Non-zero ${\Gamma }^{\theta }$ hain ${\Gamma }_{r\theta }^{\theta }={\Gamma }_{\theta r}^{\theta }=1/r$ (symmetric lower indices!). Sum ${\Gamma }_{ij}^{\theta }{\dot{x}}^i{\dot{x}}^j$ dono orderings pick karta hai: $\ddot{\theta }+\frac{1}{r}\dot{r}\dot{\theta }+\frac{1}{r}\dot{\theta }\dot{r}=\ddot{\theta }+\frac{2}{r}\dot{r}\dot{\theta }=0.$ Yeh step kyun? 2 ka factor poora point hai — kyunki $i,j$ symmetric hain, double sum $r\theta$ aur $\theta r$ alag count karta hai, $1/r$ ko double karta hai.

3. Interpret karo. $\ddot{r}-r{\dot{\theta }}^2=0$ Newton hai centripetal term ke saath; $\ddot{\theta }+\frac{2}{r}\dot{r}\dot{\theta }=0$ angular-momentum conservation hai (Coriolis term). Dono "forces" pure $\Gamma$ hain — woh rotating basis ke artefacts hain, real pushes nahi.

Verify: doosri equation ko $r^2$ se multiply karo: $r^2\ddot{\theta }+2r\dot{r}\dot{\theta }=\frac{d}{dt}(r^2\dot{\theta })=0$, isliye $r^2\dot{\theta }=$ const — conserved angular momentum, exactly woh jo ek free straight-line puck ke paas hona chahiye. ✔

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Ex 8 — Cell H: exam twist (conformal rescale + tensor trap)

Forecast: $e^{2f}$ kabhi zero nahi hota, aur uski derivative $2f^{\prime }$ neeche laati hai. Answer ki shape guess karo.

1. Inverse metric. Diagonal with entries $e^{2f}$, isliye $g^{xx}=g^{yy}=e^{-2f}$. Yeh step kyun? Diagonal matrix ka inverse reciprocals hota hai; humein contraction ke liye $g^{xx}$ chahiye.

2. Formula apply karo, $k=i=j=x$. ${\Gamma }_{xx}^x=\frac{1}{2}{g}^{xx}({\partial }_x{g}_{xx}+{\partial }_x{g}_{xx}-{\partial }_x{g}_{xx})=\frac{1}{2}{g}^{xx}{\partial }_x{g}_{xx}.$ Yeh step kyun? Jab teeno indices coincide karte hain, "do plus, ek minus" ek single derivative mein collapse ho jaata hai.

3. Evaluate karo. ${\partial }_x{g}_{xx}={\partial }_x{e}^{2f}=2f^{\prime }{e}^{2f}$, aur $g^{xx}=e^{-2f}$: ${\Gamma }_{xx}^x=\frac{1}{2}{e}^{-2f}\cdot 2f^{\prime }{e}^{2f}=\boxed{f^{\prime }(x)}.$ Yeh step kyun? Exponential cleanly cancel ho jaata hai — conformal metrics ka ek signature move.

4. Examiner ko rebut karo (Cell H trap). $f(x)=x$ lo toh ${\Gamma }_{xx}^x=1$, ek non-zero constant. Metric $e^{2x}(d{x}^2+d{y}^2)$ conformally flat hai, aur ek non-zero $\Gamma$ curvature prove nahi karta; sirf Riemann Curvature Tensor jo $\partial \Gamma +\Gamma \Gamma$ ki tarah build hota hai (ek invariant) decide karta hai. Kyunki $\Gamma$ coordinate change ke under ek non-tensorial second-derivative piece carry karta hai (dekho Tensor Transformation Laws), tum hamesha use non-zero bana sakte ho ek bure coordinate choice se. Yeh step kyun? Yahi recurring exam intent hai: "coordinate twisting" aur "genuine curvature" ko alag karo.

Verify (limiting check): agar $f$ constant hai, $f^{\prime }=0$ isliye ${\Gamma }_{xx}^x=0$ — metric $e^{2f}{\delta }_{ij}$ constant $f$ ke saath ek mere uniform rescale hai (still flat Cartesian), saare $\Gamma =0$ deta hai, Ex 1 se match karta hai. ✔

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Recall Ek-line placement drill

Naya problem diya, woh kaun si cell mein hai? Constant metric ::: Cell A/D — saare $\Gamma =0$, chahe diagonal ho ya skewed. Diagonal metric, ek entry ek coordinate par depend karti hai ::: Cell B/C — ek derivative bachti hai; sign plus vs minus slot par depend karta hai. Metric coefficient kahin vanish karta hai (pole, origin) ::: Cell E/F — ek $\Gamma$ ke diverge hone ki expect karo; yeh ek coordinate artefact hai. Koi $\Gamma \ne 0$ se curvature claim kare ::: Cell H — rebut karo; sirf Riemann decide karta hai.

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Connections

• Parent: Christoffel Symbols — Intro — woh formula aur derivation jise yeh page drill karta hai.
• Metric Tensor — yahan har example ka input.
• Geodesic Equation — jahan Ex 7 ke fictitious forces aate hain.
• Covariant Derivative — woh operator jinhe yeh symbols correct karte hain.
• Riemann Curvature Tensor — woh invariant jo ($\Gamma$ ke unlike) truly curvature detect karta hai (Ex 8).
• Polar Coordinates · Tensor Transformation Laws — Ex 2–3 aur Ex 8 trap ke peeche coordinate machinery.