Question bank — Christoffel symbols — intro

False. Flat 2D space in polar coordinates mein ${\Gamma }_{\theta \theta }^r=-r\ne 0$ hota hai phir bhi curvature zero hai — symbols yeh batate hain ki coordinate grid kaise bend karti hai, nahi ki space kaise bend karti hai.

False. Tum hamesha ek chosen point par har $\Gamma$ ko zero kar sakte ho "normal coordinates" choose karke; flatness ke liye $\Gamma$ ko poore neighbourhood mein zero rehna chahiye (ya uski curvature combination ka vanish hona zaroori hai).

False. Coordinate change ke under yeh ek inhomogeneous second-derivative term le aate hain, isliye yeh us homogeneous Tensor Transformation Laws ko violate karte hain jo ek tensor ko define karta hai.

True ek coordinate (holonomic) basis mein, kyunki ${\partial }_i{\mathbf{e}}_j={\partial }_j{\mathbf{e}}_i$ (position ke mixed partials commute karte hain); non-coordinate basis mein yeh fail ho sakta hai.

True. Boxed formula ${\Gamma }_{ij}^k=\frac{1}{2}{g}^{km}({\partial }_i{g}_{mj}+{\partial }_j{g}_{mi}-{\partial }_m{g}_{ij})$ inhe poori tarah $g$ aur uske first derivatives se build karta hai — kuch aur ki zaroorat nahi.

True. Wahan $g_{ij}={\delta }_{ij}$ constant hai, isliye har ${\partial }_m{g}_{ij}=0$ hai aur formula $\Gamma =0$ everywhere deta hai.

True. Dono mein ek hi inhomogeneous term hota hai, isliye subtract karne par non-tensorial junk cancel ho jaata hai aur baaki hissa cleanly transform karta hai.

False. $g$ se raise/lower karna kabhi transformation law fix nahi karta; extra second-derivative term wahan bhi maujood rehta hai chahe index kuch bhi ho.

Yeh error is baat ko ignore karta hai ki curvilinear coordinates mein basis vectors bhi change hote hain; sahi derivative hai ${\nabla }_j{V}^k={\partial }_j{V}^k+{\Gamma }_{ij}^k{V}^i$, aur $\Gamma$ term drop karne se jo quantity milti hai woh tensor bhi nahi hoti.

$\theta$ wala galat hai: usmein inverse metric $g^{\theta \theta }=1/r^2$ chahiye, jo deta hai $\frac{1}{2}\cdot \frac{1}{r^2}\cdot 2r=1/r$, nahi ki $r$. $g^{km}$ se contract karna bhool jaana classic galti hai.

Nahi — minus ${\partial }_m{g}_{ij}$ par hota hai, yaani dummy index $m$ ke respect se derivative par (jo $g^{km}$ se tied hai). Do cyclic terms jo free indices ke saath hain woh plus hain.

False; upper $k$ output basis vector ko name karta hai jabki lower $i,j$ yeh batate hain (kaun sa basis vector, kis direction mein move hua). Sirf dono lowers swap ho sakte hain.

Yeh blow-up ek coordinate artefact hai — origin ek bilkul smooth flat point hai; polar coordinates wahan sirf degenerate ho jaate hain kyunki $r=0$ par $\theta$ undefined hai.

Constant speed ka matlab zero acceleration nahi hai; velocity vector turn karta hai. Covariant derivative real centripetal turning ko ${\Gamma }_{\theta \theta }^r=-r$ ke zariye capture karta hai, jisse $-r{\dot{\theta }}^2\ne 0$ milta hai.

Kyunki practice mein Metric Tensor woh ek cheez hai jo hamesha available hai ($d{s}^2$ se), jabki embedding mein explicit basis vectors available nahi ho sakte — formula hume ambient space ki zaroorat se azaad karta hai.

Kyunki woh symmetry aati hai ${\partial }_i{\partial }_j(\text{position})={\partial }_j{\partial }_i(\text{position})$ se; non-coordinate frame mein basis ko kisi coordinate ko differentiate karke nahi mila hota, isliye mixed-partial argument fail ho jaata hai.

Kyunki $\Gamma$ ko coordinates se kisi bhi point par zero kiya ja sakta hai, isliye $\Gamma$ ki koi single-point value geometric nahi ho sakti; sirf woh combination jo coordinate changes survive kare — Riemann Curvature Tensor — true bending measure karta hai.

Plain derivative ki transformation mein ek non-tensorial term leak hoti hai; $\Gamma$ term ek equal and opposite non-tensorial term carry karta hai, aur unka sum junk cancel kar deta hai, ek clean tensor bacha ke.

Extra $-r{\dot{\theta }}^2$ exactly ${\Gamma }_{\theta \theta }^r{\dot{\theta }}^2$ hai; geodesic equation "koi real force nahi" ko in $\Gamma$-driven "fictitious force" terms mein translate karti hai jo rotating basis demand karta hai.

Tum $\Gamma$ ko ek point par khatam kar sakte ho, lekin curvature $\partial \Gamma$ ko paas mein nonzero rehne par majboor karti hai, isliye koi bhi coordinate system $\Gamma$ ko everywhere ek saath vanish nahi kar sakta — yahi impossibility curvature ka invariant signature hai.

Yeh diverge karta hai, lekin harmlessly: yeh flag karta hai ki polar coordinates origin par break down karte hain, geometry singular nahi hai — wahan ek Cartesian patch mein saare $\Gamma =0$ hote hain.

Nahi. Ek diagonal metric mein bhi ${\partial }_r{g}_{\theta \theta }$ jaise derivatives hoti hain, jo off-diagonal-looking symbols jaise ${\Gamma }_{r\theta }^{\theta }$ mein feed karti hain; $g$ ki diagonality $\Gamma$ ko diagonal nahi banati.

Saare zero, kyunki formula mein har term $g$ ki derivative hai; ek constant metric — chahe non-orthonormal hi kyon na ho — ek flat, $\Gamma$-free connection deta hai.

Nahi. Sphere genuinely curved hai, isliye uska Riemann tensor nonzero hai aur koi bhi coordinates globally connection ko flatten nahi kar sakti — zyaada se zyaada $\Gamma$ ek chosen point par vanish ho sakta hai.

Zaroor nahi — unhe ek constant scaling tak agree karna chahiye jo saare derivative ratios unchanged rakhta ho, kyunki sirf derivatives of $g$ (weighted by $g^{-1}$) enter hoti hain, isliye overall constant factors alag ho sakte hain.