Christoffel symbols — intro — Hinglish notes

4.10.11 · Maths › Advanced Topics (Elite Level)

Hume inki zaroorat KYUN hai?

Flat space mein Cartesian coordinates $(x, y)$ ke saath, basis vectors $\overset{e}{^}_{x}, \overset{e}{^}_{y}$ har jagah same direction mein point karte hain. Isliye ek vector field ki derivative bas uske components ki derivative hoti hai.

Lekin jis waqt tum polar coordinates $(r, θ)$ use karo — ya kisi sphere par raho — basis vectors $\overset{e}{^}_{r}, \overset{e}{^}_{θ}$ move karne par rotate karte hain. Ab agar tum kisi vector ko differentiate karo, to tumhe do cheezein account karni padti hain:

Components mein change, AUR
Basis vectors mein change.

EXACTLY ek Christoffel symbol kya hota hai?

Key facts:

Do lower indices ( $i, j$ ) ek coordinate basis mein symmetric hain: $Γ_{ij}^{k} = Γ_{j i}^{k}$ .
Ye tensor components NAHI hain (ye tensor ki tarah transform nahi hote) — ye coordinate system ko encode karte hain, sirf geometry ko nahi.

Hum metric se formula KAISE derive karte hain?

Hum $Γ$ purely metric $g_{ij} = e_{i} \cdot e_{j}$ se chahte hain, kyunki metric woh ek cheez hai jo hum hamesha jaante hain.

Step 1 — Metric ko differentiate karo. $\partial_{k} g_{ij} = (\partial_{k} e_{i}) \cdot e_{j} + e_{i} \cdot (\partial_{k} e_{j})$ Ye step kyun? $g_{ij} = e_{i} \cdot e_{j}$ par product rule lagao.

First-kind symbol $Γ_{ij k} = (\partial_{j} e_{i}) \cdot e_{k}$ define karo. Phir: $\partial_{k} g_{ij} = Γ_{ik j} + Γ_{j k i}$

Step 2 — Teen cyclic permutations likho. $\partial_{k} g_{ij} = Γ_{ik j} + Γ_{j k i}$ $\partial_{i} g_{j k} = Γ_{j ik} + Γ_{k ij}$ $\partial_{j} g_{k i} = Γ_{k j i} + Γ_{ik j}$ Ye step kyun? Teen equations se hum ek $Γ$ ko isolate kar sakte hain symmetry $Γ_{ab c} = Γ_{ba c}$ pehle do slots mein use karke... actually lower coordinate indices mein symmetry: $\partial_{j} e_{i} = \partial_{i} e_{j}$ , isliye $Γ_{ik j} = Γ_{k ij}$ etc.

Step 3 — Pehle do add karo, teesra subtract karo. Cancellation ke baad (symmetry use karke): $Γ_{k ij} = \frac{1}{2} (\partial_{i} g_{j k} + \partial_{j} g_{k i} - \partial_{k} g_{ij})$

Step 4 — Index raise karo inverse metric $g^{k m}$ se:

Worked Example 1 — Polar coordinates (the classic)

2D polar coordinates mein metric: $d s^{2} = d r^{2} + r^{2} d θ^{2}$ , isliye $g_{r r} = 1, g_{θ θ} = r^{2}, g_{r θ} = 0, g^{r r} = 1, g^{θ θ} = \frac{1}{r ^{2}} .$

$Γ_{θ θ}^{r}$ find karo. $Γ_{θ θ}^{r} = \frac{1}{2} g^{r r} (\partial_{θ} g_{r θ} + \partial_{θ} g_{r θ} - \partial_{r} g_{θ θ})$ Kyun? Boxed formula mein $k = r, i = j = θ$ set karo; inverse metric mein sirf $g^{r r}$ survive karta hai (diagonal hai). $= \frac{1}{2} (1) (0 + 0 - \partial_{r} r^{2}) = \frac{1}{2} (- 2 r) = - r$

$Γ_{r θ}^{θ}$ find karo. $Γ_{r θ}^{θ} = \frac{1}{2} g^{θ θ} (\partial_{r} g_{θ θ} + \partial_{θ} g_{θ r} - \partial_{θ} g_{r θ}) = \frac{1}{2} \cdot \frac{1}{r ^{2}} \cdot 2 r = \frac{1}{r}$ Kyun? Sirf $\partial_{r} g_{θ θ} = 2 r$ nonzero hai. Baaki saare Christoffel symbols vanish ho jaate hain.

Worked Example 2 — Verify karo ki flat case mein zero milta hai

Cartesian coordinates mein $g_{ij} = δ_{ij}$ (constant). Har $\partial_{m} g_{ij} = 0$ , isliye har $Γ = 0$ . Ye kyun important hai: Christoffel symbols flat space mein bhi nonzero ho sakte hain (polar mein!), isliye $Γ \neq = 0$ ka matlab "curved" nahi hota. Curvature $Γ$ ke derivatives mein rehti hai (Riemann tensor), $Γ$ mein khud nahi.

Covariant derivative (jahan inhe use kiya jaata hai)

Common Mistakes (Steel-manned)

Recall Feynman: ek 12-saal ke bachche ko samjhao

Socho tum ek giant globe par chal rahe ho aur do arrows pakde ho: ek "East" ki taraf, ek "North" ki taraf. Jaise jaise tum chalte ho, "East" aur "North" dheere dheere curved zameen follow karne ke liye twist karte hain. Agar tum jaanna chahte ho ki tumhara apna arrow (maan lo kisi phenki hui ball ki direction) sach mein kitna change ho raha hai, to pehle tumhe "East" aur "North" ka twisting subtract karna hoga. Christoffel symbols woh chote instruction cards hain jo kehte hain "jab tum North mein ek aur kadam lo, to East itna turn karta hai." Bas itna hi hain ye — apne direction-arrows ke liye twist-rate cards.

Active-Recall Flashcards

Christoffel symbols physically kya encode karte hain?

Basis vectors point-to-point kaise change hote hain:

\partial_{j} e_{i} = Γ_{ij}^{k} e_{k}

Γ_{ij}^{k}

ka metric formula batao.

Γ_{ij}^{k} = \frac{1}{2} g^{k m} (\partial_{i} g_{mj} + \partial_{j} g_{mi} - \partial_{m} g_{ij})

Kya Christoffel symbols tensors hain?

Nahi — ye ek extra inhomogeneous second-derivative term ke saath transform karte hain.

Coordinate-basis Christoffel symbols kin lower indices mein symmetric hote hain?

Do lower indices

i, j

mein:

Γ_{ij}^{k} = Γ_{j i}^{k}

2D polar coords mein

Γ_{θ θ}^{r}

compute karo.

- r

(from

- \frac{1}{2} \partial_{r} g_{θ θ} = - \frac{1}{2} \cdot 2 r

2D polar coords mein

Γ_{r θ}^{θ}

compute karo.

1/ r

Kya

Γ \neq = 0

curvature imply karta hai?

Nahi; flat polar coords mein nonzero

Γ

hota hai. Curvature ke liye

\partial Γ + ΓΓ

chahiye (Riemann).

V^{k}

ki covariant derivative likho.

\nabla_{j} V^{k} = \partial_{j} V^{k} + Γ_{ij}^{k} V^{i}

\nabla_{j} V^{k}

mein har term ka kya role hai?

\partial_{j} V^{k}

= components mein change;

Γ

term = basis vectors mein change.

Connections

Metric Tensor — woh source jisse saare $Γ$ compute hote hain.
Covariant Derivative — jahan Christoffel symbols use hote hain.
Riemann Curvature Tensor — $\partial Γ + ΓΓ$ se banta hai; true curvature.
Geodesic Equation — $\overset{x}{¨}^{k} + Γ_{ij}^{k} \overset{x}{˙}^{i} \overset{x}{˙}^{j} = 0$ .
Polar Coordinates — sabse clean non-trivial example.
Tensor Transformation Laws — isliye $Γ$ ek tensor nahi hai.