Kinetic energy in generalized coordinates: Question bank

2.1.3 · D5 · Physics › Analytical Mechanics › Generalized coordinates mein kinetic energy

Definition Is page pe use hone waale har symbol ki definition pehle yahan

Taaki yeh page apne aap mein complete ho, yahan poora cast of characters hai:

$q_{i}$ — ek generalized coordinate: system ki configuration describe karne wala ek independent number (ek angle, ek rail ke saath distance). $\overset{q}{˙}_{i} = d q_{i} / d t$ iski generalized velocity hai.
$r_{a} (q, t)$ — particle $a$ ki Cartesian position, coordinates aur (shayad) clock $t$ ke function ke roop mein likhi gayi.
$\partial_{q_{i}} r_{a} \equiv \frac{\partial r _{a}}{\partial q _{i}}$ — particle $a$ ki position kitni shift hoti hai jab aap coordinate $q_{i}$ ko nudge karo aur baaki sab freeze karo. Yeh ek vector hai jo us direction mein point karta hai jisme woh coordinate particle ko move karta hai.
$\partial_{t} r_{a} \equiv \frac{\partial r _{a}}{\partial t}$ — position kitni move hoti hai kyunki constraint khud clock ke saath drag ho rahi hai, saare $q_{i}$ frozen hain. Ek stationary track ke liye zero.
$T$ — kinetic energy $= \sum_{a} \frac{1}{2} m_{a} ∣ \dot{r}_{a} ∣^{2}$ , split hoti hai $T = T_{2} + T_{1} + T_{0}$ mein.
$M_{ij} = \sum_{a} m_{a} \partial_{q_{i}} r_{a} \cdot \partial_{q_{j}} r_{a}$ — mass (metric) matrix.
$a_{i} = \sum_{a} m_{a} \partial_{q_{i}} r_{a} \cdot \partial_{t} r_{a}$ — linear-term coefficient: coordinate $q_{i}$ particle ko jis direction mein push karta hai aur moving constraint us direction mein jisko drag karta hai, unka overlap (dot product). Nonzero tabhi hoga jab ek coordinate-motion aur ek clock-motion dono ho aur perpendicular na hon. Tab $T_{1} = \sum_{i} a_{i} \overset{q}{˙}_{i}$ .
$V$ — potential energy $V (q)$ : forces (gravity, springs) mein store hoti position-only energy. Isme koi velocities nahi hoti.
$L = T - V$ — Lagrangian: kinetic minus potential (dekho Lagrangian mechanics).
$p_{i} = \frac{\partial L}{\partial q ˙ _{i}}$ — $q_{i}$ ke conjugate generalized momentum (dekho Generalized momentum and conserved quantities).
$H = \sum_{i} p_{i} \overset{q}{˙}_{i} - L$ — energy function / Hamiltonian (dekho Energy function and Hamiltonian).

Shuru karne se pehle, teen anchors jo aapko apni jeb mein rakhne chahiye (sab parent note mein banaye gaye hain):

Scleronomic = position map $r_{a} (q)$ mein koi explicit $t$ nahi — "track" ko kheencha nahi ja raha. Rheonomic = $r_{a} (q, t)$ mein $t$ hai — koi constraint apne schedule pe khud move karta hai.
Split $T = T_{2} + T_{1} + T_{0}$ count karta hai kitne velocity factors $\overset{q}{˙}$ har piece carry karta hai: 2 (quadratic), 1 (linear), 0 (velocity-free).
Mass matrix $M_{ij}$ symmetric, positive-definite hai, aur generally coordinates $q$ pe depend karta hai.

Formula Non-trivial cancellation

$H = T_{2} - T_{0} + V$ — assert nahi, dikhaya gaya hai Neeche kai traps is baat pe hinge karte hain ki kyun $H = T + V$ sirf scleronomic systems ke liye hai. Yahan poori chain hai taaki aap ise kabhi faith pe na lo.

Definition se shuru karo $H = \sum_{i} p_{i} \overset{q}{˙}_{i} - L$ jahan $p_{i} = \partial L / \partial \overset{q}{˙}_{i}$ aur $L = T - V$ .
Kyunki $V = V (q)$ mein koi velocities nahi hain, $\partial L / \partial \overset{q}{˙}_{i} = \partial T / \partial \overset{q}{˙}_{i}$ , isliye $\sum_{i} p_{i} \overset{q}{˙}_{i} = \sum_{i} \overset{q}{˙}_{i} \partial T / \partial \overset{q}{˙}_{i}$ .
$T = T_{2} + T_{1} + T_{0}$ pe Euler-degree count apply karo: $T_{2}$ ka degree $\overset{q}{˙}$ mein 2 hai, $T_{1}$ ka 1, $T_{0}$ ka 0, isliye $\sum_{i} \overset{q}{˙}_{i} \frac{\partial T}{\partial q ˙ _{i}} = 2 T_{2} + 1 \cdot T_{1} + 0 \cdot T_{0} = 2 T_{2} + T_{1} .$
Isliye $H = (2 T_{2} + T_{1}) - (T_{2} + T_{1} + T_{0} - V) = T_{2} - T_{0} + V .$
Scleronomic case: $T_{1} = T_{0} = 0$ aur $T = T_{2}$ , isliye $H = T_{2} + V = T + V$ . Rheonomic case: $- T_{0}$ (aur gaayab hua $T_{1}$ ) identity ko kharaab kar deta hai, isliye generally $H \neq = T + V$ .

True or false — justify karo

Kinetic energy hamesha generalized velocities ka quadratic function hoti hai.

False. Sirf scleronomic constraints ke liye. Jab

\partial_{t} r_{a} \neq = 0

ho tab linear piece

T_{1} = \sum_{i} a_{i} \overset{q}{˙}_{i}

aur velocity-free piece

T_{0}

appear hote hain, isliye

T

\overset{q}{˙}

mein ek general degree-2 polynomial hai, homogeneous quadratic nahi.

M_{ij}

ko ek constant matrix maana ja sakta hai jab ek baar coordinates choose ho jayein.

False. Yeh

M_{ij} (q)

hai — e.g. polar coordinates mein

M_{θ θ} = m r^{2}

r

ke saath badhta hai. Yeh dependence tab differentiate karni padti hai jab aap Euler–Lagrange equation mein

\partial T / \partial q

form karte ho.

Mass matrix hamesha symmetric hoti hai.

True.

M_{ij} = \sum_{a} m_{a} \partial_{q_{i}} r_{a} \cdot \partial_{q_{j}} r_{a}

ek dot product se bana hai, jo

i

aur

j

mein symmetric hai, isliye indices swap karne pe identical expression milta hai.

Mass matrix ka ek physical system ke liye zero eigenvalue ho sakta hai.

False.

M

positive-definite hai:

\sum_{ij} M_{ij} \overset{q}{˙}_{i} \overset{q}{˙}_{j} = \sum_{a} m_{a} ∣ \dot{r}_{a} ∣^{2} \geq 0

, aur yeh zero tabhi hota hai jab har

\overset{q}{˙}_{i}

zero Cartesian velocity produce kare — matlab woh coordinate ek real degree of freedom hi nahi hai.

Ek scleronomic system ke liye, energy

T + V

ke barabar hoti hai.

True. Scleronomic ka matlab hai

T = T_{2}

ek homogeneous quadratic hai, isliye

\sum_{i} \overset{q}{˙}_{i} \partial T / \partial \overset{q}{˙}_{i} = 2 T_{2}

aur chain

H = \sum_{i} p_{i} \overset{q}{˙}_{i} - L

(boxed derivation dekho)

T_{2} + V = T + V

pe collapse ho jaati hai, jahan

p_{i} = \partial L / \partial \overset{q}{˙}_{i}

aur

L = T - V

Agar ek constraint time-dependent hai, phir bhi total energy

H

T + V

ke barabar hai.

False.

T_{1}, T_{0}

present hone par upar wali derivation deti hai

H = T_{2} - T_{0} + V \neq = T + V

generally. Moving constraint energy andar ya bahar feed kar sakta hai.

T_{1}

ka zero hona guarantee karta hai ki constraint scleronomic hai.

False. Rotating-wire bead mein

T_{1} = 0

hai lekin

T_{0} = \frac{1}{2} m r^{2} ω^{2} \neq = 0

; yeh fully rheonomic hai.

T_{1}

tab vanish hota hai jab

\partial_{q_{i}} r_{a} ⊥ \partial_{t} r_{a}

ho (unka dot product

a_{i} = 0

), jo tab bhi ho sakta hai jab time explicitly map mein ho.

T_{0}

ek constant hai (state se independent).

False.

T_{0} = \frac{1}{2} \sum_{a} m_{a} (\partial_{t} r_{a})^{2}

coordinates

q

pe depend karta hai (e.g.

\frac{1}{2} m r^{2} ω^{2}

r

ke saath badhta hai). Yeh "constant" sirf is sense mein hai ki velocities

\overset{q}{˙}

se free hai, position se nahi.

Alag generalized coordinates choose karne se ek given physical state pe

T

ki numerical value badal jaati hai.

False.

T

ek physical scalar hai (

\sum_{a} \frac{1}{2} m_{a} v_{a}^{2}

); ek real instant pe iski value coordinate-independent hai. Jo badlta hai woh formula hai — entries

M_{ij}

ek metric tensor ki tarah transform hote hain, lekin unka compute kiya hua number fixed rehta hai.

Kisi bhi constraint ko generalized coordinates ki choice mein absorb kiya ja sakta hai.

False. Sirf holonomic constraints (positions aur time ke beech relations,

f (r_{a}, t) = 0

) ko coordinates eliminate karne ke liye solve kiya ja sakta hai. Ek non-holonomic constraint — jo velocities ko baandhta hai aur ek position relation mein integrate nahi ho sakta, e.g. ek rolling coin ka

\overset{x}{˙} = R \dot{ϕ} cos ψ

— is tarah remove nahi ho sakta, aur is poore page ka clean

T (q, \overset{q}{˙})

reduction maanta hai ki yeh absent hai.

Error dhundho

"Polar coordinates mein

T = \frac{1}{2} m \overset{r}{˙}^{2} + \frac{1}{2} m \dot{θ}^{2}

θ

wala term galat hai:

M_{θ θ} = m r^{2}

hai,

m

nahi, isliye hona chahiye

\frac{1}{2} m r^{2} \dot{θ}^{2}

. Factor

r^{2}

isliye aata hai kyunki

\partial_{θ} r = (- r sin θ, r cos θ)

ki length

r

hai,

1

nahi.

"Polar

T

mein ek cross term

M_{r θ} \overset{r}{˙} \dot{θ}

hai, isliye main ise add karunga."

Pehle do basis vectors yaad karo:

r

ko nudge karne se milta hai

\partial_{r} r = (cos θ, sin θ)

(baahir point karta unit vector),

θ

ko nudge karne se milta hai

\partial_{θ} r = (- r sin θ, r cos θ)

(length

r

, sideways point karta). Unka dot product hai

M_{r θ} = m [(cos θ) (- r sin θ) + (sin θ) (r cos θ)] = 0

— baahri

⊥

sideways hai, isliye cross term genuinely vanish karta hai. Yeh sirf non-orthogonal coordinate systems mein nonzero hota hai.

"Kyunki

\sum_{i, j}

pair

(1, 2)

aur

(2, 1)

ko do baar count karta hai, isliye main off-diagonal coefficient ko 2 se multiply karta hoon jab main ise read karta hoon."

T_{2} = \frac{1}{2} \sum_{i, j} M_{ij} \overset{q}{˙}_{i} \overset{q}{˙}_{j}

mein

\frac{1}{2}

pehle se hi woh double-counting absorb kar chuka hai;

T

mein

\overset{q}{˙}_{1} \overset{q}{˙}_{2}

ka coefficient

M_{12}

hai (

\frac{1}{2} M_{12} + \frac{1}{2} M_{21} = M_{12}

se), isliye dobara 2 se multiply karna double-counting hai.

"Rotating wire pe bead ek straight line mein move karta hai, isliye

T = \frac{1}{2} m \overset{r}{˙}^{2}

Yeh

T_{0} = \frac{1}{2} m r^{2} ω^{2}

bhool jaata hai. Ek bead jo momentarily wire pe still bhi hai (

\overset{r}{˙} = 0

) wire ke imposed spin se room mein ghoom raha hota hai, isliye uski real Cartesian speed

r ω

hoti hai.

\dot{r}_{a} = \sum_{i} \partial_{q_{i}} r_{a} \overset{q}{˙}_{i}

— chain rule complete hai."

Explicit clock term

\partial_{t} r_{a}

missing hai. Yeh tab present hota hai jab map rheonomic ho; ise silently drop karna

T_{1}

aur

T_{0}

dono ko phek deta hai.

"Kyunki

M

ke units mass ke hain, iski determinant ek mass hai."

Generally yeh nahi hoti: ek

n \times n

metric ke liye iski determinant ke units hote hain

(mass)^{n}

times coordinates ke jo bhi length powers hote hain (e.g. polar mein

det M = m \cdot m r^{2} = m^{2} r^{2}

). Har entry

M_{ij}

alag length dimensions carry kar sakti hai depending on whether

q_{i}

ek length hai ya angle.

T_{0}

mein koi velocities nahi hain, isliye yeh dynamics mein koi role nahi nibhata."

T_{0} (q)

Lagrangian mein enter karta hai aur

\partial T_{0} / \partial q

Euler–Lagrange equation mein ek real force produce karta hai — precisely rotating wire ke liye centrifugal term. Ek velocity-free term dynamically inert term nahi hota.

"Linear coefficient

a_{i}

sirf

M_{ij}

ke ek component ka doosra naam hai."

Nahi —

M_{ij} = \sum_{a} m_{a} \partial_{q_{i}} r_{a} \cdot \partial_{q_{j}} r_{a}

do coordinate gradients ko pair karta hai, jabki

a_{i} = \sum_{a} m_{a} \partial_{q_{i}} r_{a} \cdot \partial_{t} r_{a}

ek coordinate gradient ko clock gradient ke saath pair karta hai.

a_{i}

tabhi exist karta hai jab constraint move kare;

M_{ij}

hamesha exist karta hai.

Why questions

Chain rule

\dot{r}_{a}

ke liye ek explicit

\partial_{t} r_{a}

term kyun acquire karta hai, aur kabhi kabhi hi kyun?

Kyunki

r_{a} (q, t)

t

pe depend kar sakta hai dono moving

q_{i} (t)

ke through aur directly clock ke through. Direct term

\partial_{t} r_{a}

tabhi survive karta hai jab ek constraint apne khud ke time schedule pe imposed ho (rheonomic); ek scleronomic map mein

\partial_{t} r_{a} = 0

hota hai.

T

\overset{q}{˙}

ki powers ke hisaab se group karna "natural" classification kyun hai?

Kyunki

\dot{r}_{a}

velocities mein linear hai plus ek velocity-free clock term, isliye ise square karne par sirf

2

1

, ya

0

velocity factors wale products aa sakte hain. Kisi aur tarah ka term possible hi nahi, aur har group ka ek alag physical meaning hai (kinetic, coupling, effective-potential).

Coefficient

a_{i}

ek "geometric overlap" kyun measure karta hai?

Kyunki

a_{i} = \sum_{a} m_{a} \partial_{q_{i}} r_{a} \cdot \partial_{t} r_{a}

ek dot product hai us direction ke beech jisme coordinate

q_{i}

particle ko push karta hai aur us direction ke beech jisme moving constraint use drag karta hai. Jab yeh perpendicular hon to dot product zero hota hai, isliye

T_{1}

disappear ho jaata hai chahe constraint move kar raha ho — exactly rotating-wire case.

Mass matrix "configuration space pe metric" ka naam deserve kyun karta hai?

Kyunki

2 T_{2} = \sum_{ij} M_{ij} \overset{q}{˙}_{i} \overset{q}{˙}_{j}

configuration point ki squared "speed" measure karta hai, exactly jaisi ek metric

g_{ij} \overset{q}{˙}^{i} \overset{q}{˙}^{j}

ek curved space pe distance-rate measure karta hai. System ki motion is metric mein geodesic-like motion hai (dekho Configuration space and the metric tensor).

Count

\sum_{i} \overset{q}{˙}_{i} \partial T / \partial \overset{q}{˙}_{i}

2 T

ki jagah

2 T_{2} + T_{1}

kyun deta hai?

Kyunki woh sum har piece ko uski

\overset{q}{˙}

mein degree se weight karta hai:

T_{2}

(degree 2) contribute karta hai

2 T_{2}

T_{1}

(degree 1) contribute karta hai

T_{1}

, aur

T_{0}

(degree 0) kuch nahi contribute karta. Sirf jab

T = T_{2}

akela ho tab yeh

2 T

ke barabar hota hai — jo exactly scleronomic condition hai

H = T + V

ke peeche.

M

merely non-negative ki jagah guaranteed positive-definite kyun hai?

Kyunki har generalized coordinate, by construction, ek independent degree of freedom hai, isliye koi nonzero velocity vector

\overset{q}{˙}

har Cartesian velocity ko zero nahi chhod sakta. Isliye

\overset{q}{˙} \neq = 0

ke liye strictly

\sum_{a} m_{a} ∣ \dot{r}_{a} ∣^{2} > 0

hota hai.

Edge cases

Cartesian coordinates mein ek free particle:

T_{2}, T_{1}, T_{0}

kya hain?

T = T_{2} = \frac{1}{2} m (\overset{x}{˙}^{2} + \overset{y}{˙}^{2} + \overset{z}{˙}^{2})

jahan

M_{ij} = m δ_{ij}

(constant, diagonal);

T_{1} = T_{0} = 0

kyunki map

r = (x, y, z)

mein koi explicit

t

nahi hai. Yeh degenerate "flat metric" case hai.

Aap kaise detect karte ho ki ek chosen coordinate redundant hai, aur ek concrete example kya hai?

Redundancy $M$ ke zero eigenvalue ke roop mein dikhti hai: koi nonzero velocity vector

\overset{q}{˙}

zero Cartesian motion produce karta hai, isliye

det M = 0

. Concretely, ek circle pe ek particle ko dono angle

θ

aur arc length

s = R θ

se describe karna

\overset{s}{˙} - R \dot{θ} = 0

deta hai jo hamesha koi independent motion produce nahi karta;

M

singular ho jaata hai. Fix hai sirf ek rakhna —

s

(ya

θ

) drop karo taaki coordinate count true degrees of freedom se match kare.

Limit

ω \to 0

mein rotating wire: kya answer sahi reduce hota hai?

Haan —

T_{0} = \frac{1}{2} m r^{2} ω^{2} \to 0

aur

T \to \frac{1}{2} m \overset{r}{˙}^{2}

, scleronomic straight-wire result. Rheonomic terms smoothly switch off ho jaate hain jab imposed motion ruk jaata hai, exactly jaisa hona chahiye.

\overset{r}{˙} = 0

wala rotating wire (wire pe momentarily pinned bead): kya

T = 0

hai?

Nahi —

T = \frac{1}{2} m r^{2} ω^{2}

T_{0}

se. Zero generalized velocity ka matlab zero physical velocity nahi hota jab constraint khud move kar raha ho; bead

r ω

speed se ghoomaya ja raha hota hai.

Ek pendulum of length

ℓ

jiska pivot

X_{p} = A cos Ω t

ke roop mein horizontally driven hai —

T_{2}, T_{1}, T_{0}

explicitly likhein.

x = A cos Ω t + ℓ sin θ, y = - ℓ cos θ

ke saath, differentiate karo:

\partial_{θ} r = (ℓ cos θ, ℓ sin θ)

aur

\partial_{t} r = (- A Ω sin Ω t, 0)

. Isliye

T_{2} = \frac{1}{2} m ℓ^{2} \dot{θ}^{2}

a_{θ} = m \partial_{θ} r \cdot \partial_{t} r = - m A Ω ℓ cos θ sin Ω t

deta hai

T_{1} = - m A Ω ℓ cos θ sin Ω t \dot{θ}

, aur

T_{0} = \frac{1}{2} m A^{2} Ω^{2} sin^{2} Ω t

. Dono survive karte hain kyunki driven pivot

t

ko explicitly

r (θ, t)

mein daalta hai aur

\partial_{θ} r

\partial_{t} r

ke perpendicular nahi hai.

Kya

T_{2}

kabhi ek moving system ke liye zero ho sakta hai?

Sirf instantaneously, jab saare

\overset{q}{˙}_{i} = 0

hon; ek form ke roop mein

T_{2}

kabhi identically zero nahi ho sakta kyunki

M

positive-definite hai. Isliye genuinely moving configuration (

\overset{q}{˙} \neq = 0

) mein hamesha

T_{2} > 0

hota hai.

Agar do independent coordinates hamesha orthogonal Cartesian gradients dete hain, toh

M

kaisa dikhta hai?

Yeh diagonal hota hai,

M_{ij} = 0

for

i \neq = j

, isliye

T_{2} = \frac{1}{2} \sum_{i} M_{ii} \overset{q}{˙}_{i}^{2}

mein koi cross terms nahi hote — polar case (

M_{r θ} = 0

) exactly yahi hai. Orthogonal coordinates metric ko diagonalize karte hain.

Kya yeh poori

T = T_{2} + T_{1} + T_{0}

machinery ek rolling wheel pe apply hoti hai jis par

\overset{x}{˙} = R \dot{ϕ}

constraint hai?

Directly nahi — yeh ek non-holonomic (velocity-level) constraint hai jo ek position relation mein integrate nahi ho sakta, isliye aap ek plain

T (q, \overset{q}{˙})

banane ke liye ek coordinate eliminate nahi kar sakte. Aapko constraint alag carry karna padega (Lagrange multipliers); yahan decomposition sirf holonomic constraints maanke chalta hai.

Recall Ek-line survival summary

"Map mein time ⇒ $T_{1}, T_{0}$ jaag jaate hain; $M$ ek position-dependent symmetric positive-definite metric hai; energy $= T + V$ sirf tab jab track move na kare." Trap detector ::: Hamesha poochho "Kya $t$ explicitly $r_{a} (q, t)$ mein hai?" — yeh akela sawaal decide karta hai scleronomic vs rheonomic, aur isliye yeh bhi ki $T_{1}, T_{0}$ aur $H = T + V$ shortcut apply hote hain ya nahi.