Relativistic momentum p = γmv: Question bank — Hinglish deep dive

2.3.31 · D5 · Physics › Modern Physics › Relativistic momentum p = γmv

Shuru karne se pehle, teen symbols jinhe hum har jagah use karte hain:

$β = v / c$ — speed ko light speed ke fraction mein likha. $β = 0$ matlab rest mein, $β = 1$ matlab light speed par chal raha hai. Yeh ek pure number hai jiska koi unit nahi.
$γ = 1/ 1 - β^{2}$ — Lorentz factor. Yeh hamesha $\geq 1$ hota hai aur $β \to 1$ hone par infinity tak blow up hota hai. Isse ek "stubbornness dial" samjho.
$m$ — invariant rest mass, particle ka ek fixed label jis par har observer agree karta hai. Yeh speed ke saath kabhi nahi badlta.

True or false — justify karo

p = γ m v

ka value

m c

se bada ho sakta hai.

True.

p = γ β m c

, aur

γ β

bina kisi limit ke badhta jaata hai jab

β \to 1

, isliye momentum ki koi upper limit nahi, chahe speed

c

par cap ho.

v = 0

par relativistic momentum, Newtonian momentum ke barabar hoti hai.

True. Rest mein

β = 0

toh

γ = 1

, aur dono formulas se

p = m v = 0

milta hai — relativity aur Newton ko agree karna hi chahiye jab kuch bhi move nahi kar raha.

Speed double karne se momentum hamesha double ho jaati hai.

False. Yeh sirf Newtonian limit mein hota hai. Kyunki

γ

khud speed ke saath badhta hai,

v

double karne par

p

high speeds par do se zyada factor se badhti hai.

Extra factor

γ

ka matlab hai particle ki mass badh gayi.

False. Mass

m

invariant rest mass hai aur fixed rehti hai;

γ

velocity/kinematics par baitha hai.

γ m

ko "mass" kehne se contradictions paida hote hain (kyunki yeh force ki direction par depend karta).

Photon ki momentum zero hoti hai kyunki uski mass zero hai.

False.

m = 0

ke liye

γ m v

use nahi kar sakte (yeh

0 \times \infty

hoga), isliye

E = p c

use karte hain — photon zero mass ke bawajood

p = E / c

momentum carry karta hai.

Kisi bahut fast particle ke liye

γ

ka value

0.5

ho sakta hai.

False. Har real speed ke liye

γ \geq 1

hota hai, kyunki

1 - β^{2} \leq 1

hai toh uska reciprocal

\geq 1

hoga. 1 se neeche ka value matlab hai tumne formula ulta laga diya.

Relativistic momentum phir bhi velocity ki direction mein point karne wala vector hai.

True.

p = γ m v

mein velocity vector ko positive scalar

γ m

se multiply kiya jaata hai, isliye yeh

v

ki same direction mein rehta hai.

γ

correction ke saath momentum har inertial frame mein conserved hoti hai.

True. Yahi poori wajah hai redefinition ki — displacement ko invariant proper time se divide karna

p

ko cleanly transform karta hai, isliye ek frame mein conservation doosre sab mein bhi guarantee hoti hai.

Error dhundo

γ = 1 - v^{2} / c^{2}

, isliye moving particles ke liye

γ < 1

hoga."

Error hai formula ulta likhna.

γ

reciprocal hai:

γ = 1/ 1 - v^{2} / c^{2} \geq 1

. Akela square root time-dilation shrink factor hai,

γ

nahi.

v = 0.9 c

ke liye hum phir bhi

p = m v

use kar sakte hain kyunki yeh kaafi close hai."

β = 0.9

par

γ \approx 2.29

hai, toh Newton momentum ko aadhe se zyada underestimate karta hai.

γ

correction ek baar

v ≳ 0.1 c

ho jaane par strongly matter karta hai.

"Kyunki

p \to \infty

jab

v \to c

, toh

c

par momentum undefined ho jaati hai."

Massive particle ke liye

v = c

simply unreachable hai —

γ

wahan pahunchne se pehle diverge ho jaata hai. Momentum har allowed speed

v < c

par finite rehti hai; ise kabhi

c

par evaluate nahi karna padta.

p = γ m v

toh

p = γ \times m \times v = 1 - β^{2} \cdot m v

Writer ne

1 - β^{2}

γ

ki jagah substitute kar diya, jo ulta hai. Sahi:

p = m v / 1 - β^{2}

, momentum badhti hai, ghatती nahi.

"Hum position ko lab time

d t

se differentiate karke 4-momentum banate hain."

Galat clock. Hum proper time

d τ

se divide karte hain (particle ke saath chal raha clock) kyunki

d τ

Lorentz-invariant hai jabki

d t

frame-dependent hai; wahi invariance momentum ko properly transform karta hai.

E = p c

sab particles ke liye valid hai."

Sirf massless particles jaise photons ke liye. Massive particles ke liye puri relation

E^{2} = (p c)^{2} + (m c^{2})^{2}

hai;

E = p c

woh special case hai jahan

(m c^{2})^{2}

term zero ho jaati hai.

"Kyunki

γ \geq 1

hai, relativistic momentum hamesha kam se kam

m c

hogi."

Nahi. Kam speed par

γ \approx 1

lekin

v

bahut choti ho sakti hai, isliye

p = γ m v

bhi bahut chhoti hogi.

γ

\geq 1

hona

p

par koi floor nahi lagata; velocity factor phir bhi zero tak shrink ho sakta hai.

Why questions

Momentum define karne ke liye lab time

t

ki jagah proper time

τ

kyun use karte hain?

Proper time sab observers ke liye same hota hai (invariant), isliye

d τ

se displacement ek proper 4-vector ki tarah transform hoti hai — yahi cheez frames ke across momentum conservation ko bachati hai.

Momentum speed ke saath linear se zyada tezi se kyun badhti hai?

Kyunki

p = γ m v

mein extra factor

γ

hota hai, jo khud

v

ke saath badhta hai; toh jab speed badhti hai,

v

aur

γ

dono milkar

p

ko upar push karte hain.

Massive particle kabhi light speed tak kyun nahi pahunch sakta?

v = c

tak pahunchne ke liye

γ \to \infty

chahiye, matlab infinite momentum aur infinite energy — physically impossible, isliye

c

ek unattainable ceiling hai.

Formula everyday speeds par

p = m v

mein kyun reduce ho jaata hai?

v ≪ c

ke liye

β \to 0

toh

γ \to 1

, aur

p \to m v

reh jaata hai. Har sahi relativistic law ko low speed par Newton ke paas collapse karna chahiye — correspondence principle.

Mass ko

γ

ke saath badhne dene ki bajaye invariant kyun rakhte hain?

Invariant

m

ek clean, frame-independent label hai jo relativity mein sab jagah consistent hai;

γ

ko velocity par rakhne se force laws mein direction-dependent "mass" ke contradictions avoid hote hain.

Relativity mein energy aur momentum ko saath kyun treat karna padta hai?

Yeh ek 4-momentum ke time aur space parts hain jiska invariant length

m c

hai; boost inhe mix karta hai, isliye ek ko doosre ke bina transform nahi kar sakte — isliye

E^{2} = (p c)^{2} + (m c^{2})^{2}

Momentum problems mein

γ β

(proper velocity) aksar sabse clean quantity kyun hoti hai?

Kyunki

p = γ β m c

hai, toh

p

m c

units mein set karne par problem sirf single combination

γ β

solve karne ki reh jaati hai, ek messy nested square root se bachte hue.

Edge cases

Case

v = 0

γ

aur

p

kya honge?

γ = 1

aur

p = 0

— stationary particle ki koi momentum nahi, bilkul Newton se match karta hai.

Case

m = 0

(photon): kya

p = γ m v

apply hoti hai?

Nahi — yeh indeterminate

0 \times \infty

deta hai. Sahi raasta energy relation hai, jis se massless particle ke liye

p = E / c

milta hai jo

c

par move karta hai.

Case

v \to c

massive particle ke liye:

p

ka limiting behaviour kya hoga?

γ \to \infty

toh

p \to \infty

. Momentum diverge hoti hai, aur yahi precise wajah hai ki light speed mass wali kisi bhi cheez ke liye hard limit hai.

Case negative velocity (

- x

direction mein jaata particle):

p

ka kya hoga?

γ

sirf

v^{2}

par depend karta hai isliye unchanged rehta hai, lekin

p = γ m v

- x

direction mein point karti hai — momentum ka sign velocity ke sign ko follow karta hai jabki magnitude unaffected rehti hai.

Case

β

thoda 1 se upar (faster than light): kya

γ

real hai?

Nahi.

1 - β^{2}

negative ho jaata hai, toh

γ

imaginary ban jaata hai — ek mathematical signal ki real particles ke liye superluminal speed physically allowed nahi hai.

Case do particles, same speed, ek ki mass doosre se double: momentum ratio kya hoga?

Exactly 2.

γ

aur

v

identical hain, isliye

p

linearly

m

ke saath scale karta hai; relativistic corrections ratio mein cancel ho jaate hain.

Recall Har trap ki ek-line summary

$γ \geq 1$ hamesha, velocity par baitha hai mass par nahi; $p$ ki koi ceiling nahi chahe $v$ ki ho; photons ke liye $γ m v$ ki jagah $E = p c$ use karo; aur sab kuch Newton mein reduce hota hai jab $v ≪ c$ .

Connections

Lorentz factor and time dilation — $γ \geq 1$ kyun hai aur $d t = γ d τ$ kahaan se aata hai
Why nothing exceeds the speed of light — $γ \to \infty$ edge case depth mein
Photon momentum and radiation pressure — $m = 0$ trap resolved
Mass-energy equivalence E = γmc² — energy partner $E^{2} = (p c)^{2} + (m c^{2})^{2}$ ke peeche
Energy-momentum four-vector — $E$ aur $p$ saath kyun chalte hain
Newtonian momentum p = mv — low-speed limit jis par yeh traps baar baar laute hain