Relativistic momentum p = γmv: Worked examples — Hinglish deep dive

2.3.31 · D3 · Physics › Modern Physics › Relativistic momentum p = γmv

Intuition Yeh page kis liye hai

> Parent note Relativistic momentum $p=\gamma m v$ ne tumhe formula aur teen examples diye. Yahan hum har tarah ki situation dhundhte hain jo formula tumhare saamne rakh sakta hai — slow speeds, near-light speeds, exact $v = 0$ aur $v \to c$ ki edges, massless photon, ek real-world accelerator problem, aur ek exam-style twist. Har ek pehle symbol se work out kiya gaya hai taaki tumhe koi aisa case na mile jo tumne pehle dekha na ho.

Shuru karne se pehle, chalte hain un do symbols ko dobara samjhte hain jinpe hum poore page mein rely karte hain:

$β$ (Greek "beta") sirf speed ka ek nickname hai jo light-speed ki units mein measure hoti hai: $β = v / c$ . Agar tum half light-speed par move kar rahe ho, $β = 0.5$ . Iska koi unit nahi — yeh $0$ aur $1$ ke beech ek pure fraction hai.
$γ$ (Greek "gamma", Lorentz factor) ek "stubbornness multiplier" hai: $γ = \frac{1}{1 - β ^{2}}$ . Rest par $γ = 1$ ; yeh sirf badhta hi hai.

Scenario matrix

$p = γ m v$ ke baare mein har problem in cells mein se kisi ek mein aata hai. Neeche ke worked examples us cell ke saath tagged hain jo unhe cover karti hai, aur milke yeh poora table fill kar dete hain.

#	Case class	Kya special hai	Example
A	Everyday / slow ( $β ≲ 0.1$ )	$γ \approx 1$ , Newton almost exact	Ex 1
B	Zero speed ( $v = 0$ )	degenerate: $γ = 1$ , $p = 0$	Ex 2
C	Moderately relativistic ( $β \sim 0.5$ – $0.9$ )	$γ$ bahut zyada matter karta hai	Ex 3
D	Limit $v \to c$	$γ \to \infty$ , $p \to \infty$	Ex 4
E	Massless ( $m = 0$ , $v = c$ )	formula $0 \cdot \infty$ — $E = p c$ chahiye	Ex 5
F	Inverse problem (given $p$ , find $v$ )	$γ β$ invert karna padega	Ex 6
G	Real-world word problem (accelerator, energy)	momentum + energy ka mix	Ex 7
H	Exam twist / sign & direction (2-D, vector, negative component)	$γ$ speed use karta hai, signed velocity nahi	Ex 8

Ex 1 — Cell A: everyday slow speed

Worked example

$2 g$ ka pellet $300 m/s$ par fire kiya Iska relativistic momentum nikalo. Mass $m = 0.002 kg$ , $v = 300 m/s$ , $c = 3 \times 1 0^{8} m/s$ .

Forecast: pehle guess karo — kya $γ$ ki value $1$ se kisi bhi measurable tarike se alag hogi?

Step 1 — $β$ nikalo. Yeh step kyun? $γ$ sirf $β = v / c$ ke through speed par depend karta hai, isliye hum pehle light-speed ka fraction compute karte hain. $β = \frac{300}{3 \times 1 0 ^{8}} = 1 \times 1 0^{- 6}$

Step 2 — $γ$ nikalo. Yeh step kyun? $γ$ correction factor hai; hum test karte hain ki kya ise rakhna worth it hai. $γ = \frac{1}{1 - ( 1 0 ^{- 6} ) ^{2}} = \frac{1}{1 - 1 0 ^{- 12}} \approx 1 + \frac{1}{2} \times 1 0^{- 12}$ Yeh $1.0000000000005$ hai — $1$ se practically indistinguishable.

Step 3 — Momentum. Yeh step kyun? $p = γ m v$ mein plug karo; kyunki $γ \approx 1$ yeh sirf Newton jaisa hai. $p = γ m v \approx (1) (0.002) (300) = 0.6 kg\cdotm/s$

Verify: relativistic correction $\sim 5 \times 1 0^{- 13}$ hai, isliye $p = 0.6 kg\cdotm/s$ 12 decimal places tak. Yeh correspondence principle kaam kar raha hai — dekho Newtonian momentum p = mv। Everyday speeds par relativity invisible hai.

Ex 2 — Cell B: degenerate case $v = 0$

Worked example Ek particle jo bilkul still baitha hai

Ek proton at rest ( $v = 0$ ). $γ$ kya hai aur $p$ kya hai?

Forecast: kya ek stationary object ka koi momentum hota hai? Kya $γ$ blow up hoga ya vanish ho jayega?

Step 1 — $β$ compute karo. Yeh step kyun? Sab kuch $β$ se aata hai; yahan yeh exactly $0$ hai. $β = \frac{0}{c} = 0$

Step 2 — $γ$ compute karo. Yeh step kyun? Hume check karna hai ki formula boundary par misbehave toh nahi kar raha. $γ = \frac{1}{1 - 0} = \frac{1}{1} = 1$ Toh $γ$ apni smallest possible value, $1$ , leta hai. Yeh kabhi neeche nahi jaata.

Step 3 — Momentum. Yeh step kyun? Plug in karo. $p = γ m v = 1 \cdot m \cdot 0 = 0$

Verify: ek resting particle zero momentum carry karta hai — exactly jaisa Newton kehta hai. $γ$ machinery kuch bhi nahi se momentum nahi banati. Yeh poori kahani ki lower edge hai: $γ \geq 1$ , equality sirf $v = 0$ par.

Ex 3 — Cell C: moderately relativistic (picture ke saath)

Worked example Electron at

$0.6 c$ $m = 9.11 \times 1 0^{- 31} kg$ , $v = 0.6 c$ . $p$ nikalo aur Newtonian guess se compare karo.

Forecast: true momentum, Newtonian $m v$ se kitne percent zyada hoga?

Step 1 — $β = 0.6$ . Yeh step kyun? directly diya hua hai.

Step 2 — $γ$ . Yeh step kyun? yahan relativity enter karti hai. $γ = \frac{1}{1 - 0.36} = \frac{1}{0.64} = \frac{1}{0.8} = 1.25$

Step 3 — Momentum. Yeh step kyun? $γ \cdot m \cdot v$ multiply karo; $v = 0.6 \times 3 \times 1 0^{8} = 1.8 \times 1 0^{8} m/s$ use karo. $p = 1.25 \times 9.11 \times 1 0^{- 31} \times 1.8 \times 1 0^{8} = 2.05 \times 1 0^{- 22} kg\cdotm/s$

Step 4 — Newtonian comparison. Yeh step kyun? correction ki size dekhne ke liye. $p_{Newton} = m v = 9.11 \times 1 0^{- 31} \times 1.8 \times 1 0^{8} = 1.64 \times 1 0^{- 22} kg\cdotm/s$ Ratio exactly $γ = 1.25$ hai, matlab 25% zyada.

Verify: figure dekho — burnt-orange curve $p = γ m v$ , teal Newton line $p = m v$ se upar khiinchti hai jaise $β$ badhta hai. $β = 0.6$ par vertical gap wahi $25%$ boost hai. $p$ ki units poore time $kg\cdotm/s$ hain. ✓

Ex 4 — Cell D: limit $v \to c$

Worked example Light-speed ki taraf push karna

$β = 0.9$ , $0.99$ , $0.999$ par $γ$ compute karo. $p$ ka kya hota hai?

Forecast: kya $γ$ kisi finite ceiling tak pahunchta hai, ya infinity ki taraf bhaag jaata hai?

Step 1 — $γ$ ki table. Yeh step kyun? trend dekhne ke liye jaise $β \to 1$ . $γ (0.9) = \frac{1}{1 - 0.81} = \frac{1}{0.19} = 2.294$ $γ (0.99) = \frac{1}{1 - 0.9801} = \frac{1}{0.0199} = 7.089$ $γ (0.999) = \frac{1}{1 - 0.998001} = \frac{1}{0.001999} = 22.37$

Step 2 — Pattern padho. Yeh step kyun? denominator $1 - β^{2} \to 0$ jaise $β \to 1$ , isliye $γ \to \infty$ .

Step 3 — Momentum consequence. Yeh step kyun? $p = γ m v$ mein $γ \to \infty$ ka matlab $p \to \infty$ hai. Koi bhi finite push infinite momentum supply nahi kar sakta, isliye ek massive particle kabhi $c$ tak nahi pahunch sakta — dekho Why nothing exceeds the speed of light।

Verify: figure mein plum curve $β = 1$ ke paas vertically shoot up karta hai (dotted asymptote). $(1 - β)$ ke har tenfold squeeze se $γ$ roughly triple hota hai — yeh ek genuine blow-up hai, koi plateau nahi. ✓

Ex 5 — Cell E: massless case ( $m = 0$ )

Worked example Photon ka momentum

Ek photon ka rest mass $m = 0$ hai aur yeh hamesha $v = c$ par travel karta hai. Iska momentum kya hai?

Forecast: formula deta hai $p = γ m v = \infty \times 0 \times c$ . Kya yeh $0$ hai, $\infty$ hai, ya undefined?

Step 1 — Trouble spot karo. Yeh step kyun? $v = c$ par, $γ = 1/ 1 - 1 = 1/0 = \infty$ ; $m = 0$ ke saath hume indeterminate product $\infty \cdot 0$ milta hai. Light ke liye formula $p = γ m v$ directly use nahi ho sakta.

Step 2 — Tool switch karo energy–momentum relation par. Yeh step kyun? parent note ka invariant $E^{2} = (p c)^{2} + (m c^{2})^{2}$ tab bhi valid rehta hai jab $m = 0$ , kyunki yeh kabhi $1 - β^{2}$ se divide nahi karta. $E^{2} = (p c)^{2} + (0)^{2} \Rightarrow E = p c \Rightarrow p = \frac{E}{c}$

Step 3 — Numbers. Yeh step kyun? ek green photon jiska $E = 3.0 \times 1 0^{- 19} J$ hai: $p = \frac{3.0 \times 1 0 ^{- 19}}{3 \times 1 0 ^{8}} = 1.0 \times 1 0^{- 27} kg\cdotm/s$

Verify: light zero mass ke bawajood momentum carry karti hai — yahi Photon momentum and radiation pressure drive karta hai. Units: $J / (m/s) = kg\cdotm^{2} s^{- 2} / (m/s) = kg\cdotm/s$ ✓. Indeterminate form resolve hua us tool ko choose karke jo finite rehta hai.

Ex 6 — Cell F: inverse problem (given $p$ , find $v$ )

Worked example Stated momentum ke liye speed nikalo

Ek particle ka $p = 3 m c$ hai. $β$ aur $γ$ nikalo.

Forecast: guess karo — kya $β$ $0.9$ se upar hai ya neeche?

Step 1 — Clean unknown $γ β$ introduce karo. Yeh step kyun? $p = γ m v = γ β m c$ likho. $m c$ se divide karne par pure-number combination $γ β$ isolate hota hai, jo solve karne ke liye sabse aasaan unknown hai. (Names ke baare mein dhyan rakho: dimensionful quantity $γ v = γ β c$ ko physicists proper velocity kehte hain; hamaara $γ β$ simply woh proper velocity hai jo $c$ ki units mein measure ki gayi hai.) $γ β = \frac{p}{m c} = 3$

Step 2 — Square karo aur $γ^{2}$ expand karo. Yeh step kyun? squaring se $γ$ ke andar ka square root hat jaata hai; $γ^{2} = 1/ (1 - β^{2})$ use karo. $γ^{2} β^{2} = 3 \Rightarrow \frac{β ^{2}}{1 - β ^{2}} = 3$

Step 3 — $β$ ke liye solve karo. Yeh step kyun? algebra. $β^{2} = 3 - 3 β^{2} \Rightarrow 4 β^{2} = 3 \Rightarrow β^{2} = \frac{3}{4}$ Square root lete hain toh $β = \pm \frac{3}{2}$ milta hai. Hum negative root discard karte hain: by definition $β = v / c$ ek speed fraction hai, isliye $β \geq 0$ . Isliye $β = \frac{3}{2} \approx 0.866$

Step 4 — $γ$ back out karo. Yeh step kyun? ab $β$ known hai. $γ = \frac{1}{1 - 3/4} = \frac{1}{1/4} = 2$ Check: $γ β = 2 \times \frac{3}{2} = 3$ ✓.

Verify: $β \approx 0.866$ ( $0.9$ se neeche, jaise tumne guess kiya hoga), $γ = 2$ . Wapas plug karne par: $p = γ β m c = 3 m c$ — diye gaye se match karta hai. ✓

Ex 7 — Cell G: real-world accelerator problem

Worked example LHC-style beam mein ek proton

Ek proton ( $m = 1.67 \times 1 0^{- 27} kg$ ) ko accelerate kiya jaata hai jab tak uski total energy $E = 7 GeV = 1.12 \times 1 0^{- 9} J$ nahi ho jaati. Uski rest energy $m c^{2} = 1.50 \times 1 0^{- 10} J$ ( $= 0.938 GeV$ ) hai. Iska momentum $p$ nikalo.

Forecast: kyunki $E$ rest energy se lagbhag $7.5 \times$ zyada hai, kya $p c$ bahut close to $E$ hoga ya usse bahut neeche?

Step 1 — Energy–momentum relation choose karo. Yeh step kyun? humein energy di gayi hai, speed nahi, isliye tool $E^{2} = (p c)^{2} + (m c^{2})^{2}$ (dekho Mass-energy equivalence E = γmc²) seedha $p$ tak pahunchta hai bina pehle $γ$ nikale. $(p c)^{2} = E^{2} - (m c^{2})^{2}$

Step 2 — Plug in karo. Yeh step kyun? diye gaye numbers ke saath arithmetic. $(p c)^{2} = (1.12 \times 1 0^{- 9})^{2} - (1.50 \times 1 0^{- 10})^{2} = 1.254 \times 1 0^{- 18} - 2.25 \times 1 0^{- 20}$ $(p c)^{2} = 1.232 \times 1 0^{- 18} J^{2} \Rightarrow p c = 1.110 \times 1 0^{- 9} J$

Step 3 — $c$ se divide karo. Yeh step kyun? $p c$ ki energy units hain; momentum pane ke liye $c$ se divide karo. $p = \frac{1.110 \times 1 0 ^{- 9}}{3 \times 1 0 ^{8}} = 3.70 \times 1 0^{- 18} kg\cdotm/s$

Verify: $p c = 1.110 \times 1 0^{- 9} J$ , $E = 1.12 \times 1 0^{- 9} J$ se sirf $\sim 0.9%$ neeche hai — toh proton ultra-relativistic hai ( $E \approx p c$ , almost photon-jaisa), exactly jaisa " $7 \times$ rest energy" ne predict kiya tha. $p$ ki units: $J / (m/s) = kg\cdotm/s$ ✓.

Ex 8 — Cell H: exam twist — direction & sign

Worked example Two-component (2-D) momentum

Ek electron velocity components $v_{x} = + 0.6 c$ , $v_{y} = - 0.6 c$ ke saath move karta hai. Iske relativistic momentum ka magnitude nikalo. Note karo $m = 9.11 \times 1 0^{- 31} kg$ .

Forecast: trap yeh hai — kya tum $γ$ sirf $v_{x}$ use karke compute karte ho, ya total speed use karke?

Step 1 — True speed nikalo (koi single component nahi). Yeh step kyun? $γ$ actual speed $v = v_{x}^{2} + v_{y}^{2}$ par depend karta hai, kyunki $γ$ measure karta hai ki clock kitni tez tick karta hai aur use sirf total motion se matlab hai, direction ya sign se nahi. $v = (0.6 c)^{2} + (0.6 c)^{2} = 0.6 c 2 = 0.8485 c$ Toh $β = 0.8485$ — dhyan do $v_{y}$ par minus sign square ho gaya; direction kabhi $γ$ mein enter nahi karti.

Step 2 — True speed se $γ$ compute karo. Yeh step kyun? yahan $β = 0.6$ use karna classic exam mistake hogi. $γ = \frac{1}{1 - 0.848 5 ^{2}} = \frac{1}{1 - 0.72} = \frac{1}{0.28} = 1.890$

Step 3 — Har momentum component build karo. Yeh step kyun? vector formula $p = γ m v$ same $γ$ ko har signed velocity component par multiply karta hai. $p_{x} = γ m v_{x} = 1.890 \times 9.11 \times 1 0^{- 31} \times (+ 1.8 \times 1 0^{8}) = + 3.10 \times 1 0^{- 22}$ $p_{y} = γ m v_{y} = 1.890 \times 9.11 \times 1 0^{- 31} \times (- 1.8 \times 1 0^{8}) = - 3.10 \times 1 0^{- 22}$ (units $kg\cdotm/s$ ; $p_{y}$ negative hai kyunki $v_{y}$ negative hai.)

Step 4 — Magnitude. Yeh step kyun? perpendicular components combine karo. $∣ p ∣ = p_{x}^{2} + p_{y}^{2} = 3.10 \times 1 0^{- 22} \times 2 = 4.38 \times 1 0^{- 22} kg\cdotm/s$

Verify: shortcut check — $∣ p ∣ = γ m v = 1.890 \times 9.11 \times 1 0^{- 31} \times (0.8485 \times 3 \times 1 0^{8}) = 4.38 \times 1 0^{- 22}$ ✓, component route se match karta hai. Single $γ$ speed se set hota hai; signs sirf components mein rehti hain. Dekho Energy-momentum four-vector ki kyun poora $p$ ek object ki tarah transform hota hai.

Recall 2-D problem mein

$γ$ kaunsi speed set karta hai? Total speed $v = v_{x}^{2} + v_{y}^{2}$ ::: koi single component nahi; $γ$ direction aur sign ignore karta hai.

Recall Photon ke liye

$p = γ m v$ kyun use nahi ho sakta? Kyunki $m = 0$ aur $v = c$ se $\infty \times 0 \times c$ milta hai (indeterminate) ::: iske bajaye $E = p c$ use karo.

Recall Given

$p = 3 m c$ , $β$ aur $γ$ kya hain? $β = 3 /2 \approx 0.866$ , $γ = 2$ ::: $γ β = 3$ solve karo.

Connections

Parent: Relativistic momentum $p=\gamma m v$
Lorentz factor and time dilation — jahan se $γ$ aata hai
Mass-energy equivalence E = γmc² — Ex 5 & 7 mein use hua
Energy-momentum four-vector — Ex 8 ka vector view
Why nothing exceeds the speed of light — Ex 4 ka $v \to c$ limit
Newtonian momentum p = mv — slow-speed cell A
Photon momentum and radiation pressure — massless cell E