Relativistic momentum p = γmv: Exercises — Hinglish deep dive

2.3.31 · D4 · Physics › Modern Physics › Relativistic momentum p = γmv

Definition Is page pe jo symbols use karte hain (yahan built hain, assumed nahi)

$p$ — the momentum vector. $p = γ m v$ motion ki direction ke saath point karta hai. In one-dimensional problems mein saari motion ek chosen axis (use $x$ kaho) ke saath hai, toh hum $p$ ko ek signed number $p$ se describe kar sakte hain, jo $+ x$ ke saath uska component hai. Ye poore page ka hamara convention hai: jab bhi symbol $p$ dekhte ho, yeh signed component $p = γ m v$ hai, jahan $v$ khud signed hai ( $v > 0$ matlab " $+ x$ direction," $v < 0$ matlab "opposite"). Isliye ek rightward particle ka $p > 0$ hoga aur leftward ka $p < 0$ . Signs sirf tab matter karte hain jab do particles ek doosre ki taraf move kar rahe hon aur hum unke momenta add karein (L5.1). Iska magnitude — direction ignore karke size — $∣ p ∣ = γ m ∣ v ∣ \geq 0$ hai; hum $∣ p ∣$ (bars ke saath) likhte hain jab magnitude mean karte hain, aur plain $p$ (signed) baaki jagah. L2–L4 mein har particle $+ x$ direction mein move karta hai, toh wahan $p = ∣ p ∣$ aur distinction silent hai.
$E$ — total energy. Total energy hai $E = γ m c^{2}$ : rest energy $m c^{2}$ plus kinetic energy. Yeh momentum se is relation se juda hai $E^{2} = (p c)^{2} + (m c^{2})^{2}$ , jo energy–momentum relation hai. Note karo is relation mein $p^{2}$ hai, toh yeh sirf magnitude dekhta hai — signs square karne par disappear ho jaate hain. Hum $E$ ab introduce kar rahe hain taaki Level 4 mein ready ho.
$K$ — kinetic energy. Sirf motion ki energy: $K = E - m c^{2} = (γ - 1) m c^{2}$ . Yeh tab zero ho jaati hai jab particle rest pe ho ( $γ = 1$ ).

Intuition Figure — exercises se pehle yeh padho

Figure kya dikhata hai. Horizontal axis

β = v / c

hai (speed light-speed ke fraction ke roop mein,

0 \to 1

chalti hai). Vertical axis natural units mein momentum hai,

p / (m c) = γ β

. Blue curve sach muchi relativistic momentum hai; dashed yellow line woh hai jo Newton predict karta hai,

p / (m c) = β

(seedhi

4 5^{\circ}

line). Dotted red vertical line

β = 1

pe light-speed wall hai. Green dots neeche solve kiye do exercises mark karte hain (L2.1 at

β = 0.6

aur L2.2 at

β \approx 0.95

). Kya notice karna hai. Chhote

β

pe blue curve yellow line se chepti hai — relativity aur Newton agree karte hain, L1.3 aur L3.3 se match karta hai. Jaise

β

\sim 0.5

se upar chadta hai, blue curve upar ki taraf peel karti hai aur red wall pe infinity ki taraf rocket karti hai: momentum linearly se zyada tezi se badhta hai (L3.1 ka point) aur

β \to 1

pe diverge karta hai (L3.2). Is page pe lagbhag har exercise is ek curve se read off hone wala ek single point hai. Direction reverse hone par (

β < 0

) poori picture origin se reflect ho jaati hai — identical shape, negative

p

Level 1 — Recognition

Goal: formula mein plug in karo aur correctly padho.

L1.1 — Relativistic momentum formula aur $γ$ ki definition batao. Phir kaho ki $p$ kya banta hai jab $v \to c$ aur jab $v \to 0$ .

Recall Solution L1.1

$p = γ m v = \frac{m v}{1 - v ^{2} / c ^{2}}, γ = \frac{1}{1 - v ^{2} / c ^{2}} \geq 1.$ $p$ ki direction motion ki direction hai; iska magnitude $∣ p ∣ = γ m ∣ v ∣$ hai.

Jab $∣ v ∣ \to c$ : $β \to \pm 1$ , toh $1 - β^{2} \to 0$ , toh $γ \to \infty$ , toh $∣ p ∣ \to \infty$ . Mass wali koi bhi cheez $c$ tak nahi pahunch sakti — dekho Why nothing exceeds the speed of light.
Jab $v \to 0$ : $γ \to 1$ , toh $p \to m v$ , jo Newtonian value hai. ✅

L1.2 — $β = 0.6$ ke liye $γ$ compute karo.

Recall Solution L1.2

$γ = \frac{1}{1 - 0. 6 ^{2}} = \frac{1}{1 - 0.36} = \frac{1}{0.64} = \frac{1}{0.8} = 1.25.$ Kya kiya: $β$ ko square kiya, 1 se subtract kiya, reciprocal square root liya. Reciprocal kyun: $γ$ $\geq 1$ hona chahiye taaki momentum badhta rahe; agar $0.8$ mila toh tumne galti se $1 - β^{2}$ le liya. Note karo $γ$ sirf $β^{2}$ pe depend karta hai, toh yeh $β = + 0.6$ aur $β = - 0.6$ ke liye same hai — travel ki direction $γ$ nahi badlti.

L1.3 — $2 kg$ ki ek ball $3 m/s$ pe roll karti hai. Kya uske momentum mein relativistic correction measurable hai? $γ$ estimate karo.

Recall Solution L1.3

$β = 3/ (3 \times 1 0^{8}) = 1 0^{- 8}$ . Phir $γ = \frac{1}{1 - 1 0 ^{- 16}} \approx 1 + \frac{1}{2} (1 0^{- 16}) = 1 + 5 \times 1 0^{- 17} .$ Correction $1 0^{- 16}$ mein ek part hai — bilkul unmeasurable. Everyday speeds Newtonian hain: $γ$ boost tabhi matter karta hai jab $β ≳ 0.1$ .

Level 2 — Application

Goal: har problem mein ek clean numeric plug-in. (Yahan har particle $+ x$ mein move karta hai, toh $p = ∣ p ∣ > 0$ .)

L2.1 — $v = 0.6 c$ pe ek electron ka momentum find karo.

Recall Solution L2.1

$γ = 1.25$ (L1.2 se). Lab velocity $v = 0.6 \times 3 \times 1 0^{8} = 1.8 \times 1 0^{8} m/s$ (ise $+ x$ direction liya gaya). $p = γ m v = 1.25 \times (9.11 \times 1 0^{- 31}) \times (1.8 \times 1 0^{8}) .$ $p = 1.25 \times 1.6398 \times 1 0^{- 22} = 2.0498 \times 1 0^{- 22} kg\cdotm/s \approx 2.05 \times 1 0^{- 22} kg\cdotm/s .$ Vector $p$ $+ x$ mein point karta hai, toh $p > 0$ ; agar electron doosri taraf move karta, toh $v = - 0.6 c$ aur $p = - 2.05 \times 1 0^{- 22}$ (same magnitude $∣ p ∣$ , opposite sign). Magnitude check: Newton deta $m v = 1.64 \times 1 0^{- 22}$ ; $\times 1.25$ boost use $25%$ lift karta hai.

L2.2 — Ek proton ka momentum magnitude $∣ p ∣ = 3 m c$ hai. $γ$ aur $β$ find karo.

Recall Solution L2.2

Likho $∣ p ∣ = γ m ∣ v ∣ = γ β m c$ (magnitude ke liye $β > 0$ use karke), toh $γ β = ∣ p ∣/ (m c) = 3$ . $γ β$ kyun: yeh dono unknowns ko ek quantity mein package karta hai jo data deta hai. Square karo: $γ^{2} β^{2} = 9$ . $γ^{2} = 1/ (1 - β^{2})$ use karo: $\frac{β ^{2}}{1 - β ^{2}} = 9 \Rightarrow β^{2} = 9 - 9 β^{2} \Rightarrow 10 β^{2} = 9 \Rightarrow β = 0.9 = 0.9487.$ Phir $γ = \frac{1}{1 - 0.9} = \frac{1}{0.1} = 3.162$ . Sanity: $γ β = 3.162 \times 0.9487 = 3.00$ . ✅ (Positive root liya; doosri taraf move karta proton signed $p = - 3 m c$ deta, same $γ$ aur same speed $∣ v ∣$ .)

L2.3 — $β = 0.9$ pe electron ka momentum Newtonian estimate se kitne factor zyada hai?

Recall Solution L2.3

Relativistic-to-Newtonian ratio exactly $γ$ hai, kyunki $p_{rel} / p_{Newton} = γ m v / (m v) = γ$ . $γ = \frac{1}{1 - 0.81} = \frac{1}{0.19} = 2.294.$ Toh sach wala momentum naive value ka $2.29 \times$ hai — Newton half se zyada se underestimate karta hai.

Level 3 — Analysis

Goal: $p$ kaise change hota hai uske baare mein reason karo, sirf uski value nahi.

L3.1 — Agar $v$ $0.1 c$ se $0.2 c$ tak badhta hai, toh $p$ kitne factor se change hota hai? Kya yeh exactly $2$ hai?

Recall Solution L3.1

$p = γ m v$ , toh factor hai $\frac{γ ( 0.2 c ) \cdot 0.2 c}{γ ( 0.1 c ) \cdot 0.1 c}$ .

$γ (0.1 c) = 1/ 1 - 0.01 = 1/ 0.99 = 1.00504.$
$γ (0.2 c) = 1/ 1 - 0.04 = 1/ 0.96 = 1.02062.$ $ratio = \frac{1.02062 \times 0.2}{1.00504 \times 0.1} = \frac{0.204124}{0.100504} = 2.031.$ Interpretation: $p$ double se thoda zyada badhta hai, kyunki extra $γ$ khud $v$ ke saath chadta hai. Relativity mein speed double karne se momentum double nahi hota.

L3.2 — Dikhao ki jab $β \to 1$ , $p$ bina bound ke badhta hai, aur woh $β$ find karo jiske liye $p = 10 m c$ .

Recall Solution L3.2

Jab $β \to 1$ , $γ β \to \frac{β}{1 - β ^{2}}$ : numerator $1$ ki taraf jaata hai lekin denominator $0$ ki taraf, toh fraction $\to \infty$ . Isliye $p = γ β m c \to \infty$ . Yahi light-speed ki wall hai. $p = 10 m c$ ke liye: $γ β = 10$ , toh $\frac{β ^{2}}{1 - β ^{2}} = 100$ , jisse $β^{2} = 100/101$ milta hai, toh $β = 100/101 = 0.99504.$ Light-speed ke $99.5%$ pe bhi, $p$ "sirf" $10 m c$ hai — speed ka aakhri zarra enormous momentum maangta hai.

L3.3 — Chhote $β$ ke liye, $γ$ expand karo aur Newtonian momentum mein pehla correction dikhao. $β = 0.1$ pe $p = m v$ kitna fractional error karta hai?

Recall Solution L3.3

Binomial expansion use karo $(1 - x)^{- 1/2} \approx 1 + \frac{1}{2} x$ chhote $x = β^{2}$ ke liye: $γ \approx 1 + \frac{1}{2} β^{2} \Rightarrow p \approx m v (1 + \frac{1}{2} β^{2}) .$ Binomial kyun: yeh calculator ke bina leading relativistic term ko isolate karta hai. $p$ ki $m v$ se upar fractional excess $γ - 1 \approx \frac{1}{2} β^{2}$ hai. $β = 0.1$ pe: fractional error $\approx \frac{1}{2} (0.01) = 0.005 = 0.5%$ . (Exact: $γ - 1 = 1.00504 - 1 = 0.504%$ .) Toh Newton ek-tenth light-speed pe pehle se half percent se off hai.

Level 4 — Synthesis

Goal: momentum ko energy aur four-vector ke saath combine karo. Top box se yaad karo: $E = γ m c^{2}$ (total energy), $K = (γ - 1) m c^{2}$ (kinetic energy), $E^{2} = (p c)^{2} + (m c^{2})^{2}$ se linked. Yahan saare particles $+ x$ mein move karte hain, toh $p = ∣ p ∣ > 0$ .

L4.1 — Ek electron ki total energy $E = 2 m c^{2}$ hai. Energy–momentum relation use karke uska momentum $p$ (units of $m c$ mein) find karo.

Recall Solution L4.1

Energy-momentum four-vector se $E^{2} = (p c)^{2} + (m c^{2})^{2}$ use karo. Rearranged: $(p c)^{2} = E^{2} - (m c^{2})^{2} = (2 m c^{2})^{2} - (m c^{2})^{2} = 4 m^{2} c^{4} - m^{2} c^{4} = 3 m^{2} c^{4} .$ $p c = 3 m c^{2} \Rightarrow p = 3 m c = 1.732 m c .$ $γ$ se cross-check: $E = γ m c^{2} = 2 m c^{2}$ matlab $γ = 2$ , toh $β = 1 - 1/ γ^{2} = 1 - 1/4 = 3 /2$ , aur $p = γ β m c = 2 \cdot \frac{3}{2} m c = 3 m c$ . ✅ Dono roads agree karte hain.

L4.2 — Ek photon aur ek electron dono ka momentum $p = m c$ hai (jahan $m$ electron mass hai). Unki total energies compare karo.

Recall Solution L4.2

Photon ( $m_{γ} = 0$ , dekho Photon momentum and radiation pressure): $E = p c = m c \cdot c = m c^{2}$ .
Electron ( $m \neq = 0$ ): $E = (p c)^{2} + (m c^{2})^{2} = (m c^{2})^{2} + (m c^{2})^{2} = 2 m c^{2} = 1.414 m c^{2}$ . Difference kyun: electron extra rest energy $m c^{2}$ carry karta hai jo massless photon mein nahi hoti. Same momentum, lekin massive particle ki total energy hamesha zyada hoti hai. Dekho Mass-energy equivalence E = γmc².

L4.3 — Kinetic energy hai $K = E - m c^{2} = (γ - 1) m c^{2}$ . Ek electron ke liye jis ka $K = m c^{2}$ hai (yaani $K$ uski rest energy ke equal hai), $γ$ , $β$ , aur $p$ find karo.

Recall Solution L4.3

$K = (γ - 1) m c^{2} = m c^{2} \Rightarrow γ - 1 = 1 \Rightarrow γ = 2$ . $β = 1 - 1/ γ^{2} = 1 - 1/4 = 3 /2 = 0.8660$ . $p = γ β m c = 2 \cdot \frac{3}{2} m c = 3 m c = 1.732 m c .$ Note: yeh L4.1 jaisi hi physical state hai — total energy $2 m c^{2}$ matlab kinetic energy $m c^{2}$ . Consistency confirm karti hai ki hamare relations ka web saath hold karta hai.

Level 5 — Mastery

Goal: multi-step problems jo conservation, frames, aur limits ko fuse karte hain. Yahan $p$ ke liye signed convention finally real kaam karta hai.

L5.1 — Rest mass $m$ ka ek particle $β = 0.8$ pe ( $+ x$ direction mein) move kar raha hai aur ek identical particle se head-on collide karke chipak jaata hai jo $- x$ direction mein $β = 0.8$ pe move kar raha hai. Combined object ka rest mass $M$ find karo. (Hint: total energy aur total momentum conserve hote hain — signs dhyan se dekho.)

Recall Solution L5.1

Momentum (signed components add hote hain): particle 1 $+ x$ mein move karta hai toh uska signed component $p_{1} = + γ m (0.8 c)$ hai; particle 2 $- x$ mein move karta hai toh uska signed component $p_{2} = - γ m (0.8 c)$ hai. Yeh cancel ho jaate hain: total $p_{tot} = p_{1} + p_{2} = 0$ . Lump rest pe hai. Isliye hum signs rakhte hain: magnitudes se hum galti se $2 γ m (0.8 c)$ add kar dete. Energy (ek scalar — hamesha add hoti hai): har particle ka $γ = 1/ 1 - 0.64 = 1/0.6 = 5/3$ hai, toh har ek ka $E = γ m c^{2} = \frac{5}{3} m c^{2}$ . Total: $E_{tot} = 2 \cdot \frac{5}{3} m c^{2} = \frac{10}{3} m c^{2} .$ Kyunki lump rest pe hai, uski saari energy rest energy hai: $E_{tot} = M c^{2}$ . Isliye $M = \frac{10}{3} m = 3.333 m .$ Lesson: combined mass $2 m$ se zyada hai! Collision ki kinetic energy rest mass ban gayi — Mass-energy equivalence E = γmc² ka seedha display. Rest mass conserve nahi hoti; total energy aur (signed) total momentum hoti hain.

L5.2 — Invariant verify karo. Single particle ke liye $β = 0.8$ pe (mass $m$ ), $E^{2} - (p c)^{2}$ compute karo aur confirm karo ki yeh $(m c^{2})^{2}$ ke equal hai chahe frame kisi bhi tarah move kare.

Recall Solution L5.2

$γ = 5/3$ , toh $E = \frac{5}{3} m c^{2}$ aur $∣ p ∣ = γ β m c = \frac{5}{3} \cdot 0.8 m c = \frac{4}{3} m c$ , jisse $p c = \frac{4}{3} m c^{2}$ milta hai. (Invariant $p^{2}$ use karta hai, toh $p$ ka sign yahan irrelevant hai — sirf $∣ p ∣$ enter karta hai.) $E^{2} - (p c)^{2} = (\frac{5}{3})^{2} (m c^{2})^{2} - (\frac{4}{3})^{2} (m c^{2})^{2} = (\frac{25}{9} - \frac{16}{9}) (m c^{2})^{2} = \frac{9}{9} (m c^{2})^{2} = (m c^{2})^{2} .$ Yeh deep kyun hai: $E^{2} - (p c)^{2}$ energy–momentum four-vector ki squared "length" hai. Alag observers alag $E$ aur $p$ measure karte hain, lekin yeh combination invariant hai — sabhi $m c^{2}$ compute karte hain. Wahi invariance ki wajah se humne pehli jagah proper time se differentiate kiya tha.

L5.3 — Dikhao ki relativistic momentum ko $p = \frac{E v}{c ^{2}}$ likha ja sakta hai, aur $β = 0.8$ electron ke liye check karo.

Recall Solution L5.3

Derivation. Hamare paas parent note mein built do facts hain: total energy $E = γ m c^{2}$ aur momentum $p = γ m v$ . Trick yeh hai ki notice karo dono mein same factor $γ m$ shared hai. Energy equation se woh shared factor solve karo: $E = γ m c^{2} \Rightarrow γ m = \frac{E}{c ^{2}} .$ Kya kiya / kyun: humne $γ m$ isolate kiya taaki use momentum formula mein substitute karein aur $γ$ aur $m$ ko single quantity $E$ ke favour mein eliminate karein. Ab $γ m = E / c^{2}$ ko $p = γ m v$ mein daalo: $p = (γ m) v = \frac{E}{c ^{2}} v \Rightarrow p = \frac{E v}{c ^{2}} .$ Equivalently, dono original equations ko component-by-component divide karne se directly wahi milta hai: $\frac{p}{E} = \frac{γ m v}{γ m c ^{2}} = \frac{v}{c ^{2}}$ . Useful kyun: yeh kehta hai momentum sirf energy hai jo velocity $v$ ke saath carried hoti hai (divided by $c^{2}$ ), aur yeh $v$ ki direction automatically inherit karta hai — toh $p$ ka sign $v$ ke sign ko follow karta hai bina extra kaam ke. Ek photon ke liye, $∣ v ∣ = c$ deta hai $∣ p ∣ = E / c$ — massless case seedha pop out ho jaata hai. Check ( $β = 0.8$ , toh $v = 0.8 c$ ): L5.2 se, $E = \frac{5}{3} m c^{2}$ . Phir $p = \frac{E v}{c ^{2}} = \frac{( \frac{5}{3} m c ^{2} ) ( 0.8 c )}{c ^{2}} = \frac{5}{3} \cdot 0.8 m c = \frac{4}{3} m c .$ Yeh L5.2 se direct computation $p = γ β m c = \frac{4}{3} m c$ se match karta hai. ✅

Connections

Lorentz factor and time dilation — upar use hone wale har $γ$ ka source.
Mass-energy equivalence E = γmc² — L4 aur L5 energy problems ko power karta hai.
Energy-momentum four-vector — L4.1 aur L5.2 ke peeche invariant $E^{2} - (p c)^{2}$ .
Why nothing exceeds the speed of light — L3.2 ki $p \to \infty$ wall.
Newtonian momentum p = mv — L1.3 aur L3.3 mein check kiya gaya low- $β$ limit.
Photon momentum and radiation pressure — L4.2 aur L5.3 mein $m = 0$ case.