Relativistic momentum p = γmv

WHAT it says

HOW we derive it (from first principles)

Step 1 — Relate $dt$ and $d\tau$ . The moving clock runs slow (time dilation): $dt = \gamma\, d\tau \quad\Rightarrow\quad \frac{dt}{d\tau} = \gamma$ Why this step? The particle's own clock ticks $d\tau$ ; the lab sees $dt = \gamma\,d\tau$ elapse. This is the only relativistic ingredient we need.

Step 2 — Define momentum with proper time instead of lab time. $\vec p \equiv m\frac{d\vec x}{d\tau}$ Why this step? Because $d\vec x$ (a 4-vector's space part) divided by the invariant $d\tau$ transforms like a proper 4-vector — so conservation in one frame guarantees it in all frames.

Step 3 — Convert to the velocity we actually measure. Use the chain rule: $\vec p = m\frac{d\vec x}{d\tau} = m\frac{d\vec x}{dt}\cdot\frac{dt}{d\tau} = m\,\vec v\,\gamma$ $\boxed{\ \vec p = \gamma m \vec v\ }$ Why this step? $d\vec x/dt = \vec v$ is the ordinary lab velocity; $dt/d\tau = \gamma$ from Step 1. The extra $\gamma$ is the entire relativistic correction.

Energy connection (one extra line of insight)

Worked examples

Worked example Example 1 — Electron at

$0.8c$ Find $p$ for an electron ( $m = 9.11\times10^{-31}$ kg) at $v = 0.8c$ .

$\gamma = 1/\sqrt{1-0.64} = 1/\sqrt{0.36} = 1/0.6 = 1.667$ — Why? plug $\beta=0.8$ .
$p = \gamma m v = 1.667 \times (9.11\times10^{-31}) \times (0.8 \times 3\times10^8)$
$p = 1.667 \times 9.11\times10^{-31}\times 2.4\times10^8 = 3.64\times10^{-22}\ \text{kg·m/s}$ .
Why bigger? Newton would give $2.19\times10^{-22}$ ; the factor $\gamma=1.667$ is the relativistic boost.

Common mistakes (Steel-man + fix)

Flashcards

What is the relativistic momentum formula?

p=\gamma m v = \dfrac{mv}{\sqrt{1-v^2/c^2}}

Define the Lorentz factor

\gamma

\gamma = 1/\sqrt{1-v^2/c^2}

, always

\ge 1

Why must momentum be redefined relativistically?

So momentum is conserved in every inertial frame after a Lorentz transformation.

What invariant time do we differentiate by to get a clean 4-momentum?

Proper time

d\tau

(clock riding with the particle).

Relation between lab time and proper time?

dt = \gamma\, d\tau

What does

p

become as

v\to c

\gamma\to\infty

, so

p\to\infty

— can't reach

c

Energy–momentum invariant?

E^2=(pc)^2+(mc^2)^2

Momentum of a photon (

m=0

p=E/c

In modern convention, what carries the

\gamma

The velocity/kinematics, not the mass (mass is invariant rest mass).

Newtonian limit of

p=\gamma m v

\gamma\to1

p\to mv

Recall Feynman: explain to a 12-year-old

Imagine pushing a toy car. Normally, push twice as hard and it goes twice as fast. But this is a magic car: the faster it already goes, the heavier-to-shove it feels, and near the speed of light it feels almost infinitely sluggish — so you can never quite reach light speed. The "magic stubbornness" is a number called $\gamma$ . Slow car: $\gamma=1$ , ordinary. Super-fast car: $\gamma$ huge. Momentum = $\gamma \times m \times v$ , so the stubbornness multiplies your normal push-amount.

Connections

Lorentz factor and time dilation — source of the $dt=\gamma d\tau$ step
Mass-energy equivalence E = γmc² — the energy partner of this momentum
Energy-momentum four-vector — why both transform together
Why nothing exceeds the speed of light — direct consequence of $\gamma\to\infty$
Newtonian momentum p = mv — the low-speed limit
Photon momentum and radiation pressure — the $m=0$ case

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, Newton ne bola tha momentum $p = mv$ — simple. Lekin jab particle bahut tez chalta hai (light speed ke paas), tab yeh formula fail ho jaata hai, kyunki alag-alag observers ke liye momentum conserve nahi hota. Isko theek karne ke liye hum ek extra factor $\gamma$ (gamma) laga dete hain: $p = \gamma m v$ , jahan $\gamma = 1/\sqrt{1-v^2/c^2}$ hota hai aur hamesha 1 se bada ya barabar hota hai.

Yeh $\gamma$ kahan se aaya? Trick yeh hai ki hum displacement ko normal lab-time $dt$ se divide nahi karte, balki proper time $d\tau$ se karte hain — yaani particle ki apni ghadi ka time, jo sabke liye same (invariant) hota hai. Time dilation se $dt = \gamma\, d\tau$ , aur chain rule lagao to seedha $p = \gamma m v$ nikal aata hai. Slow speed pe $\gamma \approx 1$ , to wahi purana Newton ka $p=mv$ wapas mil jaata hai — isiliye relativity galat nahi karti, bas zyada accurate hai.

Important baat: bahut log galti karte hain ki "mass badh jaata hai" ( $\gamma m$ ). Modern physics mein mass to constant rest mass hi rehta hai — $\gamma$ velocity ke saath chipakta hai, mass ke saath nahi. Aur jaise-jaise $v \to c$ , $\gamma \to \infty$ , isliye momentum infinite ho jaata, matlab koi bhi massive cheez light speed kabhi nahi pakad sakti. Photon ke liye mass zero hota hai, to phir $E = pc$ — light bhi momentum carry karti hai!

WHAT it says

HOW we derive it (from first principles)

Energy connection (one extra line of insight)

Worked examples

Common mistakes (Steel-man + fix)

Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Modern Physics

Connections