Newton thought time was a universal clock ticking the same for everyone, and that velocities just add: if you run at u u u v v v u + v u+v u + v c c c c + v c+v c + v Contradiction. The Lorentz transformation is the minimal repair of the Galilean transformation that keeps the speed of light equal to c c c
Two inertial frames: S S S S ′ S' S ′ x x x = = v = = ==v== == v == event has coordinates ( t , x , y , z ) (t,x,y,z) ( t , x , y , z ) S S S ( t ′ , x ′ , y ′ , z ′ ) (t',x',y',z') ( t ′ , x ′ , y ′ , z ′ ) S ′ S' S ′ x x x = = y ′ = y = = ==y'=y== == y ′ = y == = = z ′ = z = = ==z'=z== == z ′ = z ==
The Galilean guess (what Newton would write):
x ′ = x − v t , t ′ = t . x' = x - vt, \qquad t' = t. x ′ = x − v t , t ′ = t . steel-man-then-fix this.
Intuition Why must the transform be linear?Space is homogeneous (no special point) and time is uniform. A uniform-velocity (straight-line) motion in S S S S ′ S' S ′ linear maps do that. Curved maps would make a free particle appear to accelerate in one frame and not the other, breaking the principle of relativity.
So write the most general linear form (using y , z y,z y , z x ′ = A x + B t , t ′ = D x + E t . x' = A\,x + B\,t, \qquad t' = D\,x + E\,t. x ′ = A x + B t , t ′ = D x + E t . A , B , D , E A,B,D,E A , B , D , E
The point x ′ = 0 x'=0 x ′ = 0 S S S x = v t x=vt x = v t x ′ = 0 x'=0 x ′ = 0 0 = A x + B t ⇒ x = − B A t 0 = A\,x + B\,t \Rightarrow x = -\tfrac{B}{A}t 0 = A x + B t ⇒ x = − A B t v t vt v t B = − A v . B = -A v. B = − A v . x ′ = A ( x − v t ) x' = A(x - vt) x ′ = A ( x − v t )
single factor A A A v → − v v \to -v v → − v S S S − v -v − v S ′ S' S ′ same stretch factor A A A x = A ( x ′ + v t ′ ) . x = A(x' + vt'). x = A ( x ′ + v t ′ ) .
Fire a light pulse from the common origin at t = t ′ = 0 t=t'=0 t = t ′ = 0 S S S x = c t x=ct x = c t S ′ S' S ′ x ′ = c t ′ x'=ct' x ′ = c t ′
Substitute x = c t x=ct x = c t x ′ = A ( x − v t ) x' = A(x-vt) x ′ = A ( x − v t ) x ′ = A ( c t − v t ) = A t ( c − v ) . x' = A(ct - vt) = A t(c-v). x ′ = A ( c t − v t ) = A t ( c − v ) . x ′ = c t ′ x'=ct' x ′ = c t ′ x = A ( x ′ + v t ′ ) x = A(x'+vt') x = A ( x ′ + v t ′ ) c t = A ( c t ′ + v t ′ ) = A t ′ ( c + v ) . ct = A(ct' + vt') = A t'(c+v). c t = A ( c t ′ + v t ′ ) = A t ′ ( c + v ) .
Now multiply the two left sides and the two right sides. Use x ′ = c t ′ x'=ct' x ′ = c t ′ c t ′ ⏟ x ′ ⋅ c t = A 2 t t ′ ( c − v ) ( c + v ) . \underbrace{ct'}_{x'}\cdot ct = A^2\, t\,t'\,(c-v)(c+v). x ′ c t ′ ⋅ c t = A 2 t t ′ ( c − v ) ( c + v ) .
Worked example Why this multiplication step?Each equation alone has the unknowns t t t t ′ t' t ′ t t ′ t\,t' t t ′ A A A
Divide both sides by t t ′ t\,t' t t ′ c 2 = A 2 ( c 2 − v 2 ) . c^2 = A^2(c^2 - v^2). c 2 = A 2 ( c 2 − v 2 ) . A 2 = c 2 c 2 − v 2 = 1 1 − v 2 / c 2 ⇒ A = 1 1 − v 2 / c 2 ≡ γ . A^2 = \frac{c^2}{c^2 - v^2} = \frac{1}{1 - v^2/c^2} \;\Rightarrow\; A = \frac{1}{\sqrt{1-v^2/c^2}} \equiv \gamma. A 2 = c 2 − v 2 c 2 = 1 − v 2 / c 2 1 ⇒ A = 1 − v 2 / c 2 1 ≡ γ .
We have x ′ = γ ( x − v t ) x' = \gamma(x - vt) x ′ = γ ( x − v t ) x = γ ( x ′ + v t ′ ) x = \gamma(x' + vt') x = γ ( x ′ + v t ′ ) t ′ t' t ′ x = γ ( γ ( x − v t ) + v t ′ ) = γ 2 ( x − v t ) + γ v t ′ . x = \gamma\big(\gamma(x-vt) + vt'\big) = \gamma^2(x-vt) + \gamma v t'. x = γ ( γ ( x − v t ) + v t ′ ) = γ 2 ( x − v t ) + γ v t ′ . γ v t ′ \gamma v t' γ v t ′ γ v t ′ = x − γ 2 x + γ 2 v t = x ( 1 − γ 2 ) + γ 2 v t . \gamma v t' = x - \gamma^2 x + \gamma^2 v t = x(1-\gamma^2) + \gamma^2 v t. γ v t ′ = x − γ 2 x + γ 2 v t = x ( 1 − γ 2 ) + γ 2 v t . 1 − γ 2 = 1 − 1 1 − v 2 / c 2 = − v 2 / c 2 1 − v 2 / c 2 = − γ 2 v 2 c 2 1 - \gamma^2 = 1 - \frac{1}{1-v^2/c^2} = \frac{-v^2/c^2}{1-v^2/c^2} = -\gamma^2 \frac{v^2}{c^2} 1 − γ 2 = 1 − 1 − v 2 / c 2 1 = 1 − v 2 / c 2 − v 2 / c 2 = − γ 2 c 2 v 2 γ v t ′ = − γ 2 v 2 c 2 x + γ 2 v t . \gamma v t' = -\gamma^2\frac{v^2}{c^2}x + \gamma^2 v t. γ v t ′ = − γ 2 c 2 v 2 x + γ 2 v t . γ v \gamma v γ v t ′ = − γ v c 2 x + γ t = γ ( t − v x c 2 ) . t' = -\gamma\frac{v}{c^2}x + \gamma t = \gamma\!\left(t - \frac{vx}{c^2}\right). t ′ = − γ c 2 v x + γ t = γ ( t − c 2 v x ) .
Intuition What is conserved?Galileo conserved Δ x 2 + Δ y 2 + Δ z 2 \Delta x^2 + \Delta y^2 + \Delta z^2 Δ x 2 + Δ y 2 + Δ z 2 Δ t \Delta t Δ t spacetime interval :
s 2 = c 2 t 2 − x 2 . s^2 = c^2 t^2 - x^2. s 2 = c 2 t 2 − x 2 .
Check it:
c 2 t ′ 2 − x ′ 2 = c 2 γ 2 ( t − v x c 2 ) 2 − γ 2 ( x − v t ) 2 . c^2 t'^2 - x'^2 = c^2\gamma^2\!\left(t-\tfrac{vx}{c^2}\right)^2 - \gamma^2(x-vt)^2. c 2 t ′2 − x ′2 = c 2 γ 2 ( t − c 2 v x ) 2 − γ 2 ( x − v t ) 2 . − 2 v x t -2vxt − 2 v x t c 2 t 2 − x 2 c^2 t^2 - x^2 c 2 t 2 − x 2
Worked example 1. Recovering Galileo at low speedTake v = 30 v = 30 v = 30 c = 3 × 10 8 c = 3\times10^8 c = 3 × 1 0 8 v 2 / c 2 ≈ 10 − 14 v^2/c^2 \approx 10^{-14} v 2 / c 2 ≈ 1 0 − 14 γ ≈ 1 \gamma \approx 1 γ ≈ 1 v x / c 2 ≈ 0 vx/c^2 \approx 0 v x / c 2 ≈ 0 Why this step? It shows the transform x ′ = x − v t x'=x-vt x ′ = x − v t t ′ = t t'=t t ′ = t contains the old one as a limit. That's how a correct generalization must behave.
Worked example 2. Time of a distant eventIn S S S x 1 = 0 , t 1 = 0 x_1=0,t_1=0 x 1 = 0 , t 1 = 0 x 2 = 1 x_2 = 1 x 2 = 1 = c ⋅ 1 =c\cdot 1 = c ⋅ 1 t 2 = 0 t_2 = 0 t 2 = 0 S S S S ′ S' S ′ v = 0.6 c v=0.6c v = 0.6 c γ = 1 / 1 − 0.36 = 1.25 \gamma = 1/\sqrt{1-0.36}=1.25 γ = 1/ 1 − 0.36 = 1.25 t 2 ′ = γ ( t 2 − v x 2 / c 2 ) = 1.25 ( 0 − 0.6 c ⋅ c / c 2 ) = 1.25 × ( − 0.6 s ) = − 0.75 t'_2 = \gamma(t_2 - vx_2/c^2) = 1.25\,(0 - 0.6c\cdot c/c^2) = 1.25\times(-0.6\text{ s}) = -0.75 t 2 ′ = γ ( t 2 − v x 2 / c 2 ) = 1.25 ( 0 − 0.6 c ⋅ c / c 2 ) = 1.25 × ( − 0.6 s ) = − 0.75 t 1 ′ = 0 t'_1 = 0 t 1 ′ = 0 Why this step? t 2 ′ ≠ t 1 ′ t'_2 \ne t'_1 t 2 ′ = t 1 ′ S S S not simultaneous in S ′ S' S ′ relativity of simultaneity , falling directly out of the − v x / c 2 -vx/c^2 − v x / c 2 x x x
Worked example 3. Length contraction quick checkA rod at rest in S ′ S' S ′ simultaneously in S S S (Δ t = 0 \Delta t = 0 Δ t = 0 Δ x \Delta x Δ x Δ x ′ = γ ( Δ x − v Δ t ) = γ Δ x \Delta x' = \gamma(\Delta x - v\Delta t) = \gamma\,\Delta x Δ x ′ = γ ( Δ x − v Δ t ) = γ Δ x Δ x ′ = L 0 \Delta x' = L_0 Δ x ′ = L 0 L = Δ x = L 0 / γ L = \Delta x = L_0/\gamma L = Δ x = L 0 / γ Why this step? We had to set Δ t = 0 \Delta t=0 Δ t = 0 L = L 0 / γ L=L_0/\gamma L = L 0 / γ
Common mistake "Velocities just add:u ′ = u − v u' = u - v u ′ = u − v Why it feels right: It's true to amazing precision at everyday speeds, and our brains are trained on it. The fix: because t ′ ≠ t t' \ne t t ′ = t u ′ = u − v 1 − u v / c 2 u' = \dfrac{u-v}{1 - uv/c^2} u ′ = 1 − uv / c 2 u − v u = c u=c u = c u ′ = c − v 1 − v / c = c u' = \frac{c-v}{1-v/c} = c u ′ = 1 − v / c c − v = c c c c
Common mistake "Time dilation and length contraction are separate add-on rules."Why it feels right: Textbooks often quote them as standalone formulas. The fix: both are pure consequences of x ′ = γ ( x − v t ) x' = \gamma(x-vt) x ′ = γ ( x − v t ) t ′ = γ ( t − v x / c 2 ) t'=\gamma(t-vx/c^2) t ′ = γ ( t − v x / c 2 ) γ \gamma γ 1 / γ 1/\gamma 1/ γ
γ \gamma γ Why it feels right: Subtraction inside the root looks like it could go either way. The fix: 0 ≤ v < c ⇒ 0 ≤ v 2 / c 2 < 1 ⇒ 0\le v<c \Rightarrow 0 \le v^2/c^2 <1 \Rightarrow 0 ≤ v < c ⇒ 0 ≤ v 2 / c 2 < 1 ⇒ < 1 ⇒ γ > 1 <1 \Rightarrow \gamma>1 < 1 ⇒ γ > 1 slow , never fast.
Recall Feynman: explain to a 12-year-oldImagine everyone agrees on one rule: a flash of light must always look like it's going the same fast speed, no matter how fast you are running. To keep that promise, the universe has to cheat a little: when you move really fast, your ruler shrinks a tiny bit and your watch ticks a tiny bit slower — by just the right amount so the light still looks like the same speed. The Lorentz transformation is the exact recipe for how much to shrink the ruler and slow the watch. At normal speeds the cheat is so tiny you never notice it.
Mnemonic Remember the two equations"x minus v-t, t minus v-x-over-c-squared, all times gamma."
Space loses v t vt v t v x / c 2 vx/c^2 v x / c 2 gamma scales both . The 1 / c 2 1/c^2 1/ c 2 x x x
Recall Active recall — close your eyes first
Why must the transform be linear? ::: Homogeneity of space/uniformity of time → straight (uniform) motion maps to straight motion → only linear maps preserve that.
What single physical demand fixes γ \gamma γ = c =c = c
What quantity is invariant? ::: The spacetime interval c 2 t 2 − x 2 c^2t^2 - x^2 c 2 t 2 − x 2
What two postulates underlie the Lorentz transformation? (1) Laws of physics identical in all inertial frames; (2) speed of light
c c c is the same in all inertial frames.
Why must the transformation be linear? Space is homogeneous and time uniform, so uniform-velocity motion must map to uniform-velocity motion — only linear maps do this.
State the Lorentz factor. γ = 1 / 1 − v 2 / c 2 \gamma = 1/\sqrt{1 - v^2/c^2} γ = 1/ 1 − v 2 / c 2 , always
≥ 1 \ge 1 ≥ 1 .
Write the Lorentz transformation for x ′ x' x ′ t ′ t' t ′ x ′ = γ ( x − v t ) x'=\gamma(x-vt) x ′ = γ ( x − v t ) ,
t ′ = γ ( t − v x / c 2 ) t'=\gamma(t - vx/c^2) t ′ = γ ( t − v x / c 2 ) , with
y ′ = y , z ′ = z y'=y,\ z'=z y ′ = y , z ′ = z .
How is A = γ A=\gamma A = γ From
x = c t , x ′ = c t ′ x=ct,\ x'=ct' x = c t , x ′ = c t ′ giving
c 2 = A 2 ( c 2 − v 2 ) c^2 = A^2(c^2-v^2) c 2 = A 2 ( c 2 − v 2 ) .
What replaces Galilean velocity addition? u ′ = ( u − v ) / ( 1 − u v / c 2 ) u' = (u-v)/(1 - uv/c^2) u ′ = ( u − v ) / ( 1 − uv / c 2 ) ; it keeps
c c c invariant.
What spacetime quantity is invariant under Lorentz transforms? The interval
s 2 = c 2 t 2 − x 2 s^2 = c^2t^2 - x^2 s 2 = c 2 t 2 − x 2 .
Why is simultaneity relative? t ′ = γ ( t − v x / c 2 ) t' = \gamma(t - vx/c^2) t ′ = γ ( t − v x / c 2 ) : equal
t t t but different
x x x gives different
t ′ t' t ′ .
Limit of Lorentz transform as v ≪ c v\ll c v ≪ c γ → 1 \gamma\to1 γ → 1 ,
v x / c 2 → 0 vx/c^2\to0 v x / c 2 → 0 → Galilean
x ′ = x − v t , t ′ = t x'=x-vt,\ t'=t x ′ = x − v t , t ′ = t .
Derive length contraction direction. Measure both ends at same lab time (
Δ t = 0 \Delta t=0 Δ t = 0 ):
Δ x ′ = γ Δ x ⇒ L = L 0 / γ \Delta x' = \gamma\Delta x \Rightarrow L = L_0/\gamma Δ x ′ = γ Δ x ⇒ L = L 0 / γ .
Galilean transformation — the v ≪ c v\ll c v ≪ c Time dilation — direct consequence of the t ′ t' t ′ Length contraction — consequence of the x ′ x' x ′ Relativity of simultaneity — born of the − v x / c 2 -vx/c^2 − v x / c 2 Spacetime interval — the invariant that survives.Relativistic velocity addition — derived by differentiating the transform.Michelson–Morley experiment — experimental motivation.Minkowski diagram — geometric picture of these transforms.
Michelson-Morley: light always c
Galilean transform: velocities add
Maxwell: light speed is c
Homogeneous space, uniform time
Linearity: straight to straight
Light pulse x=ct and x'=ct'
Time dilation and length contraction
Intuition Hinglish mein samjhoDekho, Newton ka rule simple tha: time sabke liye same chalta hai aur velocities seedhe add ho jaati hain. Train me u u u v v v u + v u+v u + v c c c c + v c+v c + v
Derivation ka core idea: transform linear hona chahiye (kyunki seedhi line motion dono frames me seedhi dikhni chahiye), isliye x ′ = A ( x − v t ) x'=A(x-vt) x ′ = A ( x − v t ) x = c t x=ct x = c t x ′ = c t ′ x'=ct' x ′ = c t ′ A A A γ = 1 / 1 − v 2 / c 2 \gamma = 1/\sqrt{1-v^2/c^2} γ = 1/ 1 − v 2 / c 2 gamma factor hai. Time wali equation t ′ = γ ( t − v x / c 2 ) t'=\gamma(t-vx/c^2) t ′ = γ ( t − v x / c 2 )
Yaad rakhne wali baat: γ \gamma γ slow chalti hai aur chalta hua ruler chhota dikhta hai — bas itna ki light ka speed c c c t ′ t' t ′ v x / c 2 vx/c^2 v x / c 2 S S S S ′ S' S ′ nahi hote — isko relativity of simultaneity kehte hain. Low speed pe (v ≪ c v\ll c v ≪ c
