Lorentz transformation — derivation: Question bank

2.3.28 · D5 · Physics › Modern Physics › Lorentz transformation — derivation

Shuru karne se pehle, neeche har jagah use hone wale symbols:

$S$ = lab frame, $S^{'}$ = frame jo shared $x$ -axis ke along speed $v$ se move kar raha hai.
$c$ = speed of light (har frame mein same — yahi toh poora point hai).
$γ = \frac{1}{1 - v ^{2} / c ^{2}}$ = Lorentz factor, hamesha $\geq 1$ .
Ek event = spacetime mein ek point, $S$ mein coordinates $(t, x)$ aur $S^{'}$ mein $(t^{'}, x^{'})$ .

True or false — justify

Is transform

x^{'} = γ (x - v t)

ka koi bhi relative speed

v

ke liye valid hona?

Sirf

∣ v ∣ < c

ke liye.

v = c

par root

1 - v^{2} / c^{2}

zero ho jaata hai, toh

γ

blow up kar jaata hai aur formula toot jaata hai — koi bhi massive frame

c

tak nahi pahunch sakta.

Time dilation aur length contraction do alag-alag postulates hain.

False. Dono ek single pair

x^{'} = γ (x - v t), t^{'} = γ (t - v x / c^{2})

ke consequences hain; inhe derive karte hain, assume nahi karte.

Lorentz transformation, Galilean transformation mein reduce ho jaata hai jab

v ≪ c

ho.

True. Tab

v^{2} / c^{2} \to 0

toh

γ \to 1

aur term

v x / c^{2} \to 0

, bacha sirf

x^{'} = x - v t, t^{'} = t

. Purani theory ek limit ke roop mein survive karti hai.

γ

kuch speeds ke liye 1 se kam ho sakta hai.

False.

0 \leq v < c

ke liye

0 \leq v^{2} / c^{2} < 1

hai, toh root

< 1

hai aur uska reciprocal

γ

hamesha

> 1

hota hai. Moving clocks slow chalti hain, kabhi fast nahi.

S

mein do events jo simultaneous hain, har frame mein simultaneous hain.

False. Agar

t_{1} = t_{2}

lekin

x_{1} \neq = x_{2}

, toh

t^{'} = γ (t - v x / c^{2})

har event ke liye alag

t^{'}

deta hai — yahi Relativity of simultaneity hai.

Spacetime interval

s^{2} = c^{2} t^{2} - x^{2}

ki value

S

aur

S^{'}

mein same hoti hai.

True. Direct substitution se dikhta hai ki cross terms cancel ho jaate hain aur

c^{2} t^{'2} - x^{'2} = c^{2} t^{2} - x^{2}

; dekho Spacetime interval.

Transverse coordinates,

x

ke along boost ke under change ho jaate hain.

False. Motion sirf

x

ke along hai, toh

y^{'} = y

aur

z^{'} = z

— motion ke perpendicular direction mein kuch nahi squeeze hota.

Length contraction ka matlab hai ki rod physically spring ki tarah compress ho jaati hai.

False. Koi cheez usse push nahi karti; ye ek geometric result hai jab dono ends ko same lab time par measure kiya jaata hai. Rod ke apne frame mein uski poori length hoti hai.

Inverse transform, forward transform se alag formula use karta hai.

False (structure mein). Ye wahi formula hai

v \to - v

ke saath, kyunki

S^{'}

ke nazariye se

S

- v

speed se move karta hai — koi bhi frame special nahi hai.

Spot the error

Ek student likhta hai

t^{'} = t - v x / c^{2}

(

γ

bhool gaya).

Poori time equation ko

γ

se scale karna zaroori hai:

t^{'} = γ (t - v x / c^{2})

γ

drop karna light-speed condition aur interval invariance dono ko violate karta hai.

Ek student velocities add karta hai

u^{'} = u - v

ke roop mein ek light beam

u = c

ke liye.

Isse

c - v \neq = c

milega, jo postulate ko tod deta hai. Correct Relativistic velocity addition

u^{'} = \frac{u - v}{1 - uv / c ^{2}}

exactly

c

return karta hai jab

u = c

ho.

Ek student kehta hai "moving clock

S^{'}

mein fast chalti hai kyunki use

γ

factor milta hai."

Ulta hai. Moving clock slow chalti hai:

S^{'}

mein ek proper time interval

Δ τ

, lab mein

γ Δ τ > Δ τ

jaisa dikhta hai, toh har tick mein zyada lab time guzarta hai.

Ek student rod ki length alag-alag lab times par uske dono ends note karke measure karta hai.

Invalid. Length ke liye dono ends same

t

par record karna zaroori hai (

Δ t = 0

set karo). Alag times rod ko readings ke beech move karne dete hain, measurement corrupt ho jaata hai.

Ek student claim karta hai ki derivation ko data fit karne ke liye curved (nonlinear) map chahiye.

Nahi. Space ki homogeneity aur time ki uniformity map ko linear hone ke liye force karti hai — straight-line (uniform) motion ko straight-line motion mein map karna zaroori hai, jo sirf linear maps karte hain.

Ek student interval ko

s^{2} = c^{2} t^{2} + x^{2}

(plus sign ke saath) likhta hai.

Wrong sign. Invariant

c^{2} t^{2} - x^{2}

hai; minus sign hi spacetime geometry ko ordinary Euclidean distance se alag karta hai.

Ek student conclude karta hai ki

x^{'} = γ (x - v t)

ka matlab hai ki har event ke liye

x^{'} > x

General nahi hai.

x^{'}

ka sign aur size specific

x, t

par depend karta hai; bahut saare events ke liye

x^{'}

chhota ya negative hota hai.

γ

scale karta hai, lekin

(x - v t)

kisi bhi sign ka ho sakta hai.

Why questions

Transform linear kyun hona chahiye?

Kyunki space homogeneous hai aur time uniform,

S

mein uniformly move kar rahi ek free particle ko

S^{'}

mein bhi uniformly move karna chahiye; sirf linear maps hi straight lines ko straight lines mein bhejte hain.

Forward aur inverse transforms dono mein same factor

γ

kyun use hota hai?

Relativity ka principle kehta hai ki koi bhi frame privileged nahi hai, toh inverse exactly

v \to - v

ke saath identical dikhna chahiye, jo ek common stretch factor force karta hai.

Time equation mein

v x / c^{2}

term kyun aata hai aur Galileo ke equation mein nahi?

Kyunki light-speed ki demand space ko time mein mix karti hai;

v x / c^{2}

term exactly whi hai jo frames ke beech spatially-separated events mein time shift karta hai.

γ

find karne ke liye do light-cone equations ko divide kyun karte hain?

Akela har ek equation

t

aur

t^{'}

ko uljha deta hai; multiply aur divide karne se

t t^{'}

cancel ho jaate hain, aur

A^{2} = c^{2} / (c^{2} - v^{2})

cleanly isolate ho jaata hai.

Michelson–Morley experiment ne is poore rebuild ko kyun force kiya?

Usne paaya ki Earth ki motion ke bawajood light ki speed unchanged hai, jo additive-velocity picture ko khatam kar deta hai aur ek aisi transform ki demand karta hai jo

c

ko har frame mein fixed rakhe.

Factor

1/ c^{2}

t

ki jagah

x

ke saath kyun baitha hai?

Kyunki

x / c^{2}

ki units length squared per time hain... zyada precisely

v x / c^{2}

ki units time ki hain, jo ek spatial offset ko time shift mein convert karta hai jis equation ko zaroorat hai.

Edge cases

v \to 0

par

γ

ka kya hota hai?

γ \to 1

: transform Galilean

x^{'} = x, t^{'} = t

ban jaata hai — rest par koi relativistic effects nahi.

v \to c

par

γ

ka kya hota hai?

γ \to \infty

: lengths zero ki taraf contract hoti hain aur time dilation without bound badhta hai, yahi reason hai ki massive frames ke liye

v = c

unreachable hai.

Light pulse ke liye spacetime interval

s^{2}

kya hota hai?

Exactly zero: path

x = c t

par,

s^{2} = c^{2} t^{2} - (c t)^{2} = 0

. Light "light cone" par hai, har frame mein ek null interval.

Agar

v

negative ho (

- x

direction mein move karta frame), kya formulas tab bhi hold karte hain?

Haan.

γ

depend karta hai

v^{2}

par toh unchanged rehta hai, aur

(x - v t)

aur

(t - v x / c^{2})

ke andar signed

v

automatically reversed direction handle kar leta hai.

S

mein same place par do events ke liye (

Δ x = 0

Δ t^{'}

kya ban jaata hai?

Δ t^{'} = γ Δ t

— pure Time dilation bina kisi simultaneity mixing ke, kyunki

v x / c^{2}

term drop out ho jaata hai.

Kya koi aisa frame hai jahan do timelike-separated events ka order swap ho jaaye?

Nahi. Agar

c^{2} Δ t^{2} > Δ x^{2}

(timelike) ho, toh interval ka sign guarantee karta hai ki time-ordering sabhi frames mein preserved hai, cause aur effect ki raksha karte hue.

Recall Final self-test

Wo ek demand kaunsi hai jo $γ$ fix karti hai? ::: Light ki speed dono frames mein $c$ ke barabar hai.
Woh quantity kaunsi hai jo boost ke under invariant hai? ::: Spacetime interval $s^{2} = c^{2} t^{2} - x^{2}$ .
Time dilation aur length contraction kahan se aate hain? ::: Dono do Lorentz equations ke consequences hain, alag postulates nahi.