Visual walkthrough — Compton scattering — wavelength shift derivation

2.3.4 · D2 · Physics › Modern Physics › Compton scattering — wavelength shift derivation

Pehli line se pehle hum is poori story ke characters pe agree kar lete hain — sirf yahi objects hain is poori kahani mein:

Neeche sab kuch inhi par build hota hai. Chaliye collision dekhte hain.

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Step 1 — Collision ki picture khud

KYA: Hum ek photon ke ek resting electron se takraane ka "pehle" aur "baad" draw karte hain.

KYUN: Jab tak nahi pata ki kya kahan ja raha hai, kuch bhi conserve nahi kar sakte. Baad ke har arrow isi diagram par rehta hai, isliye pehle directions pin down karte hain.

PICTURE:

Figure ko left se right padhein:

• Pehle: ek photon horizontal ($x$) axis ke along andar aata hai. Electron (magenta dot) collision point par rest mein baitha hai.
• Baad: photon axis ke upar angle $\theta$ par niklta hai, ek nayi, lambi wavelength ${\lambda }^{\prime }$ ke saath (moti ripple ki tarah draw kiya gaya hai). Electron axis ke neeche angle $\varphi$ par recoil karta hai.

Har moving object do quantities carry karta hai jinhe hum track karte hain:

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Step 2 — Energy conserve karo (bank account ki height)

KYA: Hum kehte hain ki pehle ki total energy baad ki total energy ke barabar hai.

KYUN: Conservation of energy — collision mein energy na banti hai na nashi hoti hai. Photon jo khota hai, electron wahi paata hai.

PICTURE:

Energy ko do stacked bars ki height ki tarah socho. Baayan stack (pehle) aur daayan stack (baad) ko same total height tak pahunchna chahiye:

$\underset{\text{photon in}}{\underbrace{\frac{hc}{\lambda }}}+\underset{\begin{array}{c}\text{electron}\\ \text{at rest}\end{array}}{\underbrace{m{c}^2}}=\underset{\text{photon out}}{\underbrace{\frac{hc}{{\lambda }^{\prime }}}}+\underset{\begin{array}{c}\text{electron}\\ \text{moving}\end{array}}{\underbrace{E_e}}$

Term by term: photon ki energy $\frac{hc}{\lambda }$ se ghatt kar $\frac{hc}{{\lambda }^{\prime }}$ ho gayi (kyunki ${\lambda }^{\prime }>\lambda$), aur exactly woh missing chunk electron mein chali gayi. Electron ki energy solve karte hain:

$E_e=m{c}^2+hc\left(\frac{1}{\lambda }-\frac{1}{{\lambda }^{\prime }}\right)(\star )$

Bracket $\left(\frac{1}{\lambda }-\frac{1}{{\lambda }^{\prime }}\right)$ positive hai (kyunki $\lambda <{\lambda }^{\prime }$), isliye $E_e>m{c}^2$: electron apni resting energy se zyada lekar gaya. Theek hai — ab woh move kar raha hai.

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Step 3 — Momentum ko arrows ki tarah conserve karo

KYA: Hum momentum conserve karte hain, lekin momentum ek vector hai (ek arrow), isliye hum ise horizontal ($x$) aur vertical ($y$) parts mein tod'te hain.

KYUN: Conservation of momentum har direction mein alag-alag hold hota hai. Ek akela number "kaunsi taraf" capture nahi kar sakta — arrows kar sakte hain. Toh hum arrows draw karte hain aur har axis par unka "shadow" padte hain.

PICTURE:

Incoming photon ka arrow purely $x$ ke along point karta hai. Bounce ke baad do arrows hain: scattered photon (upar-daayan) aur electron (neeche-daayan). Har axis par unka combined shadow "pehle" ke shadow se match karna chahiye.

Horizontal ($x$): pehle, sirf incoming photon; baad mein, do horizontal shadows. $\underset{\text{in, all }x}{\underbrace{\frac{h}{\lambda }}}=\underset{\text{photon }x}{\underbrace{\frac{h}{{\lambda }^{\prime }}\cos \theta }}+\underset{\text{electron }x}{\underbrace{p_e\cos \varphi }}$

Vertical ($y$): pehle, kuch bhi vertically nahi chal raha, toh do vertical shadows cancel honay chahiye. $\underset{\text{in, no }y}{\underbrace{0}}=\underset{\begin{array}{c}\text{photon up}\end{array}}{\underbrace{\frac{h}{{\lambda }^{\prime }}\sin \theta }}-\underset{\begin{array}{c}\text{electron down}\end{array}}{\underbrace{p_e\sin \varphi }}$

Minus sign poora point hai: photon upar jaata hai, electron neeche jaata hai, aur woh exactly zero ho jaate hain.

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Step 4 — Electron ko isolate karo, phir uska angle delete karo

KYA: Hum electron terms ko akele ek side mein le jaate hain, phir dono equations ko square karke add kar dete hain.

KYUN: Hum nahi jaanna chahte ki electron kahan gaya ($\varphi$). Trick ${\cos }^2\varphi +{\sin }^2\varphi =1$ (ek right-triangle identity) $\varphi$ ko completely gayab kar deti hai.

PICTURE:

Step 3 ko rearrange karte hain taaki electron ke do shadows akele hon: $p_e\cos \varphi =\frac{h}{\lambda }-\frac{h}{{\lambda }^{\prime }}\cos \theta ,p_e\sin \varphi =\frac{h}{{\lambda }^{\prime }}\sin \theta$

Ab $p_e\cos \varphi$ aur $p_e\sin \varphi$ ko ek right triangle ki do legs ki tarah socho jiska hypotenuse $p_e$ hai. Pythagoras se legs squared add hokar hypotenuse squared banta hai — aur $\varphi$ collapse ho jaata hai:

$p_e^2={\left(\frac{h}{\lambda }-\frac{h}{{\lambda }^{\prime }}\cos \theta \right)}^2+{\left(\frac{h}{{\lambda }^{\prime }}\sin \theta \right)}^2$

Expand karo aur photon ke angle par ${\cos }^2\theta +{\sin }^2\theta =1$ use karo:

$p_e^2=\frac{h^2}{{\lambda }^2}-\frac{2h^2}{\lambda {\lambda }^{\prime }}\cos \theta +\frac{h^2}{{\lambda }^{\prime 2}}(\star \star )$

Term by term: do "pure" pieces $\frac{h^2}{{\lambda }^2}$ aur $\frac{h^2}{{\lambda }^{\prime 2}}$ (do photon speeds se), aur ek cross term $-\frac{2h^2}{\lambda {\lambda }^{\prime }}\cos \theta$ jo angle carry karta hai. Woh cross term hi hai jahan $\theta$ chhupa hai — yaad rakhna.

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Step 5 — Electron ke liye relativistic bridge

KYA: Hum electron ki energy $E_e$ ko uske momentum $p_e$ se relativistic relation use karke connect karte hain, phir equation $(\star )$ ko square karte hain taaki dono squared quantities ban jayein.

KYUN: Ab hamare paas $p_e^2$ hai momentum se (Step 4) aur $E_e$ hai energy se (Step 2). Inhe milane ke liye hum ek aise equation ki zaroorat hai jo particle ki energy aur momentum ko link kare — Relativistic energy-momentum relation. Hum yahi use karte hain (na ki $\frac{1}{2}m{v}^2$) kyunki recoil light-speed ke kareeb ho sakta hai.

PICTURE:

Yeh relation khud ek right triangle hai: $E_e$ hypotenuse hai, $p_{e}c$ aur $m{c}^2$ legs hain. $E_e^2=(p_{e}c{)}^2+(m{c}^2{)}^2⟹p_e^2=\frac{E_e^2-(m{c}^2{)}^2}{c^2}$

$(\star )$ ko ab kyun square karein? Upar wala bridge $E_e^2$ (energy squared) maangta hai, lekin Step 2 ne sirf $E_e$ (energy first power mein) diya tha. Toh hum $(\star )$ ko square karte hain sirf isliye ki bridge ki form se match ho sake — bas itna hi reason hai is move ka, kuch nahi: $E_e^2=(m{c}^2{)}^2+2m{c}^{2}hc\left(\frac{1}{\lambda }-\frac{1}{{\lambda }^{\prime }}\right)+h^2{c}^2{\left(\frac{1}{\lambda }-\frac{1}{{\lambda }^{\prime }}\right)}^2$

$(m{c}^2{)}^2$ subtract karo aur $c^2$ se divide karo — resting-energy piece khoobsurati se cancel ho jaata hai: $p_e^2=2mhc\left(\frac{1}{\lambda }-\frac{1}{{\lambda }^{\prime }}\right)+h^2{\left(\frac{1}{\lambda }-\frac{1}{{\lambda }^{\prime }}\right)}^2(\star \star \star )$

Ab hamare paas ek hi $p_e^2$ ke do expressions hain — ek arrows se $(\star \star )$, ek energy se $(\star \star \star )$. Inhe equal set karo.

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Step 6 — Equal set karo aur cancellation dekho

KYA: $(\star \star )=(\star \star \star )$ set karo aur har matching term cancel karo.

KYUN: Dono $p_e^2$ hain, toh identical honay chahiye. Cancellation ke baad jo bachta hai wahi answer hai.

PICTURE:

Pehle $(\star \star \star )$ ka last term expand karo: $h^2{\left(\frac{1}{\lambda }-\frac{1}{{\lambda }^{\prime }}\right)}^2=\frac{h^2}{{\lambda }^2}-\frac{2h^2}{\lambda {\lambda }^{\prime }}+\frac{h^2}{{\lambda }^{\prime 2}}$

Dono sides align karo: $\underset{\text{cancels}}{\underbrace{\frac{h^2}{{\lambda }^2}}}-\frac{2h^2\cos \theta }{\lambda {\lambda }^{\prime }}+\underset{\text{cancels}}{\underbrace{\frac{h^2}{{\lambda }^{\prime 2}}}}=2mhc\left(\frac{1}{\lambda }-\frac{1}{{\lambda }^{\prime }}\right)+\underset{\text{cancels}}{\underbrace{\frac{h^2}{{\lambda }^2}}}-\frac{2h^2}{\lambda {\lambda }^{\prime }}+\underset{\text{cancels}}{\underbrace{\frac{h^2}{{\lambda }^{\prime 2}}}}$

Do "pure" pieces dono sides se gayab ho jaate hain. Jo bachta hai woh sirf cross term aur energy term hai: $-\frac{2h^2\cos \theta }{\lambda {\lambda }^{\prime }}=2mhc\left(\frac{1}{\lambda }-\frac{1}{{\lambda }^{\prime }}\right)-\frac{2h^2}{\lambda {\lambda }^{\prime }}$

Do $h^2$ pieces ko left side par ikatha karo: $\frac{2h^2}{\lambda {\lambda }^{\prime }}(1-\cos \theta )=2mhc\left(\frac{1}{\lambda }-\frac{1}{{\lambda }^{\prime }}\right)$

Angle ab ek akele, saaf-suthre $(1-\cos \theta )$ ke andar hai. Ek step aur baaki hai.

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Step 7 — $\Delta \lambda$ mein final squeeze

KYA: Fractions clean up karo taaki ${\lambda }^{\prime }-\lambda$ dikhe.

KYUN: Hum shift chahte hain, ek single wavelength difference — isliye $\frac{1}{\lambda }-\frac{1}{{\lambda }^{\prime }}$ ko ek fraction mein likhte hain aur clutter divide out karte hain.

PICTURE:

Right-hand bracket ko common denominator par combine karo: $\frac{1}{\lambda }-\frac{1}{{\lambda }^{\prime }}=\frac{{\lambda }^{\prime }-\lambda }{\lambda {\lambda }^{\prime }}$

Substitute karo aur dono sides ko $\frac{2h}{\lambda {\lambda }^{\prime }}$ se divide karo (yeh har $\lambda {\lambda }^{\prime }$ khatam kar deta hai): $h(1-\cos \theta )=mc({\lambda }^{\prime }-\lambda )$

$mc$ se divide karo aur result nikal aata hai. Yahan hum electron mass $m$ ko $m_e$ rename karte hain ("e" subscript sirf yeh kehta hai "electron") taaki standard textbook symbol se match ho:

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Step 8 — Har case, drawn (isliye humne sab angles cover kiye)

KYA: Hum formula ko teen landmark angles aur degenerate "wrong mass" case par test karte hain.

KYUN: Contract kehta hai reader ko koi aisa scenario nahi milna chahiye jo humne dikhaya na ho. $\theta$ $0^{\circ }$ se $180^{\circ }$ tak range karta hai, isliye $(1-\cos \theta )$ $0$ se $2$ tak range karta hai. Sab dekho.

PICTURE:

Kyunki $\cos \theta$ symmetric hai ($\cos (-\theta )=\cos \theta$), "$\theta$ upar" ya "$\theta$ neeche" scatter hone par same shift milta hai — upar/neeche ka choice sirf yeh flip karta hai ki electron kis taraf recoil karta hai, wavelength nahi. Toh har physical angle is ek curve se cover ho jaata hai.

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Ek-picture summary

Yeh single figure poora safar compress karta hai: collision → do conservation laws → $p_e^2$ ke do expressions → equal set karo → cancel karo → shift. Arrows trace karo aur tumne Compton scattering re-derive kar li.

Recall Poore walkthrough ki Feynman-style retelling

Humne roshni ki ek bouncy ball ko ek still marble (electron) se takrate dekha. Rule ek: "pehle" shelf par total energy "baad" shelf se match karni chahiye — toh light ne jo brightness khoyi, marble ne wahi motion ke roop mein paaya (Step 2). Rule do: arrows kahin se appear nahi ho sakte. Light seedhi andar aayi thi, toh hit ke baad light ke upar bounce hone ka sideways push marble ke neeche kick karne se cancel hona chahiye (Step 3). Hum nahi jaanna chahte the ki marble kahan gaya, isliye humne uski do arrow-legs ko square karke add kiya — Pythagoras ne marble ka angle mita diya (Step 4). Phir humne tez marble ke liye Einstein ke energy-momentum triangle ka use kiya (Step 5). Ab marble ka momentum do tareekon se likha tha — ek arrows se, ek energy se — isliye humne unhe equal set kiya (Step 6). Jaadu: sab "plain" $\lambda$ pieces cancel ho gaye, sirf angle $(1-\cos \theta )$ ke roop mein wrapped bachaa. Ek last tidy-up (Step 7) ne $\Delta \lambda =\frac{h}{m_{e}c}(1-\cos \theta )$ diya. Zero bend = zero shift; seedha wapas = biggest shift; poore bhaari atom se takrao = almost koi shift nahi. Yahi wala real experiment mein unshifted peak hai. Yahi derivation ki poori kahani hai — bas light ke saath pool.

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Connections

• Parent note (Hinglish) — woh algebra jo yeh page illustrate karta hai.
• Photon momentum and energy — $p=h/\lambda$ supply karta hai, Step 3 mein arrows.
• Relativistic energy-momentum relation — Step 5 mein triangle.
• Conservation of energy aur Conservation of momentum — do rules jo Steps 2–3 drive karte hain.
• de Broglie wavelength — ulta idea: matter bhi ek wavelength carry karta hai.
• Photoelectric effect — iska saathi proof ki light photons mein aati hai.
• X-ray production and Bremsstrahlung — jahan se incoming X-rays aati hain.

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