Visual walkthrough — Compton scattering — wavelength shift derivation

Before line one we agree on the cast of characters — the only objects in this whole story:

Everything below builds on these. Let us watch the collision.

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Step 1 — The picture of the collision itself

WHAT: We draw the "before" and "after" of one photon hitting one resting electron.

WHY: You cannot conserve anything until you know what is moving where. Every later arrow lives on this diagram, so we pin down directions first.

PICTURE:

Read the figure left to right:

• Before: a photon flies in along the horizontal ($x$) axis. The electron (magenta dot) sits at rest at the collision point.
• After: the photon leaves at angle $\theta$ above the axis with a new, longer wavelength ${\lambda }^{\prime }$ (drawn as a fatter ripple). The electron recoils below the axis at angle $\varphi$.

Each moving object carries two quantities we track:

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Step 2 — Conserve energy (the height of the bank account)

WHAT: We say the total energy before equals the total energy after.

WHY: Conservation of energy — nothing creates or destroys energy in a collision. Whatever the photon loses, the electron gains.

PICTURE:

Think of energy as the height of two stacked bars. The left stack (before) and the right stack (after) must reach the same total height:

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

Term by term: the photon's energy shrank from $\frac{hc}{\lambda }$ to $\frac{hc}{{\lambda }^{\prime }}$ (because ${\lambda }^{\prime }>\lambda$), and exactly that missing chunk got added to the electron. Solving for the electron's energy:

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

The bracket $\left(\frac{1}{\lambda }-\frac{1}{{\lambda }^{\prime }}\right)$ is positive (since $\lambda <{\lambda }^{\prime }$), so $E_e>m{c}^2$: the electron ended up with more than its resting energy. Good — it is moving now.

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Step 3 — Conserve momentum as arrows

WHAT: We conserve momentum, but momentum is a vector (an arrow), so we split it into horizontal ($x$) and vertical ($y$) parts.

WHY: Conservation of momentum holds separately in every direction. A single number can't capture "which way" — arrows can. So we draw the arrows and read off their shadows on each axis.

PICTURE:

The incoming photon's arrow points purely along $x$. After the bounce there are two arrows: the scattered photon (up-right) and the electron (down-right). Their combined shadow on each axis must match the "before" shadow.

Horizontal ($x$): before, only the incoming photon; after, the two 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 ($y$): before, nothing moves vertically, so the two vertical shadows must cancel. $\underset{\text{in, no }y}{\underbrace{0}}=\underset{\begin{array}{c}\text{photon up}\end{array}}{\underbrace{\frac{h}{{\lambda }^{\prime }}\sin \theta }}-\underset{\begin{array}{c}\text{electron down}\end{array}}{\underbrace{p_e\sin \varphi }}$

The minus sign is the whole point: the photon goes up, the electron goes down, and they exactly balance to zero.

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Step 4 — Isolate the electron, then delete its angle

WHAT: We move the electron terms alone to one side, then square both equations and add them.

WHY: We do not care where the electron went ($\varphi$). The trick ${\cos }^2\varphi +{\sin }^2\varphi =1$ (a right-triangle identity) lets $\varphi$ vanish completely.

PICTURE:

Rearranging Step 3 to put the electron's two shadows alone: $p_e\cos \varphi =\frac{h}{\lambda }-\frac{h}{{\lambda }^{\prime }}\cos \theta ,p_e\sin \varphi =\frac{h}{{\lambda }^{\prime }}\sin \theta$

Now picture $p_e\cos \varphi$ and $p_e\sin \varphi$ as the two legs of a right triangle whose hypotenuse is $p_e$. By Pythagoras the legs squared add to the hypotenuse squared — and the $\varphi$'s collapse:

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

Expand and use ${\cos }^2\theta +{\sin }^2\theta =1$ on the photon's angle:

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

Term by term: two "pure" pieces $\frac{h^2}{{\lambda }^2}$ and $\frac{h^2}{{\lambda }^{\prime 2}}$ (from the two photon speeds), and one cross term $-\frac{2h^2}{\lambda {\lambda }^{\prime }}\cos \theta$ that carries the angle. That cross term is where $\theta$ hides — remember it.

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Step 5 — The relativistic bridge for the electron

WHAT: We connect the electron's energy $E_e$ to its momentum $p_e$ using the relativistic relation, then square equation $(\star )$ so both live as squared quantities.

WHY: We now have $p_e^2$ from momentum (Step 4) and $E_e$ from energy (Step 2). To make them meet we need the one equation linking a particle's energy and momentum — the Relativistic energy-momentum relation. We use it (not $\frac{1}{2}m{v}^2$) because the recoil can be near light-speed.

PICTURE:

The relation is itself a right triangle: $E_e$ is the hypotenuse, $p_{e}c$ and $m{c}^2$ are the legs. $E_e^2=(p_{e}c{)}^2+(m{c}^2{)}^2⟹p_e^2=\frac{E_e^2-(m{c}^2{)}^2}{c^2}$

Why square $(\star )$ now? The bridge above needs $E_e^2$ (energy squared), but Step 2 only gave us $E_e$ (energy to the first power). So we square $(\star )$ purely to match the form the bridge demands — that's the whole reason for this move, nothing more: $E_e^2=(m{c}^2{)}^2+2m{c}^{2}hc\left(\frac{1}{\lambda }-\frac{1}{{\lambda }^{\prime }}\right)+h^2{c}^2{\left(\frac{1}{\lambda }-\frac{1}{{\lambda }^{\prime }}\right)}^2$

Subtract $(m{c}^2{)}^2$ and divide by $c^2$ — the resting-energy piece cancels beautifully: $p_e^2=2mhc\left(\frac{1}{\lambda }-\frac{1}{{\lambda }^{\prime }}\right)+h^2{\left(\frac{1}{\lambda }-\frac{1}{{\lambda }^{\prime }}\right)}^2(\star \star \star )$

We now have two expressions for the same $p_e^2$ — one from arrows $(\star \star )$, one from energy $(\star \star \star )$. Set them equal.

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Step 6 — Equate and watch the cancellation

WHAT: Set $(\star \star )=(\star \star \star )$ and cancel every matching term.

WHY: Both are $p_e^2$, so they must be identical. What survives after cancellation is the answer.

PICTURE:

First expand the last term of $(\star \star \star )$: $h^2{\left(\frac{1}{\lambda }-\frac{1}{{\lambda }^{\prime }}\right)}^2=\frac{h^2}{{\lambda }^2}-\frac{2h^2}{\lambda {\lambda }^{\prime }}+\frac{h^2}{{\lambda }^{\prime 2}}$

Line up the two sides: $\underset{\text{cancels}}{\underbrace{\frac{h^2}{{\lambda }^2}}}-\frac{2h^2\cos \theta }{\lambda {\lambda }^{\prime }}+\underset{\text{cancels}}{\underbrace{\frac{h^2}{{\lambda }^{\prime 2}}}}=2mhc\left(\frac{1}{\lambda }-\frac{1}{{\lambda }^{\prime }}\right)+\underset{\text{cancels}}{\underbrace{\frac{h^2}{{\lambda }^2}}}-\frac{2h^2}{\lambda {\lambda }^{\prime }}+\underset{\text{cancels}}{\underbrace{\frac{h^2}{{\lambda }^{\prime 2}}}}$

The two "pure" pieces vanish from both sides. What remains is only the cross term and the energy term: $-\frac{2h^2\cos \theta }{\lambda {\lambda }^{\prime }}=2mhc\left(\frac{1}{\lambda }-\frac{1}{{\lambda }^{\prime }}\right)-\frac{2h^2}{\lambda {\lambda }^{\prime }}$

Gather the two $h^2$ pieces on the left: $\frac{2h^2}{\lambda {\lambda }^{\prime }}(1-\cos \theta )=2mhc\left(\frac{1}{\lambda }-\frac{1}{{\lambda }^{\prime }}\right)$

The angle now lives inside a single, tidy $(1-\cos \theta )$. One step to go.

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Step 7 — The final squeeze into $\Delta \lambda$

WHAT: Clean up the fractions to reveal ${\lambda }^{\prime }-\lambda$.

WHY: We want the shift, a single wavelength difference — so we rewrite $\frac{1}{\lambda }-\frac{1}{{\lambda }^{\prime }}$ as one fraction and divide out the clutter.

PICTURE:

Combine the right-hand bracket over a common denominator: $\frac{1}{\lambda }-\frac{1}{{\lambda }^{\prime }}=\frac{{\lambda }^{\prime }-\lambda }{\lambda {\lambda }^{\prime }}$

Substitute and divide both sides by $\frac{2h}{\lambda {\lambda }^{\prime }}$ (this kills every $\lambda {\lambda }^{\prime }$): $h(1-\cos \theta )=mc({\lambda }^{\prime }-\lambda )$

Divide by $mc$ and out drops the result. Here we rename the electron mass $m$ as $m_e$ (the subscript "e" just says "electron") to match the standard textbook symbol:

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Step 8 — Every case, drawn (this is why we covered all angles)

WHAT: We test the formula at the three landmark angles and the degenerate "wrong mass" case.

WHY: The contract says the reader must never meet a scenario we didn't show. $\theta$ ranges over $0^{\circ }$ to $180^{\circ }$, so $(1-\cos \theta )$ ranges from $0$ to $2$. Let's see all of it.

PICTURE:

Because $\cos \theta$ is symmetric ($\cos (-\theta )=\cos \theta$), scattering "up by $\theta$" or "down by $\theta$" gives the same shift — the up/down choice only flips which way the electron recoils, never the wavelength. So every physical angle is covered by this one curve.

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The one-picture summary

This single figure compresses the whole journey: collision → two conservation laws → two expressions for $p_e^2$ → set equal → cancel → shift. Trace the arrows and you have re-derived Compton scattering.

Recall Feynman retelling of the whole walkthrough

We drew a bouncy ball of light hitting a still marble (electron). Rule one: the total energy on the "before" shelf must match the "after" shelf — so whatever brightness the light loses, the marble gains as motion (Step 2). Rule two: arrows can't appear from nowhere. The light came in flat, so after the hit the sideways push of the light bouncing up must be cancelled by the marble kicking down (Step 3). We didn't care where the marble went, so we squared its two arrow-legs and added them — Pythagoras erased the marble's angle (Step 4). Then we used Einstein's energy-momentum triangle for the fast marble (Step 5). Now we had the marble's momentum written two ways — once from arrows, once from energy — so we set them equal (Step 6). Magic: all the "plain" $\lambda$ pieces cancelled, leaving only the angle wrapped up as $(1-\cos \theta )$. A last tidy-up (Step 7) gave $\Delta \lambda =\frac{h}{m_{e}c}(1-\cos \theta )$. Zero bend = zero shift; straight-back = biggest shift; hit a whole heavy atom = almost no shift. That last one is the unshifted peak in the real experiment. That's the entire story of the derivation — just pool with light.

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Connections

• Parent note (Hinglish) — the algebra this page illustrates.
• Photon momentum and energy — supplies $p=h/\lambda$, the arrows in Step 3.
• Relativistic energy-momentum relation — the triangle in Step 5.
• Conservation of energy and Conservation of momentum — the two rules driving Steps 2–3.
• de Broglie wavelength — the reverse idea: matter also carries a wavelength.
• Photoelectric effect — its partner proof that light comes in photons.
• X-ray production and Bremsstrahlung — where the incoming X-rays come from.

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