Here is one page from my general physics notes. The derivation is still beautiful to me even though so many years passed. I want to preserve these and sharpen my Markdown skill so I try to type them here.
Thanks for this post which introduces the math functions of Markdown in great detail.
Background
A photon of initial wavelength \(\lambda_i\) collides a rest electron with mass \(m_e\) at a plane. Then, we define that the angle between the initial path of the photon and its final path is \(\phi\). And, the angle between the initial path of the photon and the motion path of the electron is \(\theta\). The final wavelength of photon is \(\lambda_f\) while the frequency doesn't change. The velocity of electron is \(\vec{v}\) which implies its momentum \(\vec{p} = \frac{m_e \vec{v}}{\sqrt{1- v^2/c^2}}\) after the correction of special relativity.
Conservation of energy
For a photon of frequency \(f\), its energy \(E\): \[ E = hf = \frac{hc}{\lambda} \qquad where \quad c = f\lambda \] and \(h\) is the Planck constant.
From mass-energy equivalence, a particle of mass \(m\) has a rest energy $ E = mc^2 $.
The sum energy of initial photon and rest electron equals the sum energy of final photon and motional electron since the law of conservation of energy.
\[ E_{pi} = \frac{hc}{\lambda_i} \text{,}\quad E_{ei} = m_ec^2 \] \[ E_{pf} = \frac{hc}{\lambda_f} \text{,}\quad E_{ef} = \frac{m_ec^2}{\sqrt{1-\frac{v^2}{c^2}}} \]
Now, we have equation \(\text{(1)}\) by plusing the formers.
\[ E_{pi}+E_{ei}=E_{pf}+E_{ef} \] \[ \Downarrow \] \[\begin{equation}\label{eq1} \frac{hc}{\lambda_i} + m_ec^2 = \frac{hc}{\lambda_f} + \frac{m_ec^2}{\sqrt{1-\frac{v^2}{c^2}}} \end{equation}\]
Conservation of momentum
If we define the direction of the motion path of initial photon is x-axis, then we know that the initial momentum of photon \(\vec{p_{pi}} = \lbrace p_{pix} ,\; p_{piy} \rbrace\) where \(p_{pix} = \frac{h}{\lambda_i}\) and \(p_{piy} = 0\). The x-component of the final momentum of photon \(p_{pfx} = \frac{h}{\lambda_f} \cos{\phi}\) and its y-component \(p_{pfy} = \frac{h}{\lambda_f} \sin{\phi}\).
From the precondition, the momentum of rest electron can be presented as \[ \vec{p_{ei}} = \lbrace p_{eix} ,\; p_{eiy} \rbrace = \text{\{0, 0\}} \] and its final momentum: \[ \vec{p_{ef}} = \frac{m_e \vec{v}}{\sqrt{1- v^2/c^2}} = \lbrace p_{efx} ,\; p_{efy} \rbrace = \left\{ \frac{m_ev}{\sqrt{1-\frac{v^2}{c^2}}} \cos{\theta} ,\; -\frac{m_ev}{\sqrt{1-\frac{v^2}{c^2}}} \sin{\theta} \right\} \]
Conclude these components. We can obtain equation \(\text{(2)}\) and \(\text{(3)}\) from the law of conservation of momentum.
\[\begin{equation}\label{eq2} p_{pix} + p_{eix} = p_{pfx} + p_{efx} \quad \Rightarrow \quad \frac{h}{\lambda_i} = \frac{h}{\lambda_f} \cos{\phi} + \frac{m_ev}{\sqrt{1-\frac{v^2}{c^2}}} \cos{\theta} \end{equation}\]
\[\begin{equation}\label{eq3} p_{piy} + p_{eiy} = p_{pfy} + p_{efy} \quad \Rightarrow \quad 0 = \frac{h}{\lambda_f} \sin{\phi} - \frac{m_ev}{\sqrt{1-\frac{v^2}{c^2}}} \sin{\theta} \end{equation}\]
Derivation
Start from squaring the equation \(\text{(2)}\).
\[\begin{align} \frac{h}{\lambda_i} - \frac{h}{\lambda_f} \cos{\phi} & = \frac{m_ev}{\sqrt{1-\frac{v^2}{c^2}}} \cos{\theta} &\text{additive identity from (2)} \tag{2'}\\ \left( \frac{h}{\lambda_i} - \frac{h}{\lambda_f} \cos{\phi} \right)^2 & = \left( \frac{m_ev}{\sqrt{1-\frac{v^2}{c^2}}} \cos{\theta} \right)^2 &\text{square of (2')} \tag{4}\\ \frac{h^2}{\lambda_i^2} - 2\frac{h^2}{\lambda_i\lambda_f}\cos{\phi} + \frac{h^2}{\lambda_f^2} \cos^2{\phi} & = \frac{m_e^2 v^2}{1-\frac{v^2}{c^2}} \cos^2{\theta} &\tag{4'}\\ \end{align}\]
Square the equation \(\text{(3)}\), too.
\[\begin{align} \frac{h}{\lambda_f} \sin{\phi} & = \frac{m_ev}{\sqrt{1-\frac{v^2}{c^2}}} \sin{\theta} &\text{additive identity from (3)} \tag{3'}\\ \left( \frac{h}{\lambda_f} \sin{\phi} \right)^2 & = \left( \frac{m_ev}{\sqrt{1-\frac{v^2}{c^2}}} \sin{\theta} \right)^2 &\text{square of (3')} \tag{5}\\ \frac{h^2}{\lambda_f^2} \sin^2{\phi} & = \frac{m_e^2 v^2}{1-\frac{v^2}{c^2}} \sin^2{\theta} &\tag{5'}\\ \end{align}\]
Now, plus \(\text{(4')}\) and \(\text{(5')}\).
\[\begin{align} \left( \frac{h^2}{\lambda_i^2} - 2\frac{h^2}{\lambda_i\lambda_f}\cos{\phi} + \frac{h^2}{\lambda_f^2} \cos^2{\phi} \right) + \frac{h^2}{\lambda_f^2} \sin^2{\phi} & = \frac{m_e^2 v^2}{1-\frac{v^2}{c^2}} \cos^2{\theta} + \frac{m_e^2 v^2}{1-\frac{v^2}{c^2}} \sin^2{\theta} &\tag{6}\\ \frac{h^2}{\lambda_i^2} - 2\frac{h^2}{\lambda_i\lambda_f}\cos{\phi} + \frac{h^2}{\lambda_f^2} (\cancelto{1}{\cos^2{\phi} + \sin^2{\phi}}) & = \frac{m_e^2 v^2}{1-\frac{v^2}{c^2}} (\cancelto{1}{\cos^2{\theta} + \sin^2{\theta}}) &\tag{6'}\\ \frac{h^2}{\lambda_i^2} - 2\frac{h^2}{\lambda_i\lambda_f}\cos{\phi} + \frac{h^2}{\lambda_f^2} & = \frac{m_e^2 v^2}{1-\frac{v^2}{c^2}} &\tag{6''}\\ \end{align}\]
\(\text{(6'')}\) applies substitution of \(\sin^2\theta + \cos^2\theta = 1\) on \(\text{(6')}\).
Put \(\text{(6'')}\) in mind. It's time to handle the equation \(\text{(1)}\)!
\[\begin{align} \frac{h}{\lambda_i} + m_ec & = \frac{h}{\lambda_f} + \frac{m_ec}{\sqrt{1-\frac{v^2}{c^2}}} &\text{$(1)\div c$} \tag{1'}\\ \frac{h}{\lambda_i} + m_ec - \frac{h}{\lambda_f} & = \frac{m_ec}{\sqrt{1-\frac{v^2}{c^2}}} &\text{additive identity from (1')} \tag{1''}\\ \left( \frac{h}{\lambda_i} + m_ec - \frac{h}{\lambda_f} \right)^2 & = \left( \frac{m_ec}{\sqrt{1-\frac{v^2}{c^2}}} \right)^2 &\text{square of (1'')} \tag{7}\\ \left[ \left( \frac{h}{\lambda_i} - \frac{h}{\lambda_f} \right) + m_ec \right]^2 & = \frac{m_e^2 c^2}{1-\frac{v^2}{c^2}} &\tag{7'}\\ \end{align}\]
Then, minus \(\text{(6'')}\) from \(\text{(7')}\), which equals squared \(\text{(1)}\) minus the sum of squared \(\text{(2)}\) and squared \(\text{(3)}\), to get the equation \(\text{(8)}\).
\[\begin{equation} \tag{8} \left[ \left( \frac{h}{\lambda_i} - \frac{h}{\lambda_f} \right) + m_ec \right]^2 - \left( \frac{h^2}{\lambda_i^2} - 2\frac{h^2}{\lambda_i\lambda_f}\cos{\phi} + \frac{h^2}{\lambda_f^2} \right) = \left( \frac{m_e^2 c^2}{1-\frac{v^2}{c^2}} \right) - \left( \frac{m_e^2 v^2}{1-\frac{v^2}{c^2}} \right) \end{equation}\]
Simplify \(\text{(8)}\) from left hand side.
\[\begin{aligned} L.H.S. & = \left[ \left( \frac{h}{\lambda_i} - \frac{h}{\lambda_f} \right) + m_ec \right]^2 - \left( \frac{h^2}{\lambda_i^2} - 2\frac{h^2}{\lambda_i\lambda_f}\cos{\phi} + \frac{h^2}{\lambda_f^2} \right) \\ & = \left( \frac{h}{\lambda_i} - \frac{h}{\lambda_f} \right)^2 + 2m_ec \left( \frac{h}{\lambda_i} - \frac{h}{\lambda_f} \right) + m_e^2 c^2 - \frac{h^2}{\lambda_i^2} + \frac{2h^2}{\lambda_i\lambda_f}\cos{\phi} - \frac{h^2}{\lambda_f^2} \\ & = \cancel{\frac{h^2}{\lambda_i^2}} - \frac{2h^2}{\lambda_i\lambda_f} + \cancel{\frac{h^2}{\lambda_f^2}} + 2m_ech \frac{\lambda_f - \lambda_i}{\lambda_i \lambda_f} + m_e^2 c^2 - \cancel{\frac{h^2}{\lambda_i^2}} + \frac{2h^2}{\lambda_i\lambda_f}\cos{\phi} - \cancel{\frac{h^2}{\lambda_f^2}} \\ & = \frac{2h^2}{\lambda_i\lambda_f}\cos{\phi} - \frac{2h^2}{\lambda_i\lambda_f} + 2m_ech \frac{\lambda_f - \lambda_i}{\lambda_i\lambda_f} + m_e^2 c^2 \\ & = \frac{2h^2}{\lambda_i\lambda_f} \left( \cos\phi -1 \right) + \frac{2h^2}{\lambda_i\lambda_f} \frac{m_ec}{h} (\lambda_f - \lambda_i) + m_e^2 c^2 \\ & = \frac{2h^2}{\lambda_i\lambda_f} \left[ \left( \cos\phi -1 \right) + \frac{m_ec}{h} (\lambda_f - \lambda_i) \right] + m_e^2 c^2 \\ \end{aligned}\]Deal with the other side.
\[\begin{aligned} R.H.S. & = \left( \frac{m_e^2 c^2}{1-\frac{v^2}{c^2}} \right) - \left( \frac{m_e^2 v^2}{1-\frac{v^2}{c^2}} \right) \\ & = \frac{m_e^2 \left( c^2 - v^2 \right) }{1-\frac{v^2}{c^2}} \\ & = \frac{m_e^2 \left( c^2 - v^2 \right) }{\frac{c^2 - v^2}{c^2}} \\ & = m_e^2 \frac{\left( c^2 - v^2 \right) }{\left( c^2 - v^2 \right) } c^2 \\ & = m_e^2 c^2 \\ \end{aligned}\]And combine the both sides.
\[ \xcancel{\frac{2h^2}{\lambda_i\lambda_f}} \left[ \left( \cos\phi -1 \right) + \frac{m_ec}{h} (\lambda_f - \lambda_i) \right] + \cancel{m_e^2 c^2} = \cancel{m_e^2 c^2} \] \[ \Downarrow \] \[ \lambda_f - \lambda_i = \frac{h}{m_ec} \left( 1 - \cos\phi \right) \]
We also can define that \(\Delta \lambda \equiv \lambda_f - \lambda_i\) and obtain a clearer presentation of Compton wavelength.
\[ \Delta\lambda = \frac{h}{m_ec} \left( 1 - \cos\phi \right) \]