I have encountered a hard exercise which i cannot quite solve. Could anyone help me with it? This is the exercise:
Lets say we have a photon whose energy $W_f$ is equal to the relativistic energy of an electron $W_{e0}=m_e c^2$. What is the energy of a scaterred photon $W_f'$ if after the "collision" the electron is moving in a direction $\vartheta =40^\circ$ according to the original direction of a photon?
What i did first was to draw the image (It is in Slovenian language so dont bother about what is written):
Now i decided to first calculate the $\lambda$ of the incomming photon:
\begin{align} W_f &= W_{0e}\\ \frac{hc}{\lambda} &= m_e c^2\\ \lambda &= \frac{hc}{m_e c^2}\\ \substack{\text{this looks similar to the}\\\text{1st part of the Compton's law}} \longrightarrow \lambda &= \frac{h}{m_e c}\\ \lambda &= \frac{6.626\times 10 ^{-34} Js}{9.109\times10^{-31}\cdot 2.99\times 10^8 \tfrac{m}{s}}\\ \lambda &\approx 2.45pm \end{align}
Now if i try to get the energy of a scattered photon i write down the conservation of energy:
\begin{align} W_{before} &= W_{after}\\ \frac{hc}{\lambda} + m_ec^2 &= \frac{hc}{\lambda'} + m_ec^2 + W_{ke}\\ \frac{hc}{\lambda} &= \frac{hc}{\lambda'} + W_{ke}\\ \end{align}
This is one equation with two variables ($\lambda'$ and $W_{ke}$) so i am kinda stuck here and i need a new equation which must be a Compton's law. If i implement it i get:
\begin{align} \frac{hc}{\lambda} &= \frac{hc}{\lambda'} + W_{ke}\\ \frac{hc}{\lambda} &= \frac{hc}{\lambda + \Delta \lambda} + W_{ke}\\ \frac{hc}{\lambda} &= \frac{hc}{\lambda + \tfrac{h}{m_ec}(1-\cos \phi)} + W_{ke}\\ \end{align}
Still i notice that i have 2 variables (now they are $\phi$ and $W_{ke}$). At this point i know i need 1 more equation. I presume it is from the momentum conservation so i write down the momentum conservation for direction $y$ and $x$:
Direction $y$:
\begin{align} p_{before} &= p_{after}\\ 0 &= \frac{h}{\lambda'}\sin\phi - p_e \sin\vartheta\\ p_e &= \frac{h}{\lambda'}\frac{\sin\phi}{\sin\vartheta} \end{align}
Direction $x$: \begin{align} p_{before} &= p_{after}\\ \frac{h}{\lambda} &= \frac{h}{\lambda'}\cos\phi + p_e \cos\vartheta\leftarrow \substack{\text{here i implement what i got from the }\\\text{conserv. of momentum in direction $y$}}\\ \frac{h}{\lambda} &= \frac{h}{\lambda'}\cos\phi + \frac{h}{\lambda'}\frac{\sin\phi}{\sin\vartheta} \cos\vartheta\\ \frac{1}{\lambda} &= \frac{1}{\lambda'} \left(\cos\phi + \frac{\sin\phi}{\tan\vartheta}\right)\\ \lambda' &= \lambda \left(\cos\phi + \frac{\sin\phi}{\tan\vartheta}\right)\leftarrow\substack{\text{It seems to me that i could solve}\\\text{this for $\phi$ if i used Compton's law}}\\ \lambda + \Delta \lambda &= \lambda \left(\cos\phi + \frac{\sin\phi}{\tan\vartheta}\right)\\ \lambda + \tfrac{h}{m_e c} (1 - \cos\phi) &= \lambda \left(\cos\phi + \frac{\sin\phi}{\tan\vartheta}\right) \leftarrow \substack{\text{I got 1 equation for 1 variable $\phi$ but}\\\text{it gets complicated as you will see...}}\\ 1 + \tfrac{h}{\lambda m_e c} (1-\cos \phi) &= \frac{\cos\phi \tan\vartheta + \sin\phi}{\tan\vartheta}\\ \tan\vartheta + \tfrac{h}{\lambda m_e c}\tan\vartheta - \tfrac{h}{\lambda m_e c}\tan\vartheta \cos\phi &= \cos\phi \tan\vartheta + \sin \phi\\ \tan\vartheta \left(1 + \tfrac{h}{\lambda m_e c} \right) &= \cos\phi \tan\vartheta \left(1 + \tfrac{h}{\lambda m_e c}\right) + \sin\phi\\ \tan\vartheta \left(1 + \tfrac{h}{\lambda m_e c}\right) \left[1 - \cos\phi\right] &= \sin \phi\\ \tan^2\vartheta \left(1 + \tfrac{h}{\lambda m_e c}\right)^2 \left[1 - \cos\phi\right]^2 &= \sin^2 \phi\\ \tan^2\vartheta \left(1 + \tfrac{h}{\lambda m_e c}\right)^2 \left[1 - \cos\phi\right]^2 + \cos^2\phi&= \sin^2 \phi + \cos^2\phi\\ \underbrace{\tan^2\vartheta \left(1 + \tfrac{h}{\lambda m_e c}\right)^2}_{\equiv \mathcal{A}} \left[1 - \cos\phi\right]^2 + \cos^2\phi&= 1 \leftarrow \substack{\text{i define a new variable $\mathcal{A}$}\\\text{for easier calculations}}\\ \mathcal{A} \left[1 - 2\cos\phi + \cos^2\phi \right] + \cos^2 \phi - 1 &= 0\\ \mathcal{A} - 2\mathcal{A} \cos\phi + \mathcal{A}\cos^2\phi + \cos^2 \phi - 1 &= 0\\ (\mathcal{A}+1)\cos^2\phi - 2\mathcal{A} \cos\phi + (\mathcal{A} - 1) &= 0\leftarrow \substack{\text{in the end i get the quadratic equation}\\\text{which has a cosinus.}} \end{align}
Question: Is it possible to continue by solving this quadratic equation as a regular quadratic equation using the "completing the square method"?
I mean like this:
\begin{align} \underbrace{(\mathcal{A}+1)}_{\equiv A}\cos^2\phi + \underbrace{-2\mathcal{A}}_{\equiv B} \cos\phi + \underbrace{(\mathcal{A} - 1)}_{\equiv C} &= 0 \end{align}
and finally:
$$ \boxed{\cos \phi = \dfrac{-B \pm \sqrt{B^2 - 4AC}}{2A}}$$
Afterall if this is possible i get $\cos \phi$ and therefore $\phi$, $W_{ke}$ and finally $W_f'$.
EDIT:
I did try to solve this using the quadratic equation and i got solution:
\begin{align} \cos \phi &= \dfrac{-B \pm \sqrt{B^2 - 4AC}}{2A}\\ \cos \phi &= \frac{2\mathcal{A} \pm \sqrt{4\mathcal{A}^2 - 4(\mathcal{A}+1)(\mathcal{A} - 1)}}{2 (\mathcal{A}+1)}\\ \cos \phi &= \frac{2\mathcal{A} \pm \sqrt{4\mathcal{A}^2 - 4(\mathcal{A}^2-1)}}{2 (\mathcal{A}+1)}\\ \cos \phi &= \frac{2\mathcal{A} \pm \sqrt{4\mathcal{A}^2 - 4\mathcal{A}^2 + 4}}{2 (\mathcal{A}+1)}\\ \cos \phi &= \frac{2\mathcal{A} \pm \sqrt{4}}{2\mathcal{A}+2)}\\ \cos \phi &= \frac{2\mathcal{A} \pm \sqrt{2}}{2\mathcal{A}+2)}\\ \end{align}
So if i apply "+" i get $\cos \phi = 1$ which is impossible for a photon to hold its original direction! But if i apply "-" and insert a variable $\mathcal{A}$ i get:
\begin{align} \cos \phi = \frac{2 \cdot \tan^240^\circ \left(1 + \tfrac{6.626\times10^{-34}Js}{2.45\times10^{-12}m\cdot 9.109\times10^{-31}kg\cdot2.99\times10^{8}m/s}\right)^2 - 2}{2 \cdot \tan^240^\circ \left(1 + \tfrac{6.626\times10^{-34}Js}{2.45\times10^{-12}m\cdot 9.109\times10^{-31}kg\cdot2.99\times10^{8}m/s}\right)^2 + 2} = \frac{5.59 - 2}{5.59 + 2} = 0.47 \end{align}
Now i can calculate: \begin{align} \phi&=\cos^{-1}0.47 = 61.74^\circ\\ \Delta \lambda &= \frac{h}{m_e c} (1-\cos\phi) = 1.28pm\\ \lambda' &= \lambda + \Delta \lambda = 3.73pm\\ W_f' &= \frac{hc}{\lambda'} = 5.31\times10^{-14}J = 331.55 keV. \end{align}
And the result is correct according by my book. But this way of solving a problem is very long and in a case if i get it in my test i cannot solve it in time i think. So how can i solve it faster? In the comments it was mentioned that i should use the momentum coordinate system? How do i do that?
