Compton scatering problem - How to solve this large problem faster using the momentum system?

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?

• My initial thoughts are: I think conservation of Energy and momentum should suffice to solve this question. Also, working in the center of momentum frame, may simplify things. – Ali Jul 16 '13 at 12:38
• Oh i dodnt think of that. Could you show how would you do this? – 71GA Jul 16 '13 at 12:43
• since this is a homework, I will give my solution as a "Hint". – Ali Jul 16 '13 at 13:01
• I am interested in what manner does the momentum frame simplify things out. – 71GA Jul 16 '13 at 13:16
• The $\psi = 0$ solution always appears; it just means that missing the target also preserves momentum and energy, so it's a solution! – Ali Jul 16 '13 at 14:55

Even without going to the COM frame, we can solve this problem(Actually since one of the given parameters is an angle in the lab frame, I'm not sure if moving to COM will help in this case). Our given relations are:

• Momentum 4-vector length(for each particle, i.e. photon and electron): $$\left(\frac{E}{c} \right)^2-\vec{p}^2=m^2c^2$$ (for photons $m=0$)

• Conservation of Momentum 4-vector(3 equations): $$2m_e c^2=E_{\gamma}'+E_e' \\ m_e c = p_{\gamma}' \cos \phi+p_e' \cos \vartheta \\ 0=p_{\gamma}'\sin \phi-p_e' \sin \vartheta$$

These can be solved to give $E_\gamma'$:

$$E_\gamma'=\frac{5-3 \cos(2\theta)}{7-\cos(2 \theta)}m_e c^2\approx 0.335 MeV$$

• Oh my god. Please take a look at my edit where i got the same result as you did. But your simplification is amazing! Can you further explain please how did you solve the system of 3 equations you have written. – 71GA Jul 16 '13 at 14:48
• There are actually five equations, the other two are coming from the 4-momentum length; namely $\frac{E_{\gamma}}{c}=p_{\gamma}$ and $\left(\frac{E_e}{c}\right)^2-{p_e}^2={m_e}^2 c^2$. – Ali Jul 16 '13 at 14:59
• I understand the first two, but how did you get later 3? I mean these three: $$2m_e c^2=E_{\gamma}+E_e \\ m_e c = p_{\gamma} \cos \phi+p_e \cos \vartheta \\ 0=p_{\gamma}\sin \phi-p_e \sin \vartheta$$ – 71GA Jul 16 '13 at 15:01
• The first is just conservation of energy, and the other two are conservation of momentum. – Ali Jul 16 '13 at 15:05
• Oh i see now :) It is somewhat obvious but i am a bit dizzy after a whole day of trying to solve this 1 problem. Thank you. I think this will help me to solve my problems much faster when i get my test. Could you just fix the symbols $E_\gamma$ to $E_\gamma'$ as it is the "scattered photon". Just in case if anyone else reads this topic. – 71GA Jul 16 '13 at 15:12