# Compton scattering angle formula and drawings

The Compton formula for scattering angles $$\varphi$$ and $$\theta$$ for electron and photon, respectively, can be shown as: $$\cot\varphi = (1+\frac{\lambda_c}{\lambda_i})\tan\frac{\theta}{2} \tag{1}\label{eq1}$$ where $$\lambda_c = \frac{h}{m_e c}$$ is the Compton wavelength of the electron which is equal to $$2.43\times 10^{−12}$$m. Below is the spectrum of X-rays in question.

Xray $$\lambda_i$$ (nm) 0.5 0.1 0.02 0.01 0.005 0.0025
Photon $$E_i$$ (keV) 2.48 12.4 62.1 124 248 497
Ratio of $${\lambda_c}/{\lambda_i}$$ 0.0049 0.024 0.122 0.243 0.486 0.972

For $${\lambda_c}/{\lambda_i}\ll 1$$, Equation \eqref{eq1} reduces to: $$\cot\varphi \approx \tan\frac{\theta}{2} = \cot(\frac{\pi}{2}-\frac{\theta}{2})\tag{2}\label{eq2}$$ Hence $$\varphi = \frac{\pi}{2}-\frac{\theta}{2} \tag{3}\label{eq3}$$ The plot of $$\varphi$$ against $$\theta$$ for Equation \eqref{eq3} is a straight line with a slope of $$-\frac{1}{2}$$ and intercept of $$\frac{\pi}{2}$$ as shown in Fig.4 subplot for the lower 3 energy X-rays in the above table. The higher 3 energy X-rays in the above table are plotted below.

As can be seen in the above subplot Fig.4, they approximate to the straight line even for $${\lambda_c}/{\lambda_i} \approx 1$$.

When we scan $$0 < \theta <180^{\circ}$$ in sequence, $$0 < \varphi<90^{\circ}$$ is scanned in the opposite direction as illustrated below:

The customary Fig.2 of the picture showing $$\theta$$ and $$\varphi$$ in the 1st and fourth quadrant should never happen. Both $$\theta$$ and $$\varphi$$ are either in the same quadrant, 1 or 4 when they cross each other at $$\pm 60^\circ$$, or $$\theta$$ can go to 2 and 3, respectively. Would you agree?

• What is it precisely that you are asking for? Commented Apr 16 at 3:12
• As for the quadrants, you are misunderstanding. The angle $\varphi$ is measured positive as it descends into the 4th quadrant. i.e. when $\theta$ is a small positive value, i.e. in the 1st quadrant, then $\varphi$ is also "in the 1st quadrant", but in the way we have defined things, it appears in the 4th quadrant, as drawn in Figure 2 of the Wikipedia picture you linked to. i.e. The picture is correct. Commented Apr 16 at 3:30
• For instance, #3 has both $\theta$ and $\varphi$ in 1st quadrant, #6 has $\theta$ in 2nd quadrant and $\varphi$ in 1st quadrant. They have to obey Eq.3, i.e. $\varphi + \frac{\theta}{2} = \frac{\pi}{2}$. There is no instance when $\theta$ is in 1st quadrant and $\varphi$ in 4th quadrant as Wiki showed. From the subplot Fig.4 above, for $\varphi$ in 4th quadrant, i.e. negative $\varphi$, $\theta$ has to be either in 3rd or 4th quadrant, i.e. in [180, 360]. Commented Apr 16 at 6:15
• No, you are misunderstanding it. Wiki is correct; what you think of as $\varphi$ is $\phi=-\varphi$; your 1st quadrant $\varphi$ is actually in the 4th quadrant $\phi$ Commented Apr 16 at 6:32
• I am referring to your question on this site, which is in agreement with the convention on Wikipedia. By considering Equation (3) in your question there, it is very clear that as $\varphi$ increases a little bit from zero, it is going into 4th quadrant. You think it is in 1st quadrant. You are wrong on that. Commented Apr 16 at 6:53

As @naturallyInconsistent pointed out, I made a wrong interpretation of eq. (3). When the magnitude of $$\pi/2>\varphi>0$$, the scattered photon angle should be negated as $$(-\pi/2, 0)$$.

Another view is that, if we assume the photon was scattered into a positive angle $$\beta$$ direction, the conservation of momentum would yield:

\begin{align} p_i &= p_f \cos \theta+p \cos \beta \ \ \ (x-\mathrm{component}) \tag{2''} \\ 0 &= p_f \sin \theta + p \sin \beta \ \ \ (y-\mathrm{component})\tag{3''} \end{align}

By substituting $$\beta = -\varphi$$ into eq. \eqref{eq2} and eq. \eqref{eq3}, they would reduce to eq. (2) and eq. (3), respectively. If we continued to carry out the $$\beta$$ versus $$\theta$$ derivation from eq. \eqref{eq2} and eq. \eqref{eq3} as before, we would arrive at:
$$\cot\beta = -\left(1+\frac{h}{\lambda_i m_e c}\right)\tan\frac{\theta}{2}= -\left(1+\frac{\lambda_c}{\lambda_i}\right)\tan\frac{\theta}{2}.\tag{6''}\label{eq6}$$

Again, by substituting $$\beta = -\varphi$$ into eq. \eqref{eq6}, we would have eq. (6). For $${\lambda_c}/{\lambda_i} \ll 1$$, it reduces to $$\beta \approx -\left( \frac{\pi}{2}-\frac{\theta}{2}\right). \tag{3'}\label{eq3'}$$

Therefore, the correct scanning drawing of angles $$\beta$$ vs $$\theta$$ would be:

The two angles $$\beta$$ and $$\theta$$ are split in different quadrants, and never cross each other.

• Is this an answer or an update to the question? Commented Apr 20 at 2:16
• This is an answer to correct my errors. Commented Apr 20 at 2:17