Chapter 40, Problem 64AP

To determine

The derivation of equation for the Compton shift from equations 40.14 through 40.16.

Answer to Problem 64AP

The equation for the Compton shift is derived from equations 40.14 through 40.16.

Explanation of Solution

Write equation 40.15.

  $\frac{h}{{\lambda }_0}-\frac{h}{{\lambda }^{\prime }}\cos \theta =\gamma m_{e}u\cos \varphi$                                                                                         (I)

Here, $h$ is the Plank’s constant, ${\lambda }_0$ is the wavelength of the incident photon, ${\lambda }^{\prime }$ is the wavelength of the scattered photon, $\theta$ is the angle through which the photon is scattered, $\gamma$ is the Lorentz factor, $m_e$ is the mass of the electron, $u$ is the speed of the electron and $\varphi$ is the angle through which the electron is scattered.

Write equation 40.16.

  $\frac{h}{{\lambda }^{\prime }}\sin \theta =\gamma m_{e}u\sin \varphi$                                                                                            (II)

Take the square of equations (I) and (II) and add them together.

  $\begin{array}{c}{\left(\frac{h}{{\lambda }_0}-\frac{h}{{\lambda }^{\prime }}\cos \theta \right)}^2+{\left(\frac{h}{{\lambda }^{\prime }}\sin \theta \right)}^2={\left(\gamma m_{e}u\cos \varphi \right)}^2+{\left(\gamma m_{e}u\sin \varphi \right)}^2\\ {\gamma }^2{m}_e^2{u}^2\left({\cos }^2\varphi +{\sin }^2\varphi \right)=h^2\left(\frac{1}{{\lambda }_0^2}-\frac{2\cos \theta }{{\lambda }_0{\lambda }^{\prime }}+\frac{1}{{{\lambda }^{\prime }}^2}\right)\\ {\gamma }^2{m}_e^2{u}^2=h^2\left(\frac{1}{{\lambda }_0^2}-\frac{2\cos \theta }{{\lambda }_0{\lambda }^{\prime }}+\frac{1}{{{\lambda }^{\prime }}^2}\right)\end{array}$                              (III)

Write the expression for the Lorentz factor.

  $\gamma =\frac{1}{\sqrt{1-u^2/c^2}}$                                                                                                      (IV)

Here, $c$ is the speed of light in vacuum.

Take the square of equation (IV).

  ${\gamma }^2=\frac{1}{1-u^2/c^2}$

Put the above equation in equation (III).

  $\begin{array}{c}\left(\frac{1}{1-u^2/c^2}\right)m_e^2{u}^2=h^2\left(\frac{1}{{\lambda }_0^2}-\frac{2\cos \theta }{{\lambda }_0{\lambda }^{\prime }}+\frac{1}{{{\lambda }^{\prime }}^2}\right)\\ \frac{u^2}{1-u^2/c^2}=\frac{h^2}{m_e^2}\left(\frac{1}{{\lambda }_0^2}+\frac{1}{{{\lambda }^{\prime }}^2}-\frac{2\cos \theta }{{\lambda }_0{\lambda }^{\prime }}\right)\end{array}$

Divide both sides of the above equation by $c^2$ .

  $\frac{u^2/c^2}{1-u^2/c^2}=\frac{h^2}{m_e^2{c}^2}\left(\frac{1}{{\lambda }_0^2}+\frac{1}{{{\lambda }^{\prime }}^2}-\frac{2\cos \theta }{{\lambda }_0{\lambda }^{\prime }}\right)$                                                                    (V)

Define $b=\frac{h^2}{m_e^2{c}^2}\left(\frac{1}{{\lambda }_0^2}+\frac{1}{{{\lambda }^{\prime }}^2}-\frac{2\cos \theta }{{\lambda }_0{\lambda }^{\prime }}\right)$.

Rewrite equation (V) in terms of $b$ .

  $\frac{u^2/c^2}{1-u^2/c^2}=b$                                                                                  (VI)

Add $1$ to both sides of equation (VI).

  $\begin{array}{c}1+\frac{u^2/c^2}{1-u^2/c^2}=1+b\\ \frac{1-u^2/c^2+u^2/c^2}{1-u^2/c^2}=1+b\\ \frac{1}{1-u^2/c^2}=1+b\\ 1-u^2/c^2=\frac{1}{1+b}\end{array}$                                                                                       (VII)

Rewrite equation (VI) for $u^2/c^2$ .

  $u^2/c^2=b\left(1-u^2/c^2\right)$                                                                                            (VIII)

Put equation (VII) in equation (VIII).

  $\begin{array}{c}{u}^2/c^2=b\left(\frac{1}{1+b}\right)\\ =\frac{b}{1+b}\end{array}$                                                                                                 (IX)

Write equation (IV).

  $\frac{hc}{{\lambda }_0}=\frac{hc}{{\lambda }^{\prime }}+\left(\gamma -1\right)m_e{c}^2$

Modify the above equation.

  $\begin{array}{c}hc\left[\frac{1}{{\lambda }_0}-\frac{1}{{\lambda }^{\prime }}\right]=\left(\gamma -1\right)m_e{c}^2\\ \frac{hc}{m_e{c}^2}\left[\frac{1}{{\lambda }_0}-\frac{1}{{\lambda }^{\prime }}\right]=\gamma -1\\ 1+\frac{h}{m_{e}c}\left[\frac{1}{{\lambda }_0}-\frac{1}{{\lambda }^{\prime }}\right]=\gamma \end{array}$

Put equation (IV) in the above equation.

  $1+\frac{h}{m_{e}c}\left[\frac{1}{{\lambda }_0}-\frac{1}{{\lambda }^{\prime }}\right]=\frac{1}{\sqrt{1-u^2/c^2}}$

Put equation (IX) in the above equation.

  $\begin{array}{c}1+\frac{h}{m_{e}c}\left[\frac{1}{{\lambda }_0}-\frac{1}{{\lambda }^{\prime }}\right]=\frac{1}{\sqrt{1-\frac{b}{1+b}}}\\ ={\left(1-\frac{b}{1+b}\right)}^{-1/2}\end{array}$

Simplify the above equation.

  $\begin{array}{c}1+\frac{h}{m_{e}c}\left[\frac{1}{{\lambda }_0}-\frac{1}{{\lambda }^{\prime }}\right]={\left(\frac{1+b-b}{1+b}\right)}^{-1/2}\\ ={\left({\left(\frac{1}{1+b}\right)}^{-1}\right)}^{1/2}\\ ={\left(\frac{1+b}{1}\right)}^{1/2}\\ =\sqrt{1+b}\end{array}$

Square both sides of the above equation.

  $\begin{array}{c}{\left(1+\frac{h}{m_{e}c}\left[\frac{1}{{\lambda }_0}-\frac{1}{{\lambda }^{\prime }}\right]\right)}^2={\left(\sqrt{1+b}\right)}^2\\ 1+\frac{2h}{m_{e}c}\left[\frac{1}{{\lambda }_0}-\frac{1}{{\lambda }^{\prime }}\right]+\frac{h^2}{m_e^2{c}^2}{\left[\frac{1}{{\lambda }_0}-\frac{1}{{\lambda }^{\prime }}\right]}^2=1+b\end{array}$

Put the expression for $b$ in the above equation.

  $\begin{array}{c}1+\frac{2h}{m_{e}c}\left[\frac{1}{{\lambda }_0}-\frac{1}{{\lambda }^{\prime }}\right]+\frac{h^2}{m_e^2{c}^2}{\left[\frac{1}{{\lambda }_0}-\frac{1}{{\lambda }^{\prime }}\right]}^2=1+\frac{h^2}{m_e^2{c}^2}\left(\frac{1}{{\lambda }_0^2}+\frac{1}{{{\lambda }^{\prime }}^2}-\frac{2\cos \theta }{{\lambda }_0{\lambda }^{\prime }}\right)\\ 1+\frac{2h}{m_{e}c}\left[\frac{1}{{\lambda }_0}-\frac{1}{{\lambda }^{\prime }}\right]+\frac{h^2}{m_e^2{c}^2}\left[\frac{1}{{\lambda }_0^2}-\frac{2}{{\lambda }_0{\lambda }^{\prime }}+\frac{1}{{{\lambda }^{\prime }}^2}\right]=1+\frac{h^2}{m_e^2{c}^2}\left(\frac{1}{{\lambda }_0^2}+\frac{1}{{{\lambda }^{\prime }}^2}-\frac{2\cos \theta }{{\lambda }_0{\lambda }^{\prime }}\right)\\ \frac{2h}{m_{e}c}\left[\frac{1}{{\lambda }_0}-\frac{1}{{\lambda }^{\prime }}\right]-\frac{h^2}{m_e^2{c}^2}\left(\frac{2}{{\lambda }_0{\lambda }^{\prime }}\right)=-\frac{h^2}{m_e^2{c}^2}\left(\frac{2\cos \theta }{{\lambda }_0{\lambda }^{\prime }}\right)\\ \frac{h}{m_{e}c}\left[\frac{1}{{\lambda }_0}-\frac{1}{{\lambda }^{\prime }}\right]-\frac{h^2}{m_e^2{c}^2}\left(\frac{1}{{\lambda }_0{\lambda }^{\prime }}\right)=-\frac{h^2}{m_e^2{c}^2}\left(\frac{\cos \theta }{{\lambda }_0{\lambda }^{\prime }}\right)\end{array}$

Multiply both sides of the above equation by $\frac{m_e^2{c}^2}{h^2}$ .

  $\begin{array}{c}\left(\frac{m_e^2{c}^2}{h}\right)\frac{h}{m_{e}c}\left[\frac{1}{{\lambda }_0}-\frac{1}{{\lambda }^{\prime }}\right]-\left(\frac{m_e^2{c}^2}{h}\right)\frac{h^2}{m_e^2{c}^2}\left(\frac{1}{{\lambda }_0{\lambda }^{\prime }}\right)=-\left(\frac{m_e^2{c}^2}{h}\right)\frac{h^2}{m_e^2{c}^2}\left(\frac{\cos \theta }{{\lambda }_0{\lambda }^{\prime }}\right)\\ m_{e}c\left[\frac{1}{{\lambda }_0}-\frac{1}{{\lambda }^{\prime }}\right]-h\left(\frac{1}{{\lambda }_0{\lambda }^{\prime }}\right)=-h\left(\frac{\cos \theta }{{\lambda }_0{\lambda }^{\prime }}\right)\\ m_{e}c\left[\frac{{\lambda }^{\prime }-{\lambda }_0}{{\lambda }_0{\lambda }^{\prime }}\right]-\frac{h}{{\lambda }_0{\lambda }^{\prime }}=-h\left(\frac{\cos \theta }{{\lambda }_0{\lambda }^{\prime }}\right)\\ m_{e}c\left({\lambda }^{\prime }-{\lambda }_0\right)-h=-h\cos \theta \end{array}$               (X)

Conclusion:

Rearrange equation (X).

  $\begin{array}{c}{m}_{e}c\left({\lambda }^{\prime }-{\lambda }_0\right)=h-h\cos \theta \\ m_{e}c\left({\lambda }^{\prime }-{\lambda }_0\right)=h\left(1-\cos \theta \right)\\ {\lambda }^{\prime }-{\lambda }_0=\frac{h}{m_{e}c}\left(1-\cos \theta \right)\end{array}$

Therefore, the equation for the Compton shift is derived from equations 40.14 through 40.16.