Question

Exercise 23 discusses the threshold energy for two particles of mass m in a colliding beam accelerator to produce a final stationary mass M. If the accelerator is instead a stationary target type, more initial kinetic energy is needed to produce the same final mass. Show the threshold energy is $\left(\frac{M^2}{2m}-2m\right)c^2$.

Short answer

It’s proved that the threshold energy is $\left(\frac{M^2}{2m}-2m\right)c^2$.

Step-by-step solution

Given data

If one of the particles is stationary, choose different reference frames for before and after the collision.

The quantity ${\left(\frac{E_{\text{total}}}{c}\right)}^2-p_{\text{total}}^2$is conserved before and after the collision, even with different reference frames.

Relation between energy and momentum

Firstly, to calculate this quantity after the collision in the reference frame where the final product is stationary.

It can be expressed as:

${\left(\frac{E_{\text{total}}}{c}\right)}^2-p_{\text{total}}^2=M^2{c}^2$

Determine the relationship between energy and momentum 

Now, calculate this quantity before the collision in the lab frame where the target is stationary, which is,

${\left(\frac{E_{\text{total}}}{c}\right)}^2-p_{\text{total}}^2={\left(\frac{E_{\text{move}}+m{c}^2}{c}\right)}^2-P_{\text{move}}^2$

The relation between energy $E_{\text{move}}$ and momentum $P_{\text{move}}$can be expressed as follows:

$E_{\text{move}}^2=P_{\text{move}}^2{c}^2+m^2{c}^4$

Re-arrange the equation for $P_{\text{move}}$as,

$P_{\text{move}}^2=\frac{E_{\text{move}}^2-m^2{c}^4}{c^2}$

Thus, the equation ${\left(\frac{E_{\text{total}}}{c}\right)}^2-p_{\text{total}}^2={\left(\frac{E_{\text{move}}+m{c}^2}{c}\right)}^2-P_{\text{move}}^2$ becomes:

$\begin{array}{rcl}{\left(\frac{E_{\text{total}}}{c}\right)}^2-p_{\text{total}}^2& =& {\left(\frac{E_{\text{move}}+m{c}^2}{c}\right)}^2-P_{\text{move}}^2\\ & =& {\left(\frac{E_{\text{move}}-m{c}^2}{c}\right)}^2-\frac{E_{\text{move}}^2-m^2{c}^4}{c^2}\\ & =& \frac{2E_{\text{move}}m{c}^2+2m^2{c}^4}{c^2}\\ & & \end{array}$

Determine the threshold energy 

Substitute $M^2{c}^2$for ${\left(\frac{E_{\text{total}}}{c}\right)}^2-p_{\text{total}}^2$as:

$\begin{array}{rcl}{M}^2{c}^2& =& 2E_{\text{move}}m+2m^2{c}^2\\ E_{\text{move}}& =& \frac{M^2{c}^2-2m^2{c}^2}{2m}\\ & =& \left(\frac{M^2}{2m}-m\right)c^2\\ & & \end{array}$

Thus, the threshold energy can be expressed as,

$E_{\text{therebold}}=E_{\text{move}}-m{c}^2$

Substitute $\left(\frac{M^2}{2m}-m\right)c^2$ for $E_{\text{move}}$:

$\begin{array}{rcl}{E}_{\text{therebold}}& =& \left(\frac{M^2}{2m}-m\right)c^2-m{c}^2\\ & =& \left(\frac{M^2}{2m}-2m\right)c^2\\ & & \end{array}$

Thus, the threshold energy is $\left(\frac{M^2}{2m}-2m\right)c^2$.