Exercises 23 and 24 give the threshold energies for which two particles of mass m can produce a given mass M in collidingbeam and stationary-target accelerator. Evaluate the two for a collision in which two protons become three protons and one antiproton. How much more energy is neededfor the stationary target?

Let $m_p$be the initial mass of each proton.

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

It can be expressed as:

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

Let$m_p$be the initial mass of each proton.

After the collision, the two protons become three protons and one antiproton.

So, the total mass of the four-particle after the collision is $M=4m_p$ .

For the case both protons are moving, the threshold energy for this reaction can be expressed as follows:

$\begin{array}{rcl}\left(M-2m_P\right)c^2& =& 4m_P{c}^2-2m_P{c}^2\\ & =& 2m_P{c}^2\\ & & \end{array}$

Therefore, the energy needed for the colliding beam is $2m_P{c}^2$.

For the case of the stationary target, the threshold energy for this reaction can be expressed, as shown below:

$\begin{array}{rcl}\left(\frac{M^2}{2m_P}-2m_P\right)c^2& =& \left(\frac{{\left(4m_P\right)}^2}{2m_P}-2m_P\right)c^2\\ & =& 6m_P{c}^2\\ & & \end{array}$

Thus, the threshold energy for the stationary target is about $6m_P{c}^2$.

The difference in threshold energy is, $6m_P{c}^2-2m_P{c}^2=4m_P{c}^2$.Therefore, $4m_P{c}^2$ more energy is needed for the stationary target.The energy needed for the colliding beam is $2m_p{c}^2$.The threshold energy for the stationary target is about$6m_p{c}^2$ . $4m_p{c}^2$more energy is needed for the stationary target.