In the past, most experiments in particle physics involved stationary targets: one particle (usually a proton or an electron) was accelerated to a high energy E, and collided with a target particle at rest (Fig. 12.29a). Far higher relative energies are obtainable (with the same accelerator) if you accelerate both particles to energy E, and fire them at each other (Fig. 12.29b). Classically, the energy $\overline{\mathbf{E}}$of one particle, relative to the other, is just $\mathbf{4}\mathbf{E}$(why?) . . . not much of a gain (only a factor of $\mathbf{4}$). But relativistically the gain can be enormous. Assuming the two particles have the same mass, m, show that

$\mathbf{E}\mathbf{=}\frac{\mathbf{2}{\mathbf{E}}^{\mathbf{2}}}{{\mathbf{mc}}^{\mathbf{2}}}\mathbf{=}{\mathbf{mc}}^{\mathbf{2}}$ (12.58)

FIGURE 12.29

Suppose you use protons $\left({\mathrm{mc}}^2=1\mathrm{GeV}\right)$with $\mathbf{E}\mathbf{=}\mathbf{30}\mathbf{}\mathbf{GeV}$. What $\overline{\mathbf{E}}$do you get? What multiple of E does this amount to? $\left(1\mathrm{GeV}={10}^9\mathrm{electron}\mathrm{volts}\right)$[Because of this relativistic enhancement, most modern elementary particle experiments involve colliding beams, instead of fixed targets.]

Write the expression for the zeroth component of momentum in $\overline{\mathrm{S}}$ frame.

$$\begin{gathered} \overline{\mathrm{P}°}=\mathrm{\tau }\left(\mathrm{p}°-\mathrm{\beta }\mathrm{p}\text{'}\right) \\ \frac{\overline{\mathrm{E}}}{c}=\mathrm{\tau }\left(\frac{\mathrm{E}}{\mathrm{c}}\mathrm{\beta }\mathrm{p}\right) \end{gathered}$$

…… (1)

Here, E is the energy, c is the speed of light,$\mathrm{\beta }$is the Lorentz factor, and p is the momentum.

Write the expression for the energyE.

$\mathrm{E}={\mathrm{ymc}}^2$

Here, $\mathrm{y}$ is the Lorentz contraction, m is the mass, and c is the speed of light.

Write the expression for the momentum.

$\mathrm{p}=\mathrm{ymv}$

Here,vis the velocity.

Write the formula for the velocity in terms of the Lorentz factor.

$\mathrm{v}=\mathrm{\beta c}$

Substitute $\mathrm{E}={\mathrm{ymc}}^2,\mathrm{p}=\mathrm{ymv}\mathrm{and}\mathrm{v}=\mathrm{\beta c}$ in equation (1).

$$\begin{gathered} \frac{\overline{\mathrm{E}}}{c}=y\left(\frac{E}{c}-\beta \left(-ymv\right)\right) \\ \overline{\mathrm{E}}=\mathrm{y}\left(\frac{\mathrm{E}}{\mathrm{c}}-\mathrm{\beta }\left(-\mathrm{ym}\left(\mathrm{\beta c}\right)\right)\right)\mathrm{c} \\ \overline{\mathrm{E}}=\mathrm{y}\left(\frac{\mathrm{E}}{\mathrm{c}}+{\mathrm{ym\beta }}^2\mathrm{c}\right)\mathrm{c} \\ \overline{\mathrm{E}}=\mathrm{y}\left(\mathrm{E}+{\mathrm{ymc}}^2{\mathrm{\beta }}^2\right) \end{gathered}$$

........(2)

It is known that:

$\mathrm{y}=\frac{1}{\sqrt{1-{\displaystyle \frac{{\mathrm{v}}^2}{{\mathrm{c}}^2}}}}$

Substitute $\mathrm{\beta c}$ for$\mathrm{v}$in the above expression.

$$\begin{gathered} \mathrm{y}=\frac{1}{\sqrt{1-{\displaystyle \frac{{\left(\mathrm{\beta c}\right)}^2}{{\mathrm{c}}^2}}}} \\ \mathrm{y}=\frac{1}{\sqrt{1-{\displaystyle {\mathrm{\beta }}^2}}} \\ {\mathrm{y}}^2=\frac{1}{1-{\mathrm{\beta }}^2} \\ {\mathrm{\beta }}^2=\frac{{\mathrm{y}}^2-1}{{\mathrm{y}}^2} \end{gathered}$$

Substitute role="math" localid="1654322970711" $\frac{{\mathrm{y}}^2-1}{{\mathrm{y}}^2}$ for ${\mathrm{\beta }}^2$in equation (2).

$$\begin{gathered} \overline{\mathrm{E}}=7\left(\mathrm{E}+{\mathrm{ymc}}^2\left(\frac{{\mathrm{y}}^2-1}{{\mathrm{y}}^2}\right)\right) \\ \mathrm{E}=7\mathrm{E}-\left({\mathrm{y}}^{2}1\right){\mathrm{mc}}^2 \end{gathered}$$

Substitute $\frac{\mathrm{E}}{{\mathrm{mc}}^2}$ for $\mathrm{y}$ in the above expression.

$$\begin{gathered} \overline{\mathrm{E}}=\left(\frac{\mathrm{E}}{{\mathrm{mc}}^2}\right)\mathrm{E}+\left({\left(\frac{\mathrm{E}}{{\mathrm{mc}}^2}\right)}^2-1\right){\mathrm{mc}}^2 \\ \overline{\mathrm{E}}=\left(\frac{{\mathrm{E}}^2}{{\mathrm{mc}}^2}\right)+\left(\frac{{\mathrm{E}}^2}{{\mathrm{mc}}^2}\right)-\left({\mathrm{mc}}^2\right) \\ \overline{\mathrm{E}}=\frac{2{\mathrm{E}}^2}{{\mathrm{mc}}^2}-{\mathrm{mc}}^2 \end{gathered}$$ …… (3)

Substitute $30\mathrm{GeV}\mathrm{for}\mathrm{E}$ in equation (3).

$$\begin{gathered} \overline{\mathrm{E}}=\frac{2{\left(30\mathrm{GeV}\right)}^2}{1\mathrm{GeV}}-\left(1\mathrm{GeV}\right) \\ \overline{\mathrm{E}}=1799\mathrm{GeV} \end{gathered}$$

So, the multiple ofEto the required amount will be,

$\overline{\mathrm{E}}=60\mathrm{E}$

Therefore, the expression for $\overline{\mathrm{E}}$is $\overline{\mathrm{E}}=\frac{2{\mathrm{E}}^2}{{\mathrm{mc}}^2}-{\mathrm{mc}}^2$ and the multiple of E to the required amount is $\overline{\mathrm{E}}=60\mathrm{E}$.