91影视

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

$$\begin{gathered} \overline{\mathrm{P}掳}=\mathrm{蟿}\left(\mathrm{p}掳-\mathrm{尾}\mathrm{p}\text{'}\right) \\ \frac{\overline{\mathrm{E}}}{c}=\mathrm{蟿}\left(\frac{\mathrm{E}}{\mathrm{c}}\mathrm{尾}\mathrm{p}\right) \end{gathered}$$

鈥︹�� (1)

Here, E is the energy, c is the speed of light,$\mathrm{尾}$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.

$\mathrm{v}=\mathrm{尾肠}$

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

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

........(2)

It is known that:

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

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

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

Substitute role="math" localid="1654322970711" $\frac{{\mathrm{y}}^2-1}{{\mathrm{y}}^2}$ for ${\mathrm{尾}}^2$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.

$$\begin{gathered} \overline{\mathrm{E}}=\left(\frac{\mathrm{E}}{{\mathrm{mc}}^2}\right)\mathrm{E}+\left({\left(\frac{\mathrm{E}}{{\mathrm{mc}}^2}\right)}^2-1\right){\mathrm{mc}}^2 \\ \overline{\mathrm{E}}=\left(\frac{{\mathrm{E}}^2}{{\mathrm{mc}}^2}\right)+\left(\frac{{\mathrm{E}}^2}{{\mathrm{mc}}^2}\right)-\left({\mathrm{mc}}^2\right) \\ \overline{\mathrm{E}}=\frac{2{\mathrm{E}}^2}{{\mathrm{mc}}^2}-{\mathrm{mc}}^2 \end{gathered}$$ 鈥︹�� (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 $\overline{\mathrm{E}}=60\mathrm{E}$.