Chapter 39, Problem 9P

(a)

To determine

The length of each bundle in its own reference frame.

Explanation of Solution

Given:

In laboratory frame of reference length of each bundle is $1.0\text{cm}$ .

Diameter of each bundle in laboratory frame of refence is $10\mu \text{m}$ .

Each particle has energy of $50\text{GeV}$ .

Formula used:

Write expression for energy of beam.

  $E=\gamma m{c}^2$

Solve above expression for $\gamma$ .

  $\gamma =\frac{E}{m{c}^2}$

Write expression for proper length.

  $L_p=\gamma L$

Substitute $\frac{E}{m{c}^2}$ for $\gamma$ in above expression.

  $L_p=\frac{E}{m{c}^2}L$  ....... (1)

Calculation:

Substitute $50\text{Gev}$ for $E$ , $0.511\text{MeV}$ for $m{c}^2$ and $1.0\text{cm}$ for $L$ in equation (1).

  $\begin{array}{l}{L}_p=\frac{50\text{GeV}}{0.511\text{MeV}}\left(\frac{{10}^9\text{eV}}{1\text{GeV}}\right)\left(\frac{1\text{MeV}}{{10}^6\text{eV}}\right)\left(1.0\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)\\ L_p=978.5\text{m}\end{array}$

Conclusion:

Thus, length of each bundle is $978.5\text{m}$ .

(b)

To determine

The minimum proper length of accelerator.

Explanation of Solution

Given:

In laboratory frame of reference length of each bundle is $1.0\text{cm}$ .

Diameter of each bundle in laboratory frame of refence is $10\mu \text{m}$ .

Each particle has energy of $50\text{GeV}$ .

Formula used:

Write expression for energy of beam.

  $E=\gamma m{c}^2$

Solve above expression for $\gamma$ .

  $\gamma =\frac{E}{m{c}^2}$

Write expression for proper length.

  $L_p=\gamma L$

Substitute $\frac{E}{m{c}^2}$ for $\gamma$ in above expression.

  $L_p=\frac{E}{m{c}^2}L$  ....... (1)

Write expression for proper length of accelerator.

  $L_p=\frac{L_{p,acc}}{\gamma }$

Solve above expression for $L_{p.acc}$ .

  $L_{p,acc}=\gamma L_p$

Substitute $\frac{E}{m{c}^2}$ for $\gamma$ in above expression.

  $L_{p,acc}=\frac{E}{m{c}^2}{L}_p$  ....... (2)

Calculation:

Substitute $50\text{Gev}$ for $E$ , $0.511\text{MeV}$ for $m{c}^2$ and $1.0\text{cm}$ for $L$ in equation (1).

  $\begin{array}{l}{L}_p=\frac{50\text{GeV}}{0.511\text{MeV}}\left(\frac{{10}^9\text{eV}}{1\text{GeV}}\right)\left(\frac{1\text{MeV}}{{10}^6\text{eV}}\right)\left(1.0\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)\\ L_p=978.5\text{m}\end{array}$

Substitute $50\text{Gev}$ for $E$ , $0.511\text{MeV}$ for $m{c}^2$ and $978.5\text{m}$ for $L_p$ in equation (2).

  $\begin{array}{l}{L}_{p,acc}=\frac{50\text{GeV}}{0.511\text{MeV}}\left(\frac{{10}^9\text{eV}}{1\text{GeV}}\right)\left(\frac{1\text{MeV}}{{10}^6\text{eV}}\right)\left(978.5\text{m}\right)\\ L_{p,acc}=9.6\times {10}^7\text{m}\end{array}$

Conclusion:

Thus, the minimum proper length of accelerator is $9.6\times {10}^7\text{m}$ .

(c)

To determine

The length of positron bundle which moves in the refence of electron bundle.

Explanation of Solution

Given:

In laboratory frame of reference length of each bundle is $1.0\text{cm}$ .

Diameter of each bundle in laboratory frame of refence is $10\mu \text{m}$ .

Each particle has energy of $50\text{GeV}$ .

Formula used:

Write expression for energy of beam.

  $E=\gamma m{c}^2$

Solve above expression for $\gamma$ .

  $\gamma =\frac{E}{m{c}^2}$

Write expression for length of positron bundle.

  $L_{pos}=\frac{L}{\gamma }$

Substitute $\frac{E}{m{c}^2}$ for $\gamma$ in above expression.

  $L_{pos}=\frac{L}{\frac{E}{m{c}^2}}$  ....... (1)

Calculation:

Substitute $50\text{Gev}$ for $E$ , $0.511\text{MeV}$ for $m{c}^2$ and $1.0\text{cm}$ for $L$ in equation (1).

  $\begin{array}{l}{L}_{pos}=\frac{1.0\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)}{\frac{50\text{GeV}}{0.511\text{MeV}}\left(\frac{{10}^9\text{eV}}{1\text{GeV}}\right)\left(\frac{1\text{MeV}}{{10}^6\text{eV}}\right)}\\ L_{pos}=0.1\times {10}^{-6}\text{m}\left(\frac{1\mu \text{m}}{{10}^{-6}\text{m}}\right)\\ L_{pos}=0.1\mu \text{m}\end{array}$

Conclusion:

Thus, the length of positron is $0.1\mu \text{m}$ .