Chapter 2, Problem 16P

(a)

To determine

The momentum and energy in $S^{\prime }$ frame.

Answer to Problem 16P

The momentum and energy in $S^{\prime }$ frame are $p^{\prime }=\frac{m\left[\left(u-v\right)/\left(1-uv/c^2\right)\right]}{\sqrt{1-{\left(u-v\right)}^2/{\left(1-uv/c^2\right)}^2}}$ and $E^{\prime }=\frac{m{c}^2}{\sqrt{1-{\left(u-v\right)}^2/{\left(1-uv/c^2\right)}^2}}$.

Explanation of Solution

Write the expression for momentum in $S$ frame,

    $p=\frac{mu}{{\left(1-v^2/c^2\right)}^{1/2}}$        (I)

Here, $p$ is the momentum, $v$ is the velocity, $m$ is the mass of the particle and $c$ is the speed of light.

Write the expression for energy in $S$ frame,

    $E=\frac{m{c}^2}{{\left(1-u^2/c^2\right)}^{1/2}}$        (II)

Here, $E$ is the energy, $v$ is the velocity, $m$ is the mass of the particle and $c$ is the speed of light.

Conclusion:

Write the expression for momentum in $S^{\prime }$ frame, by substituting $\frac{u-v}{1-uv/c^2}$ for $u^{\prime }$

    $\begin{array}{c}{p}^{\prime }=\frac{m{u}^{\prime }}{\sqrt{1-{\left(u^{\prime }/c\right)}^2}}\\ =\frac{m\left[\left(u-v\right)/\left(1-uv/c^2\right)\right]}{\sqrt{1-{\left[\left(u-v\right)/\left(1-uv/c^2\right)\right]}^2\left(1/c^2\right)}}\\ =\frac{m\left[\left(u-v\right)/\left(1-uv/c^2\right)\right]}{\sqrt{1-{\left(u-v\right)}^2/{\left(1-uv/c^2\right)}^2}}\end{array}$        (III)

Write the expression for energy in $S^{\prime }$ frame by substituting $\frac{u-v}{1-uv/c^2}$ for $u^{\prime }$,

    $\begin{array}{c}{E}^{\prime }=\frac{m{c}^2}{\sqrt{1-{\left(u^{\prime }/c\right)}^2}}\\ =\frac{m{c}^2}{\sqrt{1-{\left[\left(u-v\right)/\left(1-uv/c^2\right)\right]}^2\left(1/c^2\right)}}\\ =\frac{m{c}^2}{\sqrt{1-{\left(u-v\right)}^2/{\left(1-uv/c^2\right)}^2}}\end{array}$        (II)

The momentum and energy in $S^{\prime }$ frame are $p^{\prime }=\frac{m\left[\left(u-v\right)/\left(1-uv/c^2\right)\right]}{\sqrt{1-{\left(u-v\right)}^2/{\left(1-uv/c^2\right)}^2}}$ and $E^{\prime }=\frac{m{c}^2}{\sqrt{1-{\left(u-v\right)}^2/{\left(1-uv/c^2\right)}^2}}$.

(b)

To determine

Show that $E^2-p^2{c}^2$ is equal to ${E^{\prime }}^2-{p^{\prime }}^2{c}^2$.

Answer to Problem 16P

It is proved that $E^2-p^2{c}^2$ is equal to ${E^{\prime }}^2-{p^{\prime }}^2{c}^2$.

Explanation of Solution

Write the given expression

  $E^2-p^2{c}^2$        (I)

Here, $E$ is the energy, $p$ is the momentum, and $c$ is the speed of light.

Write the given expression

  ${E^{\prime }}^2-{p^{\prime }}^2{c}^2$        (II)

Here, $E^{\prime }$ is the final energy, $p^{\prime }$ is the final momentum, and $c$ is the speed of light.

Write the expression for momentum in $S$ frame,

    $p=\frac{mu}{{\left(1-v^2/c^2\right)}^{1/2}}$        (III)

Here, $p$ is the momentum, $v$ is the velocity, $m$ is the mass of the particle and $c$ is the speed of light.

Write the expression for energy in $S$ frame,

    $E=\frac{m{c}^2}{{\left(1-u^2/c^2\right)}^{1/2}}$        (IV)

Here, $E$ is the energy, $v$ is the velocity, $m$ is the mass of the particle and $c$ is the speed of light.

Write the expression for momentum in $S^{\prime }$ frame,

    $p^{\prime }=\frac{m\left[\left(u-v\right)/\left(1-uv/c^2\right)\right]}{\sqrt{1-{\left(u-v\right)}^2/{\left(1-uv/c^2\right)}^2}}$        (V)

Here, $p^{\prime }$ is the momentum, $v$ is the velocity, $m$ is the mass of the particle and $c$ is the speed of light.

Write the expression for energy in $S^{\prime }$ frame,

    $E^{\prime }=\frac{m{c}^2}{\sqrt{1-{\left(u-v\right)}^2/{\left(1-uv/c^2\right)}^2}}$        (VI)

Here, $E^{\prime }$ is the energy, $v$ is the velocity, $m$ is the mass of the particle and $c$ is the speed of light.

Conclusion:

Substitute expression (III) and (IV) in expression (I)

  $\begin{array}{c}{E}^2-p^2{c}^2={\left(\frac{m{c}^2}{{\left(1-u^2/c^2\right)}^{1/2}}\right)}^2-{\left(\frac{mu}{{\left(1-v^2/c^2\right)}^{1/2}}\right)}^2{c}^2\\ \text{Further solving}\\ =m^2{c}^4\end{array}$

Substitute expression (V) and (VI) in expression (II)

  $\begin{array}{c}{E^{\prime }}^2-{p^{\prime }}^2{c}^2={\left(\frac{m{c}^2}{\sqrt{1-{\left(u-v\right)}^2/{\left(1-uv/c^2\right)}^2}}\right)}^2-{\left(\frac{m\left[\left(u-v\right)/\left(1-uv/c^2\right)\right]}{\sqrt{1-{\left(u-v\right)}^2/{\left(1-uv/c^2\right)}^2}}\right)}^2{c}^2\\ \text{Further solving}\\ =m^2{c}^4\end{array}$

Hence it is proved that $E^2-p^2{c}^2$ is equal to ${E^{\prime }}^2-{p^{\prime }}^2{c}^2$.