Chapter 39, Problem 55P

To determine

To Show:The momentum and energy of a particle in frame S’ are related to its momentum and energy in frame S by the transformation equations.

Answer to Problem 55P

  $p_x=\gamma \left(p_x-\frac{vE}{c^2}\right)$

  $p{\text{'}}_y=p_y$

  $p{\text{'}}_z=p_z$

  $\frac{E\text{'}}{c}=\gamma \left(\frac{E}{c}-\frac{v{p}_x}{c}\right)$ .

Explanation of Solution

Given:

A particle moving with speed u along the y-axis in frame S .

Formula Used:

Relativistic momentum, $\overrightarrow{p}=\frac{m\overrightarrow{u}}{\sqrt{1-\frac{u^2}{c^2}}}$

Relativistic energy, $E=\frac{m{c}^2}{\sqrt{1-\frac{u^2}{c^2}}}$

Where, $\overrightarrow{u}$ is the velocity of the particle and u is its speed.

Calculations:

In any inertial frame

  $\overrightarrow{p}=\frac{m\overrightarrow{u}}{\sqrt{1-\frac{u^2}{c^2}}}$

As the particle is moving in the y axis

  ${\overrightarrow{p}}_x={\overrightarrow{p}}_z=0$

  ${\overrightarrow{p}}_y=\frac{m{\overrightarrow{u}}_y}{\sqrt{1-\frac{u^2}{c^2}}}$

Substitute these values in the transformation equations:

  $p{\text{'}}_x=\gamma \left(0-\frac{vE}{c^2}\right)=-\gamma \frac{vE}{c^2}$

  $p{\text{'}}_y=p_y$

  $p{\text{'}}_z=0$

  $\frac{E\text{'}}{c}=\gamma \left(\frac{E}{c}-0\right)=\gamma \frac{E}{c}$

In S’ frame, the momentum components are

  $p_x\text{'}=\frac{m{u}_x\text{'}}{\sqrt{1-\frac{u{\text{'}}^2}{c^2}}}$

  $p_y\text{'}=\frac{m{u}_y\text{'}}{\sqrt{1-\frac{u{\text{'}}^2}{c^2}}}$

  $p_z\text{'}=\frac{m{u}_z\text{'}}{\sqrt{1-\frac{u{\text{'}}^2}{c^2}}}$

The inverse velocity transformations are

  $u_x\text{'}=\frac{u_x-v}{\sqrt{1-\frac{u_{x}v}{c^2}}}$

  $u_y\text{'}=\frac{u_y}{\sqrt{1-\frac{u_{y}v}{c^2}}}$

  $u_z\text{'}=\frac{u_z}{\sqrt{1-\frac{u_{z}v}{c^2}}}$

Substitute

  $u_x=u_z=0$

  $u_y=u$

Thus,

  $u_x\text{'}=-v$

  $u_y\text{'}=\gamma u$ and

  $u_z\text{'}=0$

  $u{\text{'}}^2=v^2+\frac{u^2}{{\gamma }^2}$

To verify that

  $p{\text{'}}_z=p_z=0$

  ${\overrightarrow{p}}_z\text{'}=\frac{m(0)}{\sqrt{1-\frac{u{\text{'}}^2}{c^2}}}=0$

Next

  $p{\text{'}}_y=p_y$

  $\begin{array}{c}{p}_y\text{'}=\frac{m{u}_y\text{'}}{\sqrt{1-\frac{u{\text{'}}^2}{c^2}}}\\ =\frac{mu}{\gamma \sqrt{1-\frac{v^2}{c^2}-\frac{u^2}{{\gamma }^2{c}^2}}}\\ =\frac{mu}{\sqrt{1-\frac{u{\text{'}}^2}{c^2}}}\frac{\sqrt{1-\frac{u{\text{'}}^2}{c^2}}}{\gamma \sqrt{1-\frac{v^2}{c^2}-\frac{u^2}{{\gamma }^2{c}^2}}}\end{array}$

  $\begin{array}{c}{p}_y\text{'}=\frac{mu}{\sqrt{1-\frac{u^2}{c^2}}}\sqrt{\frac{\left(1-\frac{u^2}{c^2}\right)\left(1-\frac{v^2}{c^2}\right)}{1-\frac{v^2}{c^2}-\frac{u^2}{c^2}\left(1-\frac{v^2}{c^2}\right)}}\\ =p_y\sqrt{\frac{1-\frac{v^2}{c^2}-\frac{u^2}{c^2}\left(1-\frac{v^2}{c^2}\right)}{1-\frac{v^2}{c^2}-\frac{u^2}{c^2}\left(1-\frac{v^2}{c^2}\right)}}\end{array}$

Hence $p{\text{'}}_y=p_y$

Next, to verify $p_x=\gamma \left(p_x-\frac{vE}{c^2}\right)$

  $\begin{array}{c}{\overrightarrow{p}}_x\text{'}=\frac{m{u}_x\text{'}}{\sqrt{1-\frac{u{\text{'}}^2}{c^2}}}\\ =\frac{-mv}{\gamma \sqrt{1-\frac{v^2}{c^2}-\frac{u^2}{{\gamma }^2{c}^2}}}\\ =-\frac{\gamma v}{c^2}\frac{m{c}^2}{\sqrt{1-\frac{u^2}{c^2}}}\frac{{\gamma }^{-1}\sqrt{1-\frac{u^2}{c^2}}}{\sqrt{1-\frac{v^2}{c^2}-\frac{u^2}{{\gamma }^2{c}^2}}}\end{array}$

  $\begin{array}{c}{p}_x\text{'}=-\frac{\gamma v}{c^2}E\sqrt{\frac{\left(1-\frac{u^2}{c^2}\right)\left(1-\frac{v^2}{c^2}\right)}{1-\frac{v^2}{c^2}-\frac{u^2}{c^2}\left(1-\frac{v^2}{c^2}\right)}}\\ =-\frac{\gamma v}{c^2}E\sqrt{\frac{1-\frac{v^2}{c^2}-\frac{u^2}{c^2}\left(1-\frac{v^2}{c^2}\right)}{1-\frac{v^2}{c^2}-\frac{u^2}{c^2}\left(1-\frac{v^2}{c^2}\right)}}\end{array}$

  $p_x\text{'}=-\frac{\gamma v}{c^2}E$

Finally

  $\begin{array}{l}\frac{E\text{'}}{c}=\gamma \left(\frac{E}{c}-\frac{v{p}_x}{c}\right)\\ =\gamma \frac{E}{c},\text{or }E\text{'}=\gamma E\end{array}$

  $\begin{array}{c}E\text{'}=\frac{m{c}^2}{\sqrt{1-\frac{u{\text{'}}^2}{c^2}}}\\ =\frac{\gamma m{c}^2}{\sqrt{1-\frac{u^2}{c^2}}}\frac{{\gamma }^{-1}\sqrt{1-\frac{u^2}{c^2}}}{\sqrt{1-\frac{u{\text{'}}^2}{c^2}}}\\ =\gamma E\frac{{\gamma }^{-1}\sqrt{1-\frac{u^2}{c^2}}}{\sqrt{1-\frac{v^2}{c^2}-\frac{u^2}{{\gamma }^2{c}^2}}}\end{array}$

  $\begin{array}{l}E\text{'}=\gamma E\sqrt{\frac{\left(1-\frac{u^2}{c^2}\right)\left(1-\frac{v^2}{c^2}\right)}{1-\frac{v^2}{c^2}-\frac{u^2}{c^2}\left(1-\frac{v^2}{c^2}\right)}}\\ =\gamma E\sqrt{\frac{1-\frac{v^2}{c^2}-\frac{u^2}{c^2}\left(1-\frac{v^2}{c^2}\right)}{1-\frac{v^2}{c^2}-\frac{u^2}{c^2}\left(1-\frac{v^2}{c^2}\right)}}\end{array}$

  $E\text{'}=\gamma E$

The x,y,z and t transformation equations are

  $x\text{'}=\gamma (x-vt),y\text{'}=y,z\text{'}=z\text{ and }t\text{'}=\gamma \left(t-\frac{vx}{c^2}\right)$

The x,y,z and ct transformation equations are

  $x\text{'}=\gamma \left(x-\frac{v}{c}ct\right),y\text{'}=y,z\text{'}=z\text{and }ct\text{'}=\gamma \left(ct-\frac{v}{c}x\right)$

The $p_x,p_y,p_z\text{ and E/c}$ transformation equations are

  ${\text{p'}}_x=\gamma \left(p_x-\frac{v}{c}\frac{E}{c}\right),p{\text{'}}_y=p_y,p{\text{'}}_z=p_z\text{ and }\frac{E\text{'}}{c}=\gamma \left(\frac{E}{c}-\frac{v}{c}{p}_x\right)$

Conclusion:

  $p_x=\gamma \left(p_x-\frac{vE}{c^2}\right)$

  $p{\text{'}}_y=p_y$

  $p{\text{'}}_z=p_z$

  $\frac{E\text{'}}{c}=\gamma \left(\frac{E}{c}-\frac{v{p}_x}{c}\right)$ .