Chapter 39, Problem 40PQ

To determine

The proof of $v_{\text{y}}=\frac{v_{\text{y}}{}^{\prime }}{\gamma \left(1+\frac{v_{\text{rel}}}{c^2}{v}_{\text{x}}{}^{\prime }\right)}$ and $v_{\text{y}}{}^{\prime }=\frac{v_{\text{y}}}{\gamma \left(1-\frac{v_{\text{rel}}}{c^2}{v}_{\text{x}}\right)}$.

Answer to Problem 40PQ

The proof of $v_{\text{y}}=\frac{v_{\text{y}}{}^{\prime }}{\gamma \left(1+\frac{v_{\text{rel}}}{c^2}{v}_{\text{x}}{}^{\prime }\right)}$ and $v_{\text{y}}{}^{\prime }=\frac{v_{\text{y}}}{\gamma \left(1-\frac{v_{\text{rel}}}{c^2}{v}_{\text{x}}\right)}$ has been determined.

Explanation of Solution

Write the Lorentz’s transformation equations.

  $y=y^{\prime }$                                                                                                                          (I)

  $t=\gamma \left(t^{\prime }+\frac{v_{\text{rel}}{x}^{\prime }}{c^2}\right)$                                                                                                        (II)

  $t^{\prime }=\gamma \left(t-\frac{v_{\text{rel}}x}{c^2}\right)$                                                                                                       (III)

Here, $y^{\prime }$,$x^{\prime }$ and $t^{\prime }$ are the position and time in prime coordinate, $\gamma$ is relativistic factor,$y$, $x$ and $t$ are the position and time in unprimed coordinate, $v_{\text{rel}}$ is the relative velocity between two coordinates and $c$ is the speed of light.

Differentiate above Lorentz’s equations to find $dy$, $dt$ and $d{t}^{\prime }$.

  $dy=d{y}^{\prime }$                                                                                                                  (IV)

  $dt=\gamma \left(d{t}^{\prime }+\frac{v_{\text{rel}}d{x}^{\prime }}{c^2}\right)$                                                                                                (V)

  $d{t}^{\prime }=\gamma \left(dt-\frac{v_{\text{rel}}dx}{c^2}\right)$                                                                                                 (VI)

Case 1:

Write the equation for the velocity of the particle in the unprimed frame.

  $v_{\text{y}}=\frac{dy}{dt}$                                                                                                                   (VII)

Substitute equation (IV) and equation (V) in above equation to find $v_{\text{y}}$.

  $\begin{array}{c}{v}_{\text{y}}=\frac{d{y}^{\prime }}{\gamma \left(d{t}^{\prime }+\frac{v_{\text{rel}}d{x}^{\prime }}{c^2}\right)}\\ =\frac{d{y}^{\prime }}{\gamma d{t}^{\prime }\left(1+\frac{v_{\text{rel}}}{c^2}\frac{d{x}^{\prime }}{d{t}^{\prime }}\right)}\\ =\frac{\frac{d{y}^{\prime }}{d{t}^{\prime }}}{\gamma \left(1+\frac{v_{\text{rel}}}{c^2}\frac{d{x}^{\prime }}{d{t}^{\prime }}\right)}\end{array}$                                                                                          (VIII)

Write the equation for the velocity of the particle in primed frame along $y$ direction.

  $v_{\text{y}}{}^{\prime }=\frac{d{y}^{\prime }}{d{t}^{\prime }}$                                                                                                                  (IX)

Write the equation for the velocity of the particle in primed frame along $x$ direction.

  $v_{\text{x}}{}^{\prime }=\frac{d{x}^{\prime }}{d{t}^{\prime }}$                                                                                                                (X)

Substitute equation (IX) and (X) in equation (VIII) to find $v_{\text{y}}$.

  $v_{\text{y}}=\frac{v_{\text{y}}{}^{\prime }}{\gamma \left(1+\frac{v_{\text{rel}}}{c^2}{v}_{\text{x}}{}^{\prime }\right)}$                                                                                                  (XI)

Case 2:

Write the equation for the velocity of the particle in the primed frame.

  $v_{\text{y}}{}^{\prime }=\frac{d{y}^{\prime }}{d{t}^{\prime }}$                                                                                                               (XII)

Substitute $dy$ for $d{y}^{\prime }$ and $\gamma \left(dt-\frac{v_{\text{rel}}dx}{c^2}\right)$ for $d{t}^{\prime }$ in above equation to find $v_{\text{y}}{}^{\prime }$.

  $\begin{array}{c}{v}_{\text{y}}{}^{\prime }=\frac{dy}{\gamma \left(dt-\frac{v_{\text{rel}}dx}{c^2}\right)}\\ =\frac{dy}{\gamma dt\left(1-\frac{v_{\text{rel}}}{c^2}\frac{dx}{dt}\right)}\\ =\frac{\frac{dy}{dt}}{\gamma \left(1-\frac{v_{\text{rel}}}{c^2}\frac{dx}{dt}\right)}\end{array}$                                                                                          (XIII)

Write the equation for the velocity of the particle in unprimed frame along $y$ direction.

  $v_{\text{y}}=\frac{dy}{dt}$                                                                                                                (XIV)

Write the equation for the velocity of the particle in unprimed frame along $x$ direction.

  $v_{\text{x}}=\frac{dx}{dt}$                                                                                                                (XV)

Substitute equation (IX) and (X) in equation (VIII) to find $v_{\text{y}}$.

  $v_{\text{y}}=\frac{v_{\text{y}}}{\gamma \left(1-\frac{v_{\text{rel}}}{c^2}{v}_{\text{x}}\right)}$                                                                                                 (XVI)

Conclusion:

Therefore, the transformation of a velocity component perpendicular to ${\overrightarrow{v}}_{\text{rel}}$ is $v_{\text{y}}=\frac{v_{\text{y}}{}^{\prime }}{\gamma \left(1+\frac{v_{\text{rel}}}{c^2}{v}_{\text{x}}{}^{\prime }\right)}$ and the inverse transformation from laboratory frame to the primed frame is $v_{\text{y}}{}^{\prime }=\frac{v_{\text{y}}}{\gamma \left(1-\frac{v_{\text{rel}}}{c^2}{v}_{\text{x}}\right)}$.