Chapter 39, Problem 14PQ

(a)

To determine

The proof for the expression $t=\gamma \left(t\text{'}+\frac{v_{\text{rel}}}{c^2}x\text{'}\right)$.

Answer to Problem 14PQ

The Lorentz transformation relating $x$ and $x\text{'}$ confirms the equation $t=\gamma \left(t\text{'}+\frac{v_{\text{rel}}}{c^2}x\text{'}\right)$.

Explanation of Solution

Write the expression to obtain the position of the object along $x$ axis in primed frame.

    $x\text{'}=\gamma \left(x-v_{\text{rel}}t\right)$                                                                                                   (I)

Here, $x$ is the position of the object along $x$ axis in laboratory frame, $\gamma$ is the Lorentz factor, $x\text{'}$ is the position of the object along $x$ axis in primed frame, $v_{\text{rel}}$ is the relative velocity of the object and $t$ is the time interval in laboratory frame.

Write the expression to obtain the position of the object along $x$ axis in laboratory frame.

    $x=\gamma \left(x\text{'}+v_{\text{rel}}t\text{'}\right)$                                                                                                    (II)

Here, $x$ is the position of the object along $x$ axis in laboratory frame, $\gamma$ is the Lorentz factor, $x\text{'}$ is the position of the object along $x$ axis in primed frame, $v_{\text{rel}}$ is the relative velocity of the object and $t\text{'}$ is the time interval in primed frame.

Substitute $\gamma \left(x\text{'}+v_{\text{rel}}t\text{'}\right)$ for $x$ in equation (I).

    $\begin{array}{c}x\text{'}=\gamma \left(\left(\gamma \left(x\text{'}+v_{\text{rel}}t\text{'}\right)\right)-v_{\text{rel}}t\right)\\ x\text{'}={\gamma }^2\left(x\text{'}+v_{\text{rel}}t\text{'}\right)-\gamma v_{\text{rel}}t\\ x\text{'}={\gamma }^{2}x\text{'}+{\gamma }^2{v}_{\text{rel}}t\text{'}-\gamma v_{\text{rel}}t\\ \gamma v_{\text{rel}}t={\gamma }^{2}x\text{'}+{\gamma }^2{v}_{\text{rel}}t\text{'}-x\text{'}\end{array}$

Further solve the above equation.

    $t=\frac{{\gamma }^2-1}{\gamma v_{\text{rel}}}x\text{'}+\gamma t\text{'}$                                                                                                 (III)

Write the expression to obtain the Lorentz factor.

    $\gamma =\frac{1}{\sqrt{1-{\left(\frac{v_{\text{rel}}}{c}\right)}^2}}$

Here, $\gamma$ is the Lorentz factor, $v_{\text{rel}}$ is the relative velocity of the object and $c$ is the speed of light.

Take square both sides in the above equation.

    ${\left(\gamma \right)}^2={\left(\frac{1}{\sqrt{1-{\left(\frac{v_{\text{rel}}}{c}\right)}^2}}\right)}^2$

Subtract $1$ from both sides in the above equation.

    $\begin{array}{c}{\left(\gamma \right)}^2-1=\frac{1}{1-{\left(\frac{v_{\text{rel}}}{c}\right)}^2}-1\\ =\frac{{\left(\frac{v_{\text{rel}}}{c}\right)}^2}{1-{\left(\frac{v_{\text{rel}}}{c}\right)}^2}\end{array}$

Divide both sides by $\gamma$ in the above equation.

    $\frac{{\left(\gamma \right)}^2-1}{\gamma }=\frac{\left[\frac{{\left(\frac{v_{\text{rel}}}{c}\right)}^2}{1-{\left(\frac{v_{\text{rel}}}{c}\right)}^2}\right]}{\gamma }$

Substitute $\frac{1}{\sqrt{1-{\left(\frac{v_{\text{rel}}}{c}\right)}^2}}$ for $\gamma$ in the above equation.

    $\begin{array}{c}\frac{{\left(\gamma \right)}^2-1}{\gamma }=\frac{\left[\frac{{\left(\frac{v_{\text{rel}}}{c}\right)}^2}{1-{\left(\frac{v_{\text{rel}}}{c}\right)}^2}\right]}{\left[\frac{1}{\sqrt{1-{\left(\frac{v_{\text{rel}}}{c}\right)}^2}}\right]}\\ =\sqrt{1-{\left(\frac{v_{\text{rel}}}{c}\right)}^2}\frac{{\left(\frac{v_{\text{rel}}}{c}\right)}^2}{1-{\left(\frac{v_{\text{rel}}}{c}\right)}^2}\\ =\frac{{\left(\frac{v_{\text{rel}}}{c}\right)}^2}{\sqrt{1-{\left(\frac{v_{\text{rel}}}{c}\right)}^2}}\end{array}$

Substitute $\frac{1}{\sqrt{1-{\left(\frac{v_{\text{rel}}}{c}\right)}^2}}$ for $\gamma$ in the above equation.

    $\frac{{\left(\gamma \right)}^2-1}{\gamma }=\gamma {\left(\frac{v_{\text{rel}}}{c}\right)}^2$                                                                                                (IV)

Substitute $\gamma {\left(\frac{v_{\text{rel}}}{c}\right)}^2$ for $\frac{{\left(\gamma \right)}^2-1}{\gamma }$ in equation (III).

    $\begin{array}{c}t=\gamma {\left(\frac{v_{\text{rel}}}{c}\right)}^2\frac{x\text{'}}{v_{\text{rel}}}+\gamma t\text{'}\\ =\gamma \left(t\text{'}+\frac{v_{\text{rel}}}{c^2}x\text{'}\right)\end{array}$

Therefore, the Lorentz transformation relating $x$ and $x\text{'}$ confirms the equation $t=\gamma \left(t\text{'}+\frac{v_{\text{rel}}}{c^2}x\text{'}\right)$.

(b)

To determine

The Lorentz transformation relating $x$ and $x\text{'}$ confirms the equation $t\text{'}=\gamma \left(t-\frac{v_{\text{rel}}}{c^2}x\right)$.

Answer to Problem 14PQ

The Lorentz transformation relating $x$ and $x\text{'}$ confirms the equation $t\text{'}=\gamma \left(t-\frac{v_{\text{rel}}}{c^2}x\right)$.

Explanation of Solution

Rewrite (II).

  $x=\gamma \left(x\text{'}+v_{\text{rel}}t\text{'}\right)$

Substitute $\gamma \left(x+v_{\text{rel}}t\right)$ for $x\text{'}$ in equation (II).

    $\begin{array}{c}x=\gamma \left(\left(\gamma \left(x+v_{\text{rel}}t\right)\right)+v_{\text{rel}}t\text{'}\right)\\ x={\gamma }^2\left(x+v_{\text{rel}}t\right)+\gamma v_{\text{rel}}t\text{'}\\ x={\gamma }^{2}x+{\gamma }^2{v}_{\text{rel}}t+\gamma v_{\text{rel}}t\text{'}\\ \gamma v_{\text{rel}}t\text{'}=x-\left({\gamma }^{2}x+{\gamma }^2{v}_{\text{rel}}t\right)\end{array}$

Further solve the above equation.

    $t\text{'}=\frac{{\gamma }^2-1}{\gamma v_{\text{rel}}}x-\gamma t$                                                                                                  (V)

Consider equation (IV).

    $\frac{{\left(\gamma \right)}^2-1}{\gamma }=-\gamma {\left(\frac{v_{\text{rel}}}{c}\right)}^2$

Substitute $-\gamma {\left(\frac{v_{\text{rel}}}{c}\right)}^2$ for $\frac{{\left(\gamma \right)}^2-1}{\gamma }$ in equation (V).

    $\begin{array}{c}t\text{'}=-\gamma {\left(\frac{v_{\text{rel}}}{c}\right)}^2\frac{x}{v_{\text{rel}}}+\gamma t\\ =\gamma \left(t-\frac{v_{\text{rel}}}{c^2}x\right)\end{array}$

Therefore, the Lorentz transformation relating $x$ and $x\text{'}$ confirms the equation $t\text{'}=\gamma \left(t-\frac{v_{\text{rel}}}{c^2}x\right)$.