Question

Derive the Lorentz transformations for $t$and $t\text{'}$.

Hint: See the comment following Equation $36.22$.

Short answer

$t^1=\gamma \left(t-\frac{v}{c^2}x\right)$

Step-by-step solution

Given Information

We have to derive the Lorentz transformations for $t$ and $t\text{'}$.

Simplify

To derive the Lorentz transformations we have the below equation:

$$\begin{gathered} \gamma =\frac{1}{\sqrt{1-{\displaystyle \frac{v^2}{c^2}}}} \\ =\frac{1}{\sqrt{1-{\beta }^2}} \end{gathered}$$

As we know that

$x=ct$and

$x\text{'}=ct\text{'}$

Then,

$$\begin{gathered} x\text{'}=\gamma (x-vy) \\ x=\gamma (x\text{'}+vt) \\ xx\text{'}={\gamma }^2(xx\text{'}+xvt\text{'}-vtx\text{'}-v^{2}tt\text{'} \\ {c}^{2}tt\text{'}={\gamma }^2(c^{2}tt\text{'}+ctvt\text{'}-ctvt\text{'}-v^{2}tt\text{'}) \\ {c}^2={\gamma }^2(c^2-v^2) \\ {\gamma }^2=\frac{c^2}{c^2-v^2} \\ {\gamma }^2=\frac{1}{1-{\displaystyle \frac{v^2}{c^2}}} \\ \gamma =\frac{1}{\sqrt{1-{\displaystyle \frac{v^2}{c^2}}}} \end{gathered}$$

Next,

$$\begin{gathered} x=\gamma (x\text{'}-(-v)t\text{'}) \\ x=\gamma (x\text{'}+vt\text{'}) \\ \frac{x}{\gamma }=x\text{'}+vt\text{'} \\ \frac{x}{\gamma }-x\text{'}=vt\text{'} \\ \frac{x}{\gamma v}-\frac{x\text{'}}{v}=t\text{'} \\ t\text{'}=\frac{x}{\gamma v}-\frac{\gamma (x-vt)}{v} \\ t\text{'}=\gamma \left(\frac{x}{{\gamma }^{2}v}-\frac{x}{v}+t\right)(Eq.1) \end{gathered}$$

To get a number for t, we need to simplify the indicated component of the equation and then return it to the same equation.

$$\begin{gathered} \frac{x}{{\gamma }^{2}v}-\frac{x}{v}=x\left(\frac{1}{{\gamma }^{2v}}-\frac{{\gamma }^2}{{\gamma }^{2}v}\right) \\ \frac{1-{\gamma }^2}{{\gamma }^{2}v}=\frac{{\displaystyle \frac{c^2-v^2}{c^2-v^2}}-{\displaystyle \frac{c^2}{c^2-v^2}}}{{\displaystyle \frac{c^{2}v}{c^2-v^2}}} \\ \frac{-v^2}{c^2-v^2}·\frac{c^2-v^2}{c^{2}v}=\frac{v}{c^2} \end{gathered}$$

To find the value for t, we'll now substitute values in equation (1).

$t\text{'}=\gamma \left(t-\frac{v}{c^2}x\right)$