Question

Solve Eqs. 12.18 for$\mathit{x}\mathbf{,}\mathit{y}\mathbf{,}\mathit{z}\mathbf{,}\mathit{t}$in terms of$\overline{\mathbf{x}}\mathbf{,}\overline{\mathbf{y}}\mathbf{,}\overline{\mathbf{z}}\mathbf{,}\overline{\mathbf{t}}$ and check that you recover Eqs. 12.19.

Short answer

The coordinates of the S frame in terms of coordinates in $\overline{S}$are found as:

$$\begin{gathered} \mathrm{x}=\mathrm{\gamma }\left(\begin{array}{c}\overline{\mathrm{x}}+\mathrm{y}\overline{t}\end{array}\right) \\ \mathrm{y}=\overline{\mathrm{y}} \\ \mathrm{z}=\overline{\mathrm{z}} \\ \mathrm{t}=\mathrm{\gamma }\left(\begin{array}{c}\overline{t}+\frac{\mathrm{v}}{{\mathrm{c}}^2}\overline{\mathrm{x}}\end{array}\right) \end{gathered}$$

Step-by-step solution

Expression for the Lorentz transformation equations:

Write the expression for the Lorentz transformation equations of coordinates $x,y,z$and tin $\overline{S}$frame.

First equation:

$\overline{\mathbf{x}}\mathbf{=}\mathbf{Y}\mathbf{(}\mathbf{x}\mathbf{-}\mathbf{vt}\mathbf{)}$ …… (1)

Second equation:

$\overline{y}=y$

Third equation:

$\overline{z}=z$

Fourth equation:

localid="1654594577273" $\overline{\mathbf{t}\mathbf{}}\mathbf{=}\mathbf{Y}\mathbf{(}\mathbf{1}\mathbf{-}\frac{\mathbf{VX}}{{\mathbf{c}}^{\mathbf{2}}}\mathbf{)}$ …… (2)

Determine the coordinate of x in terms of x:

Rearrange the equation (2) in terms of t.

$\mathrm{t}=\frac{\overline{\mathrm{t}}}{\mathrm{\gamma }}+\frac{\mathrm{vx}}{{\mathrm{c}}^2}$ ……. (3)

Substitute the value of equation (3) in equation (1).

$$\begin{gathered} \overline{\mathrm{x}}=\gamma \left(\begin{array}{c}x-v\left(\begin{array}{c}\frac{\overline{t}}{\gamma }+\frac{vx}{c^2}\end{array}\right)\end{array}\right) \\ \overline{x}=\gamma \times \left(\begin{array}{c}1-\frac{v^2}{c^2}\end{array}\right)-v\overline{t} \end{gathered}$$

Rearrange the above expression in terms of x.

$\mathrm{x}=\mathrm{\gamma }\left(\begin{array}{c}\overline{\mathrm{x}}+\mathrm{v}\overline{\mathrm{t}}\end{array}\right)$ …… (4)

Determine the coordinate of t in terms of t :

Substitute the value of equation (4) in equation (3).

$$\begin{gathered} \overline{\mathrm{t}}=\gamma \left(\begin{array}{c}1-\frac{v\left(\begin{array}{c}\gamma \left(\begin{array}{c}\overline{x}+\mathrm{v}\overline{\mathrm{t}}\end{array}\right)\end{array}\right)}{c^2}\end{array}\right) \\ \overline{t}=\gamma \left[t-\frac{v}{c^2}\left(\begin{array}{c}\frac{\overline{x}}{\gamma }+vt\end{array}\right)\right] \end{gathered}$$

Rearrange the above expression in terms of t.

$\overline{\mathrm{t}}=\gamma \left(\begin{array}{c}t-\frac{v}{c^2}\overline{x}\end{array}\right)$

Therefore, the coordinates of the S frame in terms of coordinates in are found as:

$$\begin{gathered} \mathrm{x}=\mathrm{\gamma }\left(\begin{array}{c}\overline{\mathrm{x}}+\mathrm{v}\overline{\mathrm{t}}\end{array}\right) \\ \mathrm{y}=\overline{\mathrm{y}} \\ \mathrm{z}=\overline{\mathrm{z}} \\ \mathrm{t}=\mathrm{\gamma }\left(\begin{array}{c}\overline{\mathrm{t}}+\frac{\mathrm{v}}{{\mathrm{c}}^2}\overline{\mathrm{x}}\end{array}\right) \end{gathered}$$