Verify that the special case ${\mathit{x}}^{\mathbf{\text{'}}}\mathbf{=}\mathbf{-}{\mathit{t}}^{\mathbf{\text{'}}}\mathbf{,}\mathbf{}\mathit{x}\mathbf{=}\mathbf{0}$leads to equation (2-6) when inserted in linear transformations (2-4) and that special case. ${\mathit{x}}^{\mathbf{\text{'}}}\mathbf{=}\mathbf{-}\mathit{c}{\mathit{t}}^{\mathbf{\text{'}}}\mathbf{,}\mathbf{}\mathit{x}\mathbf{=}\mathit{c}\mathit{t}$in turn leads to (2-8).

The linear transformations needed for the solution are derived from the Lorentz transformations. These are called linear transformations because the position and time between two different frames are related linearly.

The linear transformation for position is given as:

$x^{\text{'}}=Ax+Bt$

Substitute all the values in the above equation.

role="math" localid="1657544757126" $$\begin{gathered} -v{t}^{\text{'}}=A\left(0\right)+Bt \\ \frac{t^{\text{'}}}{t}=-\frac{B}{v}......\left(1\right) \end{gathered}$$

The linear transformation for time is given as:

$t^{\text{'}}=Cx+Dt$

Substitute all the values in the above equation.

role="math" localid="1657544279177" $$\begin{gathered} t^{\text{'}}=C\left(0\right)+Dt \\ \frac{t^{\text{'}}}{t}=D........\left(2\right) \end{gathered}$$

Compare equation (1) and equation (2).

$D=-\frac{B}{V}$

The expression for (2-5) is for fixed object at origin is given as:

$$\begin{gathered} B=-Av \\ A=\frac{B}{V}......\left(4\right) \end{gathered}$$

Compare equation (3) and equation (4).

$D=A$

Therefore, the linear transformations changed to equation$x^{\text{'}}=-v{t}^{\text{'}}andx=0.$

The linear transformation for position is given as:

$x^{\text{'}}=Ax+Bt$

Substitute all the values in the above equation.

$$\begin{gathered} -c{t}^{\text{'}}=A\left(ct\right)+Bt \\ {t}^{\text{'}}=-\left(A+\frac{B}{C}\right)t.........\left(5\right) \end{gathered}$$

The linear transformation for time is given as:

$t^{\text{'}}=Cx+Dt$

Substitute the magnitude of the time in the above equation.

$$\begin{gathered} \left(A+\frac{B}{c}\right)t=C\left(ct\right)+At \\ \left(A+\frac{B}{c}\right)t=(cC+A)t \\ \left(A+\frac{B}{c}\right)t=(cC+A) \end{gathered}$$

Substitute $-Av$for $B$ in the above equation.

$$\begin{gathered} \left(A+\frac{(-Av)}{c}\right)=(CC+A) \\ Ac-Av=c^{2}C+Ac \\ C=-\left(\frac{V}{c^2}\right)A \end{gathered}$$

Therefore, the linear transformations changed to equation (2-8) for $x^{\text{'}}=c{t}^{\text{'}}$and$x=ct$.