Question

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).

Short answer

Its verified that the linear transformations changed to (2-4) for $x^{\text{'}}=-v{t}^{\text{'}},x=0$ and (2-8) for $x^{\text{'}}=-c{t}^{\text{'}},x=ct$, .

Step-by-step solution

Significance of the linear transformations

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.

Determination of the expression for linear transformations for (2-6)

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$x^{\text{'}}=-v{t}^{\text{'}}andx=0.$

Determination of the expression for linear transformations for (2-8)

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$x=ct$.