Question

For the situation given in Exercise 22, find the Lorentz transformation matrix from Bob’s frame to Anna’s frame, then solve the problem via matrix multiplication.

Short answer

The Lorentz transformation matrix is $\left[\begin{array}{cccc}\frac{5}{3}& 0& 0& -\frac{4}{3}\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ -\frac{4}{3}& 0& 0& \frac{5}{3}\end{array}\right]$.

Via matrix multiplication, the value of x' is gathered as$\frac{17}{3}\mathrm{Iy}=5.67\mathrm{Iy}$ and the value of t' is gathered as $-\frac{10}{3}\mathrm{yr}=-3.33\mathrm{yr}$.

Step-by-step solution

Given data

Anna is in a spaceship moving away from Earth at v=0.8c

Significance of the Lorentz transformation

The Lorentz transformation mainly describes the time and the space coordinate of one reference frame. It also describes the relationship amongst two frames of coordinates.

Determination of the Lorentz transformation

The equation of the Lorentz transformation is expressed as:

$\left[\begin{array}{c}x\text{'}\\ y\text{'}\\ z\text{'}\\ ct\text{'}\end{array}\right]=\left[\begin{array}{cccc}{y}_v& 0& 0& -y_v\frac{v}{c}\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ -y_v\frac{v}{c}& 0& 0& y_v\end{array}\right]\left[\begin{array}{c}x\text{'}\\ y\text{'}\\ z\text{'}\\ ct\text{'}\end{array}\right]$

…(i)

Here, the value of $y_{08c}$is expressed as:

$$\begin{gathered} y_{0.8c}=\frac{1}{\sqrt{1-{\displaystyle \frac{{\left(0.8c\right)}^2}{c^2}}}} \\ =\frac{1}{\sqrt{1-{\displaystyle \frac{{\left(0.64c\right)}^2}{c^2}}}} \\ =\frac{1}{\sqrt{1-0.64}} \\ =\frac{5}{3} \end{gathered}$$

Here, the value of $-y_{0.8c}\frac{v}{c}$is expressed as:

$$\begin{gathered} -y_{0.8c}\frac{v}{c}=-\frac{5}{3}\times 0.8 \\ =-\frac{4}{3} \\ \mathrm{Substitute}\mathrm{the}\mathrm{values}\mathrm{in}\mathrm{the}\mathrm{equation}\left(\mathrm{i}\right),\mathrm{the}\mathrm{matrix}\mathrm{is}\mathrm{expressed}\mathrm{as}: \\ \left[\begin{array}{c}\mathrm{x}\text{'}\\ \mathrm{y}\text{'}\\ \mathrm{z}\text{'}\\ \mathrm{ct}\text{'}\end{array}\right]=\left[\begin{array}{cccc}\frac{5}{3}& 0& 0& -\frac{4}{3}\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ -\frac{4}{3}& 0& 0& \frac{5}{3}\end{array}\right] \\ \left[\begin{array}{cccc}\frac{5}{3}& 0& 0& -\frac{4}{3}\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ -\frac{4}{3}& 0& 0& \frac{5}{3}\end{array}\right]\left[\begin{array}{c}5\mathrm{Iy}\\ 0\\ 0\\ \mathrm{c}\left(2\mathrm{yr}\right)\end{array}\right]=\left[\begin{array}{c}\frac{25\mathrm{Iy}}{3}+\frac{-8\mathrm{Iy}}{3}\\ 0\\ 0\\ -\frac{20\mathrm{c}\mathrm{Iy}}{3}+\frac{10\mathrm{c}\mathrm{yr}}{3}\end{array}\right] \\ =\left[\begin{array}{c}\frac{17\mathrm{Iy}}{3}\\ 0\\ 0\\ -\frac{10\mathrm{c}\mathrm{yr}}{3}\end{array}\right] \\ \left[\begin{array}{c}\mathrm{x}\text{'}\\ \mathrm{y}\text{'}\\ \mathrm{z}\text{'}\\ \mathrm{ct}\text{'}\end{array}\right]=\left[\begin{array}{c}\frac{17\mathrm{Iy}}{3}\\ 0\\ 0\\ -\frac{10\mathrm{c}\mathrm{yr}}{3}\end{array}\right] \end{gathered}$$

From the above matrix, the value of x' is gathered as$\frac{17}{3}\mathrm{Iy}=5.67\mathrm{Iy}$and the value of t' is gathered as $-\frac{10}{3}\mathrm{yr}=-3.33\mathrm{yr}.$.

Thus, the Lorentz transformation matrix is $\left[\begin{array}{cccc}\frac{5}{3}& 0& 0& -\frac{4}{3}\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ -\frac{4}{3}& 0& 0& \frac{5}{3}\end{array}\right]$.

Via matrix multiplication, the value of x' is gathered as $\frac{17}{3}\mathrm{Iy}=5.67\mathrm{Iy}$and the value of t' is gathered as $-\frac{10}{3}\mathrm{yr}=-3.33\mathrm{yr}.$