Question

(a) Determine the Lorentz transformation matrix giving position and time in frame $\mathit{s}\mathbf{\text{'}}$from those in the frame $\mathit{s}$for the case $\mathit{\nu }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{5}\mathit{c}$.

(b) If frame $\mathit{s}\mathbf{\text{'}}\mathbf{\text{'}}$moves at 0.5crelative to frame , the Lorentz transformation matrix is the same as the previous one. Find the product of two matrices, which gives $\mathit{x}\mathbf{\text{'}}\mathbf{\text{'}}$and t'' from x and t .

(c) To what single speed does the transformation correspond? Explain this result.

Short answer

1. The Lorentz transformation is $\left[\begin{array}{cccc}1.15& 0& 0& -0.575\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ -0.575& 0& 0& 1.15\end{array}\right]$
2. The Lorentz transformation is $\left[\begin{array}{cccc}1.65& 0& 0& -1.32\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ -1.32& 0& 0& 0=1.65\end{array}\right]$ .
3. This corresponds to a speed of 0.8c.

Step-by-step solution

Determine Lorentz transformation for the given case

(a) The general Lorentz transformation matrix is given by

$\left[\begin{array}{c}x\text{'}\\ y\text{'}\\ z\text{'}\\ ct\text{'}\end{array}\right]=\left(\begin{array}{cccc}{\gamma }_{\nu }& 0& 0& -{\gamma }_{\nu }\frac{\nu }{c}\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ -{\gamma }_{\nu }\frac{\nu }{c}& 0& 0& {\gamma }_c\end{array}\right)\left(\begin{array}{c}x\\ y\\ z\\ ct\end{array}\right)$

for

$$\begin{gathered} \nu =0.5c,{\gamma }_{0.5c}=\frac{1}{\sqrt{1-0.5^2}} \\ =1.15 \end{gathered}$$

The resulting transformation matrix is as follows.

$\left(\begin{array}{cccc}1.15& 0& 0& -0.575\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ -0.575& 0& 0& 1.15\end{array}\right)$

Determine Lorentz’s transformation for S'' With Respect to S

(b) If another frame $s\text{'}\text{'}$moves at velocity $0.5c$with respect to $s\text{'}$, the position and time in $s\text{'}\text{'}$in terms of that of the frame $s$is as follows.

$$\begin{gathered} \left[\begin{array}{c}x\text{'}\text{'}\\ y\text{'}\text{'}\\ z\text{'}\text{'}\\ ct\text{'}\text{'}\end{array}\right]=\left[\begin{array}{cccc}1.15& 0& 0& -0.575\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ -0.575& 0& 0& 1.15\end{array}\right]\left[\begin{array}{cccc}1.15& 0& 0& -0.575\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ -0.575& 0& 0& 1.15\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\\ ct\end{array}\right] \\ =\left[\begin{array}{cccc}1.65& 0& 0& -1.32\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ -1.32& 0& 0& 1.65\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\\ ct\end{array}\right] \end{gathered}$$

(c) Now, in the result of part (b), ${\gamma }_v=1.65$one can find the velocity corresponding with it by using the formula for the Lorentz factor that will give the velocity as $\nu =0.79c$. If we have solved it classically, we would have got v = c. If the value of velocity was more than $0.5c$ for any one of the frames, classically, the resulting velocity would be more than the speed of light which violates Einstein’s second postulate that nothing can travel faster than light and the speed of light is a constant irrespective of the reference frame considered.

For relativistic resultant velocity of $s\text{'}\text{'}$with respect to the frame $s$, our result can be verified using the relativistic velocity addition formula.

$$\begin{gathered} \nu =\frac{{\nu }_1+{\nu }_2}{1+{\displaystyle \frac{{\nu }_1{\nu }_2}{c^2}}} \\ =\frac{0.5c+0.5c}{1+{\displaystyle \frac{0.5c^2}{c^2}}} \\ =0.8c \end{gathered}$$

Hence, this transformation corresponds to speed 0.8c.