Question: A 1 kg object moves at 0.8c relative to Earth.

(a) Calculate the momentum and the energy of the object.

(b) Determine the Lorentz transformation matrix from the earth’s frame to the object’s frame.

(c) Find the momentum and total energy of the object in the new frame via matrix multiplication.

Lorentz transformations are a six-parameter family of linear transformations from a coordinate frame in space-time to another frame that moves at a constant velocity relative to the first.

Let’s consider Earth’s frame and the object’s moving frame as . An object is moving along the x-axis at a speed of .

The object’s momentum and energy measured with respect to the frame are given as follow.

Energy:

$$\begin{gathered} E={\gamma }_{v}m{c}^2 \\ E=\frac{5}{3}\left(1kg\right)c^2 \\ =\frac{5}{3}{c}^2 \\ =1.5\times {10}^{17}J \end{gathered}$$

Momentum:

$$\begin{gathered} p={\gamma }_{v}mv \\ =\frac{5}{3}\left(1kg\right)\left(0.8c\right) \\ =\frac{4}{3}c \end{gathered}$$

The relationship between two coordinate frames of two frames of references that move relative to each other can be represented in the form of a matrix called as Lorentz transformation matrix and is given by,

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

For the given case the transformation matrix will be,

for 0.8c and ${\gamma }_v=\frac{5}{3}$.

$\left[\begin{array}{cccc}\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$3$}\right.& 0& 0& \raisebox{1ex}{$-4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ \raisebox{1ex}{$-4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.& 0& 0& \raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\end{array}\right]$

Assuming object moving along x-direction you can write transformation matrix for momentum and energy in frame as,

$$\begin{gathered} \left[\begin{array}{c}{p\text{'}}_x\\ {p\text{'}}_y\\ {p\text{'}}_z\\ \raisebox{1ex}{$E^{\text{'}}$}\!\left/ \!\raisebox{-1ex}{$c$}\right.\end{array}\right]=\left[\begin{array}{cccc}-{\gamma }_v& 0& 0& -{\gamma }_v\frac{v}{c}\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ -{\gamma }_v\frac{v}{c}& 0& 0& -{\gamma }_v\end{array}\right]\left[\begin{array}{c}{p}_x\\ p_y\\ p_z\\ \raisebox{1ex}{$E^{\text{'}}$}\!\left/ \!\raisebox{-1ex}{$c$}\right.\end{array}\right] \\ =\left[\begin{array}{cccc}\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$3$}\right.& 0& 0& \raisebox{1ex}{$-4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ \raisebox{1ex}{$-4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.& 0& 0& \raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\end{array}\right]\left[\begin{array}{c}0\\ 0\\ 0\\ \frac{5}{3}c\end{array}\right] \\ =\left[\begin{array}{c}\frac{5}{3}\left(\frac{4}{3}c\right)+-\frac{4}{3}\left(\frac{5}{3}c\right)\\ 0\\ 0\\ -\frac{4}{3}\left(\frac{4}{3}c\right)+\frac{5}{3}\left(\frac{5}{3}c\right)\end{array}\right] \\ =\left[\begin{array}{c}0\\ 0\\ 0\\ c\end{array}\right] \end{gathered}$$

The momentum $p{\text{'}}_x=0$ in objects frame will be zero because the object itself in its frame will be stationary. The energy will be

$$\begin{gathered} \frac{E\text{'}}{c}=c \\ E\text{'}=c^{2}J \\ E\text{'}=9\times {10}^{16}J \end{gathered}$$

The energy in Object’s frame $E\text{'}$ include only internal energy $\left(m{c}^2\right)$ and not kinetic energy for the same reason why momentum is zero i.e., the object is stationary in its own frame.