91影视

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

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

Energy:

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

Momentum:

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

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,

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

For the given case the transformation matrix will be,

for 0.8c and ${纬}_v=\frac{5}{3}$.

$\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,

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

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

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

The energy in Object鈥檚 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.