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Lei Mao

Machine Learning, Artificial Intelligence, Computer Science.

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Introduction

In my previous blog post “Special Relativity Explained”, I introduced Lorentz transformation and its key consequences. Two of the key consequences, time dilation, and length contraction, qualitatively suggest that the motion we observed from different reference frames might be different. We use velocity to describe the speed and direction of the motion.


In this blog post, I would like to derive the relativistic velocity transformation based on Lorentz transformation.

Reference Frames

In the spacetime, we have two reference frames, a reference frame $S$ and another reference frame $S’$ moving at a velocity $v$ with respect to it. So the two reference frames in this scenario are inertial reference frame. The coordinate axes in each reference frame are parallel, i.e., the $x$ and $x’$ axes are parallel, the $y$ and $y’$ axes are parallel, and the $z$ and $z’$ axes are parallel, and remain mutually perpendicular. We assume the relative motion is along the coincident $xx’$ axes. At $t = t’ = 0$, the origins of both coordinate systems are the same, $(x,y,z) = (x’,y’,z’) = (0, 0, 0)$.


An event in the time space could be observed and recorded by the observers on the two reference frames using spacetime coordinates $(t,x,y,z)$ in the reference frame $S$ and $(t’,x’,y’,z’)$ in the reference frame $S’$, respectively.


The motion is a sequence of continuous events in spacetime. It could go along any directions. In the reference frame $S$, we use $\mathbf{u} = \{u_x, u_y, u_z\}$ to describe the velocity of motion. In the reference frame $S’$, similarly, we use $\mathbf{u^{\prime}} = \{u_x^{\prime}, u_y^{\prime}, u_z^{\prime}\}$ to describe the velocity of motion. Here, $u_x$ and $u_x^{\prime}$, $u_y$ and $u_y^{\prime}$, $u_z$ and $u_z^{\prime}$, represent the portions of velocities along the $x$, $y$, $z$ axes, respectively. If the quantity is positive, it means that direction of the portion of the velocity is toward the positive axis; if the quantity is negative, it means that direction of the portion of the velocity is toward the negative axis.

Galilean Velocity Transformation

Using Galilean transformation, the transformation for the velocity is trivial.


The transformation for the velocity observed from the reference frame $S$ to the reference frame $S’$ based on Galilean transformation is

\[\begin{align} u_x^{\prime} &= u_x - v \\ u_y^{\prime} &= u_y \\ u_z^{\prime} &= u_z \\ \end{align}\]

The transformation for the velocity observed from the reference frame $S’$ to the reference frame $S$ based on Galilean transformation is

\[\begin{align} u_x &= u_x^{\prime} + v \\ u_y &= u_y^{\prime} \\ u_z &= u_z^{\prime} \\ \end{align}\]

Relativistic Velocity Transformation

Lorentz Transformation

\[\begin{align} t^{\prime} &= \gamma (t - \frac{v}{c^2}x)\\ x^{\prime} &= \gamma (x - vt)\\ y^{\prime} &= y\\ z^{\prime} &= z\\ \end{align}\]

where

\[\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}\]

and it is the Lorentz factor.

Inverse Lorentz Transformation

\[\begin{align} t &= \gamma (t^{\prime} + \frac{v}{c^2}x^{\prime})\\ x &= \gamma (x^{\prime} + vt^{\prime})\\ y &= y^{\prime}\\ z &= z^{\prime}\\ \end{align}\]

Relativistic Velocity Transformation Derivation

In Lorentz transformation, the variables are $(t,x,y,z)$ in the reference frame $S$ and $(t’,x’,y’,z’)$ in the reference frame $S’$. We take infinitesimal change on the two sides of the equations.

\[\begin{align} dt^{\prime} &= \gamma (dt - \frac{v}{c^2}dx)\\ dx^{\prime} &= \gamma (dx - vdt)\\ dy^{\prime} &= dy\\ dz^{\prime} &= dz\\ \end{align}\]

By the definition of velocities, we have

\[\begin{align} u_x^{\prime} &= \frac{dx^{\prime}}{dt^{\prime}} = \frac{\gamma (dx - vdt)}{\gamma (dt - \frac{v}{c^2}dx)} = \frac{\gamma (dx - vdt) \frac{1}{dt}}{\gamma (dt - \frac{v}{c^2}dx) \frac{1}{dt}} = \frac{\frac{dx}{dt} - v}{1 - \frac{v}{c^2}\frac{dx}{dt}} = \frac{u_x - v}{1 - \frac{v}{c^2}u_x} \\ \end{align}\] \[\begin{align} u_y^{\prime} &= \frac{dy^{\prime}}{dt^{\prime}} = \frac{dy}{\gamma (dt - \frac{v}{c^2}dx)} = \frac{dy \frac{1}{dt}}{\gamma (dt - \frac{v}{c^2}dx) \frac{1}{dt}} = \frac{\frac{dy}{dt}}{\gamma (1 - \frac{v}{c^2}\frac{dx}{dt})} = \frac{u_y}{\gamma (1 - \frac{v}{c^2}u_x)} \\ \end{align}\] \[\begin{align} u_z^{\prime} &= \frac{dz^{\prime}}{dt^{\prime}} = \frac{dz}{\gamma (dt - \frac{v}{c^2}dx)} = \frac{dz \frac{1}{dt}}{\gamma (dt - \frac{v}{c^2}dx) \frac{1}{dt}} = \frac{\frac{dz}{dt}}{\gamma (1 - \frac{v}{c^2}\frac{dx}{dt})} = \frac{u_z}{\gamma (1 - \frac{v}{c^2}u_x)} \\ \end{align}\]

Similarly, deriving from the inverse Lorentz transformation,

\[\begin{align} u_x &= \frac{dx}{dt} = \frac{\gamma (dx^{\prime} + vdt^{\prime})}{\gamma (dt^{\prime} + \frac{v}{c^2}dx^{\prime})} = \frac{\gamma (dx^{\prime} + vdt^{\prime}) \frac{1}{dt^{\prime}}}{\gamma (dt^{\prime} + \frac{v}{c^2}dx^{\prime}) \frac{1}{dt^{\prime}}} = \frac{\frac{dx^{\prime}}{dt^{\prime}} + v}{1 + \frac{v}{c^2}\frac{dx^{\prime}}{dt^{\prime}}} = \frac{u_x^{\prime} + v}{1 + \frac{v}{c^2}u_x^{\prime}} \\ \end{align}\] \[\begin{align} u_y &= \frac{dy}{dt} = \frac{dy^{\prime}}{\gamma (dt^{\prime} + \frac{v}{c^2}dx^{\prime})} = \frac{dy^{\prime} \frac{1}{dt^{\prime}}}{\gamma (dt^{\prime} + \frac{v}{c^2}dx^{\prime}) \frac{1}{dt^{\prime}}} = \frac{\frac{dy^{\prime}}{dt^{\prime}}}{\gamma (1 + \frac{v}{c^2}\frac{dx^{\prime}}{dt^{\prime}})} = \frac{u_y^{\prime}}{\gamma (1 + \frac{v}{c^2}u_x^{\prime})} \\ \end{align}\] \[\begin{align} u_z &= \frac{dz}{dt} = \frac{dz^{\prime}}{\gamma (dt^{\prime} + \frac{v}{c^2}dx^{\prime})} = \frac{dz^{\prime} \frac{1}{dt^{\prime}}}{\gamma (dt^{\prime} + \frac{v}{c^2}dx^{\prime}) \frac{1}{dt^{\prime}}} = \frac{\frac{dz^{\prime}}{dt^{\prime}}}{\gamma (1 + \frac{v}{c^2}\frac{dx^{\prime}}{dt^{\prime}})} = \frac{u_z^{\prime}}{\gamma (1 + \frac{v}{c^2}u_x^{\prime})} \\ \end{align}\]

Summary

The transformation for the relativistic velocity observed from the reference frame $S$ to the reference frame $S’$ is

\[\begin{align} u_x^{\prime} &= \frac{u_x - v}{1 - \frac{v}{c^2}u_x} \\ u_y^{\prime} &= \frac{u_y}{\gamma (1 - \frac{v}{c^2}u_x)} \\ u_z^{\prime} &= \frac{u_z}{\gamma (1 - \frac{v}{c^2}u_x)} \\ \end{align}\]

The transformation for the relativistic velocity observed from the reference frame $S’$ to the reference frame $S$ is

\[\begin{align} u_x &= \frac{u_x^{\prime} + v}{1 + \frac{v}{c^2}u_x^{\prime}} \\ u_y &= \frac{u_y^{\prime}}{\gamma (1 + \frac{v}{c^2}u_x^{\prime})} \\ u_z &= \frac{u_z^{\prime}}{\gamma (1 + \frac{v}{c^2}u_x^{\prime})} \\ \end{align}\]

Discussion

Remember the conditions of time dilation we derived in “Special Relativity Explained” which is $\Delta x = 0$. If $\Delta x = 0$ and $u_x = 0$, $u_y^{\prime} = \frac{u_y}{\gamma}$, thus $|u_y^{\prime}| \leq |u_y|$. In the same manner, $|u_z^{\prime}| \leq |u_z|$. Note that because $|\mathbf{u^{\prime}}| = \sqrt{u_x^{\prime2} + u_y^{\prime2} + u_z^{\prime2}}$ and $|\mathbf{u}| = \sqrt{u_x^2 + u_y^2 + u_z^2}$, $|\mathbf{u^{\prime}}| < |\mathbf{u}|$, the magnitude of the velocity is smaller observed from the motion reference frame $S’$ compared to the static reference frame $S$. That is why on the high speed rocket we feel the jumping up and down motion happened on the earth becomes slower.


Similarly, if $\Delta x^{\prime} = 0$ and $u_x^{\prime} = 0$, $u_y = \frac{u_y^{\prime}}{\gamma}$, thus $|u_y| \leq |u_y^{\prime}|$. In the same manner, $|u_z| \leq |u_z^{\prime}|$. We could also get $|\mathbf{u}| < |\mathbf{u^{\prime}}|$, the magnitude of the velocity is smaller observed from the motion reference frame $S$ compared to the static reference frame $S’$. That is why on the earth we feel the jumping up and down motion happened on the high speed rocket becomes slower.


However, if $u_x \neq 0$ or $u_x^{\prime} \neq 0$, the velocities along different axes are not necessary slower than “expectation”. For example, if $u_x = v$, then $u_x^{\prime} = 0$, On the high speed rocket, we feel there is no movement along the $x$ axis. However, regarding the velocity on the $yy’$ axis, we have

\[\begin{align} u_y^{\prime} &= \frac{u_y}{\gamma (1 - \frac{v}{c^2}u_x)} = \frac{u_y}{\gamma (1 - \frac{v^2}{c^2})} = \frac{u_y}{\gamma \frac{1}{\gamma^2}} = \gamma u_y \\ \end{align}\]

Therefore, $|u_y^{\prime}| \geq |u_y|$.


If we compute and compare the magnitude of the velocities observed in reference frame $S$ and $S’$, we have

\[\begin{align} \mathbf{u^{\prime2}} &= u_x^{\prime2} + u_y^{\prime2} + u_z^{\prime2}\\ &= u_y^{\prime2} + u_z^{\prime2} \\ &= \gamma^2 u_y^2 + \gamma^2 u_z^2 \end{align}\] \[\begin{align} \mathbf{u^2} &= u_x^2 + u_y^2 + u_z^2\\ &= v^2 + u_y^2 + u_z^2 \end{align}\]

We want to determine which is larger for $|\mathbf{u}|$ and $|\mathbf{u^{\prime}}|$.

\[\begin{align} \mathbf{u^{\prime2}} - \mathbf{u^2} &= \gamma^2 u_y^2 + \gamma^2 u_z^2 - v^2 - u_y^2 - u_z^2 \\ &= (\gamma^2 - 1) (u_y^2 + u_z^2) - v^2 \\ &= \bigg(\frac{1}{1 - \frac{v^2}{c^2}} - 1\bigg) (u_y^2 + u_z^2) - v^2 \\ &= \frac{v^2}{c^2 - v^2} (u_y^2 + u_z^2) - v^2 \\ &= \bigg( \frac{u_y^2 + u_z^2}{c^2 - v^2} - 1 \bigg) v^2 \\ \end{align}\]

This means that when $u_y^2 + u_z^2 > c^2 - v^2$, $|\mathbf{u^{\prime}}| > |\mathbf{u}|$; when $u_y^2 + u_z^2 < c^2 - v^2$, $|\mathbf{u^{\prime}}| < |\mathbf{u}|$.

References