## Section52.10Transformation of Velocity

We can find relation between velocity of a particle in S and S' frames by using Lorentz transformations. As before, let frame S' be moving at constant speed $V$ towards positive $x$ axis with respect to frame S. Then, we have

\begin{align*} \amp t = \gamma \left( t' + \frac{V}{c^2} x' \right)\\ \amp x = \gamma \left( x' + V t' \right)\\ \amp y = y',\ \ z = z'. \end{align*}

From these we will get the followinfg for small intervals.

\begin{align*} \amp dt = \gamma \left( dt' + \frac{V}{c^2} dx' \right)\\ \amp dx = \gamma \left( dx' + V dt' \right)\\ \amp dy = dy',\ \ dz = dz'. \end{align*}

Therefore $x$-component of velocity in S is

\begin{align*} v_x \amp = \frac{dx}{dt} = \frac{dx' + V dt' }{dt' + \frac{V}{c^2} dx'} = \frac{dx'/dt' + V }{1 + \frac{V}{c^2} dx'/dt'}= \frac{v'_x + V }{1 + \frac{V}{c^2} v'_x}, \end{align*}

where I replaced $dx'/dt'$ by $v'_x\text{,}$ which is the $x$-component of velocity in S' frame. We can work out $y$ and $z$ components similarly.

\begin{align*} v_y \amp = \frac{dy}{dt} = =\frac{dy' }{ \gamma\left(dt' + \frac{V}{c^2} dx'\right) } = \frac{v'_y /\gamma}{1 + \frac{V}{c^2} v'_x}\\ v_z\amp = \frac{v'_z/\gamma }{1 + \frac{V}{c^2} v'_x} \end{align*}

To summarize, we found that

\begin{align*} \amp v_x = \dfrac{v_x^{\prime} + V}{1 + V v_x^{\prime}/c^2},\\ \amp v_y = \dfrac{v_y^{\prime}/\gamma}{1 + V v_x^{\prime}/c^2},\\ \amp v_z = \dfrac{v_z^{\prime}/\gamma}{1 + V v_x^{\prime}/c^2}. \end{align*}

This is the velocity addition law that replaces the Galilean velocity addition laws, $v_x = v_x^{\prime} +V\text{,}$ $v_y = v_y^{\prime}\text{,}$ and $v_z = v_z^{\prime}\text{.}$ We see that when the relative velocity of the frames is much smaller than the speed of light, i.e., when $V \lt c\text{,}$ the special relativity velocity addition law reduces to the Galilean velocity law.

A spaceship moving at a speed of $0.99\:c$ with respect to Earth sends light beam in the forward direction. The light beam travels with speed $c$ with respect to the spaceship. What will be the speed of the beam as observed by an observer on Earth?

Hint

$c\text{.}$

Solution

We really need no formula to answer this question. Speed of light is same in all frames regardless of the state of motion. Let us check if this is borne out by the velocity addition formulas of this section. Let positive $x$-axis be pointed towards the forward direction of the spaceship. Then, we need to add

\begin{equation*} V = 0.99\:c\ \ \text{and}\ \ v^{\prime}_x = c. \end{equation*}

Using the velocity addition formula we get the speed of light with respect to the Earth-based observer

\begin{equation*} v_x = \dfrac{v_x^{\prime} + V}{1 + V v_x^{\prime}/c^2} = \dfrac{c + 0.99\:c}{1 + 0.99} = c. \end{equation*}

A spaceship moving at a speed of $0.9\:c$ with respect to Earth fires a rocket in the forward direction that has a speed of $0.9\:c$ with respect to the spaceship. What will be the speed of the rocket as observed by an observer on Earth?

Hint

$0.99\:c\text{.}$

Solution

Let the positive $x$-axis be pointed towards the forward direction of the spaceship. Then, we need to add

\begin{equation*} V = 0.9\:c\ \ \text{and}\ \ v^{\prime}_x = 0.9\:c. \end{equation*}

Using the velocity addition formula we get the speed of light with respect to the Earth-based observer

\begin{equation*} v_x = \dfrac{v_x^{\prime} + V}{1 + V v_x^{\prime}/c^2} = \dfrac{0.9\:c + 0.9\:c}{1 + 0.9^2} = 0.99\:c. \end{equation*}

A proton beam is shot in the forward direction from a moving spaceship. The protons move with respect to the spaceship at speed $0.5c\text{.}$ The spaceship itself moves at a speed $0.5c$ with respect to Earth. What is the speed of the proton with respect to Earth?

Hint

Straightforward application of velocity addition law.

$0.8 c\text{.}$

Solution

Using velocity addition law with positive $x$-axis in the forward direction, which is the direction directly away from the Earth we get

\begin{align*} v \amp = \frac{v_1 + v_2}{1 + \frac{v_1 v_2}{c^2}} = \frac{c}{ 1 + 0.5\times 0.5} = 0.8 c. \end{align*}

A spaceship is moving away from Earth at a speed $0.5c\text{.}$ Inside the spaceship there is an electron accelerator. The electron accelerator speeds the electrons to $0.8c$ with respect to the spaceship but the velocities of the electrons are in the direction of Earth. What will be the speed of an electron with respect to Earth?

Hint

Watch for direction.

Solution

With positive $x$-axis pointing away from Earth, we get

\begin{align*} v \amp = \frac{v_1 + v_2}{1 + \frac{v_1 v_2}{c^2}} = \frac{0.5c - 0.8c}{ 1 - 0.5\times 0.8} = - 0.5 c. \end{align*}

A pion moving at a speed of $0.9c$ with respect to the laboratory splits into two photons, one in the forward direction and the other in the backward direction, each traveling at the speed of light. What are the speeds of the forward and backward emitted photons with respect to the laboratory?

Hint

Trick question.

$c\text{.}$

Solution

Both are $c$ as required by special relativity assumptions.

Two spaceships are on a collision course moving perpendicularly to the same point as observed by an observer at rest with respect to the meeting point. Suppose the common point is the origin, spaceship A is on the negative $x$-axis and moving towards the positive $x$-axis direction with speed $0.9c\text{,}$ and spaceship B is on the negative $y$-axis and moving towards the positive $y$-axis direction with speed $0.9c\text{.}$

(a) What will be the velocity of spaceship A with respect to spaceship B? (b) If the spaceships are a distance $D=100\text{ km}$ from the origin at time $t=0\text{,}$ when will they collide in the frame of the meeting point?

Hint

(a) $0.9 c\, \hat i - 0.9 c\, \hat j\text{,}$ (b) $3.7\times 10^{-4}\text{ sec}\text{.}$

Solution

(a) To get the velocity of A with respect to B. We want to add the velocity of O with respect to B, $\vec v_{OB}$ and velocity of A with respect to O, $\cev v_{AO}\text{.}$ They are

\begin{align*} \amp \vec v_{OB} = - 0.9 c\, \hat j.\\ \amp \vec v_{AO} = 0.9 c\, \hat i. \end{align*}

Therefore, $x$ and $y$ components of the required velocity will be

\begin{align*} \amp v{AB,x} = \frac{ v_{OB,x} + v_{AO,x} }{ 1 + v_{OB,x} v_{AO,x}/c^2 } = 0.9 c.\\ \amp v{AB,y} = \frac{ v_{OB,y} + v_{AO,y} }{ 1 + v_{OB,y} v_{AO,y}/c^2 } = - 0.9 c. \end{align*}

In the frame of the meeting point, i.e., origin, A goes from its location to origin wth time $t_O\text{.}$

\begin{equation*} t_O = \frac{D}{v_{AO,x}} = \frac{100\times1000}{0.9\times 3\times 10^{8}} = 3.7\times 10^{-4}\text{ sec}. \end{equation*}